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0:00
Let's talk about Mosfets. Here's what the circuit symbols look like. Now, the one on the left is what engineers often write when they're writing a MOSFET by hand. The circuit symbol on the right is what you'll often see in circuit simulation software. You could say that it's a more accurate or more representative circuit symbol for an n-channel MOSFET, but sometimes you'll see these two symbols used interchangeably. Sometimes you'll also see a similar circuit
0:30
symbol but with little breaks there between the drain and the source. In any case, these are all examples of n-channel MOSFETS. We have the gate on the left. We have the drain up top and we have the source down at the bottom. It can be a little confusing when you look at the source to identify the source because you can't really remember, "Does the arrow go from the left to the right or the right to the left?" Anyway, you just kind of have to remember how the circuit symbol looks.
0:58
Between the gate and the drain and the source is a channel. And the way the MOSFET basically works is you apply a voltage to the gate, which then creates a conducting channel that connects the drain to the source if the voltage is high enough. Let's look at an example circuit that involves an n-channel MOSFET. So in this circuit we have a 10V power supply up at the top, a drain resistor RD and then we have a voltage source over here to give us whatever gate voltage we want.
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What I'm going to do now is to make a plot of the drain current versus the gate-source voltage. Here is our drain current right here. And what we're interested in knowing is how that drain current changes as a function of the gate voltage. One of the key parameters of a MOSFET is the threshold voltage. It's very similar to this turn-on voltage of 0.7 volts in a bipolar junction transistor. In MOSFETs, though, the threshold voltage can change from transistor to
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transistor depending upon how it was designed. In any case, the threshold voltage is a key parameter. It basically tells you whether or not the transistor is going to be on or off. If the gate voltage is lower than the threshold, it'll be off. If the gate voltage is above threshold, then the transistor starts to turn on and I'm using the word "starts" to turn on, because it's not a very linear relationship between the drain current and the gate source
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voltage. For this particular transistor, we can identify the threshold voltage right here on the curve. If our voltage is too low, no drain current is flowing. As soon as the gate voltage starts to exceed the threshold voltage, then we start to get some current flowing through the transistor. But it's not a line, is it? It has kind of a curved shape. In fact, this line is parabolic.
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The curve that relates the drain current and the gate-source voltage in a MOSFET can be modeled by a simple equation. If we just subtract the gate-source voltage and threshold voltage, square it, and multiply it times a constant, then we wind up with the drain current. We could think of the gate voltage as the input, if we're trying to build an amplifier, for example. And we could think of the drain current as the output. Normally when we're designing amplifiers,
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we want a linear relationship between the output and the input, and we don't have it here, do we? However, if we don't have a very large variation in our input, then it's almost linear. Let me show you what I mean. Let's imagine that we're going to operate our transistor somewhere around this point where the gate source voltage is about five volts. If I then put a signal like an AC signal on our gate and add it to the DC signal that's already
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there, then we might move up and down this curve a little bit. We can imagine a sinusoidal variation, for example, around our five volts. As long as our variation is not too small, then the curve kind of looks like a line. It approximates a linear relationship. For instance, let's assume that we have an input source that varies between 4.9 volts and 5.1 volts. A small 0.2 volt variation. How does it look at the output side? Well,
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let's go ahead and graph it. We actually have two curves here plotted one on top of the other. I have a blue curve that represents our input sinusoidal voltage, and I have a purple curve representing our output drain current. It's hard to see that there are two curves here, isn't it? They're overlapping each other because the output is just a scaled version of the input. Nice. It's a linear amplifier. But what happens if my input voltage swing is a little bit more?
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What happens if I go all the way down here and all the way up here? Well, then it's kind of a curve instead of a line. Is my output current swing going to be simply a scaled version of my input voltage swing? Well, no. In this case, my input is a perfect sine wave. You can see the blue curve that I have plotted here and refer to the left axis. The drain current that comes out though, is not a perfect replica of my input.
