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I don't know what op-amps have against me, but they don't like me. Yesterday, I wanted to amplify an AC signal coming from a photodiode using a non-inversing configuration mode. I had a lot of noize, so I wanted to inverse it twice.

My problem is the following: When I measure the voltage across D1, it gives me 230mV. The output of my op-amp is a non-inverted signal with an amplitude of 15mV.

NB: It is not high frequency ,the board, the op-amp, the voltage source, the oscilloscope, the resistors, the wires, everything has been changed; it is not failure related. Thank you!

schematic

simulate this circuit – Schematic created using CircuitLab

Shouldn't the op-amp try to compensate to bring the fake ground to a 0V, so output a negative signal? (at the very least)... I expected a 2V over inverted PWM signal...

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    \$\begingroup\$ What are the power rails? \$\endgroup\$ – user133493 Aug 17 '17 at 1:51
  • \$\begingroup\$ a wave generator of 5VDC coming from the oscilloscope himself \$\endgroup\$ – PyThagoras Aug 17 '17 at 1:58
  • \$\begingroup\$ You mean the positive supply is 5V and the negative supply is ground? \$\endgroup\$ – user133493 Aug 17 '17 at 1:59
  • \$\begingroup\$ Yes this is what I mean. \$\endgroup\$ – PyThagoras Aug 17 '17 at 2:04
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    \$\begingroup\$ You can't drive the output outside the power supply rails. If the negative supply is 0V then -2V on the output isn't going to happen. You could disconnect the negative supply from ground, and instead tie the noninverting input to a divider that halves the supply, but then you won't see +-2V swing anyway due to the input common mode range for the LM358 which can't come closer than about 1.5V to the positive supply. Your supply is insufficient for the desired output signal swing. \$\endgroup\$ – user133493 Aug 17 '17 at 2:09
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You can't drive the output outside the power supply rails. If the negative supply is 0V then -2V on the output isn't going to happen. You could disconnect the negative supply from ground, and instead tie the noninverting input to a divider that halves the supply, but then you won't see +-2V swing anyway due to the input common mode range for the LM358 which can't come closer than about 1.5V to the positive supply. Your supply is insufficient for the desired output signal swing.

THanks to @replete for the answer.

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Do you have the diode the correct way round in the schematic?

The way you have it shown it should produce a negative voltage from the diode and that should be amplified by the opamp. I have used that particular arrangement successfully.

Is the output of the diode negative or positive? If it is positive reverse the connections to the diode.

The circuit will not be very sensitive as it will require 100uA of photocurrent to produce 1V at the output.

Also for that particular arrangement R1 is not necessary. The opamp will act as a Transimpedance Amplifier producing 1v per 100uA of photocurrent. The voltage across the diode will be about zero. It will actually be the offset voltage of the opamp which is +/-3mV or so depending upon which particular version.

The sensitivity can be improved by increasing R2. With an LM358 the bias current will cause an error if you use more than about 1megohm.

A better amplifier with lower bias current such as a CMOS version can be more sensitive you need one with a rail-to-rail input capability. (the common mode range of the input includes ground).

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