http://imgur.com/XQJrZ

- Find the input impedance at the nodes Va and Vb to ground.

Would this just be R1 for both, respectively? Since it is going to ground, there is current going through the resistors, resulting in input impedance, correct?

- Find the output impedance at the nodes Vo to ground.

For an ideal op-amp, it would be zero, but it can't be zero in this case because of the current going from the bottom amplifier to the top amplifier to Vo, is that correct? I can't figure out how to find it.

 

I can't figure out how to find it.

How were you taught to find input and output impedance? Surely, either your teacher explained it, or it is explained in your text book. If not, ... why not? If yes, ... then make an attempt to apply the method. How do you expect to be able to figure this out with vague hand waving when it requires a solid mathematical approach, or experience from looking at similar circuits? Lack of experience is understandable when you are learning, but you should have the tools readily at hand, or you should be demanding the tools from whomever is supposed to be teaching you.

How can we even judge what level your class is and what mathematics is suitable to present to you if we don't have guide from your first attempts. Also, how are we to know what type of opamp model you have been taught to apply to problems? Are you using an ideal model, or a non-ideal model? If the latter, which one?

Anyway, the rules of this forum require you to make a reasonable attempt at answering the question, or at the very least setup the problem and identify the appropriate equations and constraints.

 

Well for input impedance, it goes Va to the resistor to ground, so the input impedance is just Vin / Iin, where Vin = Va and Iin = (Va - 0)/R1, which results in Input impedance of R1. Same for Vb.

But for output impedance, I can't figure out if it's just 0 because its an ideal op-amp and the output is connected directly to the output of the op-amp, or if I have to use KCL at the middle node:

(Vo - Va) / R3 + (Vb - Va) / R2 = 0

From here, I would be finding Zout = Vout / Iout. But I can't figure out what Iout.

 

Well for input impedance, it goes Va to the resistor to ground,

You're still handwaving. Why would you say that the resistor is going to ground? I don't see a ground there.

You need a systematic way to attack these problems. This is why I wanted you to consult a textbook or class notes. The key thing you still have not answered is which opamp model you want to use. A step by step approach might look something like this.

1. Determine the OPAMP model
2. Determine if you need separate AC and DC equivalent circuits, or just one equivalent circuit valid for both AC and DC.
3. Write equivalent circuit or circuits for the calculations of interest.
4. Apply source to input or output. You are free to use a voltage source or current source.
5. Use circuit equations to determine output of source. If voltage source is used as the input, then current from the source is the output, and vice versa. Note that even output impedance calculations require a source going into the output (very strange idea).
6. Impedance is always ratio of voltage to current.

There are also heuristic methods that allow you to see the impedances by quick inspection, but that usually takes more experience developed from using the more formal methods. But, some cases are easier than others to see.

 

Krl003,
You are completely WRONG!
The input resistance of a non-inverting opamp is very high, maybe 1M or more.
The input resistors are in series with the very high input resistance, not connected to 0V so the current in the resistors is almost nothing.

 

You're still handwaving. Why would you say that the resistor is going to ground? I don't see a ground there.

The question reads

- Find the input impedance at the nodes Va and Vb to ground.

As I said before, we calculate this Zin = Vin / Iin. Vin is the voltage drop from Va to ground, which is Va, and Iin would just be the current across R1 with a voltage Va across it, so it would just be Iin = Va/R1, resulting in Zin = R1. Is that correct?

 

resulting in Zin = R1. Is that correct?

As Audioguru said, that is wrong.

 

Repeat:
The input resistance of an opamp is very high, maybe 1M (one million) ohms or more. The input resistor is in series with the input so it increases the very high input resistance.

This is a non-inverting opamp circuit. Maybe you are thinking about an inverting opamp circuit.

 

So how do I find the input impedance?

 

So how do I find the input impedance?

I've given some guidelines above, but that's basically the question I am asking you. How have you been instructed to find the impedance? What opamp model have you been instructed to use? Somehow the description you've given doesn't seem to be correct. We can help you, but it helps to have a base to start from.

The opamp model you use can have a big effect on the answers, particularly if you want to look at frequency dependence of feedback effects. In this case, the input impedance calculation seems to be dependent on the model also. The current that flows into the + terminal of the opamp is quite small which equates to an effective high input impedance. Now, do you want to assume the impedance is infinity, or do you want to use a large but finite value? It's not clear why you want to say the input impedance is just the resistor value. That resistance is very small in comparison and does not significantly matter if you are using a very simple opamp model. All I can think (as Audioguru wonders) is perhaps you are thinking of negative feedback creating a virtual ground, as it does in the inverting amplifier configuration. However, that does not apply here.

Anyway, getting back to "how you do it" I started making a step by step guide above. How far can you get in that process?

 

So how do I find the input impedance?

The datasheet for the opamp lists it minimum input resistance (300k ohms for an old opamp) and its typical input resistance (2M for an old opamp).
Modern Jfet input opamps have an input resistance that is typically one trillion ohms for a TL07x opamp.

 

The problem says its a differential biopotential amplifier.

 

You posted an ordinary differential amplifier that has a very high input resistance.
The input impedance is the same but drops at higher frequencies.