# Homework Help: Operational Amplifiers

1. Oct 8, 2012

### ElijahRockers

1. The problem statement, all variables and given/known data

Va = 4V
Vb = 9V
Vc = 13V
Vd = 8V

The 220k resistor is replaced by a variable resistor Rf.

What value of Rf will cause the Op Amp to saturate? Note: 0 ≤ Rf ≤ ∞

When Rf obtains that value, what will be the current flowing into the output terminal of the op amp?

3. The attempt at a solution

In my notes, it says Vcc/A = Vp - Vn is the condition that will cause saturation but I'm not sure where to go with that from here.

I solved the circuit for currents and everything, but I'm not sure how to calculate what Rf should be to make Vcc/A = Vp - Vn.

Vcc/A is just 15/(10^6) as i understand it but Vp = Vn because of the virtual short...

And in this diagram, Vn = 8 right?

Sooo, I solved for Vp and got 8.000015V, then I divided that by the current that is flowing through the 220k resistor to see what the new resistance should be, and it comes out to 219981, which is basically just 220k anyway.

What am I doing wrong?

EDIT: I should show some of my calculations. Using KCL, I found that the output voltage using the 220k resistor is 14V.

Last edited: Oct 8, 2012
2. Oct 8, 2012

### Staff: Mentor

One approach would be to determine the value of Rf that would make the gain of the circuit such that, given the input conditions, it would force the output to one of rails (a "rail" being one of the op-amp supply voltages, which is the maximum extent that the output can possibly swing).

You might begin by replacing the input network (what's to the left of the "x" below) with its Thevenin equivalent to make the circuit look more like the typical basic op-amp circuit with negative feedback.

As a reasonable approximation you can assume that the gain A of the op-amp is very large, so that the gain of the circuit is set by the resistor network that surrounds it. The current that results at the Vn node from the input network must pass through the feedback resistor.

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3. Oct 10, 2012

### ElijahRockers

Well I found the thevenin resistance of the input network (i think, 12kΩ?), but I'm not sure about the voltage... they are in parallel so wouldn't that mean they'd have to be equal?

Anyway, so if I am interpreting your explanation correctly, I'm trying to find Rf such that Vo = 15 or -15?

4. Oct 10, 2012

### Staff: Mentor

12 kΩ looks fine. But the voltage sources are not in parallel -- they have resistors in series with them. One trick you might find helpful when dealing with a set of voltage+resistors in parallel is to first convert each voltage+resistor pair to its Norton equivalent. All the resulting Norton equivalents will be in parallel, and so the current sources can be summed and the resistances collected parallel-fashion. This yields a single Norton model thank can trivially be turned back to its Thevenin equivalent.
Yes. And you should be able to find a good clue as to whether that should be a + or a - 15V.

5. Oct 10, 2012

### ElijahRockers

Ohh ok. So for thevenin voltage then I get 7.668V for the input network.

but am I right in thinking that since Vn is 8V that Vp must also be 8V (virtual short)?

which would make the current flowing into Vp from the input network = (7.668-8)/12k (which means current is actually flowing into the input network)?

6. Oct 10, 2012

### ElijahRockers

And I just found in my notes that

$V_0 = -R_f(\frac{V_a}{R_a}+\frac{V_b}{R_b}+\frac{V_c}{R_c})$

So Vo should be negative 15 because Rf has to be positive?

EDIT: that forumula doesn't take into account the extra resistor in the input network so i guess it wouldn't work

7. Oct 10, 2012

### ElijahRockers

ok assuming that the voltage at Vn is 8V, I set (Vo-8)/220kΩ + (7.668-8)/12kΩ = 0

I put in 15 for Vo, and solved for Rf = 253kΩ...

Is this correct?

EDIT: It was correct! Thank you. :)

8. Oct 10, 2012

### Staff: Mentor

Heh. I knew you'd get there given a bit of time to think...