I was having a look at some guitar effect circuits and came across with a company that offers PCBs for some well known circuits. I have noticed that most of ther circuits have a 1N4001 (D1) placed between VCC and GND, without any kind of resistor in series.

This one, for example: circuit

At first I thought it was meant to avoid reverse biasing the transistors, but risking the power supply by placing it in parallel doesn't make much sense to me.

I think it would be better to place it in series to the source or, if needing to keep that parallel setup, at least put a resistor in series to the diode to avoid a direct short circuit.

I just would like to know if the circuit is wrong or I am missing something.

Thank you very much.

  • 2
    \$\begingroup\$ That -9V is wrong. It should say +9V. (Just look at the nearby polar capacitor, and the direction of the NPN transistor.) The diode is then reverse-biased. If 9V is above its reverse break-down voltage, then no current flows across the diode (except maybe some negligible surface leakage current). So it's like the diode is not there. \$\endgroup\$
    – Kaz
    Oct 1, 2012 at 3:38

5 Answers 5


I nearly always put such a 'fools diode' in my circuit, even when I am breadboarding. Reverse power will kill most of the chips I use. Over-current protection is a function for the power supply, not for the powered circuit. This makes more sense:

  • most power supplies are current limited in the first place, think of basic batteries, and 7805-like regulated power supplies (but NOT NiMh accu packs!)

  • Having the protection in the attached circuit will not protect against an error in the circuit, or the wring to the circuit, so it must still be complement by a protection in the supply

  • I build much more powered circuits than I build power supplies, so it makes more economic sense to put the current limiting in the supply.

When I use a battery (NiMh etc) that does not inherently limit its current I add a fuse, polyfuse or the like (often the battery supplier has already done this). A series resistor is often not practical because it makes the voltage the circuit sees dependent on the current it consumes.

You circuit is fed by 9V, most likely it is intended to be fed by a 9V battery, which is current limited by itself (unless maybe when it is an NiMh 9V pack? or are those protected in any way?) If you are really worried I suggest you either

  • put a fuse or polyfuse between the battery and this circuit, or

  • as your circuit probably draws very little current and probably can live with a small voltage drop, put an schottky diode in series (an 1N5819 will drop ~ 0.4V at 100mA). This circuit was probably designed long before schottky diodes became common place.

  • \$\begingroup\$ Wouldn't 1N5817 be better here? It will have lower voltage drop. \$\endgroup\$
    – AndrejaKo
    Sep 30, 2012 at 20:13
  • \$\begingroup\$ @AndrejaKo : yes, but IME the 1n5819 is a yellebean. Anyway, it was just an example. There are probaly even better ones than an 15817, but a 0.4V drop is probably OK (I assume the circuit is for a 9V battery, which drops quickly anyway). \$\endgroup\$ Sep 30, 2012 at 20:22
  • \$\begingroup\$ Fool's Diode: +1. \$\endgroup\$ Oct 1, 2012 at 13:39
  • 1
    \$\begingroup\$ @romkyns: At the cost of a diode voltage drop. \$\endgroup\$
    – nibot
    Oct 25, 2012 at 21:59
  • 1
    \$\begingroup\$ @nibot: and the loss of a common ground. \$\endgroup\$ Oct 26, 2012 at 6:45

I sometimes do this. The diode shorts out the power supply. During a short circuit, a good power supply should just trip without blowing up.


Putting it in series means you have an additional voltage drop. And for battery powered application even the 0.2V of schottky diode counts.


The diode is reverse biased when the circuit is powered up correctly. So it's for protection against reverse connection. But you'd expect some current limiting and other protection in there too.


I think it would be better to place it in series to the source

Yes Sir, You are right, its in series ........Please look at it again.

The point above D1 is a node voltage and not some power source.It leaves your relative ground at +9 V with respect to the node marked as -9 V.So D1 and C7 are parallel components which have a potential difference of 9 V.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.