Video Lectures: - http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-21/

Lecture Notes: - http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

Positive and negative feedback of op amps give the same output but positive feedback cannot possible work that way. In reality a small disturbance in the positive feedback amplifier output will cause the output to increases until it reaches saturation. Static analysis of positive feedback does indeed yield v-out equals minus R2 over R1 times v-in but better tools will be needed to analyze positive feedback.

Static analysis of positive feedback circuit is done with a dependent source model. Nodal analysis is used. This does not really work so a dynamic model is used instead. This dynamic model consists of a resistor, capacitor and a dependent source.

A positive and negative feedback is connected to the dynamic model. The dynamic model is analyzed by first finding the values of v_plus and v_minus in terms of the fraction of the voltage output. Then the node method is applied to the capacitor. What we would be left with is a differential equation which has a solution v_0 equals K e to the minus t divided by capital T.

We will see how the system behaves with respect to t, which in turn behaves on gamma minus and

gamma plus. So we first plot v_0, which is equal to K, a small disturbance. So we have to look at various situations. So let's say we start with t being positive and t will be positive when gamma minus is greater than gamma plus. This will result in a slope that is going down. If gamma minus is greater than gamma plus, negative feedback is stronger than positive feedback. In other words, net, I am feeding more of the output to the negative terminal than to the positive terminal. So if the negative feedback is stronger than my positive feedback, then what's going to happen is output, k, is

going to very quickly go down to 0. So that is my stable situation. Not surprisingly, you recall, for op amps with negative feedback, if there is a perturbation to the output, very quickly that would go to 0 because of the negative feedback.

So in this situation, if gamma plus is greater than gamma minus, in other words, if

positive feedback is stronger- then t is negative. What we will notice is that there is no minus sign

in the exponential, and so therefore, my response is going to spiral out of control leading to an

unstable situation.

And then what happens when- the natural thing to consider next is when gamma plus more

or less equal to gamma minus. In this case, t is very large and v_0 is more or less equal to K for a

long period of time. This is called neutral equilibrium.

We could build a number of interesting circuits with the op amp without negative feedback.

Again, recall when the op amp doesn't have negative feedback, it is going to move with any slight perturbation. It is going to try to hit the positive rail, Vs, or it's going to hit the negative rail, Vs.

So in this particular instance, in the op amp portal I want to show you Vs plus and minus Vs explicitly so you know that Vs and V minus Vs are the positive and negative supply. So the op amp is going to hit those values, depending on what the input is. So the first circuit is a comparator.

What is a comparator? A comparator is a circuit where, when an input is applied, the output is going to shoot to either a plus, a Vs, or a V minus, depending on whether the input is positive or negative.

In this case, what I'll do is I'll connect V minus to ground. So V minus is set at zero volts. I want to apply an input to V plus. So my input, Vi, is applied to V plus. So now, what's going to happen is that when Vi is greater than zero volts, the output is going to shoot positive. And if Vi is less than zero, the output is going to shoot to a negative, minus Vs. So this is a comparator. What it does is that the output is a high if the input is positive, and the output is a low if the input is negative. So we can build a transfer function of the comparator. We can also do a time behavior out of this.

A comparator has a lot of uses. It can be used to tell whether the input is positive or negative. And it can also used to to take the analog signal and turn that into a one bit digital value, a zero, one kind of sequence at the output, for a positive and negative going analog signal as an input. Now, one of the issues with a comparator like this is that it has the unfortunate property that small perturbations, or small noise imposed at the input, can cause weird, unexpected behavior

Let's say my input looks like there's some noise imposed on the input. If we focus on a certain part right where there isn't a clean crossing, the part where the signal is crossing the zero line, the output is going to go bouncing back and forth causing some spikes.

So we are going to use positive feedback to build what we call hysteresis not run the op amp

in the open loop mode like this. By building hysteresis into the circuit, we will make the circuit remember what happened in the past, so that it doesn't behave so flippantly as did the open loop op amp.

So here's my op amp,with v_I as in the past. And what we are going to do is apply positive feedback. In this case, a resistance R2, a resistance R1, and connect the voltage divider between R2 and R1 to the positive terminal. And this is my output v_0 taken with respect to ground. For this example, let us assume, just for fun, that R1 is equal to R2, so that, at the center here, this is v_0 divided by 2 where R1 equals R2. Let's assume, to be specific, V_S is equal to 15 volts. So let's say, because the output is at 15 volts, v plus is therefore at 7.5 volts. So as soon as vi reaches plus 7.6 volts, the output is going to swing hugely negative, and hit the negative rail, minus 15 volts.

The moment the output hits minus 15 volts, then, as the sixth step, v plus goes to minus 7.5 volts.

So v plus goes to minus 7.5 volts. So what happens then? If v plus goes to minus 7.5 volts, there is a big negative voltage between v plus and v minus. So at that point, even if the voltage vi begins to meander around 7.5 let's say, because of noise, from 7.6 it goes to 7.7. No problem. v plus minus v minus is still negative. What if at that point, v minus goes to 7.4? Amazingly enough, no change in the output. Why is that? If vi goes to 7.4 volts because of noise instantaneously, notice that, because v plus, through positive feedback, has now switched to minus 7.5 volts, this is a huge negative offset.

So let me draw you a little chart so you get a better sense of what is going on here.

So I could draw what's called a state diagram. These state diagrams tend to capture the memory property of these circuits. And notice that my op amp circuit has two states.

One state is where the output v0 is equal to plus 15 volts, and the second state is where the output

v0 is minus 15 volts. Two states. In the first state, notice that v plus is at 7.5 volts, and in the second state, v plus is at minus 7.5 volts. I have two states there. So let's use a state diagram to understand

what we've just built. So you recall I said life started out with me being the first state, v0 equals plus 15 volts, so I start off in this state. And I said I started off with vi being 0. Then what happened was vi went past 7.5 volts. So if vi became greater than 7.5 volts, what happened? When vi became greater than 7.5 volts, then my output switched to the second state. Eventually we will notice that in this circuit, we have a memory in the system. That is, this state here, the v0 equals minus 15, remembers that vi had gone above 7.5 volts. And once it went above 7.5 volts, this state says, OK,

it is not going to change until vi then goes below negative 7.5 volts. It remembers that.

We are going to plot vi versus vo in this diagram. The output is a memory property here.

The circuit remembers the past. And that is called hysteresis. This is, if you recall, many of you have played with magnets and so on, and you can magnetize some of these materials, and that's where the term is used commonly, And that property which remembers what happened during the past is called hysteresis.

Now, why is this useful? This is useful, as you will be able to get a clean wave form at the output because of the memory property.

OK, so now let's take a look at another fun circuit called the oscillator. Again, circuits like this can be used to build what we call a clock. So before we look at what clocks are useful for, let's

try to build an oscillator. An oscillator is a circuit that oscillates back and forth by itself between high voltage and a zero voltage. We'll call it high voltage and a low voltage. And there are many fun uses for a circuit like that. OK, let's build a circuit and then talk about some applications of a oscillator.

In this fun circuit, we would apply both negative and positive feedback.

,Let us start with what you've seen before and apply some positive feedback using a usual R and R voltage divider like so. Since R and R are equal, then the voltage at my positive input is simply going to V naught divided by 2. At the negative terminal, we are going to build an RC circuit.

Notice that current cannot go into the minus terminal because it's infinite resistance.And so that current, 15 volts divided by R, has nowhere to go, but it's going to start charging up the capacitor.

