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I am working on a buck converter project where I am using the TL494 as a PWM controller IC.

I am confused about the power ratings of the resistors that are used in the entire circuit.

I have added the schematic for your reference.

It would be great if anyone helps with the calculations or procedure that I should use to find the power ratings of the resistors.

enter image description here

JRE
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Suprio Barua
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  • Note that this is an example circuit that is not stable because of lack of frequency compensation. See [here](https://electronics.stackexchange.com/questions/511214/why-is-there-no-frequency-compensation-required-in-tis-tl494-example-buck-regul). – Unimportant Nov 23 '22 at 13:55

1 Answers1

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R8 and R9 basically form a load for the output voltage. So their total dissipation will be

$$ P_{dr}=\frac{V_o^2}{R8+R9} $$

During voltage regulation the voltage across R9 will be fixed by the voltage comparator's reference voltage, which is set by the 5V-REF and R3-R4 dividers, 2.5V.

So, the voltage across R9 will be 2.5V during voltage regulation (it'll be lower if the voltage regulation is lost i.e. output current limiting is dominating).

$$ P_{R9}=\frac{2.5^2}{5100}=1.25\mathrm{mW} $$

If a failure happens and Q2 gets shorted, regardless of the operation of the controller IC, output voltage will be 32V and the voltage across the divider resistors will be 16V. This makes the dissipation higher:

$$ P_{R8-fail}=P_{R9-fail}=\frac{16^2}{5100}=50\mathrm{mW} $$

Other resistors will see lower voltages during normal operation. So your best bet is to select all of the resistors except R13 with a power rating of higher than 50 mW (0805-case SMD resistors have a power rating of 125mW and 1206-case ones have that of 250 mW).

R13 has a special case. If the voltage regulation is lost and the current is limited by the controller (e.g. under an overload condition) then the voltage across this resistor will be kept at 1V, which is set by R1-R2 divider and 5V-ref. So its max dissipation can be

$$ P_{d-R13}=\frac{1^2}{0.1}=10\mathrm{W} $$

Considering the failure that I explained above (seeing 32V across the output) the load's current can be much higher. So you need to take that into account as well, and select a resistor having a power rating of way higher than 10W. You can basically use a single resistor e.g. 0.1R/20W, or use paralleled resistors e.g. 5 of 0.5R/5W resistors in parallel to get 0.1R/25W.

Rohat Kılıç
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