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In other words, this is not a linear amplifier. There's some distortion that this transistor has added to the circuit. This is one reason why MOSFETs might not be as good at making linear amplifiers as bipolar junction transistors. In bipolar transistors, the collector current is beta times the base current. That's a linear relationship. Here we don't exactly have a linear relationship unless our voltage swings are small. That's certainly something to keep in mind if
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you're considering whether to use a MOSFET versus a bipolar junction transistor. Now, MOSFETs do have certain advantages over bipolar junction transistors. One of the key advantages is that there is virtually no gate current that flows towards a MOSFET. There's an insulator between the gate and the channel of a MOSFET, so the only gate current that actually flows is leakage current, and that's usually really small. It's not like the situation in a bipolar
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transistor where you might have substantial base current flowing. Let's take a look at the typical data sheet for a MOSFET. As I mentioned, one of the key parameters that you should be on the lookout for when you're considering a MOSFET is the gate-source threshold voltage. Take a look at the variations that you can expect though. For a single MOSFET type, the data sheet tells us that the threshold current might be anywhere between 0.8 balls and three volts. That can make MOSFETs
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a little bit difficult to use. With a bipolar transistor, you know that around 0.7 volts, that transistor is going to turn on. With a MOSFET, even if you choose the same MOSFET, well, it might turn on at 0.8 volts, it might turn on at 2.1 volts, it might turn on at three volts. We don't know, it might be random. So whatever circuit we designed, we're going to have to accommodate different threshold voltages. If the threshold voltage changes, it effectively means that this
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entire curve shifts from left to the right. Our circuit has to be able to deal with it. Another parameter that we're usually interested in knowing when we think about MOSFETs is the slope of this line. For example, right here at 200 milliamps, this is almost linear and it has a particular slope. That slope is called the transconductance, and that's another key parameter that describes how a MOSFET is going to perform. On this data
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sheet, it's called the forward transconductance, and quite often the transconductance is labeled with a little g. Since the slope of this line is current divided by voltage, it means that the transconductance has units of inverse ohms or Siemens. On the datasheet, the transconductance was measured at a drain current of 200 milliamps. So it means we're taking the slope of the line at this single point. The slope of the line can vary anywhere between
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about 80 and 320 mS. If you know the threshold and the transconductance at a single operating point, you can derive that equation that I showed you just a moment ago. Let's work a few simple examples to get the hang of working with these MOSFETs. First of all, take a look at this circuit. Is this transistor on or off? Well, the voltage here is a half a volt, and we know from the datasheet that the threshold voltage is
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going to be between 0.8 and 3 volts. We're below threshold no matter what. Therefore, this transistor is always off. There's not going to be any current flowing from drain to source. Let's look at another example. What if we put 3.1 volts at the gate? Well, if we just look at the range of the threshold voltages, 0.8 to 3 volts, we say, "Oh! We're over the threshold, therefore the transistor is on". Yes, but it's just barely
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on, and you're not going to get very much current flowing down through the drain. Let's go back for just a moment and look at that curve. If the transistor is just barely on, the slope of that line or the forward transconductance is going to be really low; that's not very good. So we could say, yes, it's on, but because it's not on very much, it's not good. So let me just say "barely on." What if we put eight volts at the gate? Well,
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then it's very clear this transistor is on. If we're trying to use the transistor as a switch, then this is perfect. What about 20 volts? Okay, it's still on, isn't it? But it can't really turn "more on" if it's already on. The gate creates a conducting channel between drain and source, but once the channel is on, you can't somehow make it more on once it's already on. Having 20 volts there is probably not significantly different than having 8 volts there at the gates.
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Well, now go back to a very simple circuit involving a MOSFET and I'm going to put a voltage source at the gate and I'm going to attach the drain to another voltage source. Depending upon the voltage applied at the gate, we get different curves that relate to the drain current to the drain voltage. Imagine for a moment that I put 2.5 volts at the gate. It means that we're going to be operating along this yellow curve right here. The 2.5
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V curve. As long as you keep the voltage here on the gate maintained at 2.5 volts, then you can expect a relatively constant drain current regardless of the drain voltage that you apply to that transistor. That means that the MOSFET functions in this region almost like perfect current source. Over here at the left side though, we see that the curve kind of bends down. Therefore, we can almost identify two distinct regions of operation
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when it comes to these curves. Let's go ahead and label them. I'll kind of draw a line to divide these two regions because they have different names. The region to the left of that red line that I've just drawn is the Ohmic region. The region to the right of that red line is called saturation. For completeness. I can also label a third part of the curve, which corresponds to this green line down at the bottom: Off. We can see for that green line in
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particular that a gate-source voltage of 1.9 volts is insufficient to turn the transistor on. Therefore no current flows through the drain and we're on the Off region of the curve. Take a look at the Ohmic region for just a moment. Notice how the slope of all of these lines changes with the applied gate voltage. That's kind of interesting because it means that the transistor is almost behaving like a resistor where the resistance
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is controlled by the gate voltage. That's why it's called the Ohmic region. It's like a voltage-controlled resistor. Over here in the saturation region, that's usually where you want to be if you're actually using the transistor, though. In this region of the graph, the drain current is a function only of the gate voltage. It's not a function of the drain voltage, is it? Those curves are flat. The reason why we want to operate in this region when we're designing
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amplifiers is because we want our output or the drain current to be a function only of our input, the gate voltage. We don't want it to depend on the drain voltage, do we? Now, before I move on, I want to quickly draw a comparison of the MOSFET's curves with those of a bipolar transistor, because there is an unfortunate naming convention that many people find confusing. Notice that this region of the MOSFET curves is called "saturation", but with a bipolar transistor,
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saturation kind of corresponds to what I would call the Ohmic region for a MOSFET. That's unfortunate because sometimes it can be difficult to keep the two apart. With a bipolar transistor, we want to operate in the forward active region. That means the transistor is on. With a MOSFET, we want to operate in the saturation region. That means the transistor is on. Let's take a look at a typical datasheet for an N-channel enhancement mode MOSFET.
