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As long as the power dissipation is below 0.25W then you'll be fine. I would recommend you keep the dissipation below 1/2 the rated power. If you don't then you may need to consider the ambient temperatures.
Thanks for your help, i'm not sure if these two resistors at 0.25 watts is enough over head. As a newbie to electronics i know they set the reference voltage and output voltage on the chip but not sure if quarter watt resistors can cope with the power of there job in the circuit, it was the 1% tolerance i was looking for for a stable output voltage offered by 1% resistors.
Power is calculated as I^2.R or V^2/R If you know the current through them or the voltage across them, then knowing the resistance, calculating the power is easy.
For example the resistor between the output and the adj pin might be 120 ohms. It will have 1.25 volts across it, the power dissipates is 1.25*1.25/120 = 0.01 watts. This is way below 0.25W.
Let's assume your output voltage is 12V. So the voltage across the lower resistor will be 10.75V (and it's value will be about 1k) the power dissipated is about 0.12W -- and this is almost exactly half of 0.25W, so I probably would use a 0.25W resistor and not worry about it.
1% resistors will not necessarily give you a more precises output voltage. But it certainly won't be worse! The main variable is the LM317 which has only about 5% accuracy, so it will swamp the resistors in terms of error.
For example the resistor between the output and the adj pin might be 120 ohms. It will have 1.25 volts across it, the power dissipates is 1.25*1.25/120 = 0.01 watts. This is way below 0.25W.
Let's assume your output voltage is 12V. So the voltage across the lower resistor will be 10.75V (and it's value will be about 1k) the power dissipated is about 0.12W -- and this is almost exactly half of 0.25W, so I probably would use a 0.25W resistor and not worry about it.
1% resistors will not necessarily give you a more precises output voltage. But it certainly won't be worse! The main variable is the LM317 which has only about 5% accuracy, so it will swamp the resistors in terms of error.
Thank you for your help and the formula, thats about the voltage i need. I might as well use 0.5 watt resistors then if the chip has 5% tolerance, just wanted to do away with the trim pot as this tolerance is 20% and my voltage can be fixed so it all looks workable, thanks again for your help. 
