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Circuit for LED flashlight
- Thread starter PsySc0rpi0n
- Start date
It will drop twice as more voltage in Re than before and will halve the LEDs current!
I'm not sure but I think I can write something like the equation for the diodes, B-E junction and voltage drop at Re:
Ie = (2*Vd - Vbe - Ve) / 33
I know it's the same as with the LEDs because I'm checking with the simulator, but I can't get to that value.
Yes I do understand the constant current circuit thing and also the part that the LEDs won't have any role regarding the current control!
Well, if LEDs doesn't play any role in the current, I think nothing would happen.
The only part of the circuit that can take those changes would be at 1kΩ resistor, if I understand!
Same for the 15V supply. 1kΩ resistor will have to dissipate everything other than what is happening in the constant part of the circuit. Diodes, B-E junction, LEDs and transistor will remain constant!
hi,
Well, if LEDs doesn't play any role in the current, I think nothing would happen.
If the voltage drop across the load exceeds the supply voltage, the CC source cannot work as expected.
You must have a supply voltage that is greater than the voltage drop across the load, by say 2 or 3 volts at least.
Look up Constant Current circuit Compliance Voltage
https://www.sunpower-uk.com/glossary/what-is-compliance-voltage/
15V.
The only part of the circuit that can take those changes would be at 1kΩ resistor, if I understand!
There would be a very slight rise in the Base voltage, but the circuit would still give close to the 20mA current.
E
Well, if LEDs doesn't play any role in the current, I think nothing would happen.
If the voltage drop across the load exceeds the supply voltage, the CC source cannot work as expected.
You must have a supply voltage that is greater than the voltage drop across the load, by say 2 or 3 volts at least.
Look up Constant Current circuit Compliance Voltage
https://www.sunpower-uk.com/glossary/what-is-compliance-voltage/
15V.
The only part of the circuit that can take those changes would be at 1kΩ resistor, if I understand!
There would be a very slight rise in the Base voltage, but the circuit would still give close to the 20mA current.
E
The voltage drop across the 4 LEDs would be 4 times the voltage drop of one of them.
The voltage drop across the transistor would be 12V - 4*V_Led - Ve.
What resistor? Re? Would be 12 - 4*V_Led - Vce or maybe it's simply around 0.7V.
Hum, ok.
I'm now trying to put the circuit in a stripboard!
The voltage across the LEDs is just 4x 3.2 = 12.8V. How does that work with a 12V supply?
Bob
I may do it tomorrow if I get this done today. Because to try to assemble this as small as possible is not going to be easy. Between connecting points and interrupting conduction on these strip boards is always a challenge!hi Psy,
Let's know how it goes.
You may like to consider how you design that CC circuit using a PNP transistor.?
E
But I'm not using 4 LEDs. That was your question! I just answered it.
I'm considering to change the circuit and use 9 LEDs. 3 branches in parallel of 3 series LEDs in series. I'm considering this because this way, maybe I can make a bigger light and be easier to place components on the strip board. Trying to arrange the 3 LEDs, 2 diodes and 2 resistors and placing the LEDs in kind of in a triangle displacement isn't being easy in the smallest space possible.
And @crutschow suggestion might even be easier. Less components and maybe easier to displace in the strip board. Dinner time now. Be back later!
And @crutschow suggestion might even be easier. Less components and maybe easier to displace in the strip board. Dinner time now. Be back later!
@crutschow I just used your suggestion with the 9 LEDsI mentioned above and changed the Emitter resistor to 100Ω. But this resistor is still dissipating too much power! Around 0.3W. I think it will become too hot, no?
