# Single Ideal Op Amp Circuit

Before we start the problem, we should go over a couple things about op amps that’ll help us solve the problem and check our work. For one, notice how the resistor for Vo is connected to the negative terminal of the op amp (not directly, as it goes through the 63 k-ohm resistor, but you could draw a line between Vo and the negative terminal without making a new wire). This means that the op amp is undergoing negative feedback (NFB). And because of this, we can say that the voltages at the positive and negative terminals of the op amp are the same (So Vp = Vn). Since the op amp is ideal we can also say that the currents going into the positive and negative terminals are 0 A. Finally, since the power source, Vg, is connected to the positive terminal, this is a non-inverting op amp. Because of this Vg and Vo will have the same sign, or in other words, if Vg is positive, then Vo will also be positive. As a side note, if Vg was connected to the negative end then we would have had an inverting op amp, so if Vg was positive Vo would be negative.

The next thing to address is what it means for an op amp to operate in the linear mode, and what it means to saturate. These have to do with the 12V and -12V terminals coming out of the op amp. Those are its power source, or Vcc. So long as Vo is between -12 and 12 V inclusive, the op amp is operating in the linear mode. If you calculate Vo and its greater than 12V or less than -12V, then the op amp is saturated. Vo can never actually leave that range, so if the op amp is saturated you would set Vo equal to 12V or -12V, and you would have to recalculate what the voltage at the negative terminal of the op amp is, as the Vp = Vn only holds in the linear mode (I’ll go over this at the end once we actually solve the problem). Keep in mind that the current into the positive and negative ends are always 0 no matter what if the op amp is ideal.

This illustrates what it means to be in the linear mode or saturated.

Now to the actual problem. First, let's draw what we know based on the fact that the op amp is ideal, NFB, and we are operating in the linear mode.

Since there are 3 parts to the problem, we want to solve for Vo in a general form first (so let's not plug in Vg = 4V yet). The first thing to do in an op amp problem is usually to find either Vn or Vp. In this case, it's easier to find Vp since Vg, one of our knowns, is connected to Vp. For this, we can use the voltage divider rule, which states that the voltage across any set of resistors in series (so the same current passes through all of them) connected to a source and in a loop with no branches (so no current gets diverted in a different direction) is:

So, using voltage divider:

We can do this because both Vg and the 68 k-ohm resistor were directly connected to ground, so they are technically in a loop (as drawn above). And since the current to the positive terminal is 0, all current from Vg runs through both the 12 k-ohm and 68 k-ohm resistor, so they are in series too, thus satisfying the conditions to use the voltage divider rule.

NOTE: I didn’t write in the thousand part for the resistors since that would just cancel out anyway. It would technically be 68,000/(68,000 + 12,000).

NOTE: while the wire to the positive terminal is technically a branch, the current through it is 0, so we don’t have to count it.

Since Vp = Vn, we can now move to the negative terminal and use the fact that the same current passes through the 30 k-ohm and 63 k-ohm resistor (since again, the current to the negative terminal is 0). Unfortunately, we can’t use voltage divider again, since the two resistors are not in a loop (since once the current moves past the 63 k-ohm resistor it joins up with the current from the op amp and goes through the 27 k-ohm resistor). We instead can just use ohm's law, which states that V = IR, where V is the voltage difference across the resistor, I is the current through the resistor, and R is the resistance of the resistor.

We can plug in what we have for Vn (from solving for Vp), and solve for Vo in terms of Vg.

Note how Vo is positive, just like Vg. This is again because the op amp is non-inverting.

For part b, we want to find the range of Vg that allows the op amp to remain in the linear mode, so we want Vo to be less than or equal to 12V, and greater than or equal to -12V (since that is when the op amp will saturate). We can use the general equation we found in part a relating Vg and Vo to do this. We just set Vo equal to 12 and solve for Vg.

For part c, we want to know what R, given that R replaces the 63 k-ohm resistor in the circuit, would cause the op amp to saturate if Vg is equal to 2V. We know the op amp saturates if Vo is greater than 12V or less than -12V, so we will use one of those as a boundary case to solve the problem...but how do we know which one? Again, we use the fact that this is a non-inverting op amp. Since Vg is positive, we will use Vo = 12V.

So if R is 181.764 k-ohm or greater, the op amp would saturate.

NOTE: This is no longer part of the problem, but if you were asked to find what Vo and Vn would be if Vg = 6V (taking into account that we are using 63 k-ohm again for the resistor), this is how you would do it.

Just plugging 6V into the formula for Vo we had before:

We know this is above the saturation value of 12V. So we set Vo = 12V (as it can’t physically be greater than that), and now have to find Vn, since, due to the op amp being saturated, Vn doesn’t have to equal Vp anymore. For this we once again use that the current through the 30 k-ohm and 60 k-ohm resistor are the same.

So Vn = 3.87 V, which is different from Vp, since Vp = (17/20)Vg = (17/20)*6 = 5.1 V.