# Single supply opamp calculations

• posted

Hi, I'm doing some more opamp calculations just to get used to opamp golden rules some more. I decided to do an ideal analysis on an inverting, single-supply opamp:

After applying some golden rules, I got the following equations:

(V1 / R1) + (Vo / R2) - [ (Va * (R2 - R1) ] / (R1 * R2) = 0

Re-arranging the terms, I get:

Vo / V1 = -(R2 / R1) + (Va * V1) * [ (R2 - R1) * R1] / (R2 * R2)

Did I make a mistake in my calculation? I was expecting something like this:

Vo / V1 = -(R2 / R1) + V2

Thanks!

• posted

yes... your dimensions are not even correct.

Va = Vb = V2 for an ideal op amp, so...

I1 = (V2 - V1)/R1 I2 = (Vo - V2)/R2

since I1 = I2 for an ideal op amp we have

(V2 - V1)/R1 = (Vo - V2)/R2

or

R2/R1*(V2 - V1) + V2 = Vo

If V2 was 0 then you would just have -R2/R1*V1 = Vo which is the basic inverting op amp. In this case you have shifted by voltages by V2...

i.e.

-R2/R1*(V1 - V2) = (Vo - V2)

Jon

• posted

Looks like you're trying to sum the currents at the node labelled "Va" - but don't forget, the current through R2 is NOT Vo/R2 unless Va = 0...which cannot be the case here, given the offset voltage (V2) at the non-inverting (Vb) terminal. Nor is the current through R1 simply V1/R1, for the same reason.

There are two ways to do this - assume (Va-Vb) is zero (which basically assumes infinite gain for the op amp), in which case the currents are (V1-V2)/R1 and (Vo-V2)/R2 - OR you can assume that there is an error voltage here (Ve = Va - Vb) which is amplified by the op-amp such that Vo = AVe (where A is the gain). The latter path gives you the more accurate solution, which can then be reduced to what you expect by assuming that A gets very, very large and noting which terms will them drop out.

Bob M.

• posted

Thanks, Jon!

I also had the assumption that I1 = I2 and that Va=Vb=V2, but I think I made a mistake when I split my initial equation into the following:

I1 = - (V1 - V2) / R1 I2 = - (V2 - Vo) / R2

to this....

(V1 / R1) - (V2 / R1) = (V2 / R2) - (Vo / R2)

(V1 / R1) + (Vo / R2) - (V2 / R1) - (V2 / R2) = 0

then I used the following relationship: (A / C) +/- (A / D) = [ A * (D +/- C) ] / CD on

-(V2 / R1) - (V2 / R2) and turn it into

-[ V2 * (R2 - R1) ] / (R1 * R2)

(V1 / R1) + (Vo / R2) - (-[ V2 * (R2 - R1) ] / (R1 * R2))

... I still can't quite pick out my exact mistake, yet.. but I'm sure after much scrutiny it'll hit me like a light bulb.

• posted

I understand this portion.

I'm having a hard time grasping the last sentence in this paragraph. Sorry.

• posted

He's simply taking into account that a real op amp does not have infinite gain... a needless assumption in this case.... don't worry about it. If you can't do it for the ideal case you can't do it for the non-ideal case.

• posted

Well, yes, I was trying to get into the non-ideal case, because I think it's important that people eventually see where all of these magic formulas actually come from. If we take the usual inverting-amp configuration with the non-inverting input grounded (in other words, I'm going to ignore what was originally called "V2" here as a needless complication at this point - all it will really do is offset the output, anyway), the analysis really isn't all that difficult.

We'll call the input (at the "free" end of R1) Vin, the output Vout, and the voltage across the op-amp inputs Va. The "ideal" analysis would have you assume that Va is zero, or call the inverting input a "virtual ground," or some such, without really explaining why you do that. But again, it's really not all that hard to go through the full analysis, and what you'll wind up with is the general formula for any amplier used in this configuration, no matter how big the gain is.

We will, however, make the assumption that no current enters the inverting input (i.e., the input impedance is way bigger than anything else involved, such that any current into the input is negligible), and that makes the summing of currents at this node very simple:

(Vin - Va)/R1 = (Va - Vout)/R2

But by definition, the output voltage must be the amp gain times the voltage at the amplifier's input, which is Va, so:

If Vout = A(Va), then Va = Vout/A, and

(Vin - Vout/A)/R1 = (Vout/A - Vout)/R2

Solving for the overall gain (Vout/Vin), we get

Vout/Vin = R2 / [(1/A)*(R1/R2 + 1) - R1]

Note where the open-loop gain term "A" (i.e., the gain of the amplifier itself, without the R2 and R1 external components) winds up. If A gets very large, the 1/A term will approach zero, and take the whole first part of the denominator with it. If that's the case (and it's a reasonable assumption for an op-amp, where the open-loop gain is very commonly in the tens if not hundreds of thousands), then that first term drops out completely, and the whole thing winds up with the familiar

Vout/Vin = - (R2/R1)

Simple, no?

Bob M.

• posted

I'm following so far.

I'm understanding this portion.

This is clearer to me.

That's much better! Thanks, Bob!

• posted

- (V2 / R1) - (V2 / R2) = - [(V2 / R1) + (V2 / R2)] = -[ V2 * (R2 + R1) ] / (R1 * R2)

Regards,

Mark

• posted

Thanks, Mark!

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