Will this current limiting and current mirror LED driver circuit work?

Thread Starter

LEDgris

Joined Feb 13, 2019
2
So I'm trying to come up with a simple circuit that can provide about 800mA for each of my IR-LEDs. Also I want to use current mirroring to reduce the number of components needed. The voltage source is two 3.6V 18650 lithium batteries in parallell.
It seems that the current limiting part of my circuit functions properly but will the addition of the Q2 mosfet used for current mirroring work?ledcircuit.png
 

AlbertHall

Joined Jun 4, 2014
9,291
I doubt it will work.
How much gate voltage is necessary for the MOSFETs to pass 800mA at almost zero drain voltage?
What is the voltage drop of D1 at 800ma?
Vd1 + 2 * Vgs must be less than the supply voltage.
 

OBW0549

Joined Mar 2, 2015
3,204
I don't see anything wrong with your current mirroring scheme, provided Q2 and Q3 are a matched pair; but there is a bigger issue that will keep this circuit from functioning properly: Vcc is WAY too low.

Node Vg2 sits one Vgs (whatever Vgs turns out to be at Id = 800 mA) above ground; the top end of R1 sits at 0.8 volts above Vg2; and the drain of Q1 is going to be a few hundred mV above that. So unless you can find MOSFETs that will pass 800 mA with an extremely low Vgs, you're not going to have enough headroom for your LEDs with Vcc being only 3.6 volts.

There's also a similar issue with not having enough voltage headroom to power U4.
 

Thread Starter

LEDgris

Joined Feb 13, 2019
2
I doubt it will work.
How much gate voltage is necessary for the MOSFETs to pass 800mA at almost zero drain voltage?
What is the voltage drop of D1 at 800ma?
Vd1 + 2 * Vgs must be less than the supply voltage.
If I understand the datasheet correctly the required gate voltage is about 2.5V and the voltage drop for the diode should be about 2V so it seems that it won't work at all.
Thank you for the help.
 

Kjeldgaard

Joined Apr 7, 2016
396
If there are two IR LEDs the circuit has to drive, then I have a simple suggestion:
Put the two battery cells in series and the two LEDs also in series with one current generator.

With the higher operating voltage, it becomes easier to find operational amplifier and power MOS transistor.
 

AnalogKid

Joined Aug 1, 2013
8,305
1. At 800 mA, what is the LED Vf?

2. Vcc - Vf - 800 mV (V(R1)) = the voltage across the pass transistor. If this is a negative number, the circuit will not work.

ak
 

crutschow

Joined Mar 14, 2008
24,335
Unless you buy matched MOSFETs, the currents will likely be quite different since the Vthreshold of MOSFETs has such a wide tolerance, and those may be difficult to find.
You can use BJT's instead since, even if they aren't matched. the Vbe voltages will still be fairly well matched.
 
Last edited:

crutschow

Joined Mar 14, 2008
24,335
Here's a circuit using two power BJT's and a small BJT.
R1 and R3 help balance the current between the two LEDs.
Note that since both LED currents go through R2, its value must be halved.

upload_2019-2-15_15-47-28.png
 
Top