Op Amp circuit analysis

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I don't think this circuit can be impedance characterized only by a single number. There is an impedance from each input to ground also to consider.

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Eero-Pekka agrees with you; did you read all of what he had to say at:

He was only talking about the impedance at Vin-.

What does the impedance at Vin+ look like for some of the various applied signal combinations he described?

Eero-Pekka _is_ sort of correct for his special case...

(Ah, but that is a special condition! /rap)

Think about it, if inputs move anti-phase, R1 sees _nearly_double_ the voltage change.

But the whole discussion is picking at nits, depending on definition ;-)

For Eero-Pekka's case Iin = Ein/R1/1.9901, Ein is the movement of _just_one_side_, other side is -Ein ;-)

...Jim Thompson

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Make that a typing blunder, should be... Iin = (Ein/R1)*1.9901

...Jim Thompson

Let me see if I get this right. There's an impedance at Vin+ of simply R3+R4. Assuming the output's not pinned, Vin- sees R1 to (Vin+)*R4/(R3+R4), which you could break into a component to ground and one to Vin+ if you cared.

Clifford Heath.

I don't want to finish your calculations because I might not do what you intend.

Finish your calculations and verify or falsify Eero-Pekka's assertions for the conditions he gave.

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He had more than one case; which one are you considering (it looks like it's the first one)?

If Iin = (Ein/R1)*1.9901, then apparently Ein/Iin = R1/1.9901, which if R1=1000 and R2=100000 would give a value of 502.487 ohms. So are you agreeing that his value of 500 ohms is about right, but not exactly right?

Now, how about his other cases?

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I just did the 1K/100K case, and it's only valid if you consider one side of an anti-phase situation as if it were single-ended.

Math of the other cases is left as an exercise for the student... I need to sip a glass of wine before a conference call from Malaysia in

I try to avoid getting involved in semantic misunderstandings of what input impedance means... in the real world I would only be concerned with the DC loading of the previous stage or perhaps (in rare occasions in my line of work) balancing AC-coupling.

...Jim Thompson

There have been a few answers so far. Consider this:

Consider the first case Eero-Pekka describes, where the signal, ein, which is applied to Vin- is also applied to Vin+, but 180 out of phase. In other words, -ein is applied to Vin+ at the same time ein is applied to Vin-.

One way to do this is to apply the input (AC) signal through an ideal transformer with a center tapped secondary winding. The center tap is grounded and one end of the winding is connected to Vin- and the other end is connected to Vin+. The source driving the transformer is considered to be an ideal voltage source, with zout equal to zero.

With this hookup, the end of the transformer driving Vin- sees 500 ohms according to Eero-Pekka (assuming R1=R3=1000, R2=R4=100000 and A=infinite). The other end of the transformer, the one driving Vin+ sees 101000 ohms.

But, now imagine disconnecting the center tap of the transformer secondary from ground. Now we have a floating winding driving Vin- and Vin+ differentially. What is the differential impedance seen by this winding?

Does that seem consistent with the 500 ohms at Vin- and 101000 ohms at Vin+ when the center tap is grounded?

Come on, John Popelish! I know you can say more.

And others who want to participate, don't just offer opinions off the top of your head. Talk about results of actual analysis.

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I started wading into equations, without much of a plan and stopped when I got to 14 of them and hadn't quite come to the end. So I think a graphical solution with just a couple simple equalities is better.

1. The two opamp input s are always equal voltage by negative feedback and infinite gain.
2. The opamp inputs pass no current, so all the input current passes through all 4 of the resistors, since there is no other possibilities for it.

So, showing voltage as height (positive at top) the drops across all the current carrying components is as follows for the source applying a positive voltage to the non inverting signal input (V+ is the opamp non inverting input voltage, and V- is the opamp inverting input voltage, Cs represent the center tapped floating signal source transformer winding):

Vout=Vsignal*(R2/R1) | - | | | | | | | | | | | | R2 +------+ Vin+ | | C - | | C | | R3 - C - | V- | Vcm | V+ - C - | | R1 C | | - C | | Vin- +-----+ | | | | | | R4 | | | | | | - | 0V

Though drawn to scale for a gain of more like 4, the point is that it shows how all the resistors are in series with the input signal, and the common mode voltage Vcm) at the center of the signal source winding is always half of the output voltage, regardless of the gain.

For instance, if the gain is 100, and the input differential voltage is a 0.1 volt, the output voltage will be 10 volts, and the common mode voltage will be 5 volts.

And the differential input impedance is R1+R3, since these are in series with the input current, each to the common mode voltage, even though they are not connected together.

