Electronic – 230V AC / 460V DC Optocoupler using Led and Voltage Divider

currentoptoelectronicsvoltage divider

I am building a primitive optocoupler to sense if a bulb has blown/if current is flowing.
It receives either 230VAC or 230VDC (rectified..) as supply.

I have calculated and built a voltage divider as pictured, but I am getting no LED action on 230AC tests, would dearly love any tips on how to get it working, but still keep it "primitive" simple as a design consideration 😀
So what's wrong with this picture?!

Details:
The bulb measures 77 Ohm across the filament, and acts as a resistor (R2) in my calculations (and as the switch..).

R1 (two resistors in parralell) is 3 Ohm as measured-.
I calculate 11V@230V AC and 17V@460DC, as the voltage across the LED (but measure nothing..)

I plan to buy a 5W, 1.7 Ohm to get a more manageable 5V / 10V combo, and choose the LED resistor accordingly.

I will put a rectifying diode in series with the led I think..
Is there any advantage crossing two LEDS to get a steady light on AC, or should I just smooth the 60hz signal with cap on the LDR side?

230VandRectifiedOptocouplerandVoltageDivider

Best Answer

Basically, this won't really work - or work well. An alternative is presented at the end of this answer.

Before that, the calculations, and an explanation of why it won't work.

From Wikipedia:

The cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating.

  • Assuming the resistance of the filament was measured when the bulb was not lit, this gives a rough estimate of 77 x 15 = 1155 Ohms.
  • This implies a ~45 Watt bulb, let's go with 50 Watts for convenience. This back-calculates to a hot filament current of around I = P / V = 217 mA = ~220 mA.
  • To obtain ~5 Volts at 220 mA, R = V / I = 22.727 Ohms = ~ 22 Ohms, the nearest standard resistor value.
  • Power dissipated by this resistor: P = V x I = 1.1 Watts, so let's go with a 5 Watt resistor to be safe.
  • LED choice: Let's say red LED, 1.8 Volt forward voltage, 20 mA nominal current. Such an LED will light up quite well at 6 mA (230 Volt situation) and will also stay within current limits at 12 mA (460 Volts situation).
  • Therefore a suitable LED current limiting resistor is 560 Ohms

Using two LEDs wired in anti-parallel would be recommended, because LEDs typically have low reverse voltage ratings (often as low as 5 Volts). With two LEDs facing opposite ways, the maximum reverse voltage appearing across the non-conducting LED will be the forward voltage of the conducting one, at any point in the AC cycle. It'll work fine in DC as well, so all is good.

Now, the problems:

What happens when the bulb is just switched on, and the filament is cold?

  • Resistance = 77 Ohms, therefore current = ~ 3 Amperes! That's before adding the bleed resistor of 22 Ohms
  • Current with 22 Ohms added in series: 2.323 Amperes. Power across resistor: 119 Watts!
  • Result: Fireworks.

What happens when the bulb is supplied with 460 Volts?

  • Assuming it is a 230 Volts rated bulb: The filament burns out.
  • If it is a 460 Volts rated bulb: Not much goes wrong other than the sense resistor blowing up anyway.

Solutions:

  1. Use the light bulb to illuminate the LDR directly.
  2. Use a high wattage bulb in series with the 50 Watt bulb, for the voltage divider arrangement.
  3. Use an alternative, such as a current sensing transformer, then use that CT's output in place of the LDR for sensing. A CT can be made by winding several turns of thin enamel wire around the wire supplying current to the light-bulb.
  4. For AC and DC current sensing, use a prebuilt module such as this one ($4.35 from eBay including international shipping) that incorporates a nice "lil black IC", the Allegro ACS712, and will provide safe, nicely isolated current sensing:
    Current sensor module
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