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Resistor wattage in LAB PSU

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neazoi

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In this LAB PSU I have designed the 0.33 Ohms resistor, how many watts must it be?

Also, can I use another resistor value, or no resistor at all, and what effects will it have?
 

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W=(I^2)R so it needs to be rated at 0.33 times the maximum load current squared.
The resistor is there to create a voltage drop at the 'set' pin as load current flows so it adjusts the voltage to maintain a constant current.
I would anticipate some drift problems using opto-couplers like that, as everything is ground referenced you could directly couple the voltages from the potentiometers to the ICs.
There should be a ground on the regulator IC!

Brian.
 
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    neazoi

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W=(I^2)R so it needs to be rated at 0.33 times the maximum load current squared.
The resistor is there to create a voltage drop at the 'set' pin as load current flows so it adjusts the voltage to maintain a constant current.
I would anticipate some drift problems using opto-couplers like that, as everything is ground referenced you could directly couple the voltages from the potentiometers to the ICs.
There should be a ground on the regulator IC!

Brian.

So the the resistor must be 0.33*3.3=1.089 Watts for 3.3A max load?

I noticed the drift as well and I replaced the LDRs with pots like shown in the datasheet.
 

That seems about right. I would use a 5W rated resistor so it doesn't get as hot, any change in value will change the voltage sensed across it.

Brian.
 

So the the resistor must be 0.33*3.3=1.089 Watts for 3.3A max load?

power requires the current be squared P =I^2* R
or 0.33 ohm * (3.3 A) ^2 = 3.6 W

a 5 W resistor should be ok
 

Sorry, I missed that the current hadn't been squared in post #4 but as wwfeldman points out, 5W is adequate anyway. It is the temperature coefficient that really matters as it changes the current limit.

Brian.
 

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