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

 

[Bjtpower]

Hi,

a remark on the simulation. First: well done!

you see that I_CC (= current on VCC) carries AC while I_EE does not.
If you shift V_IN (common mode as well as differential mode) you may see the opposite (I_EE carries the AC) or both carry AC.

Klaus

Hi,
How to shift that? i did not get it
Can you guide how to that?

 

[KlausST]

Hi,
How to shift that? i did not get it
Can you guide how to that?

example:

AIN+ = 3V, AIN- = 2V --> Vdiff = +1V --> let´s say the ADC_input is 1.234V (I did´n calculate this)

now change (shift) the common mode voltage by +3V:
AIN+ = 6V, AIN- = 5V --> still Vdiff = +1V .. and still you expect the same ADCinput of 1.234V

The
* input for the first OPAMP is different
* the output of the first OPAMP does not change
* the input and output of the second OPAMP does not change.

Thus the supply current for the first OPAMP changes .. but not for the second OPAMP.

**********

Now use AIN+ = 3V, AIN- = 4V. (change differential voltage)
--> supply current of both OPAMPs changes
(and output, too)

or more general:
* differential voltage changes both OPAMP supply currents
* common mode changes first OPAMP supply current only

*******
Again you ask new questions, without answering my questions. This is time consuming because we need to guess maybe this situation, maybe that situation, a lot of guessing.
A lot of brain work, a lot of writing .... and maybe 80% of we guess and write is not relevant in your application´s situation.
Please understand that this behaviour seems to be disrespectful (regarding the time we spend) ... I know it is not meant this way, but it makes the one or the other person stop from responding.

Klaus

added:
All could be done with a single OPAMP.
But in favour of equal input_impedance and measurement precision I recommend you to use an instrumentation amplifier.

 

[danadakk]

Correction to earlier post stating Zin is different for Ain+, Ain-, its same.

What is the input impedance of a differential amplifier?

I have this circuit: How would we calculate the input impedance of the differential amplifier? I have read answers that says that the inverting impedance side will just be R1=200 ohm. I have also ...

electronics.stackexchange.com

SIM confirms this :

Regards, Dana.

 

[KlausST]

SIM confirms this :

What exactly does it confirm?

I don´t understand what the SIM actually wants to show/measure.

*****
In my statement input impedance of AIN+ is not the same of AIN-
...
Don´t you need individaul measurements for AIN+ and AIN- to confirm?

Let´s take the setup of your SIM:

1)
Connect AIN+ to GND, connect AIN- to GND.
Measure the currents at each input individually.
Both should show close to zero. Equal so far. But without current and without voltage one can not calculate resistance.

2)
now leave AIN- at GND.
apply different voltages at AIN+ and measure the current at AIN+.
1V --> 166.4uA according Ohm´s law: R = V/I = 1V/166.4uA = about 6k (V_out = 0.249V)
2V --> 332.8uA = about 6k (V_out = 0.499V)
5V--> 832uA = about 6k (V_out = 1.247V)
-5V --> -832uA = about 6k (V_out = -1.247V)

3) now simply swap channels
leave AIN+ at GND
apply different voltages at AIN- and measure the current at AIN-.
1V --> 207.9uA according Ohm´s law: R = V/I = 1V/166.4uA = about 4.8k (V_out = -0.249V)
2V --> 415.8uA = about 4.8k (V_out = -0.499V)
5V--> 1040uA = about 4.8k (V_out = -1.247V)
-5V --> --1040uA = about 4.8k (V_out = 1.247V)

My conclusion: The input impedance of AIN+ is about 6k, while the input impedance of AIN- is about 4.8k. --> not equal.
You can do the same with AC and get the same results: 6k vs 4.8k

Now you can say: But the output voltage is correct in both cases. Why care about input impedance then?
Fair enough!...As long as the source impedance is ideal zero Ohms, there is no problem. (have you ever seen anything "ideal"? ;-) )
Let´s say you have 1V at AIN+ and 1V at AIN---> difference = 0 ... thus you expect 0.000V at the output.
Now install a 1k as source impedance at AIN+ --> Vout = -0.036V
But when the 1k is in the AIN- path then --> Vout = +0.034V (the absolute value differs by about 2mV to the above result)
It isn´t even symmetric!

