Amplifying circuit question

[kuksul08]

Hey guys, I had this problem on a test recently and I have been thinking over it. I can't seem to figure it out.

http://i106.photobucket.com/albums/m267/kuksul08/opamp.jpg


At first I thought it was easy, I could just orient it differently, and it's exactly like any other problem we've done (inverting opamp), but the voltage source is on the other side of the resistor. If it doesn't matter, it's a matter of using a simple equation. If it does, well then I'm still stumped.

Here's the standard form I was thinking of:

anyone know? Thanks

 

[jtr1962]

Since the non-inverting input is at ground, then the inverting input must be also. Vb is then determined by calculating the current flowing through the 3K resistor. Since Va = 0, then the negative end of the 2V source is at -2V relative to ground. Therefore, the current through the 2K resistor is -1 mA. Since the opamp inputs are infinite impendance (at least for the purposes of this exercise), then -1 mA also flows through the 3K resistor. Because the leg of the 3K resistor connecting to the inverting input is at ground, then Vb = -1mA times 3K = -3V.

 

[Alan B]

problem statement changed, analysis invalid.

 

[kuksul08]

Thanks for the help. I guess I got it wrong . I said Va was 2V because the voltage from the source there would split, but at the same time Va is grounded.

And for Vb I said the current was the voltage over the two resistors (2k and 3k), putting .4 amps through the 3k resistor for 1.2V. meh. It sucks when there are only 3 questions on a midterm.

 

[jtr1962]

I didn't see that the output of the opamp was grounded. Therefore, Vb = 0. As for Va, since the opamp isn't in linear mode then Va is essentially floating. Both resistors are grounded at one end. This means the circuit is a loop and the sum of the voltages must add up to zero. ( 2V - I * 2K = I * 3K ). Solving for I gives us I = 0.4 mA. Va = 2V - I * 2K = 1.2V

Actually, your answer was correct. My initial response would have been correct also had the output of the opamp not been grounded.

 

[kuksul08]

I didn't see that the output of the opamp was grounded. Therefore, Vb = 0. As for Va, since the opamp isn't in linear mode then Va is essentially floating. Both resistors are grounded at one end. This means the circuit is a loop and the sum of the voltages must add up to zero. ( 2V - I * 2K = I * 3K ). Solving for I gives us I = 0.4 mA. Va = 2V - I * 2K = 1.2V

Actually, your answer was correct. My initial response would have been correct also had the output of the opamp not been grounded.

Did you switch Va and Vb just then^? If so, PHEW. That's half credit for that problem. Granted I got the others on the test I'll end up with a 12.5/15. Not as great as I'd like but... next one will be better

Thanks man :thumbsup:

 

[jtr1962]

Did you switch Va and Vb just then^?

No, I just didn't notice that the op-amp output was grounded when I gave my first response. Just goes to show you have to carefully examine everything when you're taking a test. My test taking skills are rusty as I've been out of school for over 23 years.

 

[Steve K]

Yep... 1.2v sounds right. It took me a while to recognize the 2V as a battery. I would have expected a voltage source symbol with polarity markings.

Anyway... the op-amp is a red herring, and has nothing to do with the analysis. Tricky, but not particularly relevant to the real world. Well, except that it's not unusual to have to look at someone's circuit and need to figure out what the heck is going on.

Steve K.

 

[kuksul08]

Got the test back today... 11/15 DOH!

We went over the problem and the correct solution is Va=0V and Vb=-3V. He said that since the op-amp is grounded at the input, the junction there at Va is 0, and then he did KCL at A to find Vb.

 

[UncleFester]

.....Vb=-3V. .....

Ummm.. in your drawing, the output of the opamp (Vb)is grounded... that by definition is 0V... Is the drawing you provided different than the one in the test? Is there a load resistor between the opamp output and ground ON THE TEST?

 

[kuksul08]

Ummm.. in your drawing, the output of the opamp (Vb)is grounded... that by definition is 0V... Is the drawing you provided different than the one in the test? Is there a load resistor between the opamp output and ground ON THE TEST?

This is exactly what was on the test:

I figured that the circle means it's grounded, so I just drew it directly. Is that wrong?

Also the teacher drew the 2V power source on the other side of the 2k resistor, saying it doesn't matter what side it's on. I think he left part of the puzzle piece out when teaching us the basics.

 

[asdalton]

I figured that the circle means it's grounded, so I just drew it directly. Is that wrong?

Also the teacher drew the 2V power source on the other side of the 2k resistor, saying it doesn't matter what side it's on. I think he left part of the puzzle piece out when teaching us the basics.

The circles mean that the voltage is measured at the op-amp output, relative to ground.

Va = 0 because that's what an op-amp does with negative feedback.

From the diagram, the voltage between the 2k resistor and the 2V battery must be -2 volts. The current passing through the battery and 2k resistor is 1 mA.

Since no significant current enters the op-amp input, this same current passes through the 3k resistor. Then,

Vb = 0 - (3k)*(1 mA) = - 3 V.

 

[Steve K]

I figured that the circle means it's grounded, so I just drew it directly. Is that wrong?

Also the teacher drew the 2V power source on the other side of the 2k resistor, saying it doesn't matter what side it's on. I think he left part of the puzzle piece out when teaching us the basics.

The teacher should explain symbols that are being used. When I've seen the circle used, it typically meant that this was a post that a wire could be attached to. However, there's no need to draw a circle or a little ground symbol to the right of it. The teacher would only have needed to label the point "B", and ask what the voltage was at B.

The teacher is correct that the circuit will function the same whether the 2k resistor and 2v source are swapped or not.

As drawn, the circuit is a basic inverting amplifier where the gain is the ratio of the resistors.

If you can find a copy in the library, you might want to review a book called "IC op-amp cookbook". Lots of good info about op-amp circuits.

Steve K.

 

[UncleFester]

Awright, it makes a ton more sense now. The output of the opamp is NOT grounded.
Steve K is right, it's an inverting amp with the reference,(the amp + input pin) referenced to 0V (ground).
Since it's an invertng amp, the gain is the ratio of the two resistors times negative 1. or -3/2. If you take the input voltage (2V) and multiply it by the gain, it will show -3V at the output.

(-3/2) X 2 =-3

 

[kuksul08]

Thanks again. I should have just asked what the circle meant instead of assuming, because we all know what that results in!

 

[NA8]

Fun stuff:

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/opampcon.html#c1