So as the capacitor starts charging up, it eventually charges to 7.5 volts, And when it charges to 7.5 volts and exceeds it ever so slightly, boom, I'm going to have a negative voltage on the V plus minus V minus terminal pair, and the output is going to switch negative.

And I'm going to have a negative 7.5 volts now at the plus terminal, the capacitor is at plus 7.5 and the output is at minus 15. So in this case, what's going to happen is that, because the output is at minus 15 volts, the current is now going to flow in the right-hand direction so the capacitor is going to start discharging now because this has switched from plus 15, which is charging the

capacitor, to minus 15 that will begin discharging the capacitor. So as the capacitor begins to discharge and the voltage will start going more and more negative, ultimately, the voltage gets to minus 7.5 volts, in which case this switches to plus 15. It switches to plus 15 and begins to charge up the capacitor again, and so on and so forth.

Let's now get some insight into how to compute the frequency of oscillation.

First, to find the frequency of oscillation, we are going look to find the time period of oscillation,

and a rise time.So what I'm going to do is start by trying to figure out the rise time of the capacitor voltage, that is this time.

Once the system goes into a steady state, recall that, for the rise, the capacitor voltage is going to go up from minus Vs by 2 all the way up to plus Vs by 2. So let's go ahead and try to figure out what that time is. So this is vC, the capacitor voltage. So to figure out vC, I could write down in general, for a rising capacitor voltage, the intuitive method of figuring out the rise time is the following.

So vC will be some initial voltage on capacitor, initial voltage on capacitor plus the change in voltage times 1 minus e raised to minus t over RC. The change that we're talking about is the voltage that the capacitor would have reached had it gone all the way to its high value.

OK?

Starting at Vs minus 2 and, if left to itself, the capacitor would've gone all the way to Vs. That is what we have to use for this formula. So in this formula, the initial value is simply minus Vs over two.

That's the initial value in the capacitor. So we can solve this equation for tr and that gives the rise time.

A similar method is used to find the fall time.

A clock is a square wave that's applied to digital systems. And the clock can be useful for senders and

receivers to communicate effectively.

OK, as an example, suppose I have a sender and a receiver. And the sender wants to send some values. So let's say the sender really wants to send the sequence of values, 1, 1, 0, as in the waveform shown here, OK? Also notice that the waveform is a little funny, in the sense that I have a 1 and 1.

I have a high part of the waveform out here with the 1 and 1 with the low part of the waveform showing 0, but then a part of the waveform where the signal is a bit funky. So for example, it could be that the output of the gate that is producing the waveform, because of parasitic capacitance and impedances, may have some ringing associated with it.

OK, so the receiver is sitting there trying to figure out when is the signal valid. You can say the receiver can look at the signal when the clock is high, for example. That's the example I'm using here. In other cases, with what is called edge-triggered logic, you can tell the receiver, look at the signal during a given clock edge, a rising edge or a falling edge. But in our example, I'm simply going to say that the sender needs to make sure that a valid signal is available to the receiver whenever the clock is high.

So notice here the receiver is looking at the output from the sender whenever the clock is high.

Notice that the first time the clock is high, it picks up a 1. Then it picks up another 1. Then it picks up a 0 out here, OK? So it picks up a 0 at the point where the clock is high for the third time because the receiver sees a 1, 1, 0. And the receiver is pretty happy with that. So what we have done here using the clock is another kind of discretization. It is discretizing time, rather than dealing with

continuous time, by using a clock.

Notice that discretization of time is one among another major sequence of discretizations that we do in systems to make it all work, OK? Recall, in physics, we had the discrete mass or lumped mass

discipline where we lumped matter and got the point mass simplification. Then, for the digital discipline, we discretized value, and we got the digital abstraction. And now we've talked about discretizing time, So this is a pretty cool sequence of things that we do to make systems easy to design and build.

## Sunday, September 11, 2016

## Thursday, August 25, 2016

### Circuits and Electronics - The Operational Amplifier Circuits (Lecture 24)

Video Lectures: - http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-20/

Lecture Notes: - http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

We first look at the subtractor circuit. There are five steps in the first method of obtaining v_O.

Another way is the superposition method which is the easier way. The two input voltages are shorted one at a a time, v_1 is set to zero, then v_2 is set to zero. After that the output voltages are added.

Next, we look at how to build an integrator. As voltage output is proportional to the integral of the current, we need to somehow convert voltage v_I to a current. First, we try using a resistor to do so. This can happen when v_O is very small compared to v_R..

However we will find that that this is not possible as the capacitor tends to charge up v_O to a certain value.

The better way is an op amp integrator. With KVL, v_O=v_C.

To build a differentiator as current is proportional to the differential of the input voltage , we need to somehow convert current to a voltage v_O. The circuit is somewhat similar to the integrator except the resistor and capacitor are swapped.

Filters can be built with op amps as well. The problem with passive filters is that it is difficult to get v_O equal to H of j omega v_I.

The moment a load resistance is connected to the filter circuit, the j omega of the filter has changed.

When a a source with an internal resistance R_I is connected to the filter circuit, the filter characteristic will change as well. Worse is that the filter resistance loads the source, so the source may not be able to supply current.

The other issue is that it is difficult to build an inductor in an integrated circuit. As many filter circuits require inductors , we need to look at some other approach of doing such things. The approach is to use op amps and the first approach is known as the brute force approach. In this approach a buffer is used. A buffer can be connected to the input and the output but it is a bit inefficient.

We will see how op amps work with impedances. and sketch a graph for values of low and high omega. In this case, a bandpass filter is obtained.

We can build non linear circuits with op amps. With an inverting amplifier, it is possible to build an exponentiator with the Expodweeb device.

With that in mind, it is possible to build a logarithmic amplifier.

Summing amplifiers can be used to build digital to analog converters.

Lecture Notes: - http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

We first look at the subtractor circuit. There are five steps in the first method of obtaining v_O.

Another way is the superposition method which is the easier way. The two input voltages are shorted one at a a time, v_1 is set to zero, then v_2 is set to zero. After that the output voltages are added.

Next, we look at how to build an integrator. As voltage output is proportional to the integral of the current, we need to somehow convert voltage v_I to a current. First, we try using a resistor to do so. This can happen when v_O is very small compared to v_R..

However we will find that that this is not possible as the capacitor tends to charge up v_O to a certain value.

The better way is an op amp integrator. With KVL, v_O=v_C.

To build a differentiator as current is proportional to the differential of the input voltage , we need to somehow convert current to a voltage v_O. The circuit is somewhat similar to the integrator except the resistor and capacitor are swapped.

Filters can be built with op amps as well. The problem with passive filters is that it is difficult to get v_O equal to H of j omega v_I.

The moment a load resistance is connected to the filter circuit, the j omega of the filter has changed.

When a a source with an internal resistance R_I is connected to the filter circuit, the filter characteristic will change as well. Worse is that the filter resistance loads the source, so the source may not be able to supply current.

The other issue is that it is difficult to build an inductor in an integrated circuit. As many filter circuits require inductors , we need to look at some other approach of doing such things. The approach is to use op amps and the first approach is known as the brute force approach. In this approach a buffer is used. A buffer can be connected to the input and the output but it is a bit inefficient.

We will see how op amps work with impedances. and sketch a graph for values of low and high omega. In this case, a bandpass filter is obtained.

