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Offset voltage in op-amps (Amplifiers #8) | Aaron Danner | YouTubeToText
YouTube Transcript: Offset voltage in op-amps (Amplifiers #8)
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This content explains the concept of op-amp offset voltage, its origins, its detrimental effects on circuit performance, and how to correct it using the dedicated offset null pins.
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If you look at the pinout of an op-amp, you'll see two pins labeled "offset null".
That's what I'm going to be talking about here in this video. The offset null is
used to adjust something called the offset voltage. It's an imperfection in op-amps,
and there are certain circumstances where you need to correct for it.
First, I'd like to talk about what it is and the problems it causes.
Then I'll describe where it originates, and I'll tell you exactly how you can fix it.
If we tie the two input pins of an op-amp together, we expect the output to be zero.
That's what we want from an ideal op-amp. A real op-amp, though, might not give us exactly zero
volts at the output. That's bad, of course because in an ideal differential amplifier,
you would need to have that output equal to zero if there's no difference between the voltages
at the two input pins. We're going to model this imperfection as an offset voltage. You
might notice here that I'm drawing a little voltage source there at the input pin of
an op-amp. This is just a mathematical model of an imperfection that's inside the op-amp.
What we're doing here is modeling the imperfection as an input voltage that's
required to zero the output. In other words, when you go to use the op-amp in an actual circuit,
you're going to find that the output voltage is not exactly zero, even if the two input
pins are tied to one another. However, if you were to put a little bit of voltage
between the two input pins and adjust it to the appropriate value, you might be able to
zero it out. That's what we're modeling right here. With the appropriate Vos,
we can force the output to be equal to zero. That's the definition of the offset voltage.
Now, I'm going to talk about the effect of this offset voltage on some actual circuits. First,
let's look at an inverting amplifier. This is the standard configuration,
and I'm going to first assume that the op-amp is offset-free. In other words,
it's basically an ideal op-amp. With an input voltage of 10 mV, let's find the output voltage.
If everything is perfect, you might recall that the gain is just this resistance divided
by this resistance, so we expect the gain to be 100 here. But I'm going to go through
the derivation because we won't be able to make that assumption in the next viewgraph.
Because this pin is grounded, it's at 0 volts. Because of the negative feedback,
we expect this voltage also to be 0 volts. Because the op-amp is offset-free,
I then expect this voltage also to be 0 volts. I'd like to label this current i1 and this current I2.
Although I'll later be assuming that the op-amp is not quite perfect, I'm still going to assume
that we're not eating up any of that current at the input pin of the op-amp. In other words,
I'm going to assume that i1 equals I2. We can use Ohm's law across the source
resistor to find i1. We have 10 mV at this side and we have 0 volts at the other side.
We can also use Ohm's law across the feedback resistor Rf in order to find i2. We have 0 volts
on the left side of it and we have the output voltage on the right side of it. Let's bring the
100 up here to the left half of the equation, and we're left with -Vout over there at the
right. The output voltage is just -1 V. Since the input voltage was 10 mV, I can conclude that the
gain was -100. That's exactly what we expect by looking at these two resistors in this circuit.
Of course, this circuit contained an op-amp that was offset-free. What I'm
going to do now is assume that we have the same circuit, the same amplifier,
but we have an op-amp that's not offset-free. How will that affect
the gain? We're going to assume an offset voltage of 4 millivolts. It's still a pretty good op-amp,
but it's not perfect. How will this 4 millivolts affect our gain?
Well, this pin is grounded, so it's at 0 volts. Because of the negative feedback,
this voltage should also be 0 volts. However, because the op-amp is not offset-free,
we're going to have 4 millivolts at this node. Just as I did with the last problem,
I'm going to label this current i1, and I'm going to label this current as i2. i1 is just
going to be equal to i2. And we can use Ohm's law in order to find both of these currents. We
need to apply V/R for Rs in order to find i1. We have 10 mV to the left and 4 mV to the right. So,
there's our ΔV, and our resistance here is 1k. For i2, I see that we have 4 mV to the left of
the resistor, and we have our output voltage Vout to the right. The resistance here is 100k.
Let's move the 100k to the left side of this equation. The thousand cancels out,
and we're just left with 100. We have 600 millivolts on the left side of this equation. So,
when I subtract the 4 millivolts on the right, we wind up with 596 millivolts. And that's equal
to -Vout. Since our input voltage was just 10 mV, I can conclude that our gain is 59.6.
