# Subtractor

Discussion in 'General Electronics Chat' started by SimonAB, Dec 30, 2009.

1. ### SimonAB Thread Starter Member

Jul 30, 2008
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Hi all ,

I want to know the input and output impedance of an opamp subtractor. The circuit is as shown below. I want to use this in a Weaver receiver, and thus want to match its output and input impedances to 50 ohm, for maximum power transfer.

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2. ### hgmjr Moderator

Jan 28, 2005
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Signals applied to E2 see 20K input impedance and signals applied to E1 see 10K impedance.

hgmjr

3. ### Ron H AAC Fanatic!

Apr 14, 2005
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What are the frequency ranges of e1 and e2?

4. ### thyristor Active Member

Dec 27, 2009
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Not quite correct. The impedance seen by E2 is indeed 20K (R1 + R0), but the impedance seen by E1 depends on the relative values of E1 and E2. I think you have assumed that the - input of the op-amp is a virtual earth in this circuit; it is not.

Both + and - op amp terminals will have E2/2 volts at their inputs (using the resistor values given). Running the maths reveals the impedance seen by E2 is:

Z1 = 20E1/(2E1 - E2)

so if E1 is say 10v and E2 is 6v then Z1 = 14.3 Kohms

However, if E1 = 7v and E2 = 6v then Z1 = 17.5 Kohms and so on

If (and only if) E2 = 0 in this equation does Z1 become 10K (ie: R1) as in the classic inverting amplifier

Last edited: Dec 30, 2009
5. ### steveb Senior Member

Jul 3, 2008
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I think hgmjr is implying the AC input impedance because he says that "signals" will "see" 10K impedance. Generally, with nonlinear circuits, it's implied that it is the AC "linearized" or "small-signal" impedance that is of interest.

However, you raise a good point that clarification is often needed. I think that if DC input resistance needs to be considered, it's sometimes important to talk about common-mode and differential-mode signals because the two terminals could be used to monitor a voltage difference as if it is one signal.

I'm guessing that once Ron H gets his answer about the frequency of interest, he may say that both answers are wrong because the OPAMP can not be assumed to be ideal at higher frequencies.

Basically, this question is more complex than it might seem at first.

6. ### hgmjr Moderator

Jan 28, 2005
9,030
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Yes, Thyristor, you are correct. I did assume that E2 was at ground when I indicated that the impedance for signals applied to E1 would be 10K. I should have stated that in my reply.

hgmjr

7. ### hgmjr Moderator

Jan 28, 2005
9,030
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Thyrisor,

Could you elaborate on the math? I am not quite clear on where you obtained your equation for Z1.

hgmjr

8. ### thyristor Active Member

Dec 27, 2009
94
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1. The voltage at the + input of the op amp is E2/2 since R1 and R0 are equal in value.

2. Thus, the voltage at the - op amp input must also be E2/2 because the output of the op amp settles at the point where there is no difference between the inputs.

3. Therefore, the input current (i1) drawn from the E1 source must be (by Ohm's Law) (E1 - E2/2)/R1

4. Ergo, the input impedance Z1 (which is E1/i1) = E1/{(E1 - E2/2)/R1}

This rearranges to Z1 = E1R1/(E1 - E2/2) which further simplifies to:

Z1 = 2 E1 R1/(2E1 - E2)

Since R1 = 10K

then Z1 = 20E1/(2E1 - E2) .......QED

9. ### hgmjr Moderator

Jan 28, 2005
9,030
214
That certainly makes sense from a Thevenin's Equivalent perspective.

hgmjr

10. ### SimonAB Thread Starter Member

Jul 30, 2008
13
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Thanks for your detailed information thyristor. Can you please write the equation for input and out put impedances in terms of Ro and R1?

11. ### thyristor Active Member

Dec 27, 2009
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Assumption: the input impedances of the actual + and - op amp terminals are virtually infinite so that effectively no input current is drawn by the op amp. (At least none that would make any real difference to the answers below)

Input impedance:

Let V' be the voltage at the op amp + terminal.

thus V' = (E2)R0/(R1+R0) (potential divider)

It can also be seen that the input impedance of the E2 terminal is simply R1 + R0 as they are in series to ground.

V' will also be the voltage at the op amp - terminal since the op amp output will adjust itself until the + and - terminal voltages are equal.

