Ac voltage indicator using optocoupler

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Discussion Overview

The discussion revolves around designing an AC voltage indicator using an optocoupler, focusing on calculating the appropriate resistor value (R1) for the circuit. Participants explore the requirements for the optocoupler's LED, including voltage and current specifications, and consider alternative methods for voltage dropping.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks to determine the resistor value (R1) necessary for the optocoupler, specifically the voltage across it and the minimum current required to activate the photo transistor.
  • Another participant suggests that R1 should limit the current to around 20 mA, based on typical LED characteristics, and that the forward voltage drop is approximately 2V.
  • A different participant counters that using a resistor to drop voltage from 354V to 2V would require a high resistance, leading to significant power dissipation, and proposes that the minimum current needed to activate the photo transistor is likely less than 20 mA.
  • One participant recommends using a series capacitor to achieve high reactance without power loss, suggesting that this could be a viable alternative to using resistors.
  • Another participant reiterates the need to understand the optocoupler's datasheet to determine the minimum current required for the photo transistor, emphasizing that this value can vary by part number and load current.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate current for the optocoupler and the feasibility of using resistors versus capacitors for voltage dropping. There is no consensus on the minimum current required to activate the photo transistor or the best method for designing the circuit.

Contextual Notes

Participants note the importance of consulting the datasheet for specific optocoupler models, as characteristics such as current transfer ratio can vary significantly between different parts.

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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