Total current required by OPAMP

[Bjtpower]

Hi,

I am trying to do paper calculation for current requirement for OPAMP AD8022

https://www.analog.com/media/en/technical-documentation/data-sheets/ad8022.pdf

I would like to make sure how much current required by OPAMP when it is in operation mode.
I have captured below snapshot from PAGE No: 3

I have supply voltage: +-5V
So 55 mA is the correct current which will draw from my power supply? or 55mA+11mA=66mA (Quicent current of both opamp) or something else?

I

 

[KlausST]

So how can each single input impedance can be the same but the voltage drop and current waves are different?

You got it! The input impedance can´t be the same! They are different. That´s what I´m saying all the time.

Klaus

 

[danadakk]

For the lazy ones (sorry should not have said that) here a classic diff amp sim, which is what the OP seems to have intended.

Given the use of a fast OpAmp with 1970's levels of offset and bias current galore OP concerned with DC case ?

Shown, I assume, was a diff amp structure, POST #3, with balanced R's, so here is one simplified AC sim (down to ONE HERTZ, as before) :

Now if I unbalance that circuit in any way I do get different results.

So in closing we have a case where AIN+ Z = AIN- Z, and we have case(s) where its not. If we use the 3 OpAmp
design this discussion goes away (essentially).

https://www.ti.com/lit/an/slyt310/slyt310.pdf

https://www.analog.com/media/en/technical-documentation/application-notes/AN-1026.pdf

Over and out, Dana.

 

[KlausST]

Hi,

now a differential mode setup. Why do you ignore that I always talk about "individual" impedance?
Why talking about "1970" OPAMPs .. while I refer to the general circuit, independent of Opamp. The OPAMp does not matter at all.

One input imedance is "R1" = 4.81k,
The other input impedance is "R3 + R4" = 4.81k + 1.2k = 6.01k.
Where is the OPAMP you refer to?

Using different - unsuitable - test setups. Ignoring math, ignoring simulation results.
Then why don´t you build the setup with real parts?
Connect one input (AIN+) to GND and use an OhmMeter w.r.t. GND at the other input (AIN-)
Then do the same with inputs swapped.

Scared to show the results here? Why?

Klaus

 

[danadakk]

So in closing we have a case where AIN+ Z = AIN- Z, and we have case(s) where its not.

What is it you dont understand, I made your point there are conditions where Zins are not equal.

Using different - unsuitable - test setups

I used OP circuit post #3.

while I refer to the general circuit, independent of Opamp. The OPAMp does not matter at all.

OMG Klaus, I did not know these circuits one could omit entirely the active element. Wow
what a cost saving measure.....Please write Analog Devices and TI that OpAmps and their
finite G and poles etc do not affect impedances, as they have found in their ap notes. IEEE
as well. Oh and my sim shows OpAmp impact on Z's, known now for decades.

ignoring simulation results

Did not ignore my or your results. Both have validity as I pointed out in post 42.

Connect one input (AIN+) to GND and use an OhmMeter w.r.t. GND at the other input (AIN-)

I do not use OhmMeters for wideband verification. I use a VNA.

Scared to show the results here? Why?

For the last time, my case showing conditions where AIN+ Z = Ain- Z :

Dana.

 

[KlausST]

Hi,

A circuit is not a test setup.
It´s so simple: the key is "individual".

OMG Klaus, I did not know these circuits one could omit entirely the active element. Wow
what a cost saving measure.....Please write Analog Devices and TI that OpAmps and their
finite G and poles etc do not affect impedances, as they have found in their ap notes. IEEE
as well. Oh and my sim shows OpAmp impact on Z's, known now for decades.

You're just making a fool of yourself with statements like that. I never said you could omit an OPAMP.

I do not use OhmMeters for wideband verification. I use a VNA.

Now you try to move to wideband. Why?
Just because you don´t want to see what an OhmMeter will show?
4.8k and 6k? Any hobbyist can do this test. No special equippment needed, as simple as can be.

