A question related to Operational Amplifier.

[DarkStalker]

Hello. I believe there's a flaw in my understanding of the operational amplifier. I'm getting a bizarre result that I'd really appreciate if anyone can help me with. :)

1. http://tinyurl.com/6e9ovvp
The part I'm having problem with is Q9 (b) (iv).


2. V=IR

The Attempt at a Solution

You'd have to look at my attempts in other parts to help me with the problem:

Q9 (b) (i): Going by the 'virtual earth' concept, I deduced that the potential at both X and Y is 0V, and it's the same at the junction to the left of Y. The potential at A is +1.8V, so the potential difference b/w the 6000 ohm resistor is (1.8-0)=1.8V.
Using V=IR,
1.8 = I x 6000,
I = 3 x 10^-4 = 0.3mA

Q9 (b) (ii): The junction left of Y is connected to the 4000 ohm resistor, which is being provided a -0.4V potential from one end (the point B). The potential difference b/w that junction and B is therefore [0 - (-0.4)] = 0.4V.
Again, V=IR
I = 1 x 10^-4 A = 0.1mA
This current is flowing out of the junction, towards B.

Now the question asks for current at Y, which by Kirchoff's Law is:
0.3 - 0.1 = 0.2 mA

Q9 (b) (iii): The entire current at Y would flow into the feedback resistor, so current in feedback resistor = 0.2 mA

Now here's the part that confuses me

Q9 (b) (iv): My understanding is that, the potential difference across the feedback resistor would be the same as the potential difference between the output and earth. So in order to calculate V(out), I have to calculate the p.d. across the feedback resistor. I do this by using V=IR across this resistor:

V= IR
V= 2 x 10^-4 x 200000 = 40V

Going by the direction of flow of current, V(out) = -40 V.


Now this is what I don't understand. The calculations clearly show that the p.d. across the amplifier is 40V, because a current of 0.2 mA is flowing through a resistor of 200000 ohms. The pd HAS to be this much for that amount of current to flow.

But we know that the output of the amplifier is limited by it's power supply, which cannot exceed +/- 9V. So, at best, the output voltage should be -9V. But that's not what the calculations say.

Am I wrong about my assumption of the potentials being 0V at the point Y, or have I missed some property of 'ideal amplifiers'? That's my problem.

Thanks in advance for reading through all this.

 

[DarkStalker]

Oops, this should have been in the "Advanced Physics" section.

Can a mod please move this?

 

[Jmf]

The assumption that the voltages at the inverting and non-inverting inputs are equal only holds when the op amp is working in it's usual operating region (i.e. output not saturated to +9 or -9V). That's my understanding, anyway, since we usually assume negligible output impedance and a large input impedance for operational amplifiers.

You're right in thinking that the output can't really go lower than -9V (the output of the op amp would just saturate - it's no longer in it's linear region). It seems odd that this would be the case in an exam question though. Perhaps there's an error in your calculation for the current? (though I haven't seen it yet).

EDIT: Actually, since the question states that it's an 'ideal' op amp, and it previously asks you what you can say about the minimum and maximum output voltages, perhaps the answer it's aiming at is that the output voltage is -9V?

If this were a real op amp though, I think that it would turn out that the voltage at the inverting input would actually change (not be 0V, and not be equal to the other input, but rather be less) when the op-amp saturates, in order to allow the output to be (nearly) -9V for whatever current was going through the feedback resistor.

Your numbers seem fine anyway, I can't see any errors.