The load seen by opamp in non-inverting config. is TWICE w.r.t. the inverting config.?

Load seen by opamp in non-inverting config. is TWICE w.r.t. inverting config.?

Hi! I've come across the result that the load seen by an opamp is radically different depending if it's in inverting or non-inverting configuration. I wonder if I'm making some mistake, or if it's a well known fact and I was just ignorant.
Considering the figures in attachment (resistive feedback and no explicit load are considered for simplicity):

For the inverting configuration:
Vout/Vin = -A.Rf/[Rf+Ri(1+A)] --> -Rf/Ri for large A
Iout/Vin = -(1+A)/[Rf+Ri(1+A)] --> -1/Ri for large A
Rout = Vout/Iout = Rf(A)/(1+A) --> Rf for large A

For the non-inverting configuration:
Vout/Vin = A(Ri+Rf)/[Rf+Ri(1+A)] --> 1+Rf/Ri for large A
Iout/Vin = A/[Rf+Ri(1+A)] --> 1/Ri for large A
Rout = Vout/Iout = Ri+Rf for ANY value of A!

Thus, the load seen by the opamp in the non-inverting configuration is much larger than in the inverting case (e.g. twice as much, assuming Rf=Ri). Can someone please confirm if this is correct (a reference to a book or paper showing this would be GREATLY appreciated), or if there is a mistake?

Thanks in advance!

Re: Load seen by opamp in non-inverting config. is TWICE w.r.t. inverting config.?

Hi,

you are correct.

But "DON`T PANIC". This is just the feedback path. And it should no be a problem for an OPAMP to drive it´s own feedback path.
The feedback path can/should be high ohmic compared to the real load it is able to drive.

Usually what we call "load" is the circuit right to the OPAMP output. (not shown in your circuit)

Klaus

Re: Load seen by opamp in non-inverting config. is TWICE w.r.t. inverting config.?

Thank you for your reply, Kaus. I am dealing with an inverting switched-capacitance (SC) circuit where the input, feedback and load elements are all capacitors of the same order of magnitude. A reviewer of a paper claims the load seen by the opamp should be the series of the feedback and input capacitors (plus the explicit load), but according to the analysis above the input capacitor should have no effect. What can I be missing?

Re: Load seen by opamp in non-inverting config. is TWICE w.r.t. inverting config.?

I think it's ambiguous to designate a higher load impedance as "larger load". The OP output load is actually lighter in non-inverting configuration.

In settled state when the SC charge balancing is finished, Cf has been charged to Vout. Thus in terms of charge balance, Cf is the internal load, to be summed with respective external loads to total amplifier load. But if you analyze the SC amplifier loop gain, the higher impedance of the Cf and Ci series circuit acts as amplifier load.

Re: Load seen by opamp in non-inverting config. is TWICE w.r.t. inverting config.?

FvM said:

I think it's ambiguous to designate a higher load impedance as "larger load". The OP output load is actually lighter in non-inverting configuration.

Click to expand...

Yes, I see your point! (a higher resistance means less current, therefore "less loading", right?)
Irrespective of this, my main doubt is if the input impedance counts or not as loading for the opamp.

FvM said:

In settled state when the SC charge balancing is finished, Cf has been charged to Vout. Thus in terms of charge balance, Cf is the internal load, to be summed with respective external loads to total amplifier load. But if you analyze the SC amplifier loop gain, the higher impedance of the Cf and Ci series circuit acts as amplifier load.

Click to expand...

I understand what you mean, but, if I analyzed the loop gain (after the SC charge balancing is done), shouldn't I arrive to the same conclusion as for the resistive case? Why with capacitors I should see Cin as loading, while with resistors I don't?

- - - Updated - - -

UPDATE: I just found a book [1] stating (not demonstrating) the same as the reviewer comment: the load capacitance is written as Cload_total = Cload_explicit + (Cf in series with Ci) = Cload_explicit + Cf*Ci/(Cf+Ci).
Why in the capacitive case Cin appears in the opamp load expression, while in the resistive case it doesn't? Is my definition of opamp loading wrong?
[1] Gustavsson, "CMOS Data Converters for Communications", 2002

Re: Load seen by opamp in non-inverting config. is TWICE w.r.t. inverting config.?

Why in the capacitive case Cin appears in the opamp load expression, while in the resistive case it doesn't? Is my definition of opamp loading wrong?

Click to expand...

The behavior is the same with capacitive and resistive load, but the answer depends on the exact question. In which relation are you analyzing load? Is it for derivation of transfer function, it need to consider open loop gain, feedback factor, total output load. Ci counts both for feedback factor and total load. Or do you calculate the charge respectively current absorbed by the load in closed loop operation?

Re: Load seen by opamp in non-inverting config. is TWICE w.r.t. inverting config.?

FvM said:

The behavior is the same with capacitive and resistive load, but the answer depends on the exact question. In which relation are you analyzing load? Is it for derivation of transfer function, it need to consider open loop gain, feedback factor, total output load. Ci counts both for feedback factor and total load. Or do you calculate the charge respectively current absorbed by the load in closed loop operation?

Click to expand...

I need an expression of the effective load "seen" by the opamp, to use it as an specification for its design.
You can see my actual circuit in the new attachment below; won't it behave the same as the inverting resistive case? Why in the capacitive case would Cin appear in the expression for the load seen by the opamp during the amplification phase?