Below is a circuit I put together that uses a 555 timer and an LED Driver:

555 timer + LED driver

But the strange thing I see is that the timer does not follow the 1.1*R*C equation. The led remains lit for much longer.

I tested this circuit by removing the MOSFET and triggering pin 2 manually. I get the same effect. When I completely remove the LED driver and replace it with just a resistor and LED, it works. I also replaced the 100uF with a 10uF and I see the same problem.

Is the inductor affecting the timer? Is it not a good idea to use such a boost converter + driver to light the LED? Should I replace this circuit with say a 2 transistor constant current source circuit?

I have independently tested the LED driver and it works very well - even with a 1.5 V battery. For the given circuit, I am using a 9V battery.

  • \$\begingroup\$ The way the boost converter is powered from the output of the 555 looks fishy. When the output turns off the inductor will continue to suck current from it or from Vcc of the ZXSC300. Maybe a diode to ground would help. \$\endgroup\$
    – starblue
    Mar 24, 2013 at 8:37
  • \$\begingroup\$ @starblue : Sorry, could you please explain where exactly the diode should be connected? \$\endgroup\$
    – M-V
    Mar 29, 2013 at 6:36
  • \$\begingroup\$ put a diode between pin 3 and 1. \$\endgroup\$
    – wooddy
    Jul 23, 2020 at 10:47

1 Answer 1


It's possible that switching noise from the driver is affecting the 555. Consider adding a capacitor between Vcc of IC1 and ground. I would try a value between 10 μF and 100 μF.

  • \$\begingroup\$ Even a LED with a resistor can cause a problem. Years ago I made a battery-powered op-amp guitar preamp, it worked fine but when I added an LED to it to show it was on, it got very noisy (current through LED is noisy apparently). Adding decoupling caps helped only a little. I took it out, and figured out much later that my single-rail op-amp design was bad, and it was strongly amplifying any AC signal on the V+. \$\endgroup\$
    – greggo
    Mar 23, 2013 at 17:37

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.