We can build non linear circuits with op amps. With an inverting amplifier, it is possible to build an exponentiator with the Expodweeb device.

With that in mind, it is possible to build a logarithmic amplifier.

Summing amplifiers can be used to build digital to analog converters.

## Thursday, August 11, 2016

### Circuits and Electronics - The Operational Amplifier Abstraction (Lecture 23)

Video Lectures:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-19/

Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

The Operational Amplifier an input port represented with a plus and a minus. That's where the input signal is applied. Then there is an output port.

There are two terminals for the input. So I have a pair of terminals, plus minus, and the input is applied across those two terminals. This has what is called a differential input.

It also has a power port, and here we go. The way the power port works is with both a positive and negative supply.

A more abstract representation for the op amp is one without the power supplies.

We will look at a very simple equivalent circuit for an ideal op amp. For the input, you are going to have a pair of terminals.

And let's say the input is v. So v is the voltage difference between the two terminals. And then, the output of the op amp comes from a dependent ideal, a dependent source - an ideal voltage-controlled voltage source. So let's say this is what it looks like from between ground and out here, that is v_O.

And for this dependent voltage source, its voltage is going to be Av, where A is a very large number.

Where A tends to infinity. And in practice, A is typically 10 to the 6, or thereabouts. And so notice that v is v plus minus v minus.

Av, the output v_0, can also be written as A times v plus minus v minus.

Both input terminals display an open circuit. So therefore, my current going in is going to be identically equal to 0 at all times. And this is ideal op amp so i plus equals 0 where i plus is

the current into the plus terminal. And similarly, i minus is equal to 0. So both i plus and i minus, the currents into the plus and minus terminals of the op amp are identically equal to 0. That says that there's an open circuit at the inputs of the operational amplifier.

I have a dependent voltage source, it's a voltage source at the output. So therefore, it has 0 output resistance.. What about the input resistance? Resistance looking into the plus terminal is infinity.

Similarly, resistance looking into the minus terminal is infinity. And the reason is that i plus is identically equal to 0 at all times. And so is i minus. So , infinite input resistance.

For the ideal op amp, there is no saturation - what this is saying is that the output voltage v_0 can

take on any value.

So let's very quickly summarize what we've seen so far. So for this ideal op amp, the current into the positive terminal is 0, infinite resistance. Current into the negative terminal is also 0, infinite

impedance there as well. The voltage between the plus minus terminal pair is v. And my output is modeled as a voltage-controlled voltage source. It's a dependent voltage source. And the voltage is going to be Av. It's going to be amplifying the input voltage difference between the terminal pair by A. And because it's a ideal dependent voltage source, the output resistance is 0. And then A, the gain, is virtually infinity for this ideal op amp. Op amps are the basic building block of the analog industry. Most analog designs are done with such abstractions. We may have more detailed piece of information that we use in the op amp model. Maybe make it less ideal.

So a range of voltages V_I is applied to the input of the MOSFET. And we are going to plot V_I versus V_0 for the MOSFET show the concept of saturation. So we're going to have our op amp with +12 and -12V power supplies shown as well. It's not going to be the perfectly ideal op amp, where there is no saturation. And then, for V_0, there will be a resistor, R_L.

The input voltage V_I is going to be swept across a range of values So that as I sweep the values of VI across a range of values, I want to go ahead and observe what V_0 looks like. Now, recall, from the diagram up here, recall that V_0 is equal to A times V. A might be on the order of 10 to the sixth. Given that V_0 is A times V, I'm going to have a huge, huge gain.

For a small change in V_I, there is going to be a massive change in V_0. So we will see a V_0 versus V_I with a huge slope. If A is on the order of 10 to the sixth, then for a 10-microvolt input--

OK, so in this case, I have 10 here minus 10 here. So for a 20-microvolt input, if A is 10 to the sixth, what's going to happen here is that when some point is reached -- at +12-volt and -12V, there will be saturation.

The region, where the output is truly A times V, is called the active region. And when you use the MOSFET, generally, when you build linear devices, that is where you want the MOSFET to operate.

So far, so good, but one of the issues here is that this A , even though it's about 10 to the sixth or thereabouts, it's really, really unreliable. Among other things, A could be temperature dependent.

So even though A is really large, the op amp can behave weirdly as its temperature changes.

Now, how do we get to use the op amp?

Where by using a little trick, a little trick called feedback, we are going to make the op amp behave exactly like we want it to.

So we have the abstract op amp here on the left and the equivalent circuit of the op amp on the

right-hand side.

A circuit will be built and connected to the abstract op amp. And then an equivalent circuit diagram with the same circuit, will be built so that A and other parameters can be analyzed.The circuit is a

non-inverting amplifier. It's going to be an amplifier that's going to gives some gain, a small amount of gain. And it's going to pass the input without inversion.

We will eventually end up with an equation involving v_0 and v_I and we have to make an approximation where A is very large. We will get v_0 is approximately equal to v_I

times r1 plus r2 divided by r2.

It can be seen that the non-inverting op amp provides a stable output even when heat is applied. This is due to negative feedback. To explore how this happens, we have to perturb the output voltage from 10 V of 12 V. Recall that the output of the op amp is A times v plus minus v minus. 6V is fed back to the inverting terminal, making the output decrease. Then part of the output is fed back again, this time causing the output to increase.

The key is that the stable point is when v plus equals v minus. It's not exactly equal. It's more or less equal to each other. Because notice that the moment one of them becomes greater than the other, the output tends to shoot in one direction.

From the virtual short method, it's only negative feedback that causes V-plus to be more

or less equal to V-minus. We also know a couple of other constraints, that I-plus and I-minus is 0. The method solves op amp circuits really, really quickly. And the beauty of this technique is that we don't have to deal with the A and all of that stuff anymore.

On examining the buffer circuit with the virtual short method , we can see that the voltage output follows the voltage input. Another method is putting R_1=0 and R_2= infinity.

So why is this circuit useful?

Notice that one might think that we could just take v_I and just connect it to whatever system we want to connect it to. But the problem is that this v_I may be a very sensitive sensor. It's some kind of a sensor. And it cannot provide much current. So this system that it is connected to may damage the

sensor or cause its behavior to change.

So buffers are useful in this situation, where you want to buffer the output from the input. Buffers serve as really nice isolation devices. So let's look at the properties of this buffer.

The voltage gain is simply 1.

The impedance looking in is infinity.

The output impedance is 0.

The current gain is infinity.

The power gain is also infinity.

These are pretty good statistics. That's why a buffer is very useful, to isolate sources that are quite sensitive from other uses in the rest of the system.

Next, we look at an inverting amplifier circuit. Start by grounding the V plus terminal. Notice that in all of these negative feedback circuits, some portion of the output always gets fed back to the

negative terminal. So here, output is fed back to the input with R1. And then, the input resistance, R2 and input voltage are connected.

We are going to use both the op amp equivalent circuit and the virtual short method to analyze the circuit and just show you how simple the virtual short method makes it.

The principle of superposition is employed in the op amp equivalent circuit model. So V-minus will be the sum of the components of voltage at V-minus due to V_I and V_0.

Next, we look at the input resistance of an inverting amplifier circuit. To do this we need to apply a test input v_I, and measure the current i_I going in. So then the Rin is going to be v_I divided by i_I.

To get the input resistance, we have to find i_I in terms of the resistances, v_I and v_O. Then we substitute the value of v_O. Conductance is used for the analysis. Eventually we will get a value of input resistance equal to R_2. The virtual short method is much faster for this case.

Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

The Operational Amplifier an input port represented with a plus and a minus. That's where the input signal is applied. Then there is an output port.

There are two terminals for the input. So I have a pair of terminals, plus minus, and the input is applied across those two terminals. This has what is called a differential input.

It also has a power port, and here we go. The way the power port works is with both a positive and negative supply.

A more abstract representation for the op amp is one without the power supplies.

We will look at a very simple equivalent circuit for an ideal op amp. For the input, you are going to have a pair of terminals.

And let's say the input is v. So v is the voltage difference between the two terminals. And then, the output of the op amp comes from a dependent ideal, a dependent source - an ideal voltage-controlled voltage source. So let's say this is what it looks like from between ground and out here, that is v_O.

And for this dependent voltage source, its voltage is going to be Av, where A is a very large number.

Where A tends to infinity. And in practice, A is typically 10 to the 6, or thereabouts. And so notice that v is v plus minus v minus.

Av, the output v_0, can also be written as A times v plus minus v minus.

Both input terminals display an open circuit. So therefore, my current going in is going to be identically equal to 0 at all times. And this is ideal op amp so i plus equals 0 where i plus is

the current into the plus terminal. And similarly, i minus is equal to 0. So both i plus and i minus, the currents into the plus and minus terminals of the op amp are identically equal to 0. That says that there's an open circuit at the inputs of the operational amplifier.

I have a dependent voltage source, it's a voltage source at the output. So therefore, it has 0 output resistance.. What about the input resistance? Resistance looking into the plus terminal is infinity.

Similarly, resistance looking into the minus terminal is infinity. And the reason is that i plus is identically equal to 0 at all times. And so is i minus. So , infinite input resistance.

For the ideal op amp, there is no saturation - what this is saying is that the output voltage v_0 can

take on any value.

So let's very quickly summarize what we've seen so far. So for this ideal op amp, the current into the positive terminal is 0, infinite resistance. Current into the negative terminal is also 0, infinite

impedance there as well. The voltage between the plus minus terminal pair is v. And my output is modeled as a voltage-controlled voltage source. It's a dependent voltage source. And the voltage is going to be Av. It's going to be amplifying the input voltage difference between the terminal pair by A. And because it's a ideal dependent voltage source, the output resistance is 0. And then A, the gain, is virtually infinity for this ideal op amp. Op amps are the basic building block of the analog industry. Most analog designs are done with such abstractions. We may have more detailed piece of information that we use in the op amp model. Maybe make it less ideal.

So a range of voltages V_I is applied to the input of the MOSFET. And we are going to plot V_I versus V_0 for the MOSFET show the concept of saturation. So we're going to have our op amp with +12 and -12V power supplies shown as well. It's not going to be the perfectly ideal op amp, where there is no saturation. And then, for V_0, there will be a resistor, R_L.

The input voltage V_I is going to be swept across a range of values So that as I sweep the values of VI across a range of values, I want to go ahead and observe what V_0 looks like. Now, recall, from the diagram up here, recall that V_0 is equal to A times V. A might be on the order of 10 to the sixth. Given that V_0 is A times V, I'm going to have a huge, huge gain.

For a small change in V_I, there is going to be a massive change in V_0. So we will see a V_0 versus V_I with a huge slope. If A is on the order of 10 to the sixth, then for a 10-microvolt input--

OK, so in this case, I have 10 here minus 10 here. So for a 20-microvolt input, if A is 10 to the sixth, what's going to happen here is that when some point is reached -- at +12-volt and -12V, there will be saturation.

The region, where the output is truly A times V, is called the active region. And when you use the MOSFET, generally, when you build linear devices, that is where you want the MOSFET to operate.

So far, so good, but one of the issues here is that this A , even though it's about 10 to the sixth or thereabouts, it's really, really unreliable. Among other things, A could be temperature dependent.

So even though A is really large, the op amp can behave weirdly as its temperature changes.

Now, how do we get to use the op amp?

Where by using a little trick, a little trick called feedback, we are going to make the op amp behave exactly like we want it to.

So we have the abstract op amp here on the left and the equivalent circuit of the op amp on the

right-hand side.

A circuit will be built and connected to the abstract op amp. And then an equivalent circuit diagram with the same circuit, will be built so that A and other parameters can be analyzed.The circuit is a

non-inverting amplifier. It's going to be an amplifier that's going to gives some gain, a small amount of gain. And it's going to pass the input without inversion.

We will eventually end up with an equation involving v_0 and v_I and we have to make an approximation where A is very large. We will get v_0 is approximately equal to v_I

times r1 plus r2 divided by r2.

It can be seen that the non-inverting op amp provides a stable output even when heat is applied. This is due to negative feedback. To explore how this happens, we have to perturb the output voltage from 10 V of 12 V. Recall that the output of the op amp is A times v plus minus v minus. 6V is fed back to the inverting terminal, making the output decrease. Then part of the output is fed back again, this time causing the output to increase.

The key is that the stable point is when v plus equals v minus. It's not exactly equal. It's more or less equal to each other. Because notice that the moment one of them becomes greater than the other, the output tends to shoot in one direction.

From the virtual short method, it's only negative feedback that causes V-plus to be more

or less equal to V-minus. We also know a couple of other constraints, that I-plus and I-minus is 0. The method solves op amp circuits really, really quickly. And the beauty of this technique is that we don't have to deal with the A and all of that stuff anymore.

On examining the buffer circuit with the virtual short method , we can see that the voltage output follows the voltage input. Another method is putting R_1=0 and R_2= infinity.

So why is this circuit useful?

Notice that one might think that we could just take v_I and just connect it to whatever system we want to connect it to. But the problem is that this v_I may be a very sensitive sensor. It's some kind of a sensor. And it cannot provide much current. So this system that it is connected to may damage the

sensor or cause its behavior to change.

So buffers are useful in this situation, where you want to buffer the output from the input. Buffers serve as really nice isolation devices. So let's look at the properties of this buffer.

The voltage gain is simply 1.

The impedance looking in is infinity.

The output impedance is 0.

The current gain is infinity.

The power gain is also infinity.

These are pretty good statistics. That's why a buffer is very useful, to isolate sources that are quite sensitive from other uses in the rest of the system.

Next, we look at an inverting amplifier circuit. Start by grounding the V plus terminal. Notice that in all of these negative feedback circuits, some portion of the output always gets fed back to the

negative terminal. So here, output is fed back to the input with R1. And then, the input resistance, R2 and input voltage are connected.

We are going to use both the op amp equivalent circuit and the virtual short method to analyze the circuit and just show you how simple the virtual short method makes it.

The principle of superposition is employed in the op amp equivalent circuit model. So V-minus will be the sum of the components of voltage at V-minus due to V_I and V_0.

Next, we look at the input resistance of an inverting amplifier circuit. To do this we need to apply a test input v_I, and measure the current i_I going in. So then the Rin is going to be v_I divided by i_I.

To get the input resistance, we have to find i_I in terms of the resistances, v_I and v_O. Then we substitute the value of v_O. Conductance is used for the analysis. Eventually we will get a value of input resistance equal to R_2. The virtual short method is much faster for this case.

## Sunday, July 3, 2016

### Circuits and Electronics - Time Domain Versus Frequency Domain Analysis (Lecture 22)

Video and lecture notes:- edX MITx: 6.002.2x Circuits and Electronics 3: week 3

Identifying the time domain behavior of the circuit is easy from the impedance analysis.

