I was working on a project regarding building a high power LED driver as a excitation light to excite the florescence in the plant tissue. I want to turn the LED on and off at around 200Hz~10KHz.I came up with the following circuit: enter image description here

After finish building the circuit, I want to measure the voltage across the LED. First of all, I am not exactly sure if the way I measured is correct. I put the two probes in the anode and cathode of LED separately and use math function in oscilloscope to subtract it and then I got the wave form of voltage across the LED. BTW, I don't have differential probes.

The waveform I expected to get should be when I turn the MOSFET off, the voltage across LED should be 0 and when I turn the MOSFET on, the voltage across LED should be almost like 14V.

However, I end up getting something like this: I apply the PWM signal to the gate of the MOSFET at 200Hz shown below(yellow wave)

Then the green wave is LED's cathode and the blue wave is LED's anode shown below

After using math function subtract it, I got the pink waveform like this:

enter image description here

As you can see here, the LED seems does not fully turn off. When the MOSFET is off, there are still some voltage across the LED.

Anyone familiar with why it causes this problem? Anyone has any idea how to solve this problem and to modify the circuit?Do you have any other way to drive and turn the lexuon rebel LED on and off?

I appreciate any help you can give to me!

The LED I use is luxeon Rebel Royal Blue(three of them in series )

The MOSFET I use is STP14NK50ZFP


1 Answer 1


You should measure the voltage across the 3 ohm resistor rather than across the LED. That will represent the current, and current is pretty much proportional to brightness.

Make sure that your 3\$\Omega\$ resistor is non-inductive (not wirewound) type.

Voltage is only loosely related to brightness in an LED. If you insist on measuring the voltage across the LED, put a 200\$\Omega\$ 1W non-inductive resistor across it.


simulate this circuit – Schematic created using CircuitLab

Plot without R4

enter image description here

With R4

enter image description here

  • \$\begingroup\$ Thank you! I am still confused that how to I verify the LED has been fully turned off when the MOSFET is off? \$\endgroup\$
    – haharat
    Aug 7, 2015 at 18:35
  • \$\begingroup\$ If the current is close to zero, then the LED is fully turned off. \$\endgroup\$ Aug 7, 2015 at 18:38
  • \$\begingroup\$ The voltage across the resistor is around 0V when the MOSFET is turning off. But can you explain again why measure the voltage across LED does not make sense to say that the LED has been turned off? \$\endgroup\$
    – haharat
    Aug 7, 2015 at 19:21
  • \$\begingroup\$ The MOSFET has D-S capacitance and the LED has capacitance (and your probe has a high impedance) so the LED voltage will tend to go somewhere in-between the LED Vf(on) and 0V, then slowly change to a different voltage based on leakages and input impedance (probably close to 0V). \$\endgroup\$ Aug 7, 2015 at 19:46
  • \$\begingroup\$ Please see edit above showing a simulation with and without the resistor. The 10M represents the probe impedance. You can see the LED voltage drops to only about 600mV during the 50usec off period (but the current is close to zero). With the resistor, the small amount of energy in the capacitance is quickly drained off and the LED voltage drops to close to zero very quickly. \$\endgroup\$ Aug 7, 2015 at 19:58

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