Current mirror for multiple LED driver circuit.

AI Thread Summary
The discussion revolves around designing a LED driver circuit for 12 LEDs using a 24V supply, with concerns about thermal runaway and the need for a current mirror. The user seeks guidance on selecting appropriate transistor values for the current mirror, as they have previously only used a driver for 5 LEDs. Several participants suggest alternatives, including using a higher voltage power supply or adjusting resistor values to manage current effectively. The importance of ensuring the LED voltage drop is accurately accounted for to avoid damaging the LEDs is emphasized. Ultimately, the conversation highlights the challenges of working within the constraints of the existing power supply while maintaining LED safety and performance.
collier1988
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Hello guys I wonder if you can help me?

Basically for a project I’m going to produce a led light source with 12 leds with a single small driver that runs on 24v supply. The LEDs run 3.7v typically so that would require 44.4v that’s a lot more than the led driver can supply but if I run 2x 6 LEDs in parallel this would be fine . But to overcome thermal runaway need to include a current mirror in the circuit.

Can someone please tell me how to work-out what values of transistor I need as I don’t have a clue : )
I have used this type of driver before but for only 5 LEDs so this was not a problem.

For info on the driver please see http://www.recom-international.com/pdf/Lightline/RCD-24.pdf the circuit I’m going to use is on page 4 second from top left!
It simply identifies I need 2 x NPN transistors with 2 x 0.5-2ohm resistors.

LEDs Cree XR-E Q5
http://www.dealextreme.com/p/cree-xr-e-q5-emitter-on-premium-star-228lm-at-1a-2394
 
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Why do you want a current mirror? If you have a 24V supply, it is perfect. Here is a simple circuit that power 3 LED, just parallel 3 you get 12 LED:

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I used your approx of 3.7V per LED, I don't know the real value. All the values are very approximate. It is not important. This drive about 20mA through the LEDs. I don't think you will burn the LED with say even 30mA if the voltage is way off. If you want more, lower the resistance. You know the voltage drop across the resistor is about 5V, so calculate the resistance accordingly.

But, if your value of the LED drop is way off and things start smoking...well.....I don't know you!:smile:
 
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Thanks yungman for your reply :) I’ll be using power LEDs with a max current 1000ma but using the driver I want it can only run 1200ma so over over the two parallel strings of LEDs giving a supply of 600ma. From what research I have conducted it’s apparently only advisable to run two strings of LEDs using a current mirror. The current mirror is only there to stop them going pop! I’m helping a friend make this light source and I don’t really want to pop away his LEDs. I could use a high powered driver i.e. 50v or something but the size of the driver will be a lot larger and the PSU will have to be bigger also I have used the recom RCD-24 driver before and it’s a really good bit of kit.
I don’t really want to be going below 600ma to the LED’s
 
I don't think the constant current module can work for you. If you read the spec, for 1200mA, the drop out as in p5 of the data sheet is over 2V. My understanding is if you give 24V input, it can only provide up to 21V of output. Your 24V supply is only 1.2A, 6 of the LED at 3.7V each is 22.2V, it won't work even if you don't use a current mirror. the current mirror will take up more voltage.

I don't know how accurate is the voltage across the LED. You might still be able to use my circuit. So if you have 1.8V and you want 0.6A. so the resistor is 3ohm. using my circuit with 6 LEDs. But this is dangerous, if the turn on voltage is less, you draw more current. If you are doing one experiment, you can always use a variable supply and slowly raise the voltage and look at the current and re-adjust the resistor value.

Instead of fighting these limitation, are you stuck with the power supply? Get a bigger supply so you can break into three chain and you can even use my original circuit and just scale the resistor for 600mA which is about 15 ohm. Or get a higher voltage supply.

If using the resistor is not an option and you really stuck with the 24V 1.2A supply, here is another circuit I come up:

145488[/ATTACH]"]
671jdl.jpg


I am doing something like a current mirror also, since only about 0.6V drop across the 1ohm resistor, it is not very accurate between the two chain. I use 2 separate 1K feedback resistor to the -ve input of op-amp to average the current of the two chain. Notice I use a separate transistor to do the darlington because I cannot effort to waste the limited headroom with those build in darlington transistors. There is a dimmer by adjusting the 500ohm pot on the left side even you did not mention about a dimmer, since your driver has a dimmer, so I put it in.

First take a look to see is there anything I overlooked, if you decided that my circuit is the way to go, then we can talk about what op-amp, what transistor to use in more detail. The idea is the same as your module, the difference is my design has only about 1V drop out rather than over 4V dropout including the current mirror. this is just give you an idea. Still have to look for an op-amp that has input rail to rail.

Normally, your driver being PWM, is more efficient than mine which is linear. But in this case with very little head room, the efficiency should actually be comparable.
 

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