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I wanted to create a square wave that alternates between positive and negative voltage with a full cycle time of 8ms and a 50% duty cycle. So is there a way I could make a circuit that takes in 0V and a voltage V_in (roughly in the voltage range of 1V-7V), and that outputs the square wave as well as the inverse of the square wave. The load I'm going to put on the signal is very minimal (below 10 mA).

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  • \$\begingroup\$ How accurate do your duty cycle and frequency need to be? The solution will vary widely based on this information. Also will this need to run off of a battery or is power not a consideration? \$\endgroup\$
    – Reinderien
    May 22, 2021 at 3:48
  • \$\begingroup\$ Also, do you already have a negative supply or does this need to make one? \$\endgroup\$
    – Reinderien
    May 22, 2021 at 4:26
  • \$\begingroup\$ Would an H-bridge solution work for you? electronics.stackexchange.com/questions/278645/… \$\endgroup\$
    – ErikR
    May 23, 2021 at 12:05

3 Answers 3

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You can do it with a dual opamp. The left opamp is a relaxation oscillator - can adjust the frequency with R1 - and the second opamp just inverts.

The AC-coupling at the outputs creates the bipolar output swing you wanted. The total output swing will be a volt or two less than Vin.

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Here is a simulation, with Vin = 10V:

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LATER

@Reinderien correctly pointed out this solution doesn't swing from Vin to -Vin. For that we need a voltage boost.

The following circuit will do that. It again uses an opamp for the relaxation oscillator: this should be a rail-to-rail type to get maximum voltage swing. The square wave feeds a cheap voltage inverter generating a negative voltage nearly equal to -Vin. A BAT54S does the two diodes in one cheap package. Dual comparator LM393 uses this negative voltage as a reference, the pullup resistor determines the high voltage.

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And here is the simulation (using available opamp and comparator in the library but just use an LM393 and any rail-to-rail opamp).

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A CD4047 IC has an RC oscillator, a divide by 2 stage making perfect 50% duty cycles and and direct and inverted outputs. Exactly what you need.

CD4047

CD4047 again

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  • \$\begingroup\$ Would an 8ms cycle mean 125hz, and if so, what resistance would c1 need to be? Also, I am kind of a beginner, so what exactly should I connect where in terms of 0V, V_in, signal, and inverse signal? \$\endgroup\$
    – Miguel Flores-Acton
    May 22, 2021 at 1:49
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    \$\begingroup\$ Are you sure that this is exactly what the OP needs? He is asking for a zero centered output waveform with a negative minimum. \$\endgroup\$
    – Reinderien
    May 22, 2021 at 4:36
  • \$\begingroup\$ @reinderien: with capacitors in series at the outputs, that would work, wouldn't it? \$\endgroup\$
    – mmmm
    May 22, 2021 at 8:12
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    \$\begingroup\$ @mmmm not on its own, because then the voltage maximum would be half of what's needed \$\endgroup\$
    – Reinderien
    May 22, 2021 at 13:36
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As Reinderien made a topic that is related to this, I will also share my answer here. Note that this circuit only creates an additional inverted square wave from a unipolar input square wave V1 (without a negative supply), but doesn't create the square wave itself. So it is only part of the solution.

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