Because if you recall, for our RLC circuit, Vc over Vi from the previous video was given by omega naught squared divided by s squared plus 2 alpha s plus omega naught squared. The s was equal to j omega. Now, you can get the time domain behavior completely from this equation here. OK, if you recall, at a given point in our analysis, we had substituted e to the st for a response, and when we

differentiated e to the st with respect to time, we got s e raised to st. And the e raised to st things canceled out. So each time you saw a d by dt, we ended up with an s in our equation.

So what we can do is a reverse mapping. So wherever we see an s, we can go and stick in a d by dt

there and derive the differential equation in a backward step. Wherever you see an s squared, stick in a second derivative.

However, just for kicks, let's say we care about a step input. So for example, let's try a step input.

So for a step input, the input signal will simply be some dc value starting at time t equal to 0.

There you go. So I've very quickly written my differential equation for a step input for the same circuit completely by using frequency domain analysis.

As before, even if I did the time domain analysis, my Q is the same, omega naught by 2 alpha, and

for the series RLC circuit, 2 alpha is R divided by L. So I ended up with omega naught divided by R over L, which is Q. We will look at using Q at both time domain and frequency domain behavior.

For the frequency domain, we would apply a sinusoid given by Vi cosine omega t. For the time domain, just for fun, we are going to use a unit step. And as we have seen in the past, we can do the analysis in three cases.

So one case is for overdamped behavior. So for this overdamped behavior, alpha is greater than omega naught. And in terms of Q, Q-- which is equal to omega naught by 2 alpha-- is less than 1/2.

Then for alpha less than omega naught, or Q--which is still omega naught by 2 alpha-- is greater than 1/2. this is high Q. This is underdamped.

And then, finally, a look at alpha equals omega naught. And in this case, Q equals 1/2. This is critically damped.

With the over-damped case, alpha is greater than omega naught. This is the over-damped case. In one case, a sinusoid is applied at the input and the frequency response is measured. We are going to measure the magnitude of Vc/Vi, the magnitude of the transfer function for the frequency response. For the time domain part, we are going to apply a unit step and observe for what v_C looks like.

For the frequency domain behavior, it looks like a low-pass filter. It's low-Q, no peakiness.

And at the time domain, there is a pretty sluggish circuit, which just goes up lazily and saturates at unity.

For the underdamped case, alpha is less than omega-naught. And so therefore, Q will be greater than half. Again, this is the case where I'm going to get a high Q circuit. So in this case, what's going to happen is that at omega-naught, because I have a high Q filter, I am going to get a peaky response.

The bandwidth is defined as the interval between the points in the curve where the response falls to 1 by square root 2 of the peak value. And 1 by square root 2 is 0.707. And sometimes it's also called the 0.707 frequency at which you reach the 1 by square root 2 value. Either way, it's a narrow band, and it's a very high Q filter. What happens in the time domain? Underdamped circuits gave rise to a lot of ringing. They were not sluggish. So for this capacitor's circuit, it would eventually settle down at 1.

So for instance, if I have a Q of 50, then the circuit is going to sit around ringing for about 50 rings.

And of course, notice that if I make R close to 0, if R goes close to 0, then this just becomes LC.

And recall, for an LC circuit, it will ring forever. Provide a unit step, they will simply ring forever

and ever and ever. Because Q for that circuit would be infinity when R goes to 0. So the key here is that in the time domain, Q relates to how ringy the circuit is. Q relates to ringiness. And in the frequency domain, Q relates to peakiness.

Finally, let me look at the critically damped case where alpha equals omega naught or, equivalently, Q equals half. Same circuits, input sinusoid for the frequency domain analysis and unit step for the time domain analysis. So in the case of my frequency domain analysis, let me first draw out clearly, at the very low and very high frequencies, the circuit behaves similarly. And it will look something like . . . 'bloop'. Next, let's look at the time domain behavior of the same circuit and for the step input. I just get a little, little blip. It may be seen that correlated to a flipped version of the time domain

step response but don't get confused by that.

Identifying the time domain behavior of the circuit is easy from the impedance analysis.

Because if you recall, for our RLC circuit, Vc over Vi from the previous video was given by omega naught squared divided by s squared plus 2 alpha s plus omega naught squared. The s was equal to j omega. Now, you can get the time domain behavior completely from this equation here. OK, if you recall, at a given point in our analysis, we had substituted e to the st for a response, and when we

differentiated e to the st with respect to time, we got s e raised to st. And the e raised to st things canceled out. So each time you saw a d by dt, we ended up with an s in our equation.

So what we can do is a reverse mapping. So wherever we see an s, we can go and stick in a d by dt

there and derive the differential equation in a backward step. Wherever you see an s squared, stick in a second derivative.

However, just for kicks, let's say we care about a step input. So for example, let's try a step input.

So for a step input, the input signal will simply be some dc value starting at time t equal to 0.

There you go. So I've very quickly written my differential equation for a step input for the same circuit completely by using frequency domain analysis.

As before, even if I did the time domain analysis, my Q is the same, omega naught by 2 alpha, and

for the series RLC circuit, 2 alpha is R divided by L. So I ended up with omega naught divided by R over L, which is Q. We will look at using Q at both time domain and frequency domain behavior.

For the frequency domain, we would apply a sinusoid given by Vi cosine omega t. For the time domain, just for fun, we are going to use a unit step. And as we have seen in the past, we can do the analysis in three cases.

So one case is for overdamped behavior. So for this overdamped behavior, alpha is greater than omega naught. And in terms of Q, Q-- which is equal to omega naught by 2 alpha-- is less than 1/2.

Then for alpha less than omega naught, or Q--which is still omega naught by 2 alpha-- is greater than 1/2. this is high Q. This is underdamped.

And then, finally, a look at alpha equals omega naught. And in this case, Q equals 1/2. This is critically damped.

With the over-damped case, alpha is greater than omega naught. This is the over-damped case. In one case, a sinusoid is applied at the input and the frequency response is measured. We are going to measure the magnitude of Vc/Vi, the magnitude of the transfer function for the frequency response. For the time domain part, we are going to apply a unit step and observe for what v_C looks like.

For the frequency domain behavior, it looks like a low-pass filter. It's low-Q, no peakiness.

And at the time domain, there is a pretty sluggish circuit, which just goes up lazily and saturates at unity.

For the underdamped case, alpha is less than omega-naught. And so therefore, Q will be greater than half. Again, this is the case where I'm going to get a high Q circuit. So in this case, what's going to happen is that at omega-naught, because I have a high Q filter, I am going to get a peaky response.

The bandwidth is defined as the interval between the points in the curve where the response falls to 1 by square root 2 of the peak value. And 1 by square root 2 is 0.707. And sometimes it's also called the 0.707 frequency at which you reach the 1 by square root 2 value. Either way, it's a narrow band, and it's a very high Q filter. What happens in the time domain? Underdamped circuits gave rise to a lot of ringing. They were not sluggish. So for this capacitor's circuit, it would eventually settle down at 1.

So for instance, if I have a Q of 50, then the circuit is going to sit around ringing for about 50 rings.

And of course, notice that if I make R close to 0, if R goes close to 0, then this just becomes LC.

And recall, for an LC circuit, it will ring forever. Provide a unit step, they will simply ring forever

and ever and ever. Because Q for that circuit would be infinity when R goes to 0. So the key here is that in the time domain, Q relates to how ringy the circuit is. Q relates to ringiness. And in the frequency domain, Q relates to peakiness.

