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.