Everyone can see that you are always looking for new special cases to avoid the obvious: Individual: one 4k8, the other 6k.
One can not see "individual" results when one signal source is connected to both inputs.

You simply don´t want.

One does not need to be a specialist, one does not need special knowledge, one does not need special equippment.
Every one can prove it.

Klaus

 

[danadakk]

Tiresome

Now you try to move to wideband. Why?

The OP is using a wideband OpAmp and you specifically called out "For the lazy ones (sorry had to say that) here a DC test
and an AC test." So I looked at it as a wideband differential OpAmp application. I guess in your process thats a
crime. And I did it at 1 Hz (arbitrary) that I always thought is pretty close to DC, although I do DC bias examinations
when called for.

Everyone can see that you are always looking for new special cases to avoid the obvious: Individual: one 4k8, the other 6k.
One can not see "individual" results when one signal source is connected to both inputs.

Everyone can see your specific desire to ignore both our cases are legit, as do the authors of the papers
I referenced.

Not only that you and I share a lack of OP input as to exactly what does the source signals look like,
and are they going to use it single ended (as you seem to emphasize) or differentially (as I made the point
to insure OP knew in differential operation, which he posted, there is a case for equality).

I am not a specialist, the Amatuer radio community I belong to is largely not "specialists", yet love
and use our Nano VNAs as another lovely instrument that augments our lovely DVMs. Like our lovely
scopes if thats also an issue for you.

You simply dont want.

One does not need to be a specialist, one does not need special knowledge, one does not need special equipment.
Every one can prove it.

But I am glad you now acknowledge there is a case when using a differential OpAmp configured circuit
where input Z is same, as I have done multiple times in prior posts and sim, and your case for single ended
operation shows they can be different. In fact a designer of using my case could certainly consider your case
"special", from their viewpoint. Oh and as you know, not everyone here on the forums is equipped to use
even a basic DVM, hence cannot prove it as you assert.

Over and Done, Dana.

 

[FvM]

@danadakk
I have a small question related to post #44 plots. Can you show how the measurements "Impedance @ R1" and "Impedance @ R3" are defined in SIMetrix? With the shown floating source configuration, voltages at AIN+ and AIN- have different magnitude (referenced to ground) due to asymmetrical input, as explained by KlaussST. I'd expect to see this in an impedance measurement.

 

[KlausST]

Hi,

let´s calm down.

Technical problem! Nothing personal! First things first: Define what we are talking about:
1) Let´s focus on the OP´s circuit of pos#3.
2) Let´s focus on the frequency of interest in this application.

My opinion: since there is a bandwidth limitation capacitor ....
--> I´d say the circuit is useful for frequencies 0Hz to something about 1MHz
--> and I´d say ... let´s focus on this area of frequency.

Let´s try to make it like at a wedding: If someone does not agree, please say now. And not later on claim something else. (for sure le´t accept time zone)
I don´t want to over rule the OP, thus the OP´s opinion should count the most. If the OP has a different opinion, please state it, your input is very welcome.
(Also I don´t want to forget to invite Tony and Dana to agree/disagre to the two points below)

So for the first step: does every one agree to further talk about:'
1) the circuit given by the OP in post#3
2) the frequency range 0Hz to 1MHz

... next steps later

 

[FvM]

Three cases, each showing different AIN- input impedance. You can setup a fourth case with grounded AIN+.

 

[danadakk]

@danadakk
I have a small question related to post #44 plots. Can you show how the measurements "Impedance @ R1" and "Impedance @ R3" are defined in SIMetrix? With the shown floating source configuration, voltages at AIN+ and AIN- have different magnitude (referenced to ground) due to asymmetrical input, as explained by KlaussST. I'd expect to see this in an impedance measurement.

The probe selections include "More Probe Functions" then "Impedance", one clicks on
specific part terminal. Note in case of floating signal the current thru R1 and R3 have to
be same, as it is in series with R1, R3. I also looked at virtual ground, its there until pole
start its break around 100Khz. Note I verified virtual ground intact until that break as well.