To avoid these known errors .. I recommended to use a true instrumentation amplifer that does not suffer form these problems.
This is my recommendation for "hobbyists", because it is simple and accurate. Use an INA ... and don´t care.

For sure an experienced electronics designer may say "I know what I´m doing, I know the flaws and errors of the circuit, I know how to calculte them, I know how to handle them ..".
Well, nothing against it. Maybe it saves a few cent per device.

Klaus

 

[D.A.(Tony)Stewart]

Use an INA ... and don´t care.

This OA has 130 MHz bandwidth , much more than an INA.

 

[KlausST]

This OA has 130 MHz bandwidth , much more than an INA.

True.

The OA gain is set to about 0.25 and the bandwidth is externally limited to 1.33MHz.
I didn´t do a serach, but I guess there should be INAs that satisfy this.

Klaus

 

[D.A.(Tony)Stewart]

* that AIN+ has not the same input impedance as AIN-.
*

When used as a balanced Diff Amp the Differential Rin will be 2x Rin (=4k81) since the differential input is a virtual null from negative feedback.

The Rin of AD7484 (Vin) > 10 Meg // 35 pF with 1uA bias current.

 

[KlausST]

When used as a balanced Diff Amp the Differential Rin will be 2x Rin (=4k81) since the differential input is a virtual null from negative feedback.

The Rin of AD7484 (Vin) > 10 Meg // 35 pF with 1uA bias current.

This is not what I´m talking about:

I´m referring to AIN+ and AIN- of schematic of post#3.

What I say: R_AIN+ is not the same as R_AIN-.
See my example:
Set AIN+=0 and apply a signal on AIN-. Measure voltage and current of AIN- --> calculate R_AIN-
Set AIN-=0 and apply a signal on AIN+. Measure voltage and current of AIN+ --> calculate R_AIN+
They are not the same! One is 6k the other is 4.8k.

I´m looking for an application note.

Klaus

 

[danadakk]

What exactly does it confirm?

I don´t understand what the SIM actually wants to show/measure.

*****
In my statement input impedance of AIN+ is not the same of AIN-
...
Don´t you need individaul measurements for AIN+ and AIN- to confirm?

Let´s take the setup of your SIM:

1)
Connect AIN+ to GND, connect AIN- to GND.
Measure the currents at each input individually.
Both should show close to zero. Equal so far. But without current and without voltage one can not calculate resistance.

2)
now leave AIN- at GND.
apply different voltages at AIN+ and measure the current at AIN+.
1V --> 166.4uA according Ohm´s law: R = V/I = 1V/166.4uA = about 6k (V_out = 0.249V)
2V --> 332.8uA = about 6k (V_out = 0.499V)
5V--> 832uA = about 6k (V_out = 1.247V)
-5V --> -832uA = about 6k (V_out = -1.247V)

3) now simply swap channels
leave AIN+ at GND
apply different voltages at AIN- and measure the current at AIN-.
1V --> 207.9uA according Ohm´s law: R = V/I = 1V/166.4uA = about 4.8k (V_out = -0.249V)
2V --> 415.8uA = about 4.8k (V_out = -0.499V)
5V--> 1040uA = about 4.8k (V_out = -1.247V)
-5V --> --1040uA = about 4.8k (V_out = 1.247V)

My conclusion: The input impedance of AIN+ is about 6k, while the input impedance of AIN- is about 4.8k. --> not equal.
You can do the same with AC and get the same results: 6k vs 4.8k

Now you can say: But the output voltage is correct in both cases. Why care about input impedance then?
Fair enough!...As long as the source impedance is ideal zero Ohms, there is no problem. (have you ever seen anything "ideal"? ;-) )
Let´s say you have 1V at AIN+ and 1V at AIN---> difference = 0 ... thus you expect 0.000V at the output.
Now install a 1k as source impedance at AIN+ --> Vout = -0.036V
But when the 1k is in the AIN- path then --> Vout = +0.034V (the absolute value differs by about 2mV to the above result)
It isn´t even symmetric!

