Will this circuit work for the purpose of making a high voltage LED flash on and off at 10kHz?

I have a NPN transistor to turn the LED on and off. The pad is where the LED will be soldered into as they are attached with wires.enter image description here


1 Answer 1


No, your frequency calculation is way off for 10kHz.

That will give you closer to 7Hz, more than 1000x slower. You can reducer the capacitor to about 7nF (7000pF) to get closer to 10kHz. For the bipolar 555 in particular it's important to bypass the supply. I show 100uF electrolytic in parallel with 100nF ceramic or film, but even a 10uF electrolytic will be better than nothing.

enter image description here This is the updated version enter image description here

  • \$\begingroup\$ Oh really, oooops. What values would I need? I guess I did a wrong calculation somewhere. \$\endgroup\$
    – M2T156
    Apr 24, 2019 at 14:18
  • \$\begingroup\$ The resistor values are okay, reduce the capacitor to more like 7nF (7000pF). Also 100nF in parallel with 100uF electrolytic across the power rails. \$\endgroup\$ Apr 24, 2019 at 14:21
  • \$\begingroup\$ Thank for the feedback, yes I did realise the value of C was too large. Would you mind sketching on the schematic the changes needed please. Thank you Spehro. And the NPN would work this way to turn LED on and off? \$\endgroup\$
    – M2T156
    Apr 24, 2019 at 14:25
  • \$\begingroup\$ Thank you very much and it is all very clear. I will show you my PCB once it is constructed on easyEDA. \$\endgroup\$
    – M2T156
    Apr 24, 2019 at 14:34
  • \$\begingroup\$ Spehro, while you are at correcting errors , delete R13 and Q1 which are redundant, since OUT can drive ±200 mA and R14 suggests we do not expect much current but can't be sure what LED current and Vcc is being used., but adding 100uF to Vcc is overkill \$\endgroup\$ Apr 24, 2019 at 14:38

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