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### Topic: resistor power (Read 4167 times)previous topic - next topic ##### Dec 21, 2011, 09:04 am
So on my current project, I have 95 LED's running on 5V. Each LED rated 2 volts.

28 go through one 2803A darlington. Each has 90 ohm resistor.

28 go through another 2803A darlington. Each has 90 ohm resistor.

39 are directly connected to 5V via a 150 ohm resistor.

I'd like to add a DPDT switch to make them go to half power (10mA). One half of switch sends each circuit full 5V. Other half of switch sends one pole through 150 ohm resistor, other pole through 90 ohm resistor. So each batch of LED's now has twice as much resistance, thus half the current.

My question is for these dimming resistors, what wattage they need to be.

150 ohm -> 5V2/150 = 0.167 watts

90 ohm -> 5V2/90 = 0.278 watts

But the 150 ohm resistor will have a 0.78 amp draw on it (39 LEDs). The 90 ohm will have a 1.12 amp draw on it (56 LEDs). Do I need to take this into consideration?

#### sixeyes #1
##### Dec 21, 2011, 11:20 am
Your DPDT switch won't work with the resistor your proposing because the LED resistors are in parallel. So if you tried it you wouldn't get 10mA but 20mA spread across the LEDs. Your resistor value will need to be much lower.

How are you powering the board?

At 2 amps+ wouldn't it be easier to have a regulator produce 3.3v and either switch +5v / +3.3v. #2
##### Dec 21, 2011, 07:15 pm

Your DPDT switch won't work with the resistor your proposing because the LED resistors are in parallel. So if you tried it you wouldn't get 10mA but 20mA spread across the LEDs. Your resistor value will need to be much lower.

How are you powering the board?

At 2 amps+ wouldn't it be easier to have a regulator produce 3.3v and either switch +5v / +3.3v.

The LED's get 20ma now. DPDT would switch from full power, to double resistance on each line, dropping to 10ma. 150 ohm LED's have an extra 150 ohm in the main power line, 90 ohm LED's have an extra 90ohm resistor in power line. But I don't know whether to get 3 watt resistors (1/4A @ 5V) or 10 watt resistors (1.12A @ 5V).

I'm running darlington's on the 90 ohm LED's, so 3.3V isn't an option, too much voltage drop.

#### sixeyes #3
##### Dec 21, 2011, 07:53 pm
The LED's get 20ma now. DPDT would switch from full power, to double resistance on each line, dropping to 10ma. 150 ohm LED's have an extra 150 ohm in the main power line, 90 ohm LED's have an extra 90ohm resistor in power line. But I don't know whether to get 3 watt resistors (1/4A @ 5V) or 10 watt resistors (1.12A @ 5V).
How many DPDT switches are you fitting? One per LED? If not then this "new" resistor will need to be a lot less than you think.

If I just stick to the 39 LEDs you're connecting directly to +5v, you're proposing to connect all the 150 ohm resistors to the centre terminal of a DPDT switch, with one terminal of the switch connected to +5v and the other terminal to +5v via a single 150 ohm resistor?

If so, the 150 ohm LED resistors are all in parallel but the new resistor (if it's to only drop 1.5 volts) will need to be 150 / 39 = ~ 3.9 ohms. Since you're dropping 1.5 volts across the resistor it will need to be 0.6 watts (but you could go with a 1 watt to be safe).

Iain #4
##### Dec 22, 2011, 01:19 am

The LED's get 20ma now. DPDT would switch from full power, to double resistance on each line, dropping to 10ma. 150 ohm LED's have an extra 150 ohm in the main power line, 90 ohm LED's have an extra 90ohm resistor in power line. But I don't know whether to get 3 watt resistors (1/4A @ 5V) or 10 watt resistors (1.12A @ 5V).
How many DPDT switches are you fitting? One per LED? If not then this "new" resistor will need to be a lot less than you think.

If I just stick to the 39 LEDs you're connecting directly to +5v, you're proposing to connect all the 150 ohm resistors to the centre terminal of a DPDT switch, with one terminal of the switch connected to +5v and the other terminal to +5v via a single 150 ohm resistor?

If so, the 150 ohm LED resistors are all in parallel but the new resistor (if it's to only drop 1.5 volts) will need to be 150 / 39 = ~ 3.9 ohms. Since you're dropping 1.5 volts across the resistor it will need to be 0.6 watts (but you could go with a 1 watt to be safe).

Iain

Close.

IE the 39 all in parrallel to top center post
The other 56 all in parrallel to bottom center post.

Both posts on one side have clean +5V

On other side, top post has +5V with 150 ohm resistor
Bottom post has 90 ohm resistor.