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I've been using this addition to the original circuit for analysis of the cases where Vin+ is made to be some multiple of Vin- :

R1 R2 1 ___ 3 ___ Vin- o----o-|___|-o---o------o-|___|---. | | | | | | | | |\\ | .-----. '---|-\\ | \\ m / |A >---------o--- Vout \\ / .---|+/ V | |/ | | | | | ___ | ___ Vin+ o----o-|___|-o---o------o-|___|---o 2 4 | R3 R4 o GND

There is an ideal buffer (infinite input impedance, zero output impedance, gain = m) driving Vin+. If the gain of the buffer is -1, then this is the same as the center tapped (with center tap grounded) transformer driving the original circuit.

Using this might reduce your 14 equations somewhat.

--- Redrawing: (view in Courier)

For the first case:

E2 . R1 / R2 .Vin->---[1000]--+--[100K]--+ . | | . | | . +---|-\\ | . |A >---+-->Vout . +---|+/ . | . R3 | R4 .Vin+>---[1000]--+--[100K]--GND \\ E1

With Vin+ at 1V, then the voltage at E1 will be:

(Vin+) * R4 1V * 100000R E1 = ------------- = ----------------- = 0.990V R3 + R4 1000R + 100000R

and, with E2 being forced to equal E1 with Vin- equal to -1V, Vout must move in such a direction to make E2 = E1,

When that happens, the current in R1 will be:

(Vin-) - E2 -1V - (0.99V) -1.99V I = ------------- = --------------- = -------- = -0.00199A R1 1000R 1000R

Now, as far as the source of V1 is concerned, it has to supply that

-1.99mA into a load which it's also dropping -1V across, so the impedance it sees, looking into R1, is:

E -1V R = --- = ----------- ~ 502.5 ohms I -0.00199A

For the second:

. E2 . R1 / R2 .Vin->---[1000]--+--[1000]--+ . | | . | | . +---|-\\ | . |A >---+-->Vout . +---|+/ . | . R3 | R4 .Vin+>---[1000]--+--[1000]--GND . \\ . E1

With Vin+ equal to 10 volts, E1 will be equal to:

E1 * R4 10V * 1000R E1 = --------- = --------------- = 5.0V R3 + R4 1000R + 1000R

With Vin- equal to -10V and E2 set to equal E1, then the current in R1 will be:

(Vin-) - E2 -10V - (5V) -15V I = ------------- = --------------- = -------- = -0.015A R1 1000R 1000R

and, as in the first case, the impedance the source of Vin- will be looking into will be:

E -10V R = --- = --------- = 666.666... ohms I -0.015A

So he was correct in the second case, but 2.5 ohms high in the first. ;)

JF

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If the input is connected between nodes (1) and (2), but is otherwise floating, the input impedance is simply R1 + R3....

... by eyeball, no equations required...

... because: nodes (3) and (4) have to be at the same voltage ;-)

...Jim Thompson

So would having an actual plan, instead of just diving into algebra. ;-)

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Also, should be "...he was...2.5 ohms low in the first."

You've got it just right.

Eero-Pekka says "You may simulate various conditions as I did and find out that the impedance varies vastly."

Apparently he got his numbers by simulation and not by analysis. That probably accounts for why he got 500 ohms for the first case and not the correct 502.488 ohms.

Now, if we add an ideal buffer amplifier to apply some multiple m of the voltage at Vin- to Vin+:

R1 R2 1 ___ 3 ___ Vin- o----o-|___|-o---o------o-|___|---. | | | | | | | | |\\ | .-----. '---|-\\ | \\ m / |A >---------o--- Vout \\ / .---|+/ V | |/ | | | | | ___ | ___ Vin+ o----o-|___|-o---o------o-|___|---o 2 4 | R3 R4 o GND

derive a general expression for the impedance at Vin- (in terms of R1, R2, R3, R4 and m) when A is infinite and m has some finite value.

Then, for the piece de resistance, find a general expression for the impedance at Vin- when A and m are both finite.

What if they aren't at the same voltage because the opamp gain, A, is not infinite? Then what's the differential input impedance?

What I find interesting is the difference in impedances seen as a result of grounding or not grounding the center tap of the transformer secondary.

If a transformer is not used, the differing impedances at the two inputs could cause partial conversion from common mode to differential mode. A large common mode signal would thus cause error. This has long been known, but perhap the usual analysis which says the impedance at Vin- is equal to R1 should be reconsidered; the conversion may be different than usually assumed.

For example, in Pease's column:

he has the impedance at Vin- equal to R1, and doesn't take into account what the signal at Vin+ is. That was Eero-Pekka's point.

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Picky! Picky! ;-)

Please provide the "balanced" input impedance, versus frequency, for an OpAmp with GBW=20MHz and Ao=100K... you may ignore excess phase and slew rate (for the moment ;-)

...Jim Thompson