# Help choosing resistor

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#### DMFW

##### New Member
Hi all,

I been given the circuits attached in the image, R1 and R3 need to have a load of 5W, so using P=(V^2)/R I rearranged to get R and worked out R1 and R3 need to be 28.8Ω to get 5W load. However how do I choose the resistor values of R2 and R4.

cheers.

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It depends on the voltage drop of the LED and the required forward current.

$R = \frac{V-V_F}{I_F}$

Hi all,

I been given the circuits attached in the image, R1 and R3 need to have a load of 5W, so using P=(V^2)/R I rearranged to get R and worked out R1 and R3 need to be 28.8Ω to get 5W load. However how do I choose the resistor values of R2 and R4.

cheers.

hi,
You cannot solve the resistor value until you know the 'forward voltage drop' of the LED and it required operating current.

For a general solution: R1= [12V - V led]/ I led = R1 ohms

hi,
You cannot solve the resistor value until you know the 'forward voltage drop' of the LED and it required operating current.

For a general solution: R1= [12V - V led]/ I led = R1 ohms

Thanks for the reply, the Power across R1 and R3 needs to be 5W, I'm not sure of the forward voltage drop or LED operating current, how would you get R1 and R3 to give 5W with typical led forward voltage and current.

Cheers.

Thanks for the reply, the Power across R1 and R3 needs to be 5W, I'm not sure of the forward voltage drop or LED operating current, how would you get R1 and R3 to give 5W with typical led forward voltage and current.

Cheers.

Unless it a very high powered LED, to get 5Watts of heat generation in a LED's series resistor would be unusual when powered from a 12V source.

Standard LED's are rated around 10mA to 20mA, look at this chart for typical voltage drops.

So say you had a White LED with Vfwd of 3.2V to get a 20mA current thru the LED from 12V would be [12v-3.2v]/0.02 =440R
The wattage/heat in the resistor would only be 0.176W

So to get heating of 5Watts of a resistor would mean [using a White LED] would be [8.8V^2] /5 =15.48R

So the current would be I = 8.8/15.48 = 0.55Amps

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Unless it a very high powered LED, to get 5Watts of heat generation in a LED's series resistor would be unusual when powered from a 12V source.

Standard LED's are rated around 10mA to 20mA, look at this chart for typical voltage drops.

So say you had a White LED with Vfwd of 3.2V to get a 20mA current thru the LED from 12V would be [12v-3.2v]/0.02 =440R
The wattage/heat in the resistor would only be 0.176W

So to get heating of 5Watts of a resistor would mean [using a White LED] would be [8.8V^2] /5 =15.48R

So the current would be I = 8.8/15.48 = 0.55Amps

how would you then work out the value of R2 and R4? as you have 8.8v drop across R1 and 3.2 on the LED

cheers

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how would you then work out the value of R2 and R4? as you have 8.8v drop across R1 and 3.2 on the LED

cheers

hi,

So say you had a White LED with Vfwd of 3.2V to get a 20mA current thru the LED from 12V would be [12v-3.2v]/0.02 =440R
The wattage/heat in the resistor would only be 0.176W

So to get heating of 5Watts of a resistor would mean [using a White LED] would be [8.8V^2] /5 =15.48R

ah my mistake sorry, so R1 = 15.48Ω and R2 = 440Ω?

Cheers

ah my mistake sorry, so R1 = 15.48Ω and R2 = 440Ω?

Cheers

hi,
Thats it. So the method is:

Subtract the LED forward voltage from the supply voltage, then divide the result by the required LED current,, the answer is in Ohms.

NOTE: the 15.48R at 0.55A is not a normal example.!!

For standard LED's both resistors would be 440R

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Thanks man, helped a lot.

Hi,

With this circuit the resistors are in parallel, is it possiblie to get 5W on R1 and R3 in this circuit?

cheers

heres the pic sorry.

With this circuit the resistors are in parallel, is it possiblie to get 5W on R1 and R3 in this circuit?

hi,
You have already solved that problem.????

I been given the circuits attached in the image, R1 and R3 need to have a load of 5W, so using P=(V^2)/R I rearranged to get R and worked out R1 and R3 need to be 28.8Ω to get 5W load.

so are you saying your solution works or R1 should be 28.8Ω?

cheers

so are you saying your solution works or R1 should be 28.8Ω?

cheers

hi,
The fact that those resistors are directly across the 12V supply means that the resistors with leds have no influence on R1 and R3.

OK.? ok makes sense now

cheers

Whay do you want to do this?

Sorry but it seems like a total waste of time.

Whay do you want to do this?

Sorry but it seems like a total waste of time.

hero.
If you check the Forum.... its Homework.!

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