Tthe segment needs a daylight (full power) and night (half brightness). The 56 LEDs are 5V -> 90 ohm resistor -> LED -> Darlington array --> ground

The LEDs are rated 2V 20mA. The darlington saturation voltage is 1.2.

#### sixeyes #5
##### Dec 22, 2011, 01:28 am

Close.

IE the 39 all in parrallel to top center post
The other 56 all in parrallel to bottom center post.

Both posts on one side have clean +5V

On other side, top post has +5V with 150 ohm resistor
Bottom post has 90 ohm resistor.

Tthe segment needs a daylight (full power) and night (half brightness). The 56 LEDs are 5V -> 90 ohm resistor -> LED -> Darlington array --> ground

The LEDs are rated 2V 20mA. The darlington saturation voltage is 1.2.

OK. So for the 150 ohm LEDs you'll need a 3.9 ohm @ 1 watt (0.6 watts is marginal). For the other side you'll need 90 / 56 = ~ 1.5 ohm @ 0.5 watts

Hope that helps.

Iain #6
##### Dec 22, 2011, 06:19 am
How do you figure that? I've not seen the expression ohm/(load quantity) before..?

#### sixeyes #7
##### Dec 22, 2011, 10:45 am

How do you figure that? I've not seen the expression ohm/(load quantity) before..?
If you have identical resistors in parallel, the equivalent resistance is divided by the number of resistors.

So for two identical resistors in parallel the equivalent resistance is half. For three identical resistors it's a third, for four identical resistors it's a quarter and so on. Your LED resistors are in parallel with this one resistor. If you want to halve the current you need to use the equivalent resistance to all these parallel resistors. This means you need:

150 ohm / 39 (identical resistors) = ~ 3.9 ohm
90 ohm / 56 (2 chains of 28 identical resistors, all in parallel) = ~ 1.5 ohm

For the power calculation I used P = I * I * R. Using your value of 10mA per LED I got the following:

Current through 3.9 ohm resistor is 10mA * 39 = 0.39A.
Therefore P = 0.39 * 0.39 * 3.9 = 0.6 watts. I'd get a 1 watt resistor to be safe. Current may actually be higher than 10mA.
e.g. Current is actually 11mA per LED and the power dissipated in the 3.9 ohm resistor is 0.72 watts.

Current through 1.5 ohm resistor is 10mA * 56 = 0.56A.
Therefore P = 0.56 * 0.56 * 1.5 = 0.47 watts. Again I'd get a 1 watt resistor.

If you can't get a 3.9 ohm or 1.5 ohm 1 watt resistor you can make them by combining higher value resistors in parallel, but you may not need 1 watt resistors any more. It will depend on the actual values you can buy. Worst case you could wire 39 150 ohm resistors in parallel to make your 3.9 ohm resistor. Each resistor would only dissipate 0.01 * 0.01 * 150 = 0.015 watts.

Iain #8
##### Dec 22, 2011, 11:32 am

How do you figure that? I've not seen the expression ohm/(load quantity) before..?
If you have identical resistors in parallel, the equivalent resistance is divided by the number of resistors.

So for two identical resistors in parallel the equivalent resistance is half. For three identical resistors it's a third, for four identical resistors it's a quarter and so on. Your LED resistors are in parallel with this one resistor. If you want to halve the current you need to use the equivalent resistance to all these parallel resistors. This means you need:

150 ohm / 39 (identical resistors) = ~ 3.9 ohm
90 ohm / 56 (2 chains of 28 identical resistors, all in parallel) = ~ 1.5 ohm

For the power calculation I used P = I * I * R. Using your value of 10mA per LED I got the following:

Current through 3.9 ohm resistor is 10mA * 39 = 0.39A.
Therefore P = 0.39 * 0.39 * 3.9 = 0.6 watts. I'd get a 1 watt resistor to be safe. Current may actually be higher than 10mA.
e.g. Current is actually 11mA per LED and the power dissipated in the 3.9 ohm resistor is 0.72 watts.

Current through 1.5 ohm resistor is 10mA * 56 = 0.56A.
Therefore P = 0.56 * 0.56 * 1.5 = 0.47 watts. Again I'd get a 1 watt resistor.

If you can't get a 3.9 ohm or 1.5 ohm 1 watt resistor you can make them by combining higher value resistors in parallel, but you may not need 1 watt resistors any more. It will depend on the actual values you can buy. Worst case you could wire 39 150 ohm resistors in parallel to make your 3.9 ohm resistor. Each resistor would only dissipate 0.01 * 0.01 * 150 = 0.015 watts.