If we just look at the two resistors, 100 kiloohms and 1 kiloohm, we might conclude
that the magnitude of the gain ought to be 100. But the gain that we just
found out is only 59.6. Therefore, the magnitude of the gain has been reduced,
and the reason it's been reduced is because of the presence of that imperfection in our op-amp.
That imperfection originates during the fabrication of the op-amp. The op-amp is
kind of a complicated amplifier. It doesn't just have a single transistor inside it;
it has a number of different transistors. And anytime you fabricate something lithographically
using silicon, you're going to have variations from component to component. To make a very
good differential amplifier, you need to have these components very well matched,
and that can't always be done perfectly. This is mainly due to doping variations,
lithographic errors; there can be packaging and stress issues as well.
I'm showing an example circuit on the right. This is not an op-amp;
it's just a very simple differential amplifier because I want to illustrate why the component
mismatch can cause an offset. Here, we have two MOSFETs, two resistors, and I'm assuming
that things are biased down here with a current source. Since this is a differential amplifier,
it should amplify any difference at the input pin. So, I'm just going to
tie these two input pins together so that there's no difference to multiply at all.
In this situation, the output voltage ought to be zero. Let's consider why it ought to be zero
if these two MOSFETs are well-matched. From the biasing considerations, we can see that both of
these amplifiers have the same source voltage. Because they also have the same gate voltage due
to the fact that we've tied these two input pins together, then they have the same gate-source
voltage. Because they have the same gate-source voltage, they should have the same amount of
current passing down into the drain. With the same amount of current passing down through the drain,
that current has to go through these two resistors. And if they're well matched, that is if
R1 equals R2, the output ought to be zero because the voltage here ought to equal the voltage here.
Now let's talk about what might make that not work. Why might the output voltage not
be zero? Well, if R1 is not perfectly matched to R2, the voltages are not
going to be equal to one another, because even if these two currents are the same,
the voltage drop across resistor R1 won't be exactly the same as the voltage drop across R2
unless the resistors are equal to one another. So, the mismatch in resistances can cause a problem.
But I hope you would take note that any mismatch in the MOSFETs would also result in a change
in the voltage. Even if the gate-source voltages in these two MOSFETs are the same,
some small variation in conditions from one MOSFET to another might then cause the drain
current for MOSFET M1 to be slightly different from the drain current of MOSFET M2. Therefore,
even if resistor R1 were hypothetically well-matched to resistor R2, we still might
have a nonzero output voltage due to problems inside the MOSFETs here. So, this is just meant
to give you some idea of where an offset voltage might originate in the design of an amplifier.
Of course, we hope that doesn't happen, but it can't be completely avoided. All
things being equal, the problem tends to be worse in devices using MOSFETs compared to
devices using bipolar junction transistors. So, with bipolar junction transistors,
you might end up with offset voltages on the order of a few hundred microvolts,
but with CMOS (MOSFETs), you might end up with something that's 10 to 100 times
worse. It's more difficult to match things in MOSFETs than it is to match things in BJTs.
Let's take a look at a few more circuits to get some idea of the kinds of effects
that the offset voltage can cause, and then I'll show you how to fix it.
First, let's look at two amplifiers that you're probably already familiar with:
a non-inverting amplifier and an inverting amplifier. In order to isolate the effect of
the offset voltage, let's first look at it when the input is just connected to ground. That is,
we're not going to amplify any signal at all. Under this scenario,
we would expect that the output voltage be 0 volts for both of these amplifiers.
What we're going to do next, though, is to insert the offset voltage,
or our mathematical model of it, into both of these circuits in order to see
what the output voltage would be due to the offset voltage. So, where am I going
to put it in these circuits? We can insert the offset voltage in front of either pin,
but for consistency here, I'm going to put it at the non-inverting pin in both of these circuits.
Of course, we no longer expect the output to be at 0 volts. Our job now is to find
out what it is. If you look for a moment at these two circuits,
I hope you'll realize that the circuits are now exactly the same. Look at the non-inverting pin
of the circuit here on the right. It's exactly the same as what we have connected to the pin
here for the circuit on the left. Likewise, we have resistor R2 providing feedback between
the output and the inverting pin. That's exactly what we have over here as well.
Finally, we have resistor one connected to ground. That's what we have here as well.
Since the two circuits are exactly the same, I only need to do the analysis
once. Let's go ahead and proceed with that analysis to determine the output voltage.
