# Overdriving LEDs with high currents

I'm trying to drive LEDs for the purposes of a strobe light. I'm planning on driving the LEDs at ~60 fps, and for ~100 us pulses to provide very sharp strobe images.

If the pulses are very short and infrequent, you end up needing extremely bright LEDs to make up for how infrequent the LEDs are "on" (same as if you were PWM'ing them with a 0.1" duty cycle or so). The typical solution is to overdrive the LEDs by a large factor -- I've read papers about putting 1000x the rated current through an LED, which won't damage the LED as long as it cools for sufficiently long times.

To accomplish this goal, I put together this circuit:

The LEDs are connected via a connector at P4 (labelled "LED_1"). I currently have 3x White LEDs with a forward voltage of 3.5 volts placed there.

I would expect to see (12 Volts - 3*3.5 Volts) = 1.5 Volts across the resistor R5. With a resistance of 0.02 Ohms, I would expect to see 75 Amps placed through the LED -- however, while running the circuit, I find that it's only about 1.3 Amps, with a voltage of 150 mV or so placed across the resistor.

I'm using a FET Driver to drive the MOSFET, so it doesn't seem to be an issue of it not being fully turned on.

Can anyone help me understand why this is?

thanks!

• What does your V_CAP1 voltage look like while you're strobing the LEDs? Commented Aug 9, 2017 at 21:56
• The caps charge all the way up to 12 V; the voltage drops by about 300 mV when the pulse triggers the FET, and then charges all of the way back up to 12 V before the next pulse. Commented Aug 9, 2017 at 22:08
• How is the circuit constructed? I hope you're not trying to run 75 A through a breadboard, but you never know around here. Commented Aug 9, 2017 at 22:25
• seems like there's internal resistance on the LEDs/wiring, which while normally negligible, could be enough to topple the math on your 0.02 ohm resistor... Commented Aug 9, 2017 at 23:00
• Based on Figs 1 & 2 of the IRF540N datasheet, you need more like 8-10 V on Q1's gate to get 75 mA through it. Commented Aug 10, 2017 at 3:01

Specifications for IRF540N show Rdson = 44 mOhm alone, and only at Vgs=10V control signal. So you already should see only 30% of what you expected.

If DRIVER_1 has only 5 V, the FET can barely commute 10 A of current at Vgs=5V, see specifications, Fig.1 and Fig.2, it looks more like a 100 mOhms with this control level.

To conduct 75 A of current, your wires must be short and have a solid AWG rating. If not, every wire segment will add few mOhms into the circuit.

If, God forbid, the whole test was done on a breadboard, then every junction would add 10-15 mOhms into the loop, and I counted about 7 junctions. This is another 70 mOhms.

More, the forward voltage of a LED will be quite higher at high 10X- 100X current pulse than the listed "nominal" 3.5 V, so the voltage drop across the transistor net would be quite less than 1.5 V.

Now you add all this together, and you will have the result.

• Great! This is very useful. So I'm seeing that (1) I should use a 12 V signal for DRIVER_1 in order to commute enough current and (2) I should find the lowest Rdson MOSFET I can find. Everything was done on a PCB with minimal trace lengths that were about 18 mil wide. Commented Aug 10, 2017 at 0:27
• @nathanlachenmyer, if on PCB, then your bigger limitation is likely the forward voltage of LEDs. Their forward voltage (at overcurrent) can be well above 4V, so you might need to jack-up your main supply. Also, I doubt very much that LEDs can handle 1000X overcurrent, even if pulsed. Commented Aug 10, 2017 at 0:38
• How can I tell what the forward voltage of the LED is? Since Rds is so high, I tried removing the resistor and measuring the voltage across the FET when it's triggered high -- it sits at ~360 mV; should that mean that the LED's V_f is 4.66 V? Or that it has (360 mV / 0.044 Ohms) = 8 Amps going through it? Commented Aug 10, 2017 at 18:24
• @nathanlachenmyer, no, you can't rely on 44mOhm value, it could be anything if you have Vgs=5V. The LED's V_f should be determined from manufacturer's specifications, by extrapolating their I-V curves. For example, A Cree 2-W LED has about 100 mV increase for every 400 mA in current increase, so a 2A pulse will have about 500 mV added to the nominal V_f. Commented Aug 10, 2017 at 18:45