Laser Diode Driver with MOSFET

Thread Starter


Joined Aug 12, 2022
Hi all,
I want to built laser diode driver with MOSFET. I built a driver which is attached. I need to drive laser diode up to 2A, and this current must be regulated by external voltage which varies between 0-5 volts.
I built attached circuit in real life. As pot value changes, output voltage varies between 0 to 5 volts, but when I connect multimeter in series to output pins, it measures 150 mA. I need this to be 2A. How could I make such that thing.
Thanks in advance.

Note: I used IRFZ44 MOSFET >> gives 250 mA when multimeter connected in series to output pins
IRF840 MOSFET >> gives 150 mA when multimeter connected in series to output pins



Joined Mar 10, 2019
What is the power source? If a standard 9V battery, the more current you draw, the lower the voltage will be. A lower voltage will give a lower voltage at the gate of the MOSFET, causing it to conduct less. A standard 9V battery cannot supply 2A steady state.
Those MOSFETs will drive 2A easily, provided you have at least 10V on the gate, and a power supply that can supply 2A of current.
MOSFETs are not particularily linear. That is, they conduct a bit around 2 to 4V, but have a high RDSon. That generates some heat. At 10V on the gate, they are fully "on", and the RDSon is basically what the datasheets show. With your setup, too many unknowns, and the RDSon for your setup is unknown as well, hence maybe the low currents.
PS: The IRF840 has a much higher RDSon resistance at 10V
MOSFETs do not adjust "voltage" (actually current) that well with just a pot on the gate....
Last edited:


Joined Nov 6, 2012
Using 4 or maybe 5, 9-Volt-Batteries, wired in parallel, might provide ~2-Amps for a few minutes,
otherwise, you'll need to get a large Li-Po Cell, and a "Logic-Level" FET, to provide that much Current.

You will have to have an Op-Amp, 3 Resistors, and a generic-Diode,
configured as a Current-Regulating-Circuit, with the "Logic-Level" FET,
to achieve anything remotely close to Linear-Output vs the position of the control Pot.

A 2-Amp Laser-Diode can cause instant, and permanent, damage to your Eyes.
Be careful.

Can You build a simple Perf-Board-Circuit ?


Joined Oct 7, 2019
The circuit in the first post, dose not have a 0-5V input. It is not temperature stable. Will not regulate.

I propose a circuit like this. The opamp needs to be a R-R input or at least an opamp that has an input common mode range that includes the negative rail. (some amps can not see inputs near ground) Should be a low voltage ltype.
I used a transistor not a MOSFET. It needs to have high gain and could be a Darlington type.
As the input goes from 0 to 5V the current goes from 0 to 2A.
I reduced the supply down to a USB cell phone charger. The transistor/MOSFET was getting too hot.
Supply variations from 4V to 6V do not effect the current. Temperature does not effect the current.


Joined Nov 6, 2012
A few simple changes to make the above Circuit a little simpler ...........
1) Replace R-1 with a 10K-Pot.
2) Replace R-2 with a 10K-Resistor.
3) Place a generic-Diode across the Pot to act as a 0.7-Volt Shunt-Regulator.
4) Connect the top of R-2 to the Supply-Voltage.

Now, R-4 can be used to determine the maximum Current possible.
Maximum-Current will be when there is 0.7-Volts across R-4.
.......... A maximum of 2-Amps has been requested ............
0.7V divided by 2A = 0.35 Ohms,
this is the ideal resistance-value of R-4,
and will result in a very linear Current-range of zero to 2-Amps.

The Circuit will provide equal results with a "Logic-Level" FET, or a BJT Transistor.
Use a "Rail-to-Rail" Op-Amp, such as a MCP6241 for less than a 5-Volt Supply,
and for a Power-Supply-Voltage of up to 16-Volts use a TLV2371IP .

Either type of Power-Transistor will require a Heat-Sink,
especially with higher Power-Supply-Voltages.

When using a BJT-Transistor,
the Power-Supply Voltage must exceed the Forward-Voltage of the Laser-Diode
by at least 0.7-Volts to achieve a full 2-Amps of Output-Current ,
a MOSFET doesn't have this Voltage loss, and will tend to run slightly cooler.
This only makes a significant difference near the maximum 2-Amps Output of the Circuit.


Joined Aug 21, 2008
The circuit in post #4 will work with a MOSFET.

Does "MOSFET only" mean that no other components may be used in addition to a MOSFET?

Is this homework?


Joined Nov 6, 2012
Sorry I cannot understand your words. Please draw a schematic.
Where does the "0-5V input" connect?
R-1 is replaced with a Potentiometer.
The Wiper-Terminal of the Pot is connected to the "Non-Inverting-Input" of the Op-Amp.
The Pot will have a generic-Diode across it pointing downward,
this will result in the Pot having 0.7-Volts across its Resistance-Element.
The Input to the Op-Amp will then receive anywhere from zero-Volts to 0.7-Volts.