How to incorporate potentiometer?

[kuksul08]

I need some electronics help. I will have 4AA power source connected to an AMC7135 700mA driver, connected to a cree xr-e.

I currently have a potentiometer with these specs:
1W SINGLE 5K;
Resistor element type:Variable Rotary Cermet;
Resistance, track:5kR;
Series:149;
Power rating:1W;
Temp, op. max:150(degree C);
Temp, op. min:40(degree C);
.25D by .5 in. shaft
.375D by 1.875 in. exposed threads


Can I somehow incorporate it to allow dimming of the light? If I simply put it in series, then with full batteries it will have to dissipate 1.25W which is higher than the rating.

Also if directly in series, my experience has shown that it doesn't allow much 'dimming' modulation. It's mostly dim, then suddenly increases brightness rapidly. I guess what I need is a control circuit, but I have NO idea what that is, or how to make one

Thanks for any help

 

[Blindasabat]

I can help you with one part of your problem. The dimming response.
The power you see will follow the equation:
Power = Voltage squared/Resistance

At 5k ohm , the power = 0.0045W
4k = .0057
3k = .007
2K = .01152

Assuming the pot is linear, so far you will not notice any change in light even though you are 4/5ths of the way through turning your pot.

1k = .023 Watt
500 = .046
250 = .092
125 = .18
50 = .46
25 = .92

Starting to see something...

15 = 1.54 Watts
10 = 2.3 Watts

Getting brighter!

6 = 3.84 Watts - this is where you want to be.

but you have a tiny bit of turning left.

2 = 11.52 Watts
1 = 23.04
0.5 = 46.08

Now your pot is likely non-linear, but still not steep enough to offset the abrupt ramp up of power near the end of the turn. It'll probably go from 5000 ohm to 500 in the first half turn, then more slowly from there.

Also note that the calculations show you are massively overdriving the LED at the end of the pot's turn. Luckily, the Pot and batteries have some internal resistance, but you are still overdriving your LED on 4AA's adding up to 4.8V. You should use 3AA for a max of 3.6V, then your power will drop by a factor of 1.777.

To smooth out the response curve, one thing you can so is to put a 40-50 ohm resistor in parallel with the pot. Then the Power will even out some (example with 40 Ohm):
Ohms-Watts
5000 - 0.327
4000 - 0.327
3000 - 0.328
2000 - 0.330
1000 - 0.337
500 -- 0.350
250 -- 0.376
125 -- 0.428
50 - - 0.58
25 - - 0.84
15 - - 1.19
10 - - 1.62
8 - - - 1.94
6 - - - 2.5
4 - - - 3.6
3 - - - 4.6
2 - - - 6.8
1 - -- 13.3
0.5 -- 26

This does not fix your problem of max wattage to the LED, but battery resitance will mostly fix that in real life. I'm sure you will only see around 3-5 Watts max with 3AA direct drive. Do you have a multimeter to check the amps running through it? Make sure it does not go over 1 to 1.25 Amp on fresh batteries.

In any case, you are also overdriving your pot and it will burn up if you run it full open. You need to run it a crack closed (tiny bit from open). I suggest you go to Radio shack and get a 3W or 5W one. I think I saw a 3W one there last week.
A possible solution to that is to ALSO put a resistor in series with the Pot (in addition to the parallel resistor). This will have to be a hefty 5W resistor of about 2ohm. The formulas lose accuracy at low ohm levels with real life battery & LED resistance unaccounted for. Try 1, 2, & 4 ohm (2x 2ohm in series) and measure the current. Your efficiency will drop to ~70-75% as if you were using a low quality driver, but you will save your LED from death and have actual dimming capability.

I'm not an expert in this area - having only done it on the hobby level, so if I did anything wrong, anyone please tell me

 

[kuksul08]

Thank you very much for the response

In my current design, I actually have an AMC7135 700mA driver in the circuit, so even when the pot is fully shorted (on), the LED will get 700mA. The idea was to also put the pot in series, simply adding onto the resistance from the regulator. I'm not sure if this will interfere with the current/voltage sensing of the AMC7135.

That's interesting about putting the parallel resistor with the pot. I think I can do that and still have full efficiency (of the driver) when the pot is a short.



Also, a somewhat related question: If I were to ditch the regulator and use the cheapo resistor method, would I really need a 5W resistor?

My calcs show at a maximum (fully charged battery 5.6V and especially low Vf of 3.2), I get .7A * 2.4V = 1.68W dissipated in the resistor. Using the same parameters I'd need a 3.42ohm resistor to get 700mA.
Hence, as a backup in case my regulators never get here, I was going to get some of these: http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=P3.3W-3BK-ND

..and maybe a few more just to experiment with.


Phew, once again I like the parallel resistor solution, but can you verify that stuff^^?

 

[Blindasabat]

...I actually have an AMC7135 700mA driver in the circuit, so even when the pot is fully shorted (on), the LED will get 700mA. The idea was to also put the pot in series, simply adding onto the resistance from the regulator. I'm not sure if this will interfere with the current/voltage sensing of the AMC7135.

Oh, well using a driver throws a massive wrench in the works. How did I miss that? Different drivers react somewhat differently to resistance in-line. AFAIK, the 7135's only drop voltage ~0.3 to 0.7V across themselves and do not really act like resistors even though they kind of act like a variable resistor to regulate a driver. That is as far as I understand/misunderstand that. The AMC7135 will go direct drive under ~4.5V input, so you may only need the parallel resistor and you would be good to go, but I still think you need a higher power pot as your current one will melt.

That's interesting about putting the parallel resistor with the pot. I think I can do that and still have full efficiency (of the driver) when the pot is a short.

Yes, you will have full efficiency in that case. Best of both worlds!

Also, a somewhat related question: If I were to ditch the regulator and use the cheapo resistor method, would I really need a 5W resistor?

I was guessing on that power since I don't have my jump drive with my trusty LED-resistor calculator program on it with me.
I found this one on line http://home.cogeco.ca/~rpaisley4/LEDcalc.html
and it says only 1.12W dissipated by the resistor. I found another one that had slightly different results.

My calcs show at a maximum (fully charged battery 5.6V and especially low Vf of 3.2), I get .7A * 2.4V = 1.68W dissipated in the resistor. Using the same parameters I'd need a 3.42ohm resistor to get 700mA.
Hence, as a backup in case my regulators never get here, I was going to get some of these: http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=P3.3W-3BK-ND
..and maybe a few more just to experiment with.

By all means - Experiment! The batteries will droop to about ~1.2V/cell under load, so you only need to accound for 4.8-5V for a 4-cell pack for 2.7ohm with 1.3W min rating (2W for safety sake). Otherwise you are on track IMHO.

Phew, once again I like the parallel resistor solution...

Yeah, I'm proud of myself for that one.

Power of pot with a parallel resistor = V squared / (1/((1/R pot) + (1/R parallel)))