DC Amplification using OpAmp

1. DC Amplification using OpAmp

I have a small signal of -10 mV. I need to amplify it to 4 V which need an amplification factor of 400. I would like to use inverting amplifier whose gain is R2/R1, where R2 is a feedback resistor. I am wondering what is the best practice to get the amplification by a factor of 400.

1- Use one OpAmp to get gain 400. R1 = 100 , R2 = 40 K.
2- Use two OpAmp in series. Each with a gain of 20. R1 = 500, R2 = 10 K.

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2. Re: DC Amplification using OpAmp

Hi,

One Opamp. Because itīs DC. A second OPAMP just adds more errors.

The two OPAMP solution is good for AC, when you get problems with to the gain-bandwidth limit of a single OPAMP.

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your calculation is true if the source resistance is way below 100 Ohms.
For exact overall gain calculation you need to add the source_resistance to the input_resistance.

Example:
If your source resistance is 10 Ohms, then combined with your 100R and 40k this gives just a gain of 40k/(10R + 100R) = 363.

If you want a gain of 400 then you need to subtract the source resistance from the (ideally) calculated input resistance. 100R - 10R = 90R

Klaus

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3. Re: DC Amplification using OpAmp

As an alternative, you have mentioned two inverting opamps in serties. From this, I derive that the sign of the amplification (plus or mins) does not matter, correct?
In this case, why not use a single non-invering opamp stage?
The advantage: Because of the very high input resistance the internal resistance of the signal source does not matter at all.

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4. Re: DC Amplification using OpAmp

Originally Posted by LvW
In this case, why not use a single non-invering opamp stage?
The advantage: Because of the very high input resistance the internal resistance of the signal source does not matter at all.
What about the biasing of the OP Amp. Non inverting input's resistance should be the same value as the parallel of the feedback resistance and the inverting input's one.

5. Re: DC Amplification using OpAmp

Originally Posted by CataM
What about the biasing of the OP Amp. Non inverting input's resistance should be the same value as the parallel of the feedback resistance and the inverting input's one.
I know what you mean - you are concerned about offset correction.
However, this does not change the high-resistive input for a non-inverting configuration.
1 Megohm plus 1 kohm gives 1.001 Megohm.

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6. Re: DC Amplification using OpAmp

Hi,

I am not sure if I understood source resistance. I guess it is the resistance of the source, for example signal generator, right ? Usually the standard signal generator has 50 Ohm output characteristic impedance. Is this what you mean source resistance ?

Is it true for all OpAmps that the input impedance is high and the output impedance is low ?

7. Re: DC Amplification using OpAmp

Hi,

Is this what you mean source resistance ?
Yes.

Is it true for all OpAmps that the input impedance is high and the output impedance is low ?
Yes. For the Opamp.
But not for the circuit. Especially the inverting circuit. It's input resistance is the 100 ohms resistor.

Klaus

8. Re: DC Amplification using OpAmp

Originally Posted by CataM
What about the biasing of the OP Amp. Non inverting input's resistance should be the same value as the parallel of the feedback resistance and the inverting input's one.
That's only if the opamp's input bias current is sufficient to generate a significant offset voltage through that parallel resistance value.

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