Resistor calculation questions for a 2-stage switch

[kosPap]

Hello all! I have some questions on how to calculate the resistor of a 2-stage switch for a LED module

Now, I know how to use Ohm's Law to calculate the resistor value if I know the working current and batt voltage but how can I figure both?

I know resitors actually create voltage drop, don't they? Then this creates a different current draw from the board.

But I do not have a adjustable power supply to tinker with voltage, unless i jury-rig something with a 4xAA battery case, measure total volts and current draw then take battery/dropin combination to my light box for lux measurement.

What I need is a low output at about 25 lumens using an 1A board and a Cree Q5WC LED with a combination of 2xCR123s. From previous posts I would say a 65 milliOhm resistor is good for it..

Thanks Kostas

 

[Mr Happy]

Re: Resistor calculator questions for a 2-stage switch

If you are driving the LED with a regulator board then I think there is no simple and reliable way to calculate or use a resistor for low output. Unless the board is already designed with a low output mode your only choice will be to place a resistor between the batteries and the input to the board, and you would have to find a suitable value for that resistor by trial and error.

If you want to calculate what value resistor to use without trial and error, you would need to know how the specific board responds to low input voltages and how the current draw varies in response. Those details will vary with the particular board design. It's really not simple.

 

[kosPap]

Re: Resistor calculator questions for a 2-stage switch

Well I got myself a 4AA case and rigged a Derrelight 1SM2 module

Here is what I got:

             V noload    V underload     Current   Lux
2xAA Lithium 2.97        2.93            0.06 A    55
3xAA         3.76        3.15            0.11 A    113
4xAA         5.13        3.57            0.29 A    264

Now I like the 55 lux figure most, so with an assumed 5.6 V of 2xCR123 under load I will need a drop in voltage of 2.6 Volts, ergo

V = I*R => 2.6 = 0.06*R => R = 43 Ohm

Am I correct?

 

[Mr Happy]

Re: Resistor calculator questions for a 2-stage switch

Well I got myself a 4AA case and rigged a Derrelight 1SM2 module

Here is what I got:

             V noload    V underload     Current   Lux
2xAA Lithium 2.97        2.93            0.06 A    55
3xAA         3.76        3.15            0.11 A    113
4xAA         5.13        3.57            0.29 A    264

Now I like the 55 lux figure most, so with an assumed 5.6 V of 2xCR123 under load I will need a drop in voltage of 2.6 Volts, ergo

V = I*R => 2.6 = 0.06*R => R = 43 Ohm

Am I correct?

I think so, except that the 2xCR123 will only have a load of 0.06 A under these conditions, so the voltage may be a little higher than 5.6 V when they are freshly installed. Maybe you want to plan on dropping 3.0 V through the resistor rather than 2.6 V? In practice, you may want to try values of either 40 Ohm or 50 Ohm and see which one you prefer...

 

[kosPap]

Re: Resistor calculator questions for a 2-stage switch

indeed... unhappily I have not made a rig to put 2 or 3 CR123 batts and try again....

Thnaks, kostas

 

[HarryN]

I really like resistored flashlights, and an always happy when people make them.

It might not be necessary to worry about the difference between 40 vs 50 ohms as optimum - as the resistor heats up, its resistance will vary a little anyway.

Watch your power rating on the resistors - watts = volts x amps so you are in the range of 3 volts x .1 watts = ,3 watts. You would be surprised how many resistors are not rated for this.