Finally, let me look at the critically damped case where alpha equals omega naught or, equivalently, Q equals half. Same circuits, input sinusoid for the frequency domain analysis and unit step for the time domain analysis. So in the case of my frequency domain analysis, let me first draw out clearly, at the very low and very high frequencies, the circuit behaves similarly. And it will look something like . . . 'bloop'. Next, let's look at the time domain behavior of the same circuit and for the step input. I just get a little, little blip. It may be seen that correlated to a flipped version of the time domain

step response but don't get confused by that.

## Thursday, June 30, 2016

### Circuits and Electronics - Filters (Lecture 21)

Video Lectures:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-18/

Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

We often build special circuits known as filters to eliminate that specific frequency of 50 to 60 Hz from the signal as AC in most countries are at 50 or 60 Hz. Otherwise we end up with a very low hum.

First, we need to plot the graphs of impedance against omega for capacitors, resistors and inductors.

For very low frequencies, the impedance of the capacitor is very, very high. The capacitor looks like an open circuit but for very high frequencies, the capacitor looks like a short circuit. So for the capacitor, we will be able to see the curve that shows its impedance.

For a resistor R, its behavior is frequency independent. So its impedance is simply R. And so it is

going to be a constant.

For the inductor, notice that for very low values of omega, the impedance is very low, and for high values of omega, the impedance is very high. Not surprisingly, the inductor looks like an open circuit to high frequencies and for DC and very low frequencies, the inductor looks very much like a short circuit.

So let's start with a CR circuit. If we plot the magnitude of H of omega of the transfer function for this filter, what does it look like?

So let's start with the high frequencies just for fun. For very high frequencies, note that the capacitor is a short circuit. And the resistor, of course, has a resistance R. So the capacitor is a short circuit and so Vi will appear directly at Vr. So for very high frequencies, my filter will give me Vi at Vr. So Vr divided by Vi will simply be unity.

What about for very low frequencies? At very low frequencies, the capacitor behaves like an open circuit. And if that behaves like an open circuit, then most of the voltage drop will be across the capacitor and very little will drop across the resistor and so I'm going to get a very low value. So low value for low frequencies and high values for high frequencies. So what we will get is a high pass filter.

These same principles apply for an RC circuit. So for an RC circuit, we will get a low pass filter.

For an RL circuit, at low frequencies, the inductor behaves like a short circuit so H of omega will have a low value. At high frequencies, the inductor behaves like an open circuit, so H of omega will have a high value. The RL circuit will be a high pass filter.

For an RLC circuit with the output voltage measured across the resistor, a bandpass filter is produced. At resonance, omega equals omega0 which is 1 by square root of LC and H of omega will be equal to 1.

For an RLC circuit with the output voltage measured across the capacitor and inductor, a bandstop filter is produced. At resonance, the output voltage is zero.

For an RLC circuit with the output voltage measured across the capacitor and inductor in parallel, a bandpass filter is produced. At resonance, the output voltage is zero. A major application of bandpass filters is in AM radio. The capacitor is made variable so that the bandpass filter can tune to various frequencies.

How does the radio work?

On the x axis of the graph is the frequency of the signal. On the y-axis, is the signal strength.

So the way radio system work is that different radio stations would be transmitting at different frequencies. So, for example, in the Boston area, there are a bunch of radio stations that transmit the following frequency. So for example, one of the radio stations is 1030 in the Amplitude Modulation band, And then,there may be other signals being transmitted by other radio stations. So for example, there may be a station transmitting at 1020, another at 1010 KHz and it goes on and on.

So when stations transmit, they try to maximize its signal strength, say around the 1030 frequency, ,it tries to maximize it there. And then it tries to make sure that the signal strengthdoesn't encroach too much into neighboring bands.So notice that each of these stations gets a roughly 10 kilohertz band.

So let's say, if I want to listen to WBZ News Radio, then I have to tune my capacitor here such that

the band-pass filter have its passing band focused where I care the most.

So when this is passed through, this band-pass filter will allow the 1030 range to mostly get through, and it will attenuate everything else that's further away but it is not perfect and so it will let through a little bit of the neighboring bands. That's why with AM radio you always get some interference from the sides of the two neighboring bands.

Fundamentally, that is a small value. Mostly the 1030 will come through. So this was the filter, the band-pass filter

Notice here that selectivity is important.

What is selectivity? Selectivity has to do with trying to capture the signal in a given range.

In the next video, we will see that the selectivity relates to something that you've seen before.

It will be shown how a RLC circuit, gives voltage values that are much higher than voltage values that are input. We have to look at Vc divide by Vi.. If the graph of Vc divide by Vi against omega is plotted, we will get a low pass filter.

We can do the maths with Vc over Vi using the voltage divider relation. We can write the impedance of the capacitor and divide it by the sum of all of the other impedance.

And for the magnitude, we can simply get the square of the real part plus the square of the

imaginary part square rooted.

You've taken it on faith from me that between these two points the function looks like the graph but that is not true.

Next we need to find out mathematically what happens at resonance. After some mathematical analysis, it can be seen that V_c is equal to Q times V_i. If Q is very large, V_c is more than V_i.

Next, we look at RLC circuit with V_r divided by V_i. We will get a bandpass filter.

So selectivity, we saw this in the context of radios. Selectivity has to do with how selective my filter is. Recall this is a band pass filter And selectivity says, how sharp is this band? How selective am I?

So one way of figuring out selectivity is to look at this ratio omega 0 by delta 0. Where delta 0 is called the bandwidth, it's measured at the point where the various parts of the curve reaches 1

by square root 2 of its highest value.

So higher omega 0 by delta omega, higher the selectivity. So the reason is that as delta omega becomes narrower, my filter becomes more selective.

For the next part, we need to find out what is delta omega. Absolute magnitude of V_r divided by V_i is one by square root 2. Eventually we will get a result of delta omega equal to R/L.

As 2 alpha is equal to R divided by L so omega 0 by delta omega is equal to Q. The lower the R, the sharper the peak, the more selective the filter. So high q implies high selectivity. From Q, you can tell all kinds of things about that circuit.

Q can also be defined as 2 pi times the energy stored in the circuit divided by energy lost per cycle in the circuit.

Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

We often build special circuits known as filters to eliminate that specific frequency of 50 to 60 Hz from the signal as AC in most countries are at 50 or 60 Hz. Otherwise we end up with a very low hum.

First, we need to plot the graphs of impedance against omega for capacitors, resistors and inductors.

For very low frequencies, the impedance of the capacitor is very, very high. The capacitor looks like an open circuit but for very high frequencies, the capacitor looks like a short circuit. So for the capacitor, we will be able to see the curve that shows its impedance.

For a resistor R, its behavior is frequency independent. So its impedance is simply R. And so it is

going to be a constant.

For the inductor, notice that for very low values of omega, the impedance is very low, and for high values of omega, the impedance is very high. Not surprisingly, the inductor looks like an open circuit to high frequencies and for DC and very low frequencies, the inductor looks very much like a short circuit.

So let's start with a CR circuit. If we plot the magnitude of H of omega of the transfer function for this filter, what does it look like?

So let's start with the high frequencies just for fun. For very high frequencies, note that the capacitor is a short circuit. And the resistor, of course, has a resistance R. So the capacitor is a short circuit and so Vi will appear directly at Vr. So for very high frequencies, my filter will give me Vi at Vr. So Vr divided by Vi will simply be unity.