Regards, Dana.

 

[KlausST]

Hi,

Thanks FvM.

The results are almost as expected. (only one value worries me. But to this later.)
Naming the resistors:
FvM --> OP (post#3)
R1 --> R25
R2 --> R27
R3 --> R26
R4 --> R28
(To avoid scrolling through 45 posts I now use FvM´s naming)

So what happens at this circuit? Functionally (not by errors)
I want to "visualize" it, so how to differentiate between the two inputs. Colors don´t work, so let´s use two different frequencies.
Let´s choose
* AIN+ = 1kHz with 1V amplitude
* AIN- = 10kHz with 1V amplitude
So it´s easy to recognize on a scope or a simulation.

The non inverting input side is a simple voltage divider R3 and R4 --> 4.81k and 1.2k.
We will get an input current of 1V / (4.81k Ohms + 1.2kOhms ) = 166uA amplitude (@ 1kHz) .
And we will also see the voltage at the center (which is IN+ of the OPAMP): IN+ = AIN+ * 1.2k / (4.81k + 1.2k) = 200mV Amplitude (@ 1kHz)
The input resistance well be (4.81k Ohms + 1.2k Ohms) = 6.01kOhms
And all this is functionally not influenced by AIN-.

Now a feedbacked OPAMP circuit works that it regulates it´s output in a way that IN- becomes the same as IN+. So IN- = IN+. This is important for the next step:

Let´s look at the inverting side:
What we know: we have 10kHz @ 1V amplitude at AIN- and we have IN- = IN+ = 200mV @ 1kHz (from above).
Thus we know the voltage across the 4.81k resistor: It is: V_R = (10kHz @ 1V) - (200mV @ 1kHz).
Note: now we have not only the 10kHz for the input, but also the 1kHz for the other input.
Thus the R1 current is the I_R = I_AIN- = V_R / R1 = ((10kHz @ 1V) + (1kHz @ 200mV)) / 4.81k Ohms. So a combination of two currents!

And that´s a rather special case that happens not that often. We have two identical looking inputs.
But the current of the one is only influenced by it´s own input voltage, while the other depends on BOTH input voltages.
The one input influences the other, but not the other way round.

And this makes this circuit rather complicated to analyze.

*****
Here the green trace shows the current of AIN-. It clearly is a combination of both 1kHz and 10kHz.
Hopefully what I mean with "independent view" is visualized with the two frequencies.
The two inputs look equal, but the currents, signal and impedances are different!

Klaus

 

[danadakk]

@danadakk
I have a small question related to post #44 plots. Can you show how the measurements "Impedance @ R1" and "Impedance @ R3" are defined in SIMetrix? With the shown floating source configuration, voltages at AIN+ and AIN- have different magnitude (referenced to ground) due to asymmetrical input, as explained by KlaussST. I'd expect to see this in an impedance measurement.

Here is the common mode case (as opposed to my earlier diff sims) :

Again Zin both inputs same.


Regards, Dana.

 

[FvM]

Sorry, I copied resistor identifiers from danadakk, correspondence to original post #3 circuit seemed obvious to me.

Your simulation is a different approach to visualize how AIN- input current and impedance is affected by AIN+ input voltage. That's the essence of single OP differential amplifier's asymmetrical input impedance.

You say something is unexpected in the simulation results. What is it?

I'm aware that the discussion has diverged from original "maximal OP supply current" problem. I believe, this question has been almost answered by the OP in post #19. Then KlausST mentioned that supply current is also affected by common mode voltage, which triggered the input impedance controverse.

I'm not sure if there are open questions left.

 

[danadakk]

No open left.

Regards, Dana.

 

[KlausST]

You say something is unexpected in the simulation results. What is it?

It´s the 4kOhms Zin at AIN- of the first simulation.

I don´t see the frequency of the AC input sources.
The problem is that the input current at AIN+ depends on AIN+ voltage as well as AIN- voltage.

