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Constant current source for high brightness LEDs


Chris.Elston
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I guess that people here are probably tired of LED questions, but I'll give these a shot anyway.

1) LED forward drops are specified over a smallish range, e.g.: 2.0 - 2.2V.  Does this indicate a dynamic range, or is it that any one specimen of that type of LED will have a fixed drop somewhere in that range.

2) I want to drive a high brightness red LED @ 350mA from a 5V source, assuming a 2.2V forward drop across the LED.  I calculate the current limiting resistor as 8R, power dissipation across it as ~1W.  So far, so good (I hope  :) ).  The nearest preferred value resistor available is 8R2, but not many vendors carry 1W 8R2.  So basically, the question is, what are my options beyond ordering the aforementioned resistor? I know that there are some ICs out there that do constant current, but I was wondering if there was anything I could do with good old 0.25W resistors to spread the power dissipation across multiple ones.

3) How critical is the current anyway? I know that 350mA is the recommended current, and that they will do 400mA pulsed.  I plan on controlling the brightness via PWM switching, so how does, for example a 50% duty cycle affect the current through the LED.  My guess would be that you get 50% at full current, then 50% at none - and that the off periods allow the LED to cool.  This effectively cooler running means that you can drive it at 400mA in the on phase?

Cheers,

Chris.

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I was nearly finished calculating things for the constant current sink when Microsoft told me, "Paint encountered a problem and needs to close. Tell us about it.", so it's gone.

Use 82 ohms for the upper resistor. It supplies base current to the upper transistor (use a TIP31) to turn it on. The bottom transistor can be a 2N3904 and is turned on when the current in the lower resistor (use 2.2 ohms) reaches 327mA and begins to turn off the upper transistor at this current level. If the current in the LED and in the 2.2 ohm resistor tries to increase then the voltage across the 2.2 ohm resistor rises and the 2N3904 gets turned on more to turn off the TIP31 transistor more. Self-balancing.

The TIP31 will dissipate 1W if the LED is only 2.0V and the supply is high at 5.25V. It won't need a heatsink but will be very warm.

I haven't sim'd it nor tried it so good luck. ;D

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I was nearly finished calculating things for the constant current sink when Microsoft told me, "Paint encountered a problem and needs to close. Tell us about it.", so it's gone.

:(

Use 82 ohms for the upper resistor. It supplies base current to the upper transistor (use a TIP31) to turn it on. The bottom transistor can be a 2N3904 and is turned on when the current in the lower resistor (use 2.2 ohms) reaches 327mA and begins to turn off the upper transistor at this current level.

Because it 'robs' current from the base of the top transistor? Cool, I think I get this now.

As I mentioned before, I'm planning to dim the LED via PWM.
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Maybe spoke too soon :)


The bottom transistor can be a 2N3904 and is turned on when the current in the lower resistor (use 2.2 ohms) reaches 327mA and begins to turn off the upper transistor at this current level. If the current in the LED and in the 2.2 ohm resistor tries to increase then the voltage across the 2.2 ohm resistor rises and the 2N3904 gets turned on more to turn off the TIP31 transistor more. Self-balancing.


Can you explain what effect turns on the bottom transistor?  How did you arrive at the 2.2R value  (I understand the 0.65V/R, but how is this derived)  Is it that the resistor limits any further current, so the current flows into the base? I understand the action of the two transistors together to turn each other on/off to maintain the required current, but I don't understand how that reference current is set.
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The 2N3904 transistor has a base-emitter voltage of 0.72V when it conducts 7mA, as shown on its datasheet. It takes a current of 327mA to make a voltage drop of 0.72V across the 2.2 ohm resistor.

If the current in the 2.2 ohm resistor tries to increase then the 2N3904 gets a higher base voltage and current which causes it to turn on harder and steal base current away from the TIP31 transistor.

If the current in the 2.2 ohm resistor tries to decrease then the 2N3904 transistor gets less base voltage and current and turns off, allowing the TIP31 transistor to turn on more.

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Thanks, I understand that now.  So if I read right, the 2.2R is a function of Vbe and the desired current, and in no way related to the forward voltage drop of the load or the supply voltage?  Presumably if you increase the supply voltage the power dissipation and therefore the component ratings need to go up as well?

Cheers,

Chris.

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This is a very simple constant current sink circuit so it is far from perfect. Its load current will rise a little when the supply voltage increases, mainly because the current in the upper resistor increases, therefore the current in the lower transistor will increase, causing its base-emitter voltage to rise a little which determines the current in the 2.2 ohm resistor.

If the LED voltage is changed then its current barely changes.

Of course the power ratings of the power transistor and upper resistor must be increased if the supply voltage is increased.

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The base-emitter voltage of the little transistor is 0.72V. When the 2.2 ohm resistor has 2.2V across it then its current is 327mA.

5V supply minus 2V for the LED and minus 0.72V for the base-emitter voltage of the little transistor = 2.28V. 320mA of current x 2.28V= 730mW.

5.25V supply minus the 2.0V and 0.72V above = 2.53V. 320mA of current x 2.53= 810mW.

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