Where to buy the correct resistor?

[Lumencraft (Matt)]

Hi all,

I need to find a place to buy some resistors. Radio shack in my area only carries ridiculously high value ones.

What I need to do is take the voltage of a CxxxI bin emitter to an acceptable level. The light is DD on 3 AA nimh cells, and I used a resistor calculator I found online to estimate the resistor value needed. It claims in order to get the current down to 2000am I need a 1 Ohm resistor with 4 wat capable.

Where could I find a resistor like this?

Will it be a large resistor?

 

[Mr Happy]

I think you have something wrong with the calculations. A 1 Ω resistor passing 2 A will drop 2 V. Of the 3.6 V from the battery that will only leave 1.6 V for the LED.

 

[PhotoWiz]

I think what you need to ask for is a "power resistor" — Radio Shack does have those in low resistance designations. They aren't huge, but significantly larger than the small cylindrical ones with the colored banding codes on them.

If you put two or more in parallel, you can drop the resistance further, if you can't find exactly what you need.

 

[Lumencraft (Matt)]

I think you have something wrong with the calculations. A 1 Ω resistor passing 2 A will drop 2 V. Of the 3.6 V from the battery that will only leave 1.6 V for the LED.

Hmmm... here is the calculator I used. I'm not sure if this was user error, or if the calculator needs to go back to the drawing board.

What resistor would you recommend Mr Happy?

I think what you need to ask for is a "power resistor" — Radio Shack does have those in low resistance designations. They aren't huge, but significantly larger than the small cylindrical ones with the colored banding codes on them.

If you put two or more in parallel, you can drop the resistance further, if you can't find exactly what you need.

Do you think I could fit one in the tail cap of a 3 AA minimag?

 

[inbred_weasel]

I don't think that calculator is quite right. It does seem to recommend a 1 ohm/4 watt resistor for your situation.

I think the calculation you need to do is something like this:
Vb = 3.6 volts (battery voltage)
Vf = 3.3 volts (LED forward voltage)
If = 2 amps (LED current)
Ir = If = 2 amps (resistor current is the same as LED current)

Vr = Vb-Vf = 0.3 volts (the voltage dropped across the resistor)
Rr = Vr/Ir = 0.3/2 = 0.15 ohms (resistor's resistance)

Pr = Vr*Ir = 0.3*2 = 0.6 watts (power dissipated in resistor)

So in this example you'd need a 0.15 ohm resistor rated for at least 0.6 watts.

 

[Mr Happy]

I think what you need to ask for is a "power resistor" — Radio Shack does have those in low resistance designations. They aren't huge...

I have some of the 1 Ω, 10 W resistors that Radio Shack sells. They are about 2 inches long and 1/3 inch wide. I think I would call them huge myself.

Do you think I could fit one in the tail cap of a 3 AA minimag?

No, not that kind of resistor. It's much too big.

 

[Mr Happy]

What resistor would you recommend Mr Happy?

I don't know. There is not enough information to go on.

Do you think I could fit one in the tail cap of a 3 AA minimag?

If you are considering running an LED at 2 A it is going to generate 6 or 7 W of heat that must be dispersed. That will require a massive heat sink. Can you fit the LED and the required heat sink in a minimag body?

 

[Lion251]

Hmmm... here is the calculator I used. I'm not sure if this was user error, or if the calculator needs to go back to the drawing board.

The latter.
I tested the calculator. When the actual resistor value would be below 1 Ohm, the calculator rounds it up to 1 Ohm, but still supposes that the full current will flow, and bases the required wattage of the resistor on that value.
This resistor calculator is only suitable for low-power LEDs with higher supply voltage.
If you want to drive a red LED with Vf of 1.6V at a current of 20 mA from a supply of 5V, it gives the correct result.

But, as Mr Happy already said about your plans: dissipating 6 or 7 Watts in a minimag will give you very warm light, or burnt out components.

 

[HKJ]

A more advanced calculator can be found here, but it must be downloaded and run form your computer. It can handle many different led circuits.

 

[Lumencraft (Matt)]

I don't know. There is not enough information to go on.


If you are considering running an LED at 2 A it is going to generate 6 or 7 W of heat that must be dispersed. That will require a massive heat sink. Can you fit the LED and the required heat sink in a minimag body?

I guess I should have added a bit more info.

Here is where I am at. I have made a brass heat sink the approximate shape and size of the minimag led pill. So relatively speaking I the sinking is good, and it will transfer heat to the body. However there is still a current issue. I ran this DD on 3 rested nimh cells with some degree of success. The led is soldered directly to the brass pill, but even with the cells rested I suspect that the current is about 3.8 amp. In a big light this would not be a huge problem, but in the mini it caused the loss of one of the dies of the led after only a few minutes run time.

I feel like cutting the current to around 2 amp should be a sufficient enough drop to allow use of the light until it starts to burn your hand.