In other words: if you apply voltage at AIN+ you will also see current at AIN-. The other way round it´s not the case.

That´s the difficulty with this circuit.

What do you say?
When we apply a signal on AIN+ with 1kHz.
Then we automatically get current in AIN- (@1kHZ)
We apply additionally apply a signal on AIN- with 10kHz --> then we get a mixture of two sine signals (shown above)

Now how does the toool treat this mixture signal when calcuating impedance. The applied voltage is 10kHz only, but the current is 10kHz and 1kHz.

Klaus

 

[D.A.(Tony)Stewart]

You got it! The input impedance can´t be the same! They are different. That´s what I´m saying all the time.

Klaus

That still contradicts theory. Since the negative feedback changes the current with voltage to null the differential input according to the gain and signal level, but the input impedances are identical.

You got it! The input impedance can´t be the same! They are different. That´s what I´m saying all the time.

Klaus

Yes but you still do not. both the voltage and currents are different but the input impedance is constant and only equal to the series input R.

Both the currents and voltages are different in each resistor because the inverting input has two active sources In and output feedback while the non-inverting shunt resistor goes to a DC reference.

The ratio of V/I is just one series input R which is constant and it terminates on a null differential voltage. in the classic balanced Differential. Amplifier.

To illustrate. https://tinyurl.com/yot4pnm2

--- Updated Dec 16, 2023 ---

The power dissipations can be different but the single-ended input impedances are matched.

When dealing with very small uV EEG signals and large mV grid stray noise, this balanced impedance is critical to reject the common mode voltage or current induced in cables. R tolerances upset the high CMRR of Op Amps and thus triple OpAmp INstrument Amps are needed to improve on the 34 dB CMRR due to 1% R mismatch. even if the Op Amp had 100 dB CM Rejection Ratio.

e.g. https://tinyurl.com/yuabxydw

Even twisted pairs are not perfectly twisted and on a long cables. The high grid voltages induced as CM can be far more than than Op Amp. is capable of rejecting. Then Balun transfomers are needed to raise the common mode source impedance to re-BALANCE the cable single ended impedance and extract the differential signal from CM noise. well above the wire resistance

My original answer several pages ago was correctly demonstrating how to measure input impedance by adding external matched R's. However, you saw that as just trying to measure CMRR with a common input. That was not my intent. I was showing the waveforms were attenuating equally.

I'd like to see this satisfies your curiosity now.

.

 

[FvM]

both the voltage and currents are different but the input impedance is constant and only equal to the series input R.

Did you review the input impedance cases presented in post #49. They clearly show that AIN+ and AIN- impedances of the discussed single OP differential amplifier are unequal and also depend on relation of common and differential mode input voltage.

That's nothing new, any designer dealing with this circuit should be aware of.

If your result is different, it's most likely due to a different definition of input impedance. I'm looking at the impedance seen "into" both input terminals. A circuit is symmetrical if you see equal impedance of both inputs for all drive cases. That's obviously no the case for the given circuit.

Now how does the tool treat this mixture signal when calcuating impedance. The applied voltage is 10kHz only, but the current is 10kHz and 1kHz.

You can't calculate impedance in frequency domain for a superposition of orthogonal signal components. It can be only calculated for each component separately.

Post #49 simulated "measurements" are performed in AC analysis mode, means a single frequency signal is applied at a time, frequency is swept stepwise. Respectively impedance can be calculated for each frequency, but stays almost constant over a wide frequency range.

The annotated impedance values are picked from the curves, valid for DC to about 100 kHz.

--- Updated Dec 16, 2023 ---

How can we represent the interdependency of terminal input voltages and currents in a more general way? You can setup a 2x2 Z or Y matrix, it will include the cases shown in post #49 or other special cases extracted along the thread discussion.