I've simplified the circuit drawing just a little bit to get us started,
but this circuit again represents both the inverting and the non-inverting amplifier when
we're isolating the effect of the offset voltage. This pin is grounded, so it's at 0 volts. The
voltage here is then Vos. Because of the negative feedback, and the amplifier being well-behaved,
we would also expect this voltage to be Vos. Let's label this current i1,
and this current i2. We know that i1 ought to equal i2 and we can use Ohm's law to find both
of these currents. We have 0 V to the left, and Vos to the right. There's V / R for i1.
Now let's do V / R for i2. We have Vos to the left and Vo to the right. Let's get rid of the
0 and move the R2 over to the left side of the equation. Let's now solve this for the
output voltage. What we've just found is that the output voltage is some constant value times the
offset voltage. And if you look at the constant here that's multiplied by the offset voltage,
you'll see that it's 1 + R2 / R1, so it's always a number that's greater than one.
That might not be a problem if the signal you're trying to amplify when you use the op-amp is not
a very small signal. But that's not always true. Sometimes the reason why you use an
op-amp in a circuit is precisely because you want to amplify a small signal. If the
signal you're trying to amplify when you use the op-amp is a fairly large signal to start with,
then this may not be much of a problem. But quite often the reason why you've chosen an op-amp - a
differential op-amp that is - to start with is precisely because you're trying to amplify a
very small signal. So if your small signal is very close in magnitude to the offset voltage itself,
then you could have a problem. We'll see how to fix that here in just a moment.
But first, let's look at another circuit. I'm showing both the differentiator and
the integrator circuit, and I've talked about both
of these circuits in another video. If we wanted to find the effect of the offset
voltage on the performance of either of these circuits, how would I do it?
Well, the first thing I would do is include the mathematical model of the offset voltage
here in our circuit diagrams. There we go - I've put it right in front of the non-inverting pin.
In order to isolate the effect of the offset voltage on the output voltage,
I would ground our input. Since I'm modeling the offset voltage here as a perfect voltage source,
and because that's a linear circuit element, the results that I'm calculating now are entirely
accurate, even if I have something connected to the input of either of these two amplifiers.
You might recall that you can use superposition when you're dealing with
linear sources. Let's now determine the effect of the offset voltage on
the integrator. We can leave the differentiator for another time.
I've redrawn the circuit here just a little bit, but basically, we just need to find the
output voltage in terms of Vos. Because this voltage is Vos, we also know that
this voltage is going to be Vos as well. Let's label the voltage across the capacitor as V,
and I'm going to label the current here as i1 and the current through the capacitor as i2.
Let's recall how current and voltage are related in a capacitor: the current equals C times the
derivative of the voltage across the capacitor. In this particular circuit, the voltage V is just the
offset voltage minus the output voltage. As we've done in previous derivations, I'm going to say i1
equals i2 because I'm neglecting any current that might be going into the non-inverting pin here.
Applying Ohm's law across R, I find 0 volts to the left and Vos to the right,
and we have a resistance of R. For i2, I'm going to copy the equation that I've just derived. Let's
now multiply both sides of this equation by -R. Since I'm trying to solve for the output voltage,
and it's inside of a derivative, I'm going to integrate with respect to time in order
to get it out of there. If we assume that the offset voltage is constant,
the integral here on the left is just Vos times t. R*C is just a constant
here on the right. Let's divide both sides of this equation by R*C. And finally I can
solve for the output voltage Vo. You're very quickly going to understand why I've
chosen the integrating circuit in particular to demonstrate the effect of the offset voltage.
It's really bad. Take a look at this equation. You can see that the output voltage equals some
constant plus the offset voltage times time. As the time increases, the output voltage just gets
higher and higher and higher - there's no bound to it. Therefore, a small problem: the offset voltage
can grow over time with the integrating circuit. It could quickly cause the amplifier to saturate
in other words. You might say that if you need to use the integrating circuit, you might need
to give more attention to the offset voltage than you would in other types of circuits.
That now brings us almost to the conclusion of the video. How do you fix the problem? Well,
the problem itself is not very mysterious, and we know what causes it. So, when they put these
op-amps on chips, they actually present pins in order to fix the problem. These two pins,
as I mentioned at the start of the video, are called the offset null pins. There's one of
them on pin 1, and there's another one here on pin 5. If you're just using the op-amp and you
don't care about the offset null or you don't care about being so precise, then you can just ignore
these pins, and there's no need to hook them up to anything. But if you want to fix the problem with
the offset null, you can do the following: you can hook a potentiometer or a variable resistor
in between pin 1 and pin 5, then hook the wiper to the minus VCC. Monitor the output voltage while
applying zero volts at the input and adjust the wiper of the potentiometer until the output reads
0 volts. That will give you a very easy and clear way to fix the problems with the offset voltage.
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