Moderators: Dave Moll, Andrew Davie, Steve Anderson
Steve Anderson wrote:I don't know how to phrase this nicely, but this is typically how the manufacturers in Asia would do it. You're right, in the majority of cases where the LEDs come from the same batch the forward voltage drop will be close, very close. However it goes against the grain of my thinking and training, conservative I may be, yes, but I'm I would be putting my reputation on the line, this is to be a long-term display item. For the sake of a few resistors...
Steve Anderson wrote:Andrew, this is all you need, no more, no less...
Steve A.
Andrew Davie wrote:Firstly, in that original diagram the 6V was calculated ........
Klaas Robers wrote:Why other would you have linked us to that original diagram? This is logically not a PWM driven circuit.
This current source circuit, so the transistor and the resistor from emitter to ground, has the advantage that variations in the supply votage have no influence on the current through the LEDs and hence no influence on the light output of the LED cluster. On the other hand, it only works in a proper way if the voltage on the collector of the transistor is 6 volt or more above gound level.
Klaas Robers wrote:That is true. If you don't use the current source configuration, you don't need the 6 volt at the collector. Then you drive the transistor in saturation, which is about 0.1 volt from collector to emitter (ground). Then you can use the balancing resistor immediately as a current limiting resistor. So:
- 3 LEDs in series 3 x 3.4 volts = 10.2 volts over the 3 LEDs,
- The transistor needs 0.1 volt, so 10.3 volts in total.
- Remains 12 - 10.3 = 1.7 volt for the current limiting resistors,
- which are 85 ohm (1.7 volt / 0.02 ampère) per branch.
So mount a resistor of 82 ohm (standard value) in each branch. Such a branch (3 LEDs and one 82 ohm) should run directly on 12 volts. Each branch will do this in the same way.
The need of balancing resistors is because if one branch gets hotter than other branches, this branch is drawing more current. A warm LED gets a lower "resistance". LEDs and diodes in general have a negative temperature coëfficient. Thus it is drawing even more current.... and so on. The balancing resistors equalize the 10 currents more or less. The higher the resistance of the resistors, the better this works. So a resistor in each branch and no resistor in the combined path is optimal.
Dissipation of the resistors is low: 1.7 V x 0.02 A = 0.034 watt = 34 mW.
Steve Anderson wrote:One thing I would like to give a 'heads-up' on... you need to make sure the 12V supply is truly 12V. With these unregulated 'bricks' it may not be.
Steve Anderson wrote:If you re-do Klaas's calculations where the supply is say 13V you end up with a higher current...how much? I'll leave you to do that, you should know how to do that now (I am a part-time teacher). Do it again for 14V, and as a matter of interest, 11V.
Steve Anderson wrote:You could also do a +/-1V variation the the circuit I posted above, at 18, 19 and 20V. Notice any change?
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