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I hope that this question has not been asked earlier and is wasting anyones time. I don't particularly remember my ECE101 classes that well - and seek your help!

I own my grandfather's hifi amplifier that has power light indicator that recently burnt out. The bulb is proving too hard to find right now. It was a tiny filament bulb and was powered by a 8V AC current (Schematic says bulb rated 8V 40mA).

I just wanted to check if my math is still alright here.

AC 8V RMS = 8V x Square Root of 2 = 11.3V for the DC LED.

So by using a 6V 30mA led I would take 11.3v - 6V = 5.3V extra current across the circuit.

Then I would then just use V = I x R; so 5.3V/30mA = 176 ohms resistance.

By adding the 6V 30mA LED with a 180 ohm resistor would be a workable solution?

I was also told I may need a rectifier diode and a capacitor to keep a smooth glow? Are either mandatory or just optional?

Finally can I use a full wave rectifier to avoid using the cap? Any idea on how to wire this. Sorry for my basic question but every bit of help would be useful!

Thanks in advance :)

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Your easiest wiring would be to do this:

schematic

simulate this circuit – Schematic created using CircuitLab

which gives full-wave lighting, but it'll be rather bright. It can be adjusted by raising the R1 value. White LED forward voltages are about 3V, red around 1.2V, so the LED color matters. Except for some very bright lamps intended for bicycles and lanterns, a "6V" LED is rather unusual (and probably expensive). Flickering at 120 Hz is unlikely to be a problem, so a capacitor is not necessary.

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    \$\begingroup\$ Flicker at that rate is only a problem to a small percentage of the population. I am part of that small percentage. Come Christmas, I can't go into some decorated shopping malls because of their 'it only flickers at 100Hz/120Hz' LED strings. Thank goodness we've done away with the flickery CRT computer monitors and only use slow LCDs now. \$\endgroup\$ – Neil_UK Feb 19 '17 at 10:55
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    \$\begingroup\$ I'm mainly concerned about flicker in work lights near rotating machinery: if it isn't steady light, there's potential for illusions regarding moving blades... \$\endgroup\$ – Whit3rd Feb 19 '17 at 11:14
  • \$\begingroup\$ Thanks Whit3rd. In your schematic I notice 2 diodes. Is one the LED and one a regular diode? Where does the LED go in that diagram please? \$\endgroup\$ – Gaurang Patkar Feb 20 '17 at 4:36
  • \$\begingroup\$ @GaurangPatkar: The intent was to use two LEDs, and that generates the 120Hz light modulation: with one LED and one other diode, it works the same (but my habit is to try to draw AC current symmetrically). The one-LED/one-diode variant blinks at 60 Hz. \$\endgroup\$ – Whit3rd Feb 20 '17 at 5:41
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There are three options:

  1. Use a bridge rectifier, like Passerby suggested. Without a smoothing capacitor, the LED will flicker at 100 or 120 Hz, which isn't noticeable, but it will also glow dimmer without a cap. If the drop on the bridge is 1.4 V, you can reduce the resistor to just 150 ohms. This solution requires the most components and space, so it might not be the best.

  2. Connect the LED with that 180 ohm resistor in series directly to AC. It will flicker at 50 or 60 Hz, but I'm quite sure that's not noticeable for most people. Make sure the LED will survive more than 12V reverse voltage.

  3. Connect the LED to a DC rail instead. The amplifier is most likely running on DC, so you could use the DC rails instead. Don't forget to recalculate the resistor. If the voltage is really high, you may be wasting relatively much power at 30 mA through the resistor.

  4. A 9V incandescent bulb that fits will works just fine, but be a bit dimmer. A 6V bulb should also work, but lifetime will probably be reduced.

The efficiency of a LED is about 10 times as high as for an incandescent bulb. This means that at 30 mA, the LED will be (6V*30mA*10)/(8V*40mA) = ~6 times as bright as the original bulb. So you may want to multiply the series resistor by 5.

1 Without cap: relative_current(t) = 0 if sin(t)*8*sqrt(2) < 7.4 else (sin(t)*8*sqrt(2)-7.4)/(8*sqrt(2)-7.4). And the integral of the relative current from 0 to pi is 1.129... divided by pi gives 0.114 which is already quite close to a sixth. The resistance could actually be decreased in this case, but not to anything below 131 ohms.

1 With a large enough cap: This is always at peak voltage, so the resistor could be made 5 to 6 times as large to compensate for the efficiency of a LED. 820 Ohms.

2: relative_current_when_forward_biased(t) = 0 if sin(t)*8*sqrt(2) < 6 else (sin(t)*8*sqrt(2)-6)/(8*sqrt(2)-6). The integral of that from 0 to pi divided by two pi equals to 2.1. 390 Ohms

3: 5-6 mA through the LED.

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  • \$\begingroup\$ Wow - Thanks for the info. I will try and make sense of the work now! \$\endgroup\$ – Gaurang Patkar Feb 21 '17 at 6:23
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That seems about right. The capacitor is for smoothing the output. Even with a full wave rectifier bridge, the DC output will vary widely without a capacitor.

enter image description here

The capacitor will remove the 30 or 60 hz flicker that is inherent in most AC signals. The circuit above is how you would wire it. The load is the led plus the appropriate resistor, which is the next closest standard value to 176 ohm, as you properly calculated.

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    \$\begingroup\$ It would be appropriate to point put that the full wave rectifier may require the AC source to be floating away from other circuit components. Depending what it is connected to elsewhere in the unit it may not be feasible to use the bridge. \$\endgroup\$ – Michael Karas Feb 19 '17 at 10:57
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    \$\begingroup\$ @MichaelKaras, It would not be appropriate to point that out, as the AC source does not need to be floating away from other components. \$\endgroup\$ – Tustique Feb 19 '17 at 16:39
  • \$\begingroup\$ Note that this can have quite huge startup loads on the AC source due to the capacitor. \$\endgroup\$ – yo' Feb 19 '17 at 21:53
  • \$\begingroup\$ @user38168 - You be wrong. As I said it depends upon how the rest of the circuit is hooked up. I give one example where the bridge will not produce 8V output. See the answer block I made just show you this type of example. \$\endgroup\$ – Michael Karas Feb 20 '17 at 0:08
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In comments to the answer given by Passerby @user38168 thinks that the 8V output would always work despite what other things are hooked to the 8Vrms transformer. The below circuit shows an existing center tapped transformer simulation where a full bridge across the outside winding connections of the transformer will not give a full output.

enter image description here

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