Thus the input current drawn from the E1 source (i1) will be: (Ohms law)

i1 = (E1 - V')/R1

The input impedance at the E1 terminal will be Z1 = E1/i1

Thus Z1 = E1R1/(E1 -V')

Substituting for V' = (E2)R0/(R1+R0) gives

Z1 = E1R1/{E1 - E2R0/(R1 + R0)}

Multiplying top and bottom by (R1 + R0) and dividing top and bottom by E1 reveals

Z1 = R1(R1 + R0)/(R1 + R0 - R0E2/E1)

Note therefore that Z1 is dependent on the ratio of E2/E1 and is not just a function of the resistances used.

Note also (as a check) that if we substitute R0 = R1 we obtain

Z1 = 2(E1)(R1)/(2E1 - E2) as in my earlier post

Note further that if E2 = 0 then

Z1 = R1 ((ie: a classic inverting amplifier's input impedance)

....and, finally, if E1 = E2, then Z1 = (R1 + R0)

Output impedance:

The output impedance will be the stated output impedance of the particular op amp from its data sheet. This will be very low typically and will be in parallel with any load resistor connected between the op amp's output and ground.

Last edited: Dec 31, 2009
12. ### Ron H AAC Fanatic!

Apr 14, 2005
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The math is probably correct, as far as it goes (I didn't check it), but the circuit won't work for frequencies above a few Megahertz, due to stray capacitances, and possibly op amp bandwidth, depending on the op amp. Scaling resistor values down would help.

I asked SimonAB about the frequencies involved, but he has not replied.

13. ### steveb Senior Member

Jul 3, 2008
2,433
469
An important point must be made here. Thyristor is essentially correct (I assume, I didn't check it either) in the mathematics and statements he has presented. However, this viewpoint is difficult to use in practice, because it results in nonlinear equations. By nonlinear, I mean that you can not use superposition to find independent transfer functions Z1=E1/I1 and Z2=E2/I2, which is one of the important statements Thyristor made.

Of course the circuit is nonlinear, so it is not surprising that the input impedance of one of the inputs depends on the voltage at the other input. However, the piece-wise linear nature of these types of circuits (ignoring rail limits) allows them to be treated based on considering common-mode signals and differential-mode signals. This allows linearization of the mathematics, and this is the traditional way to talk about input impedance for this circuit. That is, you separately specify the common-mode input impedance and the differential mode input impedance.

The basic idea is that any combination of E1 and E2 can be resolved into an equivalent common mode signal plus differential mode signal. Each component will see a different input impedance. The input impedance for the common mode portion of the signal will depend on both R1 and R0, while the differential mode signal will depend only on R1. This treatment is very similar (but not the same) to doing a separate AC and DC analysis, but there is no requirement for a "small-signal" analysis. The ideas work for large AC signals, as long as the outputs are not hitting the rails.

I'm deliberately not presenting the mathematics because it requires careful presentation, and it's not clear that this is interesting to anyone. If there is an interest I can look through my old notebooks and books to see if I can post the traditional approach. It's not hard to rederive it. I believe the traditional approach begins by defining common mode and differential mode signals as follows.

Differential Mode Signal: Vd=E2-E1
Common Mode Signal: Vc=(E1+E2)/2

and from this it is clear that

E2=Vc+Vd/2 and E1=Vc-Vd/2

Analysis of the circuit with these definitions, makes the math easier and allows the traditional powerful linear analysis tools to be applied.

14. ### thyristor Active Member

Dec 27, 2009
94
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I agree with your equations for E1 & E2 expressed in terms of the difference mode and common mode signals

Ergo, E2/E1 = (2Vc + Vd)/(2Vc - Vd)

Substituting this relationship into my equation for Z1, viz:

Z1 = R1(R1 + R0)/(R1 + R0 - R0E2/E1)

and putting Vc = 0 in the first instance and then putting Vd = 0 in the second instance, to determine Zd (the difference mode impedance) and Zc (the common mode impedance) reveals:

Zd = R1(R1 + R0)/(R1 + 2R0)

and

Zc = (R1 + R0)

which are linear equations independent of the actual voltage levels

15. ### steveb Senior Member

Jul 3, 2008
2,433
469
That looks right. I actually came up with a factor of 2 from your Zd, but this just results from different starting definitions for Zd. (I used Zd=Vd/I). It looks like my previous statement, that Zd depends only on R1, is wrong.

Last edited: Jan 1, 2010
16. ### kdillinger Active Member

Jul 26, 2009
141
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This should be written into an article format.

If you check datasheets from TI and ADI, they will specify a common mode and differential mode input impedance for their difference amplifiers.