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Diode protection

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Adam2014

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Hi
I'm asking for the protection diode work in this circuit?
 

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Are you asking will the protection diode work?
Based on the schematic I see, it will protect the opto-isolator and LED input from receiving a voltage greater than about -0.7 volts, which is generally lower than the reverse voltage maximum of LEDs, so yes it looks like it will protect. Not really sure what part you are looking to protect though.

The only thing that confuses me in the 4.png file is that it shows an input on both sides of the drawing. The direction of the isolator indicates that the signal goes from left to right, so is the signal source on the left really "Output from PLC"?
 
thanks
i'm know asking for interpretation to the AC input unit for PLC
why i get this signals ? how this circuit work ?
 

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  • signals.PNG
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Something doesn't look right to me here. You show the green wave, a sine wave input going into the bridge rectifier. What comes out of that should be both sides of the sinewave in a positive direction - sort of a double-hump waveform that looks kinda like "mmmm". Each of those double positive pulses should turn on the opto's diode, which in turn should turn on the phototransistor inside the optocoupler at that same doubled rate.

I expect to see the output - what you called signal2 - to have double the frequency of signal1. The red waveform appears to be just a clipped version of the sinewave at the same frequency. I would expect to see that somewhere on the other side of the circuit if a diode was shorted and burned out.

However, if the frequency was double the input, then that red waveform would appear more normal, in that the phototransistor is only able to generate a voltage and conduct in one direction and not the other. Still, I don't see any charge held over from the 100nF cap. Are you sure that signal2 is really not from the other side of the circuit? It just doesn't look right to me.
 
Hello,

Bridge rectifiers present special challenges for seeing the simulation work right.

To properly view the output, make the following changes to the circuit:
1. Disconnect the ground from the bridge rectifier, connect that point of the rectifier to the other side of the AC input.
2. Connect a 10 megohm resistor from that AC input side to ground.
3. Ground the cathode of the opto coupler internal LED.
4. Ground the bottom of C1.
5. Do not try to view the input sine wave, instead view the output of the bridge rectifier at the junction of the two cathodes (left side of R1). Also view the output at the top of C1 on the second channel.

Problems occur with bridge rectifiers because the AC input can not be common to ground when trying to view another point in the circuit because the second point is implicitly common to ground also. This isnt a problem with half wave rectifiers.

So what you should see is a full wave rectified sine on channel 1 and something similar on channel 2.

Depending on the simulation model it may also work to simply ground the lower end of C1.
 
Mr Al gives good advice.
What you should be seeing is the opto-coupler switch off (signal 2 go high) briefly as the AC passes through zero - this happens twice per cycle. This kind of circuit is often used to syncronise electronic circuits to the mains power.
 
Yea, that looks more like the real world to me. Funny, I didn't realize we had a simulation going here - I thought it was real broken hardware!
 
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