I am trying to use a potentiometer with a capacitive supply and microcontroller. As digital-to-analog output current is too low, a buffer is needed. Buffer output is the Vcc at potentiometer. Problem is the supply, while op-amp is connected, supply voltage is 1.2V at most. From datasheet current draw is ~1mA, so that shouldn't be a reason for such high voltage drop on supply. What causes this behaviour ?
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\$\begingroup\$ using a bjt as switch for op-amp supply, keeping it disconnected until voltage reaches normal level to power the mcu and DAC to function, then activate supply to op-amp, would be a solution ? my thought is 0v at non-inverted input causes the issue. \$\endgroup\$– johngerMay 22, 2018 at 8:51
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\$\begingroup\$ Could it have anything to do with applying 226V (\$160V\cdot \sqrt{2}\$) to the circuit and blowing something up? \$\endgroup\$– Sven BMay 22, 2018 at 11:26
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1\$\begingroup\$ What I am trying to say is that your problem may not be the load, but the supply. I just checked the 1N4148 and it has a peak reverse voltage of 100V, while you are applying more than double that amount. Don't you think it'll break? Without D2, the output could never reach 5.1V as a working D2 is needed to charge C2. I also do not advise working with these high voltages unless you know what you're doing. Would it not be possible to use a 5V adapter instead? \$\endgroup\$– Sven BMay 22, 2018 at 11:51
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1\$\begingroup\$ Then I think we need more info/measurements. Maybe a component broke anyway, maybe the opamp is broken and shorts too much current to ground, maybe you accidentally put the potentiometer to 0% shorting the output, maybe there is a practical issue going wrong with your setup that can't be seen on the schematic, ... \$\endgroup\$– Sven BMay 22, 2018 at 12:26
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1\$\begingroup\$ @SvenB You were right from the beginning, the high transitory current were the cause for the C1 fail. \$\endgroup\$– DorianMay 25, 2018 at 7:45
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1 Answer
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Credits to @SvenB for being the first that pointed to the right direction.
Even on steady sinusoidal AC source the current through power supply components does not go higher than hundreds of milliampere when connecting the AC power the current might go up to 2 amps.
The worst case scenario is when the AC power was disconnected with C1 voltage at a maximum (+220V) and later connected while AC power was at a maximum opposite voltage (-225V).
This will put almost 450V on R1 , a peak current of 3A for a very short time (0.3ms).
That's why components in the path R1, R6, C1, C2 D1 and D2 must tolerate a peak 3 amps current
Tough, the continuous current might be as low as few hundreds mA.
From the comments it shows that C1 was dead first but any other could fail, usually the diodes and unpolarized capacitors get shorted, and the resistors are interrupted or their resistance grows. This time C1 capacity was altered to a lower value.
Don't forget to add a discharge resistor ( ~100k ) in parallel with the AC input to protect the operator from the remaining C1 voltage
