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I am trying to simulate a voltage multiplier circuit for positive DC input using logarithmic amplifiers and inverting adder implemented by op-amps(uA-741) and non-ideal diodes. The first stage takes the natural log of two inputs, which is summed by the inverting adder in the second stage, inverted using a unity gain inverting stage, and finally sent to an antilog stage to get the product of two signals. I have no problem with the temperature-dependency of the configuration and so am trying to keep it simple.

The circuit seems fine until the last stage, where the forward biased diode forces a voltage differential(about 433 mV) between the two input terminals of the op-amp thus saturating the output. This is obviously happening because the input to the exponential stage is too high.

I am aware there was a similar problem posed in the following link: Analog multiplier using logarithmic and anti-logarithmic opamp issue

However, the poster could not provide sufficient information about his inputs, component models etc. to get a proper answer. Someone suggested raising the resistances which for me has failed to solve the problem. Thanks in advance. pspice simulation of analog multiplier

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  • \$\begingroup\$ Having used log/antilog in log-sweep-generator, I recall the need to provide a temperature-compensated bias voltage to the final opamp. \$\endgroup\$ Oct 20, 2017 at 4:48
  • \$\begingroup\$ Related information which may be of use: If you try to build a log/antilog circuit with diodes or bjt (trans-diode), you must be carefull about choice of devices (from the same lot or paired in characteristics ... AND temperature !) And you have to maintain them at the SAME temperature (because of drifting characteristics) by "thermically coupling" ... Or use diodes in the same package ... Some circuits were made with a "isolated housing" maintained at 80°C. \$\endgroup\$
    – Antonio51
    Jul 27, 2021 at 7:54

2 Answers 2

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The problem is that to get 1 V across the diode (D10?) requires about 2 A through it. And 2 A through the 10 kohm feedback resistor R11 would require 20 kV at the output of the final stage amplifier. (And of course in the real world these voltages and currents would quickly burn out these devices)

So you want to reduce the gain of the preceding stages.

You can't do that by just blindly increasing all resistor values. You have to adjust them appropriately to lower the gain. That would mean doing one or more of:

  • Lower R7 or R12
  • Increase R13
  • Increase R8 and R9
  • Increase R2 and R6

(Or, to get your simulation working, use much lower input values, like 0.5 and 0.2 V)

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  • \$\begingroup\$ I've tried this. Reducing the gain of the previous stages doesn't give me the output I want. for example halving the gain gives me an output which is the square root of the product. But I want the output to be PROPORTIONAL to the input \$\endgroup\$ Nov 12, 2017 at 10:10
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    \$\begingroup\$ @MdAjwaadZamanQuashef, the thing is, the value "1 V" isn't especially important to the diode. You're trying to multiply 2 V x 5 V and get 10 V, but the circuit might think you're multiplying 2000 mV x 5000 mV and want to get 10,000,000 mV. You need to work out how to re-scale the output of the log amp to get the proper scaling so that you're multiplying values in volts, not millivolts (or some other arbitrary scale factor). \$\endgroup\$
    – The Photon
    Nov 12, 2017 at 13:55
  • \$\begingroup\$ The place to do this is probably by adjusting R2, R6, and R11 (equally). \$\endgroup\$
    – The Photon
    Nov 12, 2017 at 13:57
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After hours trying to make it works, I found it. The input voltages should be really low, I used a fixed 0.2V (200mV) and a ramp up from zero to 0.3V (300mV). R11 should be 770Ω, or the output stage amplifiers to much. All other resistors as 10kΩ, including the input ones. The output goes from zero to +6V... yeah, the output results is 100 times higher than the input voltage multiplication result. 0.2 x 0.3 = 0.06, but it outputs 6V. You can use a voltage divider (resistive) at input, but remember the series 10k at input, the divider must be way lower than that.

May be exists a way to adjust it more correctly, but it is working very nice like that, I just need to divide the input voltage by 10 in order to obtain the correct output. I wanting to multiply 2V x 3V, use a resistor divider to push down to 200 and 300mV. Thinking about this, your output can goes up to 14.8V with a 15V power supply, your input must not goes over the board, for example 2x7 will do, but 3x7 can not, not also 4x4.

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