Ac voltage indicator using optocoupler

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SUMMARY

The discussion focuses on designing an AC voltage indicator using the PC-217 optocoupler. Key considerations include calculating the resistor R1 to ensure proper voltage and current levels for the optocoupler's LED. It is established that the typical forward voltage for standard LEDs is approximately 2V with a maximum current of 20mA. However, the participants emphasize that the minimum current required to activate the photo transistor is significantly lower than 20mA, and suggest using a series capacitor to manage voltage drop without excessive power dissipation.

PREREQUISITES
  • Understanding of optocoupler functionality, specifically the PC-217 model.
  • Basic knowledge of resistor and capacitor configurations in AC circuits.
  • Familiarity with current transfer ratio concepts in optocouplers.
  • Ability to interpret electronic component datasheets.
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  • Research the current transfer ratio for various optocouplers, including the HCPL-073A and 4N25.
  • Learn how to calculate resistor values for LED current limiting in AC applications.
  • Explore circuit designs for AC voltage indicators using capacitors and resistors.
  • Investigate the implications of using different optocoupler models in voltage indicator circuits.
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Electronics enthusiasts, circuit designers, and engineers involved in AC voltage sensing and optocoupler applications.

hisham.i
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Hello,

I want to design an AC voltage indicator. So i designed the circuit attached, but now i am trying to calculate the value of the resistor R1.

I have the following problem:


1- The voltage should be given at the terminals of the optocoupler.
2- The minimum current that should flow through the optocoupler in order to turn on the photo transistor.
3- How can i find these info. in the datasheet i tried to open the data sheet for PC-217 opto coupler but i wasn't able to find them.

Thanks
 

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hi there hisham

just remember that the LED in the opto-coupler is pretty much a standard LED
your R1 will need to provide voltage dropping and current limiting to keep the current through the diode below maximum
most average LEDs are ~ 2V forward voltage and ~ 20 mA. So depending on what the max AC voltage you are applying will determine the value of R1 to keep a max of 2V drop across the LED and a max of 20mA through it

cheers
Dave
 
Hi davenn,

But drawing droping the voltage through the resistor from 354V to 2 V at the terminals for the optocoupler will result is a high resistance required (This is not a problem i can use many resistors in series), but if you take 20 mA current the power for the resistor to be selected will be very huge. So i don't think this is the case & that the minimum current that can be used in order to open the photo transistor is much less than 20 mA.
 
use a series capacitor for a high reactance without power dissipation. Google images show a few circuits. search for "AC line voltage indicator LED"
 
No i want to use the resistor & optocoupler configuration, but the value of 20 mA is not the minimum current that can be drawn through the optocoupler to open the photo transistor.

My question is what is the minimum current that can be drawn through the optocoupler to open the photo transistor (its a general question not nessary to this project).
 
You are better off using a capacitor and resistor to drop the voltage. Your sims will show how well that works. I'll let you play with the capacitor value. There is no good reason not to do it that way. The examples show an LED, but an optocoupler is essentially an LED driving a transistor. Just replace the LED with the optocoupler.

You need to read and understand the data sheet to determine the current transfer ratio for the part you are using. Is is different for every part number. And it varies with load current. This part http://www.avagotech.com/pages/en/o...gton_transistor_output_optocoupler/hcpl-073a/ is very different than a 4n25.
 

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