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I want to create a simple circuit to flash a LED with one op-amp. I already simulated the circuit (astable multivibrator) but there is one problem with the supply. I don't know what to change so that it works for single supply. However, when I use dual-supply it will burn my LED when placed like on the circuit below. The output will oscillate between 10V and -10. Unfortunately, -10V is below the max. allowed voltage for an LED (which is maybe 5V?).

enter image description here

I have two questions now:

  1. What would I have to change when I want it to work with dual-supply (currently it would kill my led)?
  2. How do I need to change the circuit to get it work with single supply?

Thank you!

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  • \$\begingroup\$ You want to know how to get it to work for both dual and single supply while it currently works for neither? \$\endgroup\$ – Mast May 26 at 5:38
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What would I have to change when I want it to work with dual-supply (currently it would kill my led)?

Put a reverse protection diode across the LED (or maybe a reversed LED of a different colour to get a nice flashing two-colour effect).

How do I need to change the circuit to get it work with single supply?

Try this: -

enter image description here

The frequency might need a bit of an adjustment if it's critical.

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  • \$\begingroup\$ Great, thanks. For the dual-supply case: would it also be possible to connect the LED to VCC (instead of GND) + increase the resistor in front of it? For your solution I would need a special diode which doesn't break on -10V, right? \$\endgroup\$ – Spacey3 May 22 at 15:25
  • \$\begingroup\$ @Spacey3 - " I would need a special diode which doesn't break on -10V, right?" Nope. The new diode will "point" in the opposite direction as the LED When the LED is on, there will only be 2 - 3 volts across it (since the LED is on) which will not stress any diode you can buy. When the op amp is putting out -10 volts the new diode will be turned on, and will only have about 0.7 volts across it. What you do have to worry about is the op amp. a 400 ohm resistor will draw more than 20 mA, and not many general-purpose op amps will be happy trying to do that. \$\endgroup\$ – WhatRoughBeast May 22 at 15:51
  • \$\begingroup\$ @Spacey3 you can connect the single LED load to the positive rail and not worry about the reverse protection. Make sure LED anode is towards the positive end. \$\endgroup\$ – Andy aka May 22 at 17:01
  • \$\begingroup\$ I still haven't figured out how the reverse diode should work... do you mean a second diode parallel to the LED? Because when I add another diode above or below (as a serial connection) the led wouldn't light anymore at all because each diode blocks one "wave" \$\endgroup\$ – Spacey3 May 22 at 18:34
  • \$\begingroup\$ Yes, a parallel diode. Sorry I wasn’t clearer. Having said that I did say “across” and that implies parallel lol. \$\endgroup\$ – Andy aka May 22 at 18:55

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