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In a moment, we're going to be working an example and I want to use this particular transistor in my example. So for now, we're going to take note that we could expect with this MOSFET, a gate-source threshold voltage that varies between about 1 volt and 2.5 volts.
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We can also take note of the minimum and typical forward transconductance of this transistor. And on another part of the datasheet, it's mentioned that the maximum drain current for this transistor is about 280 milliamps. We'll keep that in mind, even though I'm not showing it here. Let's now work an example. We're going to be biasing this transistor so that it's definitely in the saturation region with no danger of burning out.
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First, let's label our gate-source voltage on the circuit diagram. We know that this transistor is going to be on if the gate source voltage exceeds the threshold voltage. We know that the threshold voltage might be as high as 2.5 volts. It means that we should make sure that our gate voltage is definitively higher than 2.5 volts to make sure that transistor is on. The example problems says that we want to make sure that the transistor
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is operating in the saturation region. Let's take a look at these curves from the datasheet and identify the saturation region. Well, this is the Ohmic region to the left, isn't it? So we don't want to be over here. Saturation is on the right. That's what we want. That's our second condition. For our third condition, let's make sure that the transistor has no danger of burning out. For that, I'm going to say that the drain current
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should be 200 milliamps. I want to make sure that it's on, but I don't want to exceed the maximum allowable drain current of 280 mA. It means we're going to be operating somewhere along this line. That's where the drain current equals 200 milliamps. Four: Let's make an observation. If we're trying to make sure we're in the saturation region rather than the Ohmic region, a higher drain voltage is better. Let's start plugging in some numbers and see what comes out.
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Let's start off with a trial gate voltage of five volts. That's definitely higher than 2.5 volts, so hopefully that will turn everything on. With a gate voltage of five, we can assume that the transistor is on and I'm going to write the threshold voltage. If the gate source voltage is higher than the threshold voltage, then a lot of drain current will flow and then the source voltage will get higher until it's very nearly equal to
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the threshold voltage again. Let's label the source voltage right here because if we take 5 and subtract 2.5, then we wind up with 2.5 volts there at the source. To maintain our drain current of 200 milliamps, we're going to have to choose a source resistance of 2.5 volts divided by 0.2. That gives us 12.5 ohms. However, the calculations that I just carried out assume that the threshold voltage was 2.5 volts. What happens
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if the threshold voltage of that particular transistor happened to be 1 volt? I'm not going to know ahead of time because these transistors have a lot of component-to-component variation. If the threshold voltage is 1 volt, then our drain current is going to be a little bit different. It's going to be the new source voltage divided by our source resistance, which we've already chosen. The source voltage is just going to be my gate voltage minus the threshold voltage. Our
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gate voltage is 5 and our threshold voltage is 1. We've already picked this resistor of 12.5 Ohms. This gives us a drain current of 320 milliamps. Unfortunately, that exceeds the maximum allowable drain current for this transistor. The transistor would burn out. That's bad. What did we do wrong? Well, we started by assuming that the threshold voltage is 2.5 volts. Instead, we should have started by assuming that the threshold voltage was smaller and then checked
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the large case. Let's erase and rework it. Once again. let's start with a gate voltage of five volts, which I can very easily arrange by picking appropriate values for R1 and R1. For example, I could make them each a megaohm. If the threshold voltage is one volt, then our source voltage is going to be very nearly equal to threshold when the transistor is on. This makes our source voltage about five minus one, or four volts. Once again,
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assume that our drain current is going to be 200 milliamps at that point. We can calculate our source resistor. It's going to be Vs over the drain current. In this case that comes out to be 20 ohms. Let's now find our drain voltage. Our train voltage is 10-Id*10. This is eight volts. Our drain-source voltage is just our drain voltage minus our source voltage. Eight minus four equals four volts. We're operating over here well into the saturation region, so that's good.