To avoid these known errors .. I recommended to use a true instrumentation amplifer that does not suffer form these problems.
This is my recommendation for "hobbyists", because it is simple and accurate. Use an INA ... and don´t care.

For sure an experienced electronics designer may say "I know what I´m doing, I know the flaws and errors of the circuit, I know how to calculte them, I know how to handle them ..".
Well, nothing against it. Maybe it saves a few cent per device.

Klaus

The sim show's the current thru the input R1 and R3. Shows it AC wise to be the same.
So if the input V is same to both, and same current is delivered thru the input Rs, also
both same value, that tells me, and the link I posted of another site showing Z equality,
that input Z of Ain+ = An-

Of course if you change the input values the Z changes, but the point was counter to your
earlier post.
In single stage OpAmp then yes, errors in R matching, in your case adding 1 K, totally
hose the IA behaviour. If we go to the 3 OpAmp IA, with its inputs buffered, and throw
in your 1k, we get an error V that can or can not be of concern. But input Z there is
essentially same as well, no current flows into the buffers (OK femto, pico, nano depending
on part).

One does not need to excite the inputs separately, assuming they are driven by V sources
which sim does.

Nothing is ideal in our designs, this OpAmp, like all others, has many flaws. But with matched
R's the input z of the OpAmp configed as a diff amp is same for both inputs.

Dana.

 

[D.A.(Tony)Stewart]

This is not what I´m talking about:

I´m referring to AIN+ and AIN- of schematic of post#3.

What I say: R_AIN+ is not the same as R_AIN-.
See my example:
Set AIN+=0 and apply a signal on AIN-. Measure voltage and current of AIN- --> calculate R_AIN-
Set AIN-=0 and apply a signal on AIN+. Measure voltage and current of AIN+ --> calculate R_AIN+
They are not the same! One is 6k the other is 4.8k.

I´m looking for an application note.

Klaus

Post #3 is a balanced differential input for both stages. Not unbalanced. Input currents are matched within Iio limits.
If R28 =1k2 was replaced by two 2k4 resistors (to 2.5 and 0V) then the 2nd amplifier is redundant if +/- input labels are swapped if one wanted to use single supply 5V or similar bias optimization with balanced impedance.

 

[KlausST]

Hi,

I´m not talking about common mode impedance.
I`m not talking about differential mode impedacne.

I´m taking about the individual impedances of AIN+ vs AIN-.

***
I now understand that your SIM shows the common mode impedance measuremetn setup, where AIN+ is connected to AIN- ...
I guess you agree that this situatin never happens in a real application, since the output is always expected to be zero in this special case.

***

I gave simple example how to measure the impedance of AIN+.
I gave simple example how to measure the impedance of AIN-.
They show different results!

Of course if you change the input values the Z changes, but the point was counter to your
earlier post.

I did not change the part value for the measurements above.
I set input voltages that are realsitic in an application.

But then I added "source impedance" to show that now you clearly see voltage error at the output caused by the different input impedances (previousely calcualted).
This just proved that there is a flaw in the given circuit.

Klaus

--- Updated Dec 15, 2023 ---

Post #3 is a balanced differential input for both stages. Not unbalanced. Input currents are matched within Iio limits.

Prove it by gounding one input and applying 1V to the other input and mesure the input current.
If you show me a simulation that gives the same current for AIN+ as for AIN-.

Don´t talk just what you "think". Give facts.
I´ve given my math.

Application note: https://www.ti.com/lit/pdf/snoa621
Last sentence of chapter 6 - The Difference Amplifier:
"Care must be exercised in applying this circuit since input impedances are not equal for minimum bias current error."