 

[Lumencraft (Matt)]

I don't think that calculator is quite right. It does seem to recommend a 1 ohm/4 watt resistor for your situation.

I think the calculation you need to do is something like this:
Vb = 3.6 volts (battery voltage)
Vf = 3.3 volts (LED forward voltage)
If = 2 amps (LED current)
Ir = If = 2 amps (resistor current is the same as LED current)

Vr = Vb-Vf = 0.3 volts (the voltage dropped across the resistor)
Rr = Vr/Ir = 0.3/2 = 0.15 ohms (resistor's resistance)

Pr = Vr*Ir = 0.3*2 = 0.6 watts (power dissipated in resistor)

So in this example you'd need a 0.15 ohm resistor rated for at least 0.6 watts.

Welcome to CPF inbread_weasel.

I think that I am with you. Lets see if I understand what you are saying.

For example. If my VB was 5 and led VF is 3.3 I go 5-3.3= 1.7v I need to drop across the resistor. Then 1.7V/2A=.85 ohm resistor needed?
Then 1.7vr x 2a = 3.4 watt rated capacity?

 

[Lumencraft (Matt)]

I still need to know where to get these resistors too guys.

 

[HarryN]

The challenge of obtaining high power, low value resistors is that few people make them. When I was making the "breeze" light in my sig line, it needed some 2 ohm high power resistors. (as well as some in 10 ohm) Since you need 1 ohm, that is 2 in parallel.

Most small resistors are rated for 1/8 watt, and I only know of one company that makes really high power ones, but they didn't offer them in the resistance I needed. They did agree to make a custom batch for me (1206 SMT size, 2 ohm and 10 ohm) and have some left from that project. These are about 2 x 3mm and perhaps 1mm thick / standard 1206 size.

Don't worry about the power rating of these - you will never hit it, but you do need a thermal path for moving any heat away from them, just like an LED. (your 3.4 watts, very similar to a typical power led)

If you have actually used and soldered 1206 size parts before, feel free to send me a pm and we can work something out. I have a few left, but not huge numbers.

 

[Mr Happy]

It is also possible to make fractional ohm resistors yourself using plain iron wire such as you might get from a hardware or craft store, or more specialized nickel-chrome resistance wire obtained from an electronics supplier or cannibalized from a radiant heater.

You make up an appropriate length, insulate it with heat shrink tubing (or heat resistant tubing if it might get really hot), and coil it up to fit where it needs to go. You might for instance coil it up inside the tail cap spring of a light.

 

[Stereodude]

Mouser has a lot of resistors. link You should focus on the two categories of Power Resistors.

However, you may have better luck replacing your P7 with a DxxxJ bin part that has a higher Vf that will consume less current than your CxxxI does from 3 AA's and likely won't need a current limiting resistor.

If you plan to stick with the current P7 emitter have you measured the Vf of your CxxxI P7 at 2.0A of current to calculate the exact voltage drop you need in a resistor?

 

[Th232]

Ditto with Farnell, they even appear to have 0.82 and 0.91 ohm resistors:

http://au.farnell.com/welwyn/wp4s-r82ja2/resistor-r82-4-watt-5/dp/1219251

http://au.farnell.com/irc-tt-electronics/caw5r820jlf/resistor-wirewound-0-82ohm-5w-5/dp/1457954

Either that or I'm really missing something.

 

[VegasF6]

Try this eBay seller?
http://cgi.ebay.com/0-15-15-Ohm-5-W...tem&pt=LH_DefaultDomain_0&hash=item518f19065f

 

[HarryN]

Yes, but those resistors have a body 5mm dia x 13mm long, plus the leads. That is pretty large for the required space.

Mr Happy - making your own from a wire is a good idea. Do you have a source for that kind of wire?

 

[Lumencraft (Matt)]

Ok its been a while since I have done this so before I go spending money lets see if I know whats going on.

My
VS= 4.2v
VL= 3.6v
I = 1.1a

So 4.2-3.6=.6/1.1= .545 ohm resistor needed correct? How then do I know how many watts the resistor needs to be?

 

[Lynx_Arc]

Ok its been a while since I have done this so before I go spending money lets see if I know whats going on.

My
VS= 4.2v
VL= 3.6v
I = 1.1a

So 4.2-3.6=.6/1.1= .545 ohm resistor needed correct? How then do I know how many watts the resistor needs to be?

A half ohm resistor is correct, as E = I x R or R = E / I which is the formula you used.

P= I x E Or Power (in watts) = I Current (in amps) x E Voltage
we have 0.6v drop at 1.1A multiplying the two together yields 0.66 watts. You should almost always round up on wattage on resistors in case the current is higher than expected so I would recommend 0.5 ohm 1 watt resistor or perhaps easier to get would be 2 x 1 ohm half watt resistors, which you can wire in parallel.