In UC3707, VIN = Vc = 40V.
When the input signal is 3.3V / 40KHz / Square wave, the device heats up considerably. It seems to be about 50-60 ° C.
Is this normal?
Do you have data on thermal characteristics?
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This may have something to do with what you are driving.
1) What are you driving? If they are FETs, what is the gate charge (Qg)?
For each channel, the gate drive power can be calculated by: P_Drive = Qg x Vc x Fsw (Multiply by 2 for total power)
If the power dissipated is high and you are using no gate resistors; consider using low value resistors (~ 1 Ohm to ~ 5 Ohm) to help dissipate some of the power. This will increase turn-on and turn-off times slightly as a consequence.
Regarding thermal characteristics; I do not believe they are available.
I look forward to hearing back.
Understood. Looking at the schematic, the AC coupling capacitor (0.1uF) is a large load. This is very likely why your part is getting hot...
The AC coupling capacitor probably will not be seen completely as a load, so for a rough calculation you can reduce / calculate this number more accurately by knowing how the load affects it (you will need to know the capacitance of the load too).
Using the Gate Power Equation below, we can see why the gate driver is getting so hot.
P_g = (0.1 uF)*((40 V)^2)*(40 kHz) = 6.4 W
Even if we half the capacitance, we can see that the driver must dissipate 3.2 W.
I hope this makes sense as to why the gate driver is getting so hot. If this answered your question, please click the green button, otherwise feel free to follow up!