Bootstrapping an Opamp to get a higher input impedance

[philwinder]

op amp input impedance

Hi again,
Further to my previous topic, I am looking to increase the input impedance of my opamp.

The first stumbling block is that I am using it in an inverting configuration (Charge Amplifier) and I do not beleive that I can bootstrap from the feedback because the inverted voltage applied back into the bias resistors would effectively decrease the input impedance. Is there any way to get around this? (Other than inverting again or using a non-inverting amp)

Also, I understand bootstrapping also increases the noise, but how does this compare to having a low input impedance to a high impedance source? I.e. Would the improved input impedance outweigh the increased noise?

Many thanks,
Phil Winder

 

[FvM]

bootstrap op amp

I can hardly believe, that bootstrapping (although the term is rather vague and may designate different concepts) will help you to design a low noise amplifier.

Very simple, to increase the input impedance, use a higher impedance amplifier, or a different circuit. Unfortunately, neither the imagined circuit nor the source properties, particularly it's impedance have been clearly said in your previous posts. I think, it's time to disclose some information now.

 

[philwinder]

bootstrapped charge amplifier design

Well in this case bootstrapping regards placing a certain amount of positive feedback through the basing/termination resistors to effectively multiply the value of said resistors.
The source is a hydrophone, i.e. Source resistance in the order of tens of Mohm's.
The topology is as stated before, a charge amplifier.

Cheers,
Phil

 

[FvM]

bootstrap operational amplifier

I don't see the purpose of bootstrapping with a charge amplifier. You have Cf and Rf, they are dimensioned according to the intended transfer function. Which resistor shall be multiplied?

 

[philwinder]

bootstrapping op amp

Which resistor shall be multiplied?

placing a certain amount of positive feedback through the basing/termination resistors to effectively multiply the value of said resistors.

The bias/termination resistors. But that is my question. Will it? I'm not sure it will because I do not have any positive feedback available.

Phil

 

[FvM]

op amp bootstrapping

In a charge amplifier, Rf is providing the bias, no other bias resistors are needed, also a charge amplifier has no termination resistor by operation principle. The Rf value is choosen according to the intended transfer characteristics. Why bootstrap?

Obviously, the amplifier has to be select for a sufficient low input current.

If you are thinking of a circuit, that involves additional resistors, you should show a schematic for clarification.

 

[philwinder]

designing charge amplifier

no other bias resistors are needed

Really? What about a DC path to ground? I will give it a shot, but I think that after a while the inputs will shoot off the rails due to the input bias current.

My circuit is as described with a large valued resistor to ground on both inputs, to provide a path to ground.

Cheers,
Phil

 

[FvM]

bootstrap amplifier using op-amp

Sounds as you are speaking of something different than a charge amplifier.

 

[philwinder]

boot strapping of biasing resistors

Sounds as you are speaking of something different than a charge amplifier.

Honest I am! However I have tested the bias resistors theory and you are right in the fact that they are not needed with the charge amplifier topology which must be due to the fact that the feedback path provides adequate bias current.

Thanks!