EEVblog Electronics Community Forum
Electronics => Repair => Topic started by: atgeek on November 22, 2016, 04:43:51 pm
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I've a thermostat board that is drived by a mcu. The power part consist in a transformerless power supply, the main 220v pass through a 1.5uF capacitor, then from a half wave rectifier diode and finally there are 2 resistor that provide power for a 78l05 voltage regulator. The 2 resistor were burnt, what could be the appropriate resistor value?
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So you are injecting unlimited voltage into 78L05
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Why? There are the resistors in series with 78l05 input
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Why? There are the resistors in series with 78l05 input
So what? Remove load from the output of 78L05 and you will should have way above it's maximum input voltage. Limited only by current consumption of 78L05 itself (and voltage drop of those resistors you are trying to add). If resistance of those resistors is low enough to provide low enough voltage drop to make any sense, 78L05 will blow up. If those resistors have high enough resistance to protect 78L05 in such situation, they will burn themselves (unless they are high power). What is resistance of those unnamed resistors?
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I don't know the resistance since they are burnt
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Assuming you have max 170Vdc in the divider's input you can place a 13k 3W and a 3k 1W. This will give about 32V max in the input of 7805 when it sinks no current and it will give about 7.5V when it sinks about 8,2mA. The current drawn from the source would be about 10,7mA which is 1,5W of power in the 13k resistor and about 0.35W (max) in the 3k resistor.
If you power your 7805 from a bench power supply and you determine that you don't need 8mA of current, you can increase the resistor's values accordingly.