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.
Thursday, August 25, 2016
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.
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