1
\$\begingroup\$

Every op-amp has error. We all know that. So logically, the more op-amp used, the less accuracy we will have, right?

In the circuit ,there are 2 parts: first is the current to voltage converter, second is the voltage amplifier. So let start by assuming that the current through LED as detector is -1 nA, variable resistor is 10 kOhm, so the final output should be 10V.

But I wonder: does that really need for the second op-amp? I mean, if we replace 1MOhm resistor with 10Mohm resistor (so that we wont need the second Voltage amplifier part) we will get the same value, and because fewer elements are used, the accuracy is better. Is that correct?

Also, I tried to read a lot but I still don't understand. The capacitor here acts as a filter or in order to stablize? What if we don't have capacitor?

| improve this question | | | | |
\$\endgroup\$
  • 2
    \$\begingroup\$ Final output will be 10 mV and non inverted. A single stage would be inverted. That might be important. \$\endgroup\$ – Andy aka May 1 '17 at 8:40
5
\$\begingroup\$

A perfect op amp has infinite open-loop gain, zero input current and offset voltage, infinite bandwidth, and zero output impedance. Of course real op amps have none of those attributes, but approach them as more negative feedback is applied.

If the open-loop gain is much higher than the closed loop gain then gain accuracy is determined by the external components in the feedback loop. Therefore 2 op amps set for low gain should be more accurate than a single op amp set for high gain.

Input offset voltage and current is mostly a function of the first op amp. The second op amp may have the same figures, but they are much smaller relative to the signal. The second op amp acts as a buffer which isolates the first op amp from load impedance variations.

Then there are practical considerations. A 10MΩ pot (if such a thing is available) may be less stable and harder to shield from EMI. In your circuit both op amps are inverting the signal, so the result is a non-inverted output. If that is what you want then you need two op amps anyway.

| improve this answer | | | | |
\$\endgroup\$
2
\$\begingroup\$

It does not necessarily follow that more op-amps = more error. For example, in your circuit the gain is split between two op-amps which means that there is more gain available - which can be a factor since compensated op-amps start to roll off gain at about 10Hz.

In your example again, say the op-amps are the same and have input bias current of 100pA. The error contribution due to input bias current in the first amplifier is 10% of your full scale signal. The error contribution due to input bias current of the second amplifier is less than 0.000001% of full scale. So if it makes the circuit more convenient (and it does) to have an adjustment after the first stage there is no significant disadvantage.

A frequently true (but not always) situation is that most of the error occurs in the first stage of amplification and you can pretty much ignore errors in following stages. That's because the signal is often much smaller at the input.

You really have to run the numbers to know where the errors are and how they fit into your error budget.

The capacitor is to compensate for the photodiode capacitance so that the amplifier does not oscillate. It is likely more important with a good photodiode which will have a large area of silicon and thus high capacitance. So it might work without the capacitor

| improve this answer | | | | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.