I was trying to build a simple RC circuit that causes an LED to fade in and out. However, I have electrolytic capacitors, which means that they have polarity.

If I connect a circuit as follows , is the circuit capable of damaging an LED due to back voltage? (The capacitor is supposed to have polarity, but I didn't know how to put the symbol in CircuitLab.)

I think not, because the other LED has very little resistance when the voltage is at any substantial level.


simulate this circuit – Schematic created using CircuitLab

When making the circuit in CircuitLab, I left the component values (e.g. 100 ohms or 1V) as default, as I'm mainly interested in the generic circuit, not a specific one.

  • \$\begingroup\$ If so, then millions of guitar distortion stomp-boxes based on overdriving back-to-back LEDs (and regular diodes) are in deep trouble. \$\endgroup\$
    – Kaz
    Commented May 21, 2013 at 3:01
  • \$\begingroup\$ One (1) Volt is not enough to light a LED. \$\endgroup\$
    – Turbo J
    Commented May 21, 2013 at 7:59
  • 1
    \$\begingroup\$ @TurboJ OP was pretty clear that his schematic values weren't set appropriately. \$\endgroup\$ Commented May 21, 2013 at 12:35
  • \$\begingroup\$ @TurboJ Sorry for any confusion. Madmanguruman is correct--I didn't set any of the values on the schematic, but left them at the default... But, yes, I can see that 1V wouldn't reach the "turn-on" voltage of an LED. ;) \$\endgroup\$
    – apnorton
    Commented May 21, 2013 at 13:39

2 Answers 2


Note that each diode shunts the other's reverse voltage. Neither diode is permitted to have more reverse drop across it than that which is established by the forward drop of the other. This kind of circuit basically shows a symmetric voltage drop in either polarity.

Back-to-back diodes are used (for instance in op-amp feedback) to create symmetric clipping (before the signal swings to the power rails). The clipping can be softened with resistors.

In your circuit, when the switch is in the battery position, and the capacitor is initially empty, the capacitor initially looks like a short. This is when the most current flows. The conducting diode is protected by the 100 ohm resistor, and the reverse diode is off. As the capacitor fills, the current dwindles down to zero. When the switch is flipped to the ground position, the capacitor empties. This time the other diode conducts and the first diode is off. Initially, the same current flows which flowed during the other diode at the start of charging. The current eventually dwindles to zero.

The resistor protects the diode stack by dropping most of the voltage, and this works in either direction.

To notate a polar capacitor in CircuitLab, right click on the part, click Edit Parameters, and then use the combo box next to the DISP: field to choose polar.

  • \$\begingroup\$ Resistor doesn’t “protect” from reverse voltage. Fortunately 1 V is low enough and won’t break most semiconductor devices in either polarity, but for 12 V DC power (and higher resistance) such circuit could operate on correct polarity but wouldn’t be reverse-proof. \$\endgroup\$ Commented Aug 24, 2016 at 16:08

LEDs have maximum rated reverse voltage, which is listed in the datasheet. For LTL-307EE which is indicated in the schematic, the max rated reverse voltage is 5.0V. See p.2 in the datasheet.

  • 3
    \$\begingroup\$ The maximum rated reverse voltage would not come into play at all, unless it is actually lower than the forward voltage of the opposing diode. \$\endgroup\$ Commented May 21, 2013 at 3:22
  • \$\begingroup\$ @AnindoGhosh, I'd like to clarify terms. What would not come into play is reverse breakdown of the LED. This is because the maximum rated reverse voltage is larger than the forward voltage of the opposing LED. For (hypothetical) LEDs with a 2V forward drop and 1V maximum rated reverse voltage, the reverse-biased LED >>would<< be subject to reverse breakdown and you would have to use a different circuit. \$\endgroup\$ Commented Jul 31, 2014 at 2:16

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