# Circuit for LED flashlight

#### ericgibbs

Joined Jan 29, 2010
19,118
hiPsy,
Ok, I see you changed Re = 33Ω. Any reason for this?

What voltage is 33R* 20mA= ..............................

Also, what is voltage on the Base.??

E

#### PsySc0rpi0n

Joined Mar 4, 2014
1,777
hiPsy,
Ok, I see you changed Re = 33Ω. Any reason for this?

What voltage is 33R* 20mA= 0.66V

Also, what is voltage on the Base.?? Around 2 * 0.66V = 1.32V

E

Wait, I think at the base I need to check it better.
Edited!

#### ericgibbs

Joined Jan 29, 2010
19,118
Hi Psy,
So let's assume you change the emitter resistor from 33R to say 66R, what effect will that have on the LED current.?
And
Why.?
E

Hint: The Base voltage is fixed at approx 1.35V, due to the two diodes, so it does not change

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#### PsySc0rpi0n

Joined Mar 4, 2014
1,777
Ho Psy,
So let's assume you change the emitter resistor from 33R to say 66R, what effect will that have on the LED current.?
And
Why.?
E
It will drop twice as more voltage in Re than before and will halve the LEDs current!

#### ericgibbs

Joined Jan 29, 2010
19,118
hi Psy,
That is correct , because the Base is fixed at 1.35V by those diodes.

So put the 33R back and remove the 66R.
Now short out the LED's with a wire link, what will be current through the emitter resistor of 33R. be now??
E

#### PsySc0rpi0n

Joined Mar 4, 2014
1,777
hi Psy,
That is correct , because the Base is fixed at 1.35V by those diodes.

So put the 33R back and remove the 66R.
Now short out the LED's with a wire link, what will be current through the emitter resistor of 33R. be now??
E
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.

#### ericgibbs

Joined Jan 29, 2010
19,118
hi Psy,
Remember, it is a Limited or Constant current circuit.
The Base and Emitter voltages control and set the level of the current through the transistor.

The LED's play no part in controlling the current.

E

#### PsySc0rpi0n

Joined Mar 4, 2014
1,777
hi Psy,
Remember, it is a Limited or Constant current circuit.
The Base and Emitter voltages control and set the level of the current through the transistor.

The LED's play no part in controlling the current.

E
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!

#### ericgibbs

Joined Jan 29, 2010
19,118
hi,
OK,
A couple of extra points.
What if I added another LED in series, so we have 4 LED's what effect would that have.?
also
what would change if the 12V supply was, say 15V.?
E

#### PsySc0rpi0n

Joined Mar 4, 2014
1,777
hi,
OK,
A couple of extra points.
What if I added another LED in series, so we have 4 LED's what effect would that have.?
also
what would change if the 12V supply was, say 15V.?
E
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!

#### BobTPH

Joined Jun 5, 2013
9,282
What would be the voltage across those 4 LEDs?
And across the transistor and resistor?

Bob

#### ericgibbs

Joined Jan 29, 2010
19,118
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

#### PsySc0rpi0n

Joined Mar 4, 2014
1,777
What would be the voltage across those 4 LEDs?
And across the transistor and resistor?

Bob
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.

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
Hum, ok.

I'm now trying to put the circuit in a stripboard!

#### ericgibbs

Joined Jan 29, 2010
19,118
hi Psy,
Let's know how it goes.
You may like to consider how you design that CC circuit using a PNP transistor.?
E

#### BobTPH

Joined Jun 5, 2013
9,282
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

#### PsySc0rpi0n

Joined Mar 4, 2014
1,777
hi Psy,
Let's know how it goes.
You may like to consider how you design that CC circuit using a PNP transistor.?
E
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!

The voltage across the LEDs is just 4x 3.2 = 12.8V. How does that work with a 12V supply?

Bob
But I'm not using 4 LEDs. That was your question! I just answered it.

#### crutschow

Joined Mar 14, 2008
34,843
Below is the LTspice simulation of a simple constant-current circuit that uses two transistors with a nominal 3.3V drop per LED.
The current is ≈0.6V/R3.

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#### PsySc0rpi0n

Joined Mar 4, 2014
1,777
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!

#### PsySc0rpi0n

Joined Mar 4, 2014
1,777
@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?

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#### BobTPH

Joined Jun 5, 2013
9,282
0.3W is nothing spread over all those components. It won’t even be noticeably warm.

Bob