simulate this circuit – Schematic created using CircuitLab

I found the above circuit at:

with a slight variant at:

and finally with the capacitor polarity swapped (and different R1 resistance, Q1/Q2 BJT type) at:

According to my undersanding and the simulation I've done, the polarity of C1 should be reversed according to the last circuit version above. Otherwise C1 is reverse biased when Q2 is open pulling C1's minus pin close to 7V, while C1's positive pin is close to 0.7V. If this isn't true, could someone explain how the curcuit is supposed to work?

Also if the above is true and C1's polarity needs to be swapped, C1 would still be reverse biased when Q2 is closed, albeit at a smaller absolute voltage (~0.7V vs. ~6.3V). If so, is this 0.7V reverse bias acceptable considering that this is the normal way of operation, that is it's not an occasional event? If so, is there an ideal type of capacitor for this purpose?

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    \$\begingroup\$ I agree with you. The small 0.7V reverse bias is better than the alternative. If the capacitor had significant leakage in the reverse direction the circuit wouldn't oscillate though. I don't expect that would be the case though. \$\endgroup\$ Mar 17, 2017 at 4:06
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    \$\begingroup\$ Please also post the correct picture. Posting only the incorrect circuit diagram will confuse future readers. \$\endgroup\$
    – skvery
    Mar 17, 2017 at 7:06

1 Answer 1


Stripping the oxygen from the aluminum oxide dielectric in an aluminum electrolytic capacitor requires a difference of over 1.6V. The bias of 0.7V should be acceptable, even in extended temperature conditions.

  • \$\begingroup\$ Thanks. Other places mention a 0.5V-1.5V range for acceptable reverse voltage and only for short durations. I suppose using a non-polarized cap would be safer. \$\endgroup\$
    – Imre Deák
    Mar 18, 2017 at 23:26

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