What about for very low frequencies? At very low frequencies, the capacitor behaves like an open circuit. And if that behaves like an open circuit, then most of the voltage drop will be across the capacitor and very little will drop across the resistor and so I'm going to get a very low value. So low value for low frequencies and high values for high frequencies. So what we will get is a high pass filter.

These same principles apply for an RC circuit. So for an RC circuit, we will get a low pass filter.

For an RL circuit, at low frequencies, the inductor behaves like a short circuit so H of omega will have a low value. At high frequencies, the inductor behaves like an open circuit, so H of omega will have a high value. The RL circuit will be a high pass filter.

For an RLC circuit with the output voltage measured across the resistor, a bandpass filter is produced. At resonance, omega equals omega0 which is 1 by square root of LC and H of omega will be equal to 1.

For an RLC circuit with the output voltage measured across the capacitor and inductor, a bandstop filter is produced. At resonance, the output voltage is zero.

For an RLC circuit with the output voltage measured across the capacitor and inductor in parallel, a bandpass filter is produced. At resonance, the output voltage is zero. A major application of bandpass filters is in AM radio. The capacitor is made variable so that the bandpass filter can tune to various frequencies.

How does the radio work?

On the x axis of the graph is the frequency of the signal. On the y-axis, is the signal strength.

So the way radio system work is that different radio stations would be transmitting at different frequencies. So, for example, in the Boston area, there are a bunch of radio stations that transmit the following frequency. So for example, one of the radio stations is 1030 in the Amplitude Modulation band, And then,there may be other signals being transmitted by other radio stations. So for example, there may be a station transmitting at 1020, another at 1010 KHz and it goes on and on.

So when stations transmit, they try to maximize its signal strength, say around the 1030 frequency, ,it tries to maximize it there. And then it tries to make sure that the signal strengthdoesn't encroach too much into neighboring bands.So notice that each of these stations gets a roughly 10 kilohertz band.

So let's say, if I want to listen to WBZ News Radio, then I have to tune my capacitor here such that

the band-pass filter have its passing band focused where I care the most.

So when this is passed through, this band-pass filter will allow the 1030 range to mostly get through, and it will attenuate everything else that's further away but it is not perfect and so it will let through a little bit of the neighboring bands. That's why with AM radio you always get some interference from the sides of the two neighboring bands.

Fundamentally, that is a small value. Mostly the 1030 will come through. So this was the filter, the band-pass filter

Notice here that selectivity is important.

What is selectivity? Selectivity has to do with trying to capture the signal in a given range.

In the next video, we will see that the selectivity relates to something that you've seen before.

It will be shown how a RLC circuit, gives voltage values that are much higher than voltage values that are input. We have to look at Vc divide by Vi.. If the graph of Vc divide by Vi against omega is plotted, we will get a low pass filter.

We can do the maths with Vc over Vi using the voltage divider relation. We can write the impedance of the capacitor and divide it by the sum of all of the other impedance.

And for the magnitude, we can simply get the square of the real part plus the square of the

imaginary part square rooted.

You've taken it on faith from me that between these two points the function looks like the graph but that is not true.

Next we need to find out mathematically what happens at resonance. After some mathematical analysis, it can be seen that V_c is equal to Q times V_i. If Q is very large, V_c is more than V_i.

Next, we look at RLC circuit with V_r divided by V_i. We will get a bandpass filter.

So selectivity, we saw this in the context of radios. Selectivity has to do with how selective my filter is. Recall this is a band pass filter And selectivity says, how sharp is this band? How selective am I?

So one way of figuring out selectivity is to look at this ratio omega 0 by delta 0. Where delta 0 is called the bandwidth, it's measured at the point where the various parts of the curve reaches 1

by square root 2 of its highest value.

So higher omega 0 by delta omega, higher the selectivity. So the reason is that as delta omega becomes narrower, my filter becomes more selective.

For the next part, we need to find out what is delta omega. Absolute magnitude of V_r divided by V_i is one by square root 2. Eventually we will get a result of delta omega equal to R/L.

As 2 alpha is equal to R divided by L so omega 0 by delta omega is equal to Q. The lower the R, the sharper the peak, the more selective the filter. So high q implies high selectivity. From Q, you can tell all kinds of things about that circuit.

Q can also be defined as 2 pi times the energy stored in the circuit divided by energy lost per cycle in the circuit.

## Monday, June 13, 2016

### Circuits and Electronics - The Impedance Model (Lecture 20)

Video Lectures:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-17/

Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

A simpler way to get V_p is explored in this lecture. First divide the numerator and denominator by sC, and what is obtained is something that looks like a voltage divider relationship. 1/sC can be replaced with Z_c which will make it look more like one.

Next looking at the impedance models of the resistor, capacitor and inductor, we can see how the impedance is part of Ohm's law where Z_c and Z_L are the impedances for a capacitor and an inductor. For a drive of the form Vi e raised to st, the complex amplitude, Vc of the response, is related to the complex amplitude Ic algebraically by a generalization of Ohm's Law.

Looking at the RC circuit, and replacing the capacitor with the impedance model Z_c, we will see something interesting when finding V_c. We will find that V_c is the famed complex amplitude V_p that we have been trying to derive except the method now is much simpler. All circuit methods can be used to further analyze the circuit.

Signal notation is discussed-there are four of them including the complex amplitude notation.

Here is a summary of the impedance method: -

(1) First step we replace the sinusoidal sources by their complex or real amplitudes.

(2) As a second step, we replace circuit elements by boxes. These boxes are the impedance boxes, or impedances.

(3) Determine the complex amplitudes of the voltages and currents at the various node points and the branches.

(4) The fourth step is not really necessary - obtain the time variables from the complex amplitudes.

For the series RLC circuit, we have to find out what is V_r.

Next we try to get the magnitude of the transfer function of V_r/V_i. To get the frequency response, sketch the graph of magnitude against omega. The way to do this is to figure out the asymptotes. We will need to find out the values of the magnitude for small and high values of omega. We also need to look at the value of the magnitude where a certain part of the equation goes to zero.

So when omega is very small, what happens? So when omega is very small, , and similarly, an omega squared values can be ignored. So what we are left with is approximately omega RC.

Now what does omega RC look like? Of course, at omega equals 0, it will be 0 at first. But for very low values of omega, as we increase it, it begins to go up in a linear manner.

Next, what happens for large values of omega? So when omega is very large, then 1 can be ignored in relation to omega squared LC. So we get approximately R divided by omega L for very large values of omega.

.

So what happens at that omega equals 1 by square root of LC? We will get a value of 1.

When we sketch the graph, we will get some kind of bandpass filter which allow signals to pass within a certain band of frequencies.

Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

A simpler way to get V_p is explored in this lecture. First divide the numerator and denominator by sC, and what is obtained is something that looks like a voltage divider relationship. 1/sC can be replaced with Z_c which will make it look more like one.

Next looking at the impedance models of the resistor, capacitor and inductor, we can see how the impedance is part of Ohm's law where Z_c and Z_L are the impedances for a capacitor and an inductor. For a drive of the form Vi e raised to st, the complex amplitude, Vc of the response, is related to the complex amplitude Ic algebraically by a generalization of Ohm's Law.

Looking at the RC circuit, and replacing the capacitor with the impedance model Z_c, we will see something interesting when finding V_c. We will find that V_c is the famed complex amplitude V_p that we have been trying to derive except the method now is much simpler. All circuit methods can be used to further analyze the circuit.

