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Topic: 230V => 5V smart Solution (Read 13497 times) previous topic - next topic

michinyon

Quote
Because of that series diode, it is not conducting and so there is no circuit.


So the LED is also a diode,  and it is also blocking current flow when the A/C circuit is positive at the lower connection,   and you would have the full reverse A/C voltage being blocked by the two diodes, in that path (  the diode,  and the led ),  which are in series and pointing in the same direction.

I am still very unclear how you are sure that the reverse voltage on the LED will be less than the alleged 5V breakdown limit.

Duriing the times when -230 volts is present on the A/C terminal of that circuit,  how can you be sure that 225 V of it will be present across the terminals of the external diode,   and only 5V or less across the terminals of the LED ?

bigred1212


What is the problem with that?


Mostly the problem is my lack of understanding. 


jremington

#17
Mar 07, 2014, 05:01 pm Last Edit: Mar 07, 2014, 05:19 pm by jremington Reason: 1
Quote
Duriing the times when -230 volts is present on the A/C terminal of that circuit,  how can you be sure that 225 V of it will be present across the terminals of the external diode,   and only 5V or less across the terminals of the LED ?

That is a good question, and the answer is that you can't. The actual reverse current flow in the LED during the "wrong" half cycle is most likely insufficient to destroy it, but the design with the antiparallel diode is much preferred and is in common use.

Furthermore, if a resistor, rather than a capacitor, is used to limit the diode forward current, it will generate a lot of heat (a couple of watts for 220 V @ 10 mA) and has to be sized accordingly.  I've been using several versions of the circuit I posted for many years to monitor 220VAC pump operation, with no failures.

michinyon

Quote
Mostly the problem is my lack of understanding.   


Nothing wrong with making sure you understand,   particularly when mains electricity is involved.   

To make sure that the led in the optocoupler doesn't blow up,  it would appear that the diagram of reply #8,  with the reverse bypass diode,  would be preferable to the arrangement of reply #5.

Grumpy_Mike


Quote
Because of that series diode, it is not conducting and so there is no circuit.


So the LED is also a diode,  and it is also blocking current flow when the A/C circuit is positive at the lower connection,   and you would have the full reverse A/C voltage being blocked by the two diodes, in that path (  the diode,  and the led ),  which are in series and pointing in the same direction.

I am still very unclear how you are sure that the reverse voltage on the LED will be less than the alleged 5V breakdown limit.

Duriing the times when -230 volts is present on the A/C terminal of that circuit,  how can you be sure that 225 V of it will be present across the terminals of the external diode,   and only 5V or less across the terminals of the LED ?

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