Iain

No, it's in series.... #9
##### Dec 22, 2011, 11:45 amLast Edit: Dec 22, 2011, 11:47 am by magnethead794 Reason: 1
see if this helps clarify any?

The R's in the top batch are 150 ohm, the bottom 2 batches (going through the 2803's) are 90 ohms.

#### sixeyes #10
##### Dec 22, 2011, 12:26 pm

Your LED resistors are in parallel with this one resistor.

No, it's in series....
If you don't like my terminology, think about the current. You need 10mA per LED. That's 390mA through a single resistor that needs to drop the same voltage as a 150 ohm resistor drops at 10mA. The voltage dropped by the 150 ohm @ 10 10mA is 1.5 volts. Using R = V/I you get 1.5 / 0.39 which yields 3.9 ohms. You get exactly the same result, so you can consider the LED resistors to be in parallel (from an analysis point of view).

The end result is that the resistor values I've quoted are correct and if you're still having a hard time believing it, use a 150 ohm resistor instead. You won't burn it out but each LED will get about 0.5mA each. They might glow dimly but it won't be 10mA per LED that you're hoping for.

Iain #11
##### Dec 22, 2011, 12:52 pm

Your LED resistors are in parallel with this one resistor.

No, it's in series....
If you don't like my terminology, think about the current. You need 10mA per LED. That's 390mA through a single resistor that needs to drop the same voltage as a 150 ohm resistor drops at 10mA. The voltage dropped by the 150 ohm @ 10 10mA is 1.5 volts. Using R = V/I you get 1.5 / 0.39 which yields 3.9 ohms. You get exactly the same result, so you can consider the LED resistors to be in parallel (from an analysis point of view).

The end result is that the resistor values I've quoted are correct and if you're still having a hard time believing it, use a 150 ohm resistor instead. You won't burn it out but each LED will get about 0.5mA each. They might glow dimly but it won't be 10mA per LED that you're hoping for.

Iain

Then I may need less than half current...I just put a 150 on the power line, and it's the exact brightness I need.

When I say direct link, it implies that the primary resistor is there, but not the secondary dimming resistor.

first is 150 ohm high light -> http://i1105.photobucket.com/albums/h355/magnethead494/IMAG2105.jpg

fourth is 150 ohm low light (what I want) (but the resistor got hot to the touch) -> http://i1105.photobucket.com/albums/h355/magnethead494/IMAG2108.jpg

#### sixeyes #12
##### Dec 22, 2011, 01:34 pm
Then I may need less than half current...I just put a 150 on the power line, and it's the exact brightness I need.
In which case you're not passing 10mA through each LED. Have you done a similar check for the other LEDs. Does it matter if the 39 LEDs have a different brightness to the two chains of 28?

As to the power of your single resistor for the 39 LEDs:

The 39 150 ohm resistors are in parallel, so 1/39th the resistance of the single 150 ohm resistor. So 39/40th's of the voltage will be dropped across the single dimming resistor. Volt drop will be 0.975 * (5 - 2) = 2.9v. P = (V*V) / R and this gives 50mW. So there shouldn't be a problem. If it's getting really warm there's something funny going on.

Even putting a 150 ohm resistor directly across the 5v supply would only dissipate 166mW. Are you using 1/8 watt resistors? With the LEDs in place the maximum power from a 150 ohm resistor would be 60mW. I can't see how your resistor is getting so warm.

Iain #13
##### Dec 22, 2011, 08:05 pm
So if it's dropping 2.9 volts, I is 19.3 mA , passing 2.1 volts to the LED resistor, which is then dropping current to 14mA. So how many volts are being dropped by the second resistor/how many volts are the LED's seeing?

It would look better for the other 2 arrays so be similar brightness. But I haven't hooked up the darlington's yet so I can't evaluate those ones.

#### sixeyes #14
##### Dec 22, 2011, 08:49 pm

So if it's dropping 2.9 volts, I is 19.3 mA ,
Can you verify that this is the voltage being dropped, using a multimeter?

Quote
Passing 2.1 volts to the LED resistor, which is then dropping current to 14mA. So how many volts are being dropped by the second resistor/how many volts are the LED's seeing?
The LEDs should drop about 2 volts but that may not be the case at such small currents. I don't know what LED you're using and even if I did the manufacturer may not test it for such low currents.

Can you measure the voltage across one of the LEDs when it's in the dim condition and the voltage across "its" resistor?

If you're actually getting 19.3mA through the single "dimming" resistor, this will be split between all 39 LEDs and their resistors, so it should be about 0.5mA each. This would give a volt drop across the 39 LED resistors of about 75mV. Again any actually voltages you can measure would help.

Iain

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