Klaus

 

[D.A.(Tony)Stewart]

e.g. Note that I have injected +/-5V of common mode 60 Hz on the balanced impedance.

 

[KlausST]

e.g. Note that I have injected +/-5V of common mode 60 Hz on the balanced impedance.

Again: this is not what I´m taking about.

 

[danadakk]

Again: this is not what I´m taking about.

The AIN's correspond to my sim R1, R3 inputs.

If you alter the schematic as show in POST #3 by adding your 1K of course the circuit
behaves differently.

@OP, lets state clearly for all, if you unbalance the input source Z to the circuit in POST #3
then you have in fact changed effective input Z of AIN's. But if your source Z is either a V
source, or a non ideal V source with some of its own Z, but both AINs , eg. their sources,
have equal Z, then the circuit Zin is same for both inputs. 3 OpAmp IAs (diff amps) solve that
problem by buffering the inputs such that (for most purposes) source signal Z becomes
irrelative.

Dana.

 

[KlausST]

Hi,

If you alter the schematic as show in POST #3 by adding your 1K of course the circuit
behaves differently.

Please read my posts! I didn´t alter by adding "1k" for the impedance measurements!

For the lazy ones (sorry had to say that) here a DC test and an AC test.
If you don´t see that the input currents and thus the input impedances are different .. then .. I´m speechless

And yes - I also feel sorry for the OP. As every one can see: I did not add anything. Twice the absolutely identical circuit.
Once the input signal is fed into AIN+ and at AIN-. Nothing fake, just the truth.

Klaus

 

[D.A.(Tony)Stewart]

Klaus,
We are discussing the same input impedance property, single or differential. Note the 50% reduction is matched.

 

[KlausST]

We are discussing the same input impedance property, single or differential. Note the 50% reduction is matched.

No!
Would you recommend the OP to connect AIN+ with AIN- in his application... as you do?
Such a setup is useless for a real signal measurement!
It´s nonsens!

Your setup is useful for common mode suppression measurement.
It´s not an individual input impedance measurement ... R_AIN+ is not equal R_AIN-
My math shows, my simulation shows.

I mean: You are free to do your CMRR measurement. Fine. But it has nothing to do with my input impedance statement.


Klaus

 

[D.A.(Tony)Stewart]

OK

 

[KlausST]

Hi,

You say "individual" input .. but you connected them. How can this be individual?

You say "in my example" (your example). Yes. True. Correct. Your example. Yours. But it does not show the individual input impedances. It does not follow my statement.
You may prove me that snow is yellow .. but has nothing to do with individaul input impedances.

Why don´t you show a simple simulation that mesures AIN+ input impedance, while the other input is GNDed?
Then do the same with swapped inputs .. to measure AIN- input impedance.

No - no assumptions. Math, physics, simulation... all show that the individual input impedances are different.

To all others following this thread:
Im not saying the measurement results of Tony´s simuation is wrong. They are correct. Absolutely.
The thing is he is referring to a special case where "both inputs are connected".
While I refer to "individual" input situation.

And it´s not that I want to be disrespectful against anyone. It´s not a personal problem at all, it´s a technical question that needs to be elaborated.

I know both paths look so "equal": input --> 4k8 --> (OPAMP input) --> 1k2.

The difference is: in one case (AIN+) the center node is "passive". It follows the expectable resistive voltage divider rule.
In the other case (AIN-) the center node is "regulated", it acts like a low impedance node. Force current into it´s node and no voltage move will happen.
What happens is that the OPAMP regulates it´s output in a way to compensat the "introduced" current to maintain the voltage.

In one path the GND is at the end of the string.
In the other path the "virtual GND" is at the center of the string.

Klaus

 

[D.A.(Tony)Stewart]

The non-inverting side has one source.
The inverting side has two sources (input & output via Rf)
Yet both resistors share a common virtual null.

So how can each single input impedance can be the same but the voltage drop and current waves are different?
I leave this for you to resolve.