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I'm trying to build a analog multiplier that takes in two voltages and produces the product of them. I have tried to simulate a circuit with finds the log of two voltages, sum them and then find the antilog of the result. It's required that I use this methods as it is part of the objectives in the project. However the antilog op amp stage saturates after the summing amp stage. This is the basic layout of the circuit: enter image description here The test voltages at the input was at around 10mv at each input and the supply voltages of the op amps are 6 and -6 volts respectively.

Why does the anti-log amplifier saturate and how do I fix it?

Edit: I added a differential stage which I take the log of a separate voltage and now it works. But I still can't figure out why the original circuit is advertised as the basic log-antilog multiplier.

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  • \$\begingroup\$ Are you aware that the circuit you've linked can only multiply signals that remain positive - it's not a true "4-quadrant" analogue multiplier \$\endgroup\$
    – Andy aka
    Mar 8, 2014 at 14:37
  • \$\begingroup\$ Are you also aware that the output voltage will be negative? \$\endgroup\$
    – Dave Tweed
    Mar 8, 2014 at 15:42
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    \$\begingroup\$ yes am aware of the two things above. what i am solely concerned about is the magnitude of the currents as i'm to use it in a calculator. the signs can be worked around. \$\endgroup\$
    – user38411
    Mar 10, 2014 at 10:23

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The antilog amplifier saturates because it cannot drive the output voltage to the level required to make the current through the feedback resistor match the current through the diode.

To fix this, you either need to raise the supply voltage, or scale the currents down by raising the values of all of the resistors.

Are you simulating with "ideal" opamps or models of real opamps? In the latter case, you may have offsets and bias current issues you need to address as well.

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  • \$\begingroup\$ hmm... i think this is this might be the case. i have to try this out when possible. \$\endgroup\$
    – user38411
    Mar 10, 2014 at 10:29
  • \$\begingroup\$ I tried it out but it did not work. I raised the resistances all the way up to 10 Mega ohms. No difference \$\endgroup\$
    – user38411
    Mar 12, 2014 at 10:47
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    \$\begingroup\$ In that case, you need to give us more information. For example, instead of showing us a generic circuit, show us the actual circuit you're using. Show us the simulator data. Until you do these things, we're more in the dark than you are. \$\endgroup\$
    – Dave Tweed
    Mar 12, 2014 at 11:52

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