Signal notation is discussed-there are four of them including the complex amplitude notation.

Here is a summary of the impedance method: -

(1) First step we replace the sinusoidal sources by their complex or real amplitudes.

(2) As a second step, we replace circuit elements by boxes. These boxes are the impedance boxes, or impedances.

(3) Determine the complex amplitudes of the voltages and currents at the various node points and the branches.

(4) The fourth step is not really necessary - obtain the time variables from the complex amplitudes.

For the series RLC circuit, we have to find out what is V_r.

Next we try to get the magnitude of the transfer function of V_r/V_i. To get the frequency response, sketch the graph of magnitude against omega. The way to do this is to figure out the asymptotes. We will need to find out the values of the magnitude for small and high values of omega. We also need to look at the value of the magnitude where a certain part of the equation goes to zero.

So when omega is very small, what happens? So when omega is very small, , and similarly, an omega squared values can be ignored. So what we are left with is approximately omega RC.

Now what does omega RC look like? Of course, at omega equals 0, it will be 0 at first. But for very low values of omega, as we increase it, it begins to go up in a linear manner.

Next, what happens for large values of omega? So when omega is very large, then 1 can be ignored in relation to omega squared LC. So we get approximately R divided by omega L for very large values of omega.

.

So what happens at that omega equals 1 by square root of LC? We will get a value of 1.

When we sketch the graph, we will get some kind of bandpass filter which allow signals to pass within a certain band of frequencies.

## Sunday, June 5, 2016

### Circuits and Electronics - Sinusoidal Steady State (Lecture 19)

Video Lectures:- ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-16/

Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

One of the most important reasons why we care about sinusoids is that signals can be

represented as sums of sinusoids. The technique that can transform any wave form into a

sum of sinusoids representation is called Fourier analysis. Fourier series analysis can show that

it can be represented as a sum of sinusoids. Circuits have to be linear for this to happen.

Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/

One of the most important reasons why we care about sinusoids is that signals can be

represented as sums of sinusoids. The technique that can transform any wave form into a

sum of sinusoids representation is called Fourier analysis. Fourier series analysis can show that

it can be represented as a sum of sinusoids. Circuits have to be linear for this to happen.

The response of circuits to sinusoids as a function of frequency is called the frequency

response of the circuit. As the input frequency of the amplifier increases, the amplitude of the output will decrease. Not only that, the phase will also change.

A very simple circuit example with a sinusoidal input is used - an RC network- a series connection of a resistor and capacitor. This mimics the input of the RC circuit that is part of the amplifier. So the amplifier looks like this- it has a gate capacitance, CGS. GS, and a the resistor, R. which is some parasitic resistance of the wires.

The input voltage is equal to some amplitude VI cosine of omega t, and for t greater than or equal to 0. There will be three ways of approaching the problem - the first being the most difficult - using differential equations. The third being the easiest - almost totally done by inspection.

So the first step, as is our usual practice, is to set up the differential equation by the node method.

So the current leaving the node in this direction is v minus v_I divided by R. And the current heading down this direction is C dv/dt. And the currents must sum to 0 by the node method.

So multiply the whole thing, both sides by R, and shuffle things around. So we get RC dv dt. plus v minus v_I equals 0. So what I want to do is move v_I to the right-hand side and write it like this.

And then I've been given that v_I is V_i cosine of omega t.

So now I am ready for the second step, which is to find the particular solution to the sinusoid, v_P.

This is where things become messy and is not a path to be taken. Instead we will try an exponential input as it gives rise to a nice solution. However we are not sure how this would relate to the answer at the moment.

So for that sneaky input V_i e raised to st let's go back and do our usual thing. We'll try a solution of v_PS ( where s stands for sneaky) given by V_p e raised to st. We will eventually get a solution where V_p is equal to V_i divided by 1 plus sRC. Replacing s with j omega, v_PS will be V_i divide by 1 plus j omega RC times e raised to j omega t. V_i divide by 1 plus j omega, RC is the complex amplitude.

Based on Euler Relations, the real part of the sneaky input is equal to the the input, V_i cosine of omega t. Then based on the inverse superposition argument the real output can be found by

taking the real part of the sneaky output. So V_p would be V_i divide by 1 plus j omega RC. But first we will try to work out the magnitude and phase of the complex number expression in the bracket. With sound knowledge of complex numbers, the particular solution can be found.

Next would be to find the homogeneous solution which would be v_H equals A e raised to minus t

divided by RC.

The fourth step would be to find the total solution which would be v_P + v_H. With the initial conditions A can be found.

The particular solution is the sinusoidal steady state which matters more than the transient state. Steps 3 and 4 were not relevant.

A block diagram of the approach is shown and then after that a summary but there is a simpler approach by inspection.

A magnitude plot and a phase plot can be drawn for the output. This is the frequency response.

Finally a summary of lecture 19 and a preview of what is coming up next.

A very simple circuit example with a sinusoidal input is used - an RC network- a series connection of a resistor and capacitor. This mimics the input of the RC circuit that is part of the amplifier. So the amplifier looks like this- it has a gate capacitance, CGS. GS, and a the resistor, R. which is some parasitic resistance of the wires.

The input voltage is equal to some amplitude VI cosine of omega t, and for t greater than or equal to 0. There will be three ways of approaching the problem - the first being the most difficult - using differential equations. The third being the easiest - almost totally done by inspection.

So the first step, as is our usual practice, is to set up the differential equation by the node method.

So the current leaving the node in this direction is v minus v_I divided by R. And the current heading down this direction is C dv/dt. And the currents must sum to 0 by the node method.

So multiply the whole thing, both sides by R, and shuffle things around. So we get RC dv dt. plus v minus v_I equals 0. So what I want to do is move v_I to the right-hand side and write it like this.

And then I've been given that v_I is V_i cosine of omega t.

So now I am ready for the second step, which is to find the particular solution to the sinusoid, v_P.

This is where things become messy and is not a path to be taken. Instead we will try an exponential input as it gives rise to a nice solution. However we are not sure how this would relate to the answer at the moment.

So for that sneaky input V_i e raised to st let's go back and do our usual thing. We'll try a solution of v_PS ( where s stands for sneaky) given by V_p e raised to st. We will eventually get a solution where V_p is equal to V_i divided by 1 plus sRC. Replacing s with j omega, v_PS will be V_i divide by 1 plus j omega RC times e raised to j omega t. V_i divide by 1 plus j omega, RC is the complex amplitude.

Based on Euler Relations, the real part of the sneaky input is equal to the the input, V_i cosine of omega t. Then based on the inverse superposition argument the real output can be found by

taking the real part of the sneaky output. So V_p would be V_i divide by 1 plus j omega RC. But first we will try to work out the magnitude and phase of the complex number expression in the bracket. With sound knowledge of complex numbers, the particular solution can be found.

Next would be to find the homogeneous solution which would be v_H equals A e raised to minus t

divided by RC.

The fourth step would be to find the total solution which would be v_P + v_H. With the initial conditions A can be found.

The particular solution is the sinusoidal steady state which matters more than the transient state. Steps 3 and 4 were not relevant.

A block diagram of the approach is shown and then after that a summary but there is a simpler approach by inspection.

A magnitude plot and a phase plot can be drawn for the output. This is the frequency response.

Finally a summary of lecture 19 and a preview of what is coming up next.

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