Differential Amplifier Calculations

[strahd_von_zarovich]

Hi Everyone,

I am trying to understand what did I do wrong. When I set all the capacitors the same value and impedance of R2||C5 to R1 , the output should be V2-V1 and if they are also equal then the output should be zero. However, in LTSpice, I cant get the same result.

 

[barry]

Not sure, but you might need a DC path to ground on the (-) input.

 

[danadakk]

Not sure, but you might need a DC path to ground on the (-) input.

There is thru the fdbk R.

Regards, Dana.

 

[KlausST]

Hi,

the shown circuit has different input impedance for V1 and V2.
Thus a signal at V1 generates not necessarily the same output (ignoring the sign) as a signal at V2.
Source imedance also has impact.
If you need equal input impedance (for equal behaviour) you need an instrumentation amplifier topology.

Klaus

 

[danadakk]

The input Z appears to be the same as measured at OpAmp inputs :

Which makes sense due to - fdbk forcing the inputs to see the 79 K R due to
virtual ground.


Regards, Dana.

 

[KlausST]

Hi,

@OP:
Z is not R. Even if you calculate the same value. You ignore the phase shift.
I´m not sure what you want to achieve.
But if you want a true "symmetric difference amplifier" this topology won´t work.
Read documents about
* your single OPAMP difference amplifier circuit, it´s benefits and drawbacks
* true instrumentation amplifier

@dana:
I talked about V1 and V2. Not the very OPAMP inputs.
Do a test:
* put V1 to GND, input 1V RMS sine to V2, measure the current (or the impedance) at V2
now do it the other way round:
* put V2 to GND, input 1V RMS sine to V1, measure the current (or the impedance) at V1

Is the result the same?

Klaus

 

[danadakk]

Yes, both V1 and V2 exhibit the same Z. Looks like :

When I did it at OpAmp inputs since they are at the same V that produces the same
result as in that case V1 and V2 are at Z = 0 due to driving V source, and the circuit,
bing driven by effectively a V source to set the OpAmp inputs at same V shows same
behaviour, eg. passive (effectively) two port with forward and reverse transmission same......

Regards, Dana.

 

[KlausST]

Hi,

This is not the test I was asking for.

And I only see C3, so only one input.

It does not compare the one with the other.

***

Overlooking the obvious: the input impedance of a difference amplifier

Monolithic <a href="http://www.ti.com/hpa-pa-opamp-diffamps-thehub-20150814-lp-differamps-wwe">difference amplifiers</a> are integrated circuits that incorporate an operational amplifier (op amp) and four or more precision resistors in the same package. They are incredibly useful building blocks...

e2e.ti.com

below Figure 1 it says:
Although most designers are comfortable with this simple building block, I find that one aspect of their use is commonly overlooked: the two inputs of a difference amplifier have different effective input resistances. By “effective input resistance,” I mean the input resistance resulting from both the internal resistor values and the op amp’s operation.

Klaus

 

[Easy peasy]

R1 needs another C5 across it

 

[FvM]

Yes, obviously the equivalent of C5 is missing to give zero output with V1=V2. I must confess that I didn't go through the lengthy calculations, but the symmetry condition can be seen without calculation.

 

[danadakk]

Hi,

This is not the test I was asking for.

And I only see C3, so only one input.

It does not compare the one with the other.

***

Overlooking the obvious: the input impedance of a difference amplifier

Monolithic <a href="http://www.ti.com/hpa-pa-opamp-diffamps-thehub-20150814-lp-differamps-wwe">difference amplifiers</a> are integrated circuits that incorporate an operational amplifier (op amp) and four or more precision resistors in the same package. They are incredibly useful building blocks...

e2e.ti.com

below Figure 1 it says:
Although most designers are comfortable with this simple building block, I find that one aspect of their use is commonly overlooked: the two inputs of a difference amplifier have different effective input resistances. By “effective input resistance,” I mean the input resistance resulting from both the internal resistor values and the op amp’s operation.

Klaus

I grounded one input, drove the other, as you requested. Then switched the driven and grounded
input to do the other side. But just posted one graph as both graphs/results the same.

So then ran it, not using Z probe, but just monitoring current into input caps, and
there is a difference, one sees NI input 80 uA at 1 Mhz, the other INV input 109 uA a
1 Mhz. Note because of input Cap sizes, down <= 10 Khz plots the same all the way
down to DC. Eg. INV nd NI Zin essentially the same. That I believe is because of resolution
of graphs and cursor tools to read out graph values.

So conclusion is INV Zin < NI Zin as Klaus posted. Note this would be affected by
OpAmp used, and its internal comp, but just the values found, not the relationship
that INV Zin < NI Zin


Regards, Dana.

--- Updated Aug 10, 2023 ---

Correction, just re-read the 1 Hz values and they are 125 pA NI input, 129 pA INV input.

So inconsistency with TI ap note, but is SPICE model of OpAmp the issue here....?


Regards, Dana

--- Updated Aug 10, 2023 ---

Here is Easy Peasy add 20 pF more to C across R1 on the NI path.

Regards, Dana.