Why TL081 and 2N3906 may get very hot?
The simulation of the TL081/2N3906 shows the normal working temperature, but I measure +62 ºC
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1\$\begingroup\$ At first glance I'd say it's probably because you don't have a resistor between the opamp and the base of the transistor. Why not? \$\endgroup\$– brhansJun 18, 2018 at 22:04
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2\$\begingroup\$ I don't think an LTspice simulation will try to calculate the device temperature based on power consumption...the temperature is a value provided by the user. You need to use the power consumption values along with the device datasheets to calculate the die temperature. \$\endgroup\$– Elliot AldersonJun 18, 2018 at 22:28
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3 Answers
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The TL081 is designed for a load of >2k ohms. What is your load? 10%*hFE*4R7? bad design.
The PNP will conduct Vce*Ice=Pd x Rja(+115'C/W) What is your temp rise.?
These problems can be solved from proper impedance design of a regulator. PD=V^2/R or I^2*R times thermal resistance often 120'C/W to 250'C for tiny packages with no heatsink.
WHat is your next idea to regulate your load?
answered Apr 11, 2019 at 14:26
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Your LTspice power calculation looks correct for zero load. If your real circuit has a load (especially something as innocent-looking as a bypass capacitor) your op-amp is probably oscillating. Put a scope probe on the output of the op-amp and monitor the voltage. The solution would be some compensation if that is what is happening.
You should add a diode from emitter to base on the transistor (and get a new transistor while you are at it) to protect against reverse breakdown of the B-E junction.
answered Apr 11, 2019 at 14:33
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Assuming that your current draw is the same, you probably don't have the right package:
The following is for a dip package of the TL081: The thermal junction is 115C/W which is how much heat can exit the part with a 0.5W power dissipation in the part, the part will get 115C/W*0.5W = 57.5C hotter than ambient (see how the units cancel out?).
If you have a 25C ambient temp, this will be 82.5C which is hot and exceeds the operating temp range
You can do the same calculation by finding the thermal junction resistance of your transistor package (they come in different flavors so I'm not going to guess which one you have 250mW is high for some packages).
(thermal resistance)(power dissipation) + (ambient temp) = temperature of part.
Make sure the package temperature calculation doesn't' get above the operating temperature range of the part, or you will violate an absolute maximum rating, at which point you can now longer expect the part to function or remain undamaged.
answered Jun 18, 2018 at 23:02