MOSFET for switching single LED

Question

I try to design a cost-effective but still "reasonably engineered" way of driving a single white LED with a micocontroller with the following "specs":

• For the single white LED no extra boost converter should be necessary (as a side note: I found that most simple led boost converters only work from >=2LEDs in series, so even If I used such as boost converter I would need to use two LEDs even if i only need one)
• The supply voltge is nominal 3.3V, but could be set higher to about 3.6V
• The single white LED should be run on a rather small but fixed current, let´s say somewhere between 3 - 5mA (across different PCB charges)
• The white LED is available in binned forward voltage groups and one group limits VF to be in between 2.8 and 3.2 volts (the unbinned range is up to 3.6V where I see no way of achieving what I want without a dedicated boost converter)
• Due to various reasons I'd prefer not to drive/sink the LED current directly from the uC pin, since even if 5mA are small, the driving results in some "ground offset" on the uC pin which could disturb another pin).

This seems pretty simple but is challenging for me because after some number crunching I seem to miss something.

I assume the most simple form of driving a LED with constant current, i.e. a series resistance, the LED and a low-threshold MOSFET as low-side switch. As an example, I assume using the https://www.mouser.com/datasheet/2/302/NX3008NBKW-842194.pdf

To achieve a specific, constant led current (I assume 4mA as target) I think I need to use the MOSET operating in saturation(current not dependent on VDS but settable via VGS and I assume the VGS set by the uC is constant).

I assume the uC can drive the gate of the MOSFET with 3.0V. To operate in saturation I need to keep VDS > (VGS - Vth). If I understand correctly, in this case, I actually want the minimum Vgs,th to be actually rather high, so that the saturation condition is fulfilled and the VDS voltage (voltage drop across the MOSFET) is still low.

According to datasheet I see a Vth,max = 1.1V, and Vth,min of about 0.6V, so let´s take 0.5V. Would mean that VDS must be greater than VGS-Vth = 3.0-0.5 = 2.5V. And there starts the crisis. Even with a forward voltage binned between 2.8 and 3.2V that would mean I need a minimum VDD of 2.5+3.2V = 5.7V, not even accounted for a series resistor to the LED?

There must be obviously something wrong in my thinking. I am pretty sure of having seen single blue LEDs (which are to my knowledge the same as white LEDs in terms of forward voltage) driven with the described topology. Is my assumption wrong that I need the MOSFET in saturation for setting a constant current? Driving a single LED with a few mA can not be that complicated, can it? ;-)

Answer 1

Why not use a charge pump?

V1 is a square wave source, for example the output of a PWM, can be buffered by a logic gate.

"X1" is a dual diode, make sure the two Vf in series reduce output voltage enouhgh so the LED is off when it needs to.

This is a somewhat constant current driver. If R2 and C2 are large enough to smooth the LED current well, then it will have a constant voltage output at VCC minus the two diode Vf. In this case R2 sets the current across LED Vf dispersion.

However if we remove C2 and set R2 so that C1 is discharged at each cycle then each cycle pumps an amount of charge equal to C1 times (Vcc minus one diode drop). Charge times frequency is current, so it makes a constant current drive. It's a bit more efficient than the one above, but the LED will blink at PWM frequency.