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Now let's consider what would happen if our threshold voltage were not one volt, but rather 2.5 volts. If the threshold is 2.5 volts, then when the transistor is on, the gate-source voltage is going to be nearly 2.5 volts as well. With our gate voltage of five, we would then expect a source voltage of 2.5 volts. Let's find our drain currents. Our drain current is going to be our source voltage divided by our source resistor, which we've already chosen.
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We wind up with 125 milliamps. That's lower than the maximum allowable drain current. So we're okay here. Are we still in saturation? Well, our drain voltage is going to be ten minus the drain current, which is now 0.125 amperes times our ten ohm resistor; we get 8.75 volts here for the drain voltage. And if we subtract it from our source of 2.5 V, then we wind up with 6.25 volts for our drain source voltage. We're off the curve over here on the right.
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So we're definitely still in the saturation region. We've now met all of the requirements of the example. Let's move on to example two. If these are single device measured curves, then which drain resistor will result in a particular drain-source voltage and drain current? Then we're going to draw the load line. So we're now assuming that these are the actual curves measured for this particular MOSFET. What you don't want to do is to look at the
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data sheet - and you'll always see example curves like this - you don't want to look at a data sheet and then assume that those curves apply to every single transistor that you might pull out of a drawer in the lab. These curves here in this example though, are for that particular transistor. So we know that they're accurate. Since we see here in the example that the drain source voltage is 5.5 and the drain current is 60, we know we're operating right here
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at this point on the graph. Let's apply Ohm's Law across the drain resistor. We have 8 volts minus our drain-source voltage drop and that's going to be I*R. Let's now solve for the drain current. I've written the equation this way because it's just the equation of a line. The line has a slope and an intercept, but it's very easy to find it. If the transistor is fully off, then the drain-source voltage ought to be 8 volts. This means that no current
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is flowing and the full power supply voltage appears across the MOSFET. We can connect these two points together to give us our load line. The equation that I've just written down is in fact the equation of this line. We're told on the problem that the drain current is 60 milliamps when the drain-source voltage is 5.5 V. If we plug those two numbers into this equation, then we can just calculate what RD is. Solving this
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equation yields a drain resistor of 41.7 ohms. That value can then be substituted into the equation for our line to find other values of the drain current if we were to hypothetically change the drain-source voltage. If we wanted to operate this transistor as an amplifier, I could add an AC component to my gate voltage. This means that I would be sliding up and down that load line. Some change in the input of our gate voltage would wind up as a change in the drain current,
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or equivalently, a change in the drain voltage. Let's move on to example three, which is very nearly the same circuit. The only difference here is that we're including a large load resistance of 40 ohms. Of course, the DC characteristics of this circuit are more or less unchanged. Let me go ahead and draw on the load line that we previously calculated. That's the DC load line. You'll notice now that I've actually indicated the AC component that we
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plan to add to our DC source here. If we're going to use the transistor as an amplifier, then the AC component of the source would show up at the gate, cause some change in the gate voltage, which should then show up as a change in the drain voltage and the drain current. The slope of our DC curve is just minus one over RD; that's minus one over 40. However, now we're going to have two resistors that can only be seen by the AC component. From the perspective
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of the AC signal, there's two resistors attached to the drain of that MOSFET. There's one going to an AC ground, and then there's another going to an AC ground as well. They show up as resistors in parallel, and if they're both 40 ohms, then the slope of the new load line - the AC load line - is going to be minus one over 20. The line should be centered at the DC operating point. Let's go ahead and draw the new load line. What this load line means is
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that a small change in the gate voltage - as long as the frequency is high enough so that that capacitor doesn't behave like a high pass filter... Then a small change in the gate voltage will drive us up and down that gray line rather than up and down the black line. This is how a MOSFET can be used as an amplifier. We've just shown that small changes in the gate voltage can lead to more substantial changes in the drain current or the drain voltage.
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