I'm trying to figure out how to determine the forward voltage and forward current for overdriven LEDs, and need a bit of help understanding what are the dependent variables and which are independent variables.

For a normal 3mm LED, I understand that I'd set the forward current by looking at the typical forward voltage (usually between 2.0 and 3.5 volts or so), the typical forward current (usually 20-25 mA), and the supply voltage. I'd calculate the remaining voltage and then divide by the desired current to get a resistance value; usually I find these to be like 100-200 ohms or so with a 5V supply.

The problem that I'm working with now is that I'm attempting to overdrive LEDs by a factor of 50x-100x (so, running up to 2 amps through an LED in 100 us pulses, for a duty cycle of 0.01%). I've seen papers suggesting that this should be possible, but I'm having trouble figuring out how to compute the resistance values.

I'm using a circuit, as shown below, to drive the LEDs:

enter image description here

The supply voltage is around 12 Volts, so the capacitors charge up to 12 volts, and then dump current through the LEDs. I want a resistor R2 to regulate the current and provide me a way to measure the current going through the LEDs.

I'm having trouble figuring out what the resistor value should be, because the forward voltage and forward current are both related -- if I try to drive them beyond their "typical" current, the forward voltage should comparatively increase.

The I-V curve for these LEDs is as shown below:

enter image description here

I feel like I'm in a bit of a chicken and egg scenario, where I don't know how to find the forward voltage without knowing the current, or the current without knowing the forward voltage.

How should I approach deciding the value of resistor R2?

  • \$\begingroup\$ Repeat - until, or something approximately. With formualting the equation you would solve the problem, but when you will measure in real world you'll get a slight difference, so don't worry be happy. \$\endgroup\$ Commented Aug 24, 2017 at 19:32
  • \$\begingroup\$ I'd doubt you could put 2A through the diode indicated at 12V, but whatever. If you notice the slope of I/V is pretty linear, as such you can use that slope as the dV/dI as the approximate resistance of the LED when it is on. Then you should be able to figure the rest out... but it will always be a fudgy approximation. \$\endgroup\$
    – Trevor_G
    Commented Aug 24, 2017 at 19:45
  • 1
    \$\begingroup\$ You're getting about 50mA per V after about 3.5V. Considering that 10mA @3.5V are negligible for the calculation, to get 2A you should give 40V+3.5V... \$\endgroup\$
    – next-hack
    Commented Aug 24, 2017 at 20:00
  • \$\begingroup\$ Possible duplicate, electronics.stackexchange.com/a/323200/117785 \$\endgroup\$ Commented Aug 24, 2017 at 20:04
  • 1
    \$\begingroup\$ Why do you wish to do this? Much of the suppled voltage will be dropped by the LED internal resistance and won't contribute to the light output so the overall efficiency will be low. if you just wish to drive it from the 12v supply there are other proven methods that are much better. \$\endgroup\$ Commented Aug 24, 2017 at 20:51

3 Answers 3


We can model our LED as a resistor in series with a diode.

$$V_{LED} \approx V_R+V_D$$

For a diode

$$I=I_\mathrm{S} \left( e^\frac{V_\text{D}}{n V_\text{T}} - 1 \right)$$

$$\frac{I}{I_\mathrm{S}} = e^\frac{V_\text{D}}{n V_\text{T}}-1 $$

Asssuming \$V_D >> n V_\text{T}\$

$$\ln\frac{I}{I_\mathrm{S}}\approx \frac{V_\text{D}}{n V_\text{T}}$$

$$V_\text{D} \approx n V_\text{T} \ln\frac{I}{I_\mathrm{S}} = n V_\text{T} \ln I - n V_\text{T} \ln I_\mathrm{S}$$

For a resistor

$$V_R = IR$$


$$ V_{LED} \approx n V_\text{T} \ln I - n V_\text{T} \ln I_\mathrm{S} + IR$$

The middle term is a constant (assuming constant temperate).

For small currents the first term dominantes the resposense, but for large currents the last term dominates the response, the graph from your data sheet clearly shows the last term dominating, that is the graph is roughly linear over the range shown.

A very approximate reading of your graph gives us a linear equation.

$$I = 40 (V - 3)$$

So at 12V I would expect "only" 360ma.


Peter Green's answer provides a good theoretical analysis. However, whenever you drive a component beyond the manufacturer's data sheet limits, theory must take a back seat to experimentation, and that includes taking the risk that you'll destroy one or more experimental units.

First off, though, you're going to have to change your circuit. You must operate the FET with the source grounded, or the voltage induced in R2 during pulse operation will starve the FET of gate drive.


simulate this circuit – Schematic created using CircuitLab

So, try this. Eliminate VR2. Replace R2 with a 10 ohm resistor. Note that this should be pretty safe, since the worst-case current will be less than about 1/2 amp. Now apply your gate drive and measure the voltage across R2. Since R2 is no longer grounded, this means that you'll have to do a differential pulse measurement, but you can do this by measuring the two sides of R2 on successive pulses and subtracting.

Now reduce R2 to 1 ohm and repeat. If that does not give you your 2 amps (and it won't), you'll need to increase your power supply voltage.


You shouldn't rely on forward voltage of a LED, it will vary with temperature, and LEDs heat up significantly. You should use strictly current control up to natural limit of the power supply. This kind of power supplies are called "LED drivers" these days.

enter image description here

Use of limiting resistors is the most primitive form of current control. They waste power, and are very sub-optimal.

  • \$\begingroup\$ can you use such drivers with pulse-on setups like OP's? i would think the driver would try to push the same current if the "pwm" was on as off, leading to some "wobble". am i off-base? \$\endgroup\$
    – dandavis
    Commented Aug 24, 2017 at 21:30
  • \$\begingroup\$ @dandavis, I can't attest for the quality of this particular sample of engineering, but PWM dimming control is a common method in LED lighting industry. More, I can't endorse the idea of 100X overcurrent of regular LEDs, I haven't seen any manufacturing data to support high-pulsed mode. However, a special class of LED seems to exist, designed for photo flash applications. \$\endgroup\$ Commented Aug 24, 2017 at 21:41
  • \$\begingroup\$ Since the OP specifically refers to 100 usec pulses, I'd suggest that thermal runaway is not a concern, at least as long as the PRF is low enough. \$\endgroup\$ Commented Jan 19, 2019 at 16:20
  • \$\begingroup\$ @Ale..chenski - Sorry. That responded to a comment which was then deleted, not yours. \$\endgroup\$ Commented Jan 19, 2019 at 22:01
  • \$\begingroup\$ I agree that using such short duration pulses will dim the brightness. Such high current will probably fuse open the bonding wires in the LED. \$\endgroup\$
    – Audioguru
    Commented Feb 26, 2020 at 21:19

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