Single transistor constant current power led driver.

I've seen a lot of constant current led driver circuits, however i've never seen anyone using just one transistor.

With the right resistor at the base of a transistor, you can determin the current flowing through it, right? Why does it not seem possible to make a constant current power led driver this way?

heres a circuit which I have used, if you use an LED in place of the 2 diodes, the temperature coefficient of the transistor tends to cancel out. You just adjust the value of the emitter res to set the current through the LEDs which can be run from an unregulated supply

Does the "33R" get the entire current from the led's through it?

If so, i'll need a high wattage resistor for that, right?

All you need is a votlage reference - that can be diodes, other transistors, zeners, true voltage references, or even a resistive divider (not as good).

Sometimes people use leds as voltage references too.

To be honest, i don’t quite get what the schematic, and the voltage reference does, could you explain it to me?

Inevitableavoidance: To be honest, i don't quite get what the schematic, and the voltage reference does, could you explain it to me?

Let's see if I get it (I'm not the best at electronics so I might be wrong...):

The transistor's resistance depends on the difference between base and emitter (Vbe). In this circuit the emitter isn't connected to ground so the voltage difference depends on the ratio of LED resistance to emitter resistor (they make a voltage divider). ie. Vbe isn't 1.4.

If I raise the supply voltage, the voltage at the transistor's emitter goes up, Vbe goes down. The transistor's resistance increases, the LED current goes down.

If I lower the supply voltage, the voltage at the transistor's emitter goes down, Vbe goes up. The transistor's resistance decreases, the LED current goes up.

i don't quite get what the schematic,

The two diodes function as a voltage clamp. When drive goes h igh (sufficient for the diode to conduct), the base is held at approximately 1.4v (forward voltage drop for the diodes). So the voltage on the 33r resistor is 1.4v - .7v(the transistor's Vbe) = 0.7v. So the current goes through the transistor and the led must be 0.7v/33 = 20ma. Give or take a few.

Essentially it is using the diodes as a voltage reference. As such, you can replace the diodes with other voltage references, like a zener, or a led, etc.

That worked well.

I made a little experiment: I used 2 LED's on the base and a 330Ohm resistor on the emitter. With a LED on the colecctor I measured 0.9 mA With a light bulb (4,8V 0,5A) I measured 1.0 mA

Inevitableavoidance: Does the "33R" get the entire current from the led's through it?

If so, i'll need a high wattage resistor for that, right?

Yes it does, you will need to work out how much power the resistor dissipates depending on the current, I squared R

i don't quite get what the schematic,

Think of it like this. The 33R resistor drops a voltage according to how much current is flowing. This in turn raises up the voltage on the emitter and throttles back the base current until an equilibrium is reached. If more current flows through the LEDs, because they heat up, then there will be more current through the 33R and the voltage across it would like to rise but this turns off the transistor a bit more and so in effect nothing happens. The resistor is acting as a feed back mechanism for the transistor and thus supplying a constant current in the light of any changes in the load. The value of this constant current is controlled by the size of this resistor. You want more current, you make that resistor smaller so more current has to flow through it in order to develop sufficient voltage to throttle off the transistor.

Grumpy_Mike: The resistor is acting as a feed back mechanism for the transistor and thus supplying a constant current in the light of any changes in the load. The value of this constant current is controlled by the size of this resistor. You want more current, you make that resistor smaller so more current has to flow through it in order to develop sufficient voltage to throttle off the transistor.

So...if I know what the 'drive' voltage is (eg. 5V for an Arduino output pin) I don't need the diodes between the transistor base and ground. I can replace them with whatever resistor gives me 1.4V at the transistor base?

That would give me a constant current with three resistors and a transistor.

You could but it would not have the degree of stability that the diodes have with the voltage being independent (ish) of the current through them.

I have one more question to Boffin1's schematic. I have tried with a emitter resistor at 5 Ohm and different supply voltages, and I get these results

Supply Voltage Collector current mA 3.8 280 4.5 310 7.6 450

I use a BC337 and to LED's on basis

Are these values as expected ? I was hoping for a constant current independent of voltage to drive my 1W LED's

Erni: I use a BC337 and to LED's on basis

Are these values as expected ? I was hoping for a constant current independent of voltage to drive my 1W LED's

One Watt?

You probably need a lower value resistor between 'drive' and 'base' or the transistor can't turn fully on.

One Watt?

Yes 350 mA - 3V

You probably need a lower value resistor between 'drive' and 'base' or the transistor can't turn fully on.

Ok, I wil try. In the above example my base current is 1.6, 2, 5 mA, with a 1kOhm resistor, so I am in OutputPin safe area

Thanks for your answer

I use a BC337 and to LED's on basis

Not really sure what that means.

Are these values as expected ?

No. If you have the circuit properly configured, you should see very little variance on the collector current.

Not really sure what that means.

a missing W

I use a BC337 and two LED’s on basis (from base to ground)

No. If you have the circuit properly configured, you should see very little variance on the collector current.

That was what I expected too

Not sure what leds those are: they can render the circuit not working, depending what they are.

If you look into how the circuit works, it requires that those diodes (or leds) are sufficiently conducting when the drive goes high. That means that combined, those diodes cannot drop more than 5v (from a 5v arduino), and the resistor should be sufficiently small to allow the conduction.

It also requires that the collector sits higher than the base so the supply voltage - voltage on your load is higher than the base.

You will find that lower voltage drop over the diodes works better. I would have used just one led - two leds add no value in their circuit.

You probably need a lower value resistor between 'drive' and 'base' or the transistor can't turn fully on.

Ok, I wil try. In the above example my base current is 1.6, 2, 5 mA, with a 1kOhm resistor, so I am in OutputPin safe area

Thanks for your answer [/quote]

Also...grab a multimeter and check that the voltages on collector, base and emitter all make sense.

dhenry: It also requires that the collector sits higher than the base so the supply voltage - voltage on your load is higher than the base.

Doesn't that make it useless for 5V supplies and most LEDs?

eg. If a LED needs 3.6V (blue/green/white) then there's only 1.4V left over for the transistor/resistor...not enough!

OK I tried: Diodes on base: 2 x 1N4148 Resistor on base: 330 Ohm Resistor on emitter: 10 Ohm

This setup give a collector current almost independent of the suply voltage:

Supply Voltage /Collector current

3.8V 78 mA 7,6V 89 mA 11,4V 91 mA

The base current is 9,5 mA