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I found this circuit on another thread on this site and I believe it is close to what I need for my own application (Short duration (approx 128 us), low duty cycle (<1%) high power LED pulses). However as I am not so strong on electronics there are a few questions I have to help me understand it?

  1. Why are there 2 capacitors in it?
  2. The forward voltage of the stated LED (4V) seems to be a lot less than the 30V of the PSU. I know to drive it at higher currents the forward voltage needs to be increased but cannot quite see how this is happening in this circuit. I'm assuming the 10M Ohm resistor is a pull down resistor for the MOSFET.

thanks in advance, Simon

enter image description here

Antonio51
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simon mcloughlin
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    Looks like a recipe to burn out the LED very quickly with the initial current peak. Where did you get this one from? – winny Feb 15 '22 at 17:15
  • @winny, here is the original post https://electronics.stackexchange.com/questions/222569/high-power-pulsed-led-driving-circuit – simon mcloughlin Feb 17 '22 at 11:41
  • Thanks. Please calculate peak current and compare with the datasheet. – winny Feb 17 '22 at 12:39

3 Answers3

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Why are there 2 capacitors in it?

These capacitors, initially are 0V. When power is applied, it flows through the 10Ω resistor and charges these. If left for long enough, they will reach 30V, same as the supply.

There are two, and of differing values, because:

  • Most 100nF capacitors have very low ESR or equivalent series resistance. This means the capacitance inside has very little resistance associated with it, so can generate very high current spikes.
  • Most 10µF capacitors have moderate ESR.

So the usage of both is implying that one is good at giving a really strong, very short pulse to get the LED lit quickly, while the other will help extend the pulse length.

The forward voltage of the stated LED (4V) seems to be a lot less than the 30V of the PSU. I know to drive it at higher currents the forward voltage needs to be increased but cannot quite see how this is happening in this circuit.

Assume the capacitors are fully charged. When the MOSFET gate is "fired" or raised up to 10V, the drain (top) and source (bottom) pins essentially become a low-value resistor, perhaps 0.1Ω. Thus, the charged capacitors dump their charge, very quickly, into the LED, through the MOSFET, through the 10mΩ resistor, and back the the capacitors.

Then the capacitors will be nearly depleted, so will recharge again, from the battery and 10Ω resistor.

I'm assuming the 10M Ohm resistor is a pull down resistor for the MOSFET.

This is not a 10 MEG Ohm (10,000,000Ω), but a 10 milli-Ohm (.01Ω) resistor. Capital M is always millions. This is here for two reasons:

  • Limit the maximum current that can go through the LED
  • Measure the voltage across the resistor during the pulse, and from this, the current (and then power) can be calculated.

Note I'd consider this a fairly dangerous circuit. There is little preventing the LED from being damaged; if the supply voltage is too high or capacitor values too large, the current spike could destroy the LED.

rdtsc
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  • I think the question about the capacitors is why is there 2 rather than 1. – Henry Gomersall Feb 15 '22 at 16:35
  • @rdtsc thanks for the detailed response. The datasheet for the IRF520 says the rds on is 0.27Ohms, so with this I should be able to calculate the voltage at the LED? – simon mcloughlin Feb 17 '22 at 11:44
  • Well I'm afraid it isn't that simple. LEDs are non-linear devices (for example, one could appear "open" when powered from 1.5V, slightly conductive at 1.8V, and almost a dead short at 2.5V.) Since they are a diode, they have an exponential current with applied voltage; 1.5V=1µA, 1.8V=10mA, 2.5V=5A etc.) These caps are attempting to force 30V into it - waaay more than it could handle continuously. However the capacitors are small (not much charge) so will *very* quickly discharge, leading to a very strong and very short pulse. So the "voltage" on the LED will never get to 30V. – rdtsc Feb 17 '22 at 13:21
  • @rdtsc, ok thanks. So for that very strong, short pulse, is it valid to assume the voltage on the LED is circa 30V? – simon mcloughlin Feb 17 '22 at 18:33
  • No, it will rise, perhaps above 2.5V, but likely not much more than that before the caps run out of charge. This happens on the order of nanoseconds and microseconds - very fast. If you have an oscilloscope, connect it across that 10mΩ resistor, and set the vertical scale to say, 500mV/division. If the pulse appears one division high, then 0.5V was dropped across 0.010Ω. I=E/R = 0.5V/0.01Ω = 50A peak pulse current. – rdtsc Feb 17 '22 at 22:07
  • ok thanks, I understand it a lot more than I did and feel this answer has been the most helpful so have accepted it – simon mcloughlin Feb 18 '22 at 12:06
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The resistor is 10 milli-Ohms. The small capacitor is for high frequency smoothing and the larger one seems to be a charge reservoir. The worst case current is 3A through the 10 Ohm current limiting resistor.

I am assuming the specified current of the LED is around 300mA? So this dumps 10x that current. If the driver stays ON it will burn out the LED, so be careful

Dirk Bruere
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1

cannot quite see how this is happening in this circuit

To well seeing what happens, make always a simulation.
The "energy" of the capacitor is discharged into the LED, the recharged full.

enter image description here

Note that in your case, it should be this ... enter image description here

Antonio51
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  • thanks for the response and simulation. Does the LED in the simulation have the same characteristics as the one in the original circuit (Luminus SST-90). The peak current in the simulation seems very high (100+A). The original poster suggested around 50 amps was measured through the 10.0m ohm resistor. – simon mcloughlin Feb 17 '22 at 16:11
  • Sorry, I don't have SST-90 in my database. It is only a generic "LED". I changed rs=100 mOhm and BV=50 parameters, Level 1. – Antonio51 Feb 17 '22 at 16:17