Optocoupler AC input circuit

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dswartz2

Joined May 19, 2009
31
I am trying to design a circuit that will take any AC voltage up to 230V. I need my output to always be 24Vdc.

I am using optocouplers to get the change from whatever signal input it is, to the 24V signal. This is shown in the attached circuit. My load is a low active load, so when the opto is turned on, the 24V are taken to ground and gives my load a low signal.

the spec for the optocouplers are attached as well.
I am using the PS2532-2

I am wondering what kind of changes need to be done to my circuit to achieve what I am trying to do. I think I have a pretty decent start/idea, but I would like some direction on what to do now?

Thanks for any help
 

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beenthere

Joined Apr 20, 2004
15,819
The two statements -
I need my output to always be 24Vdc
- and -
the 24V are taken to ground and gives my load a low signal
- seem to be in conflict.

Your AC detector is probably not going to work as you wish. With 230 VAC applied, the current through the optocoupler emitter will be 230/22K, or about 10 ma. That is just barely enough to affect the detector, and probably not enough to pull the output to ground - certainly not with 2.2K to 24 volts.

What is the purpose of the zener diode, and the diode in parallel with it? Is it acceptable for the output to pulse at the line frequency?

It would be very useful for you to explain why you designed the circuit the way you did. Can you give us the theory of operation?
 

Thread Starter

dswartz2

Joined May 19, 2009
31
OK.

I am trying to design a board for an interface that will take any AC input signal, and still have a 24Vdc output signal.

My load is called an IDD. It is a door drive that runs on 24Vdc. It is a low active IDD, which means it is at a constant 24V and when it drops to zero, it reads it as an incoming signal, which will then operate for whatever input signal was called. When the optocoupler is off, the constant 24V is being drawn in to the IDD. When the optocoupler turns on, it brings the 24V to ground, which gives the IDD the low signal.

For example: The "open door" push-button is pushed. This sends the AC signal to the optocoupler. The optocoupler then turns on, and brings the constant 24Vdc on the collector to ground. This, in turn, gives the IDD a low signal, which would then trigger the door to open.


The output cannot pulse because it has to keep the constant low signal until the door is all the way open.

I put the diode in parallel with the zener because I was hoping that the diode in the optocoupler and the one right outside it on the input would handle both the positive and negative ripples of the AC input. This was the main part that I wanted to have some answers to....

I hope this explains it a little better.
 

beenthere

Joined Apr 20, 2004
15,819
You need to make the AC into DC for a constant signal. You need to decide on the lowest level AC voltage the circuit can respond to. You will need to regulate the rectified AC so the current through the optocoupler is constant over the range of AC input voltages.

You may need to use a larger pullup resistor on the optocoupler collector.
 

Thread Starter

dswartz2

Joined May 19, 2009
31
bertus,

I'm sorry, i never noticed that you suggested that circuit to me



I did change the types of optocouplers i was using, they accept up to 80mA and a forward voltage from 1.2 up to 1.4 V. This makes it easier to be able to use a lot wider range of inputs.

So, this circuit you have shown can take both DC and AC voltage inputs? And what ranges were they?
 

bertus

Joined Apr 5, 2008
22,270
Hello,

The 15K resistor limits the current at high input voltages.
The bridge rectifier turns the AC into DC.
The bridge rectifiers also can be used for DC input (polarity does not matter).
The capacitor fills the gaps in the AC pulses at the output of the bridge rectfier in case of AC input.
The Zener diode limits the voltage.
The 820 Ohm resistor limits the current through the led to ca. 15 mA.

I do not know the sensitivity of the optocoupler at low input voltages.
perhaps you will have to "play" with the values of the reisitors.

Greetings,
Bertus
 

Thread Starter

dswartz2

Joined May 19, 2009
31
hello,

thanks for the explaination of the circuit.

What I don't understand, is how can I use a 35V rated cap. if i'm inputing 230Vac in to the rectifier? Just at 110Vac, the positive end of the rectifier is 50Vdc!

What is the problem here?
 

bertus

Joined Apr 5, 2008
22,270
Hello,

The zener diode will top the voltage at 15 Volts.
I always take 2X the voltage used for the capacitor 35 Volts is the closest value.

Greetings,
Bertus
 

Thread Starter

dswartz2

Joined May 19, 2009
31
alright... instead of a 15V zener, I used a 51V because I had it laying around. I changed the resistor on the input of the opto to work for this.

The circuit works, and I was just wondering something. When I applied 110Vac, the voltage going on the positive end of the rectifier stayed around 10Vdc.... why is it such a drop in voltage? I'm not complaining, because this makes it work pretty simple, but I just wanted to know why this is happening.

thanks
 

bertus

Joined Apr 5, 2008
22,270
Hello,

When you use a 51 Volts zener, you should use a 100 Volts capacitor.
The 820 Ohm resistor should be higher, about 2k2 Ohms.

Greetings,
Bertus
 

Thread Starter

dswartz2

Joined May 19, 2009
31
yes, like i said above, i replaced the 820 resistor with a 2.2k ohm resistor. I am using a 50 volts capacitor, and i havent gotten bast 26V on the output of the rectifier, so I think that it will be ok.
 

Thread Starter

dswartz2

Joined May 19, 2009
31
ok thanks for everything.

I've tested it for all voltages for DC and AC and it works for all of them.

NOW.....


I have a signal coming back from my IDD. The signal coming back is using 24Vdc. The control that the signal is going back to has to be the voltage that it sent out (the input voltage from before...Ex. 230Vac)

Attached is how I have been doing it with the DC voltages. How would I modify this circuit to get an AC voltage back?

24Vdc is going in to the opto through the 2.2kohm resistor. On the collector of the opto, is a resistor that is connected to the source voltage that the signal needs to be when it goes back.

Any ideas?
 

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bertus

Joined Apr 5, 2008
22,270
Hello,

The 24 Volts DC is input.
What should come on the output?

Greetings,
Bertus

PS when posting circuits you can attach PNG drawings, those are less lossy as JPG.
A DOC file needs to load a wordprocessor to read is (Open Office in my case, I am using LINUX).
 

Thread Starter

dswartz2

Joined May 19, 2009
31
When the 24Vdc is on the input, the output will be anything from 24Vdc-134Vdc, or 100Vac-230Vac.

There cannot be any voltage drop. So I thought about using a 2.2kohm resistor, but it would have to be a 40 or 50 W rated resistor. Is there any other way this could be done?

Also, If I did want the AC output, how would I apply the positive and negative leads from the source?
 
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