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When working with a LM324N opamp, what is the proper way to configure it to accept a bipolar signal, and output it with a positive offset so that it can be inputted into a unipolar ADC? From my understanding, the LM324N is a bipolar opamp, so I was wanting to power the opamp with a single positive supply, and still achieve this.

In addition to this, I would like the input impedance to be 15 megaohms, and the output impedance be 1 megaohm while achieving a gain of 10.

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  • \$\begingroup\$ Probably need to configure it as a summing amplifier. It won’t work by directly using a single supply. \$\endgroup\$ Commented Jan 21 at 5:41
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    \$\begingroup\$ The LM324 has too high an input bias current to provide a 15 megaohm input impedance. You need to use one with a very low input current, such as a CMOS or JFET type. \$\endgroup\$ Commented Jan 21 at 6:33
  • \$\begingroup\$ For a complete answer, we will need to know the exact details of the input signal voltage range and frequency, and desired output signal range. It is strange to ask for a 1MΩ output impedance, usually you want as low an output impedance as possible. Why do you need 1MΩ? \$\endgroup\$ Commented Jan 21 at 7:16

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You can do something like this, using a RRIO CMOS-input op-amp, not an LM324.

schematic

simulate this circuit – Schematic created using CircuitLab

This accepts -5V to +5V in and produces (almost) 0V to 5V out, with a 15MΩ input resistance.

An LM324 would not work well, even with a balancing resistor on the other input, the offset could be more than +/-500mV (and would likely vary a lot with temperature).

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OA1 is the input buffer and gain of 10, OA3 is the reference voltage/2 buffer with gain of 1, OA2 is the adder of two signals.

schematic

simulate this circuit – Schematic created using CircuitLab

EDIT: Perhaps replacing 100k resistors with 20k and R2 with 2.2k would give better results.

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