--- Updated Dec 16, 2023 ---

See https://en.m.wikipedia.org/wiki/Impedance_parameters

--- Updated Dec 16, 2023 ---

If we associate port 1 with AIN+ and port 2 with AIN-, we get the Z matrix

6.01k 0
1.2k 4.81k

 

[D.A.(Tony)Stewart]

Yes I just reviewed it now and the 1st plot is the same as I have been asserting all along.

The 2nd and 3rd plots do not compute input impedance since the feedback current is shared giving a false sense of V/I(R1)= Zin..

 

[KlausST]

Hi,

@tony_lth.
Thanks for still analyzing the problem. I really appreciate this.
Still I can be wrong, you can be wrong, or there simply is a misunderstanding. But we don´t discuss about you and me, we discuss about electronics.
So the lines below are written just to show my view of the things, my calculations and my worries. Not emotional at all. Again, thanks.
***

In the second sim you wrote "Thus false impedance appears lower".
So my worries are: wht is false here?

Let´s go a step back and simplify the circuit. Replacing the two resistors at the IN+ by connecting IN+ to GND.

What we get is the "standard inverting OPAMP circuit" like in text books.
R1 = 4.81k and R2 = 1.2k.
They are connected in series exacly like in the difference amplifier circuit.

The textbooks say the "inverting OPAMP input impedance is R1".
So are they "false", too?
I´m totally with you, that the string has two sources.
That the input current is influenced by the OPAMP´s output. In numbers, input impedance is:
* 4.81k when OPAMP output is connected to R2
* infinite when R2 is diconnected
* 6.01k when R2 is grounded.
So in my opinion there are three values, and each value is "true" for the given situation. Still same two resistors in a string.

Thus I don´t think it`s "false" to say "in circuit XYZ the input impedance is 4.81k"

*******
Usually when we measure input resistance/impedance. We measure the applied voltage, measure the current and divide them.
This goes very well with most circuits.
We always get: I = V / Z.
And according this formula: when V = 0 .... --> I becomes 0 in any case.

When we look at the difference amplifier circuit.
And apply voltage to AIN+: the current follows above formula: I_AIN+ = V_AIN+ / (4.81k + 1.2k). Always. Always... So far so good.
The same is not always true for the AIN- input. It does NOT follow above formula in ANY case. (Still there are cases when it does)
Setup: AIN+ = 1V. (or any other voltage than 0V)
And now you get the "anomaly" of this circuit: When you apply 0V to AIN- --> you don´t get 0mA.
You get 0mA when you apply about 200mV.

This does not follow Ohm´s law. (But still is physically as well as mathematically explainable).
An input usually is a node that only "draws" power, but now this node "delivers" power ... back to the sensor.

I used this circuit many times. And - since most of my applications are focussed on precision - I encountered problems. And wondered myself where they com from.
It MAY introduce errors that are not that obvious.

And this information that this circuit is somehow "tricky" I wanted to give to the OP - and any other readers of this threa - when I wrote the "input impedance statement"

Klaus

 

[FvM]

The 2nd and 3rd plots do not compute input impedance since the feedback current is shared giving a false sense of V/I(R1)= Zin..

Apparently we are having different definitions of amplifier input impedance. For me it's Voltage at input terminal/Current into input terminal. Doesn't matter where the current flows.

Consider the amplifier as a black box with terminals. For the source, only terminal voltages and current matters. It doesn't care what's in the box, if there's feedback etc.

As derived in post #57, differential amplifier input impedance can be better analyzed as two port with Z parameter matrix to describe mutual interdependence of inputs.

Impedance matrix is assuming ideal OP and abstracting from filter capacitors, valid for 0...100 KHz range.

--- Updated Dec 17, 2023 ---

LTspice is capable of calculating Z-matrix for you using .net command

Result:
Z11(vz1): mag: 6010.01 phase: -8.69652e-006° impedance
Z21(vz1): mag: 1200.09 phase: -4.41657e-005° impedance
Z12(vz1): mag: 0.051964 phase: 0.000948551° impedance
Z22(vz1): mag: 4810.61 phase: 1.24365e-007° impedance