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Connecting a Red LED to 230V: Power Indicator Bulb Circuit, Safety & Modification Tips

mateusz konrad 6978 13
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  • #1 16712674
    mateusz konrad
    Level 7  
    Posts: 7
    Does such a scheme make sense? The red LED is to indicate power to the bulb, i.e. if the bulb is on, the diode will also. If it is bad I would ask for some modification suggestions. The entire system is connected to a 230v socket.
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    • Connecting a Red LED to 230V: Power Indicator Bulb Circuit, Safety & Modification Tips 15060134763211681755646.jpg (2.98 MB) You must be logged in to download this attachment.
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    #2 16713599
    wniedzie
    Level 14  
    Posts: 113
    Help: 9
    Rate: 66
    does not make sense.
    I understand that you want to detect the current flowing in the main circuit. But in your schematic, both legs of the diode are connected at the same point in the main circuit, so there will be no voltage drop there. Theoretically, you could put a small 1 Ohm resistor in the main circuit, which would produce a voltage capable of lighting a diode. The problem is that depending on the power of the bulb, this resistor would have to have a lot of power, because it would heat up cruelly.
    Anyway, with the right choice of this resistor, you do not need any additional elements, you just connect the LED diode parallel to the resistor.
    But that's all theory. If you enter the bulb power, we can try to calculate something specifically.
  • #3 16713607
    Jarosx9
    Level 35  
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    Move this your chip above the bulb (connected to both sides), the voltage drop will be on the bulb. And choose the right resistor for the bulb power.
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    #4 16713610
    wniedzie
    Level 14  
    Posts: 113
    Help: 9
    Rate: 66
    Jarosx9 wrote:
    Put your layout above the light bulb


    There will always be 230V on the bulb, regardless of whether the bulb is on or not (burned out).
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    #5 16713615
    jdubowski
    Tube devices specialist  
    Posts: 21595
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    mateusz konrad wrote:
    Does such a scheme make sense?


    It makes no sense, and you need to connect the system in parallel to the bulb.
    Secondly - with such a resistance, the capacitor is not needed there. If you want to give a capacitor with the calculated Xc, reduce the resistance to 1k.

    Added after 1 [minutes]:

    mateusz konrad wrote:
    i.e. if the bulb is on, the diode is also


    And if the bulb is blown, what is going to happen?
  • #6 16713649
    _jta_
    Electronics specialist  
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    When powered through a bridge, the 0.22uF capacitor (necessarily for a voltage of at least 275V AC, it's good if it has the X2 mark) should be just right for a typical LED; with one anti-parallel diode it needs 0.47uF as you want full brightness. A resistor with not too low power (otherwise it can burn out when turned on), and add a resistor with high resistance (1M or more) parallel to the capacitor so that the capacitor discharges after disconnecting from the mains - otherwise you have a chance to discharge it through your hand and it's not pleasant.
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  • #7 16713773
    jarek_lnx
    Level 43  
    Posts: 22535
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    You can make a system that will light the LED in series with the bulb, three or four rectifier diodes will give the appropriate voltage drop to power the LED
    Connecting a Red LED to 230V: Power Indicator Bulb Circuit, Safety & Modification Tips
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  • #8 16714275
    mateusz konrad
    Level 7  
    Posts: 7
    And if I connect the entire system to 12 V (12v 1-6W power supply) 3W bulb power, which scheme is better? 1 or 2
    and won't the bulb shine dimmer? What resistor would be appropriate?
    Attachments:
    • Connecting a Red LED to 230V: Power Indicator Bulb Circuit, Safety & Modification Tips 15060895120091717737011.jpg (3.27 MB) You must be logged in to download this attachment.
  • #9 16714321
    kortyleski
    Level 43  
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    You want 230 AC or 12 DC. At 230V you need to put a small rectifying diode with the led and a 470k ohm resistor in series with the set. connect as shown in diagram 2
  • #10 16714366
    Freddy
    Level 43  
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    System no.1 has no right to work.
  • #11 16714462
    _jta_
    Electronics specialist  
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    And what is the power of this power supply? If 1W, that's not enough for a 3W bulb. And you need a bulb for these 12V (maybe from the car), not 230V.
  • #12 16714573
    mateusz konrad
    Level 7  
    Posts: 7
    _jta_ wrote:
    And what is the power of this power supply? If 1W, that's not enough for a 3W bulb. And you need a bulb for these 12V (maybe from the car), not 230V.

    the power of the power supply is up to 6W and the bulb is adapted to 12v
  • #13 16714765
    Marian B
    Level 38  
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    And I will ask "sillyfrant" why the signaling of a glowing light bulb? After all, you can see that it shines if it is a lighting bulb, e.g. in a corridor or in a room.
    Sooner, the LED in the switch would be useful when the light is off. Then it is mounted on the switch contacts. Not directly, of course.
  • #14 16715170
    _jta_
    Electronics specialist  
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    I once installed a LED next to the switch showing that the light is on in the next room - my grandmother sometimes forgot to turn it off, and when the door was closed, it was not visible. It is true that it was possible to drill a hole in the door and see if the bulb was lit or not ...
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Topic summary

✨ The discussion revolves around connecting a red LED as a power indicator in a circuit powered by 230V. Initial responses indicate that the proposed schematic is flawed, as both legs of the LED are connected at the same point, resulting in no voltage drop. Suggestions include using a small resistor in series to create a voltage drop sufficient to power the LED, while ensuring the resistor can handle the power without overheating. Other responses recommend connecting the LED in parallel with the bulb and using a capacitor for stability. Concerns are raised about safety, particularly regarding capacitor discharge and the implications of a blown bulb. The conversation also touches on alternative setups for lower voltage systems (12V) and the practicality of using an LED indicator in various applications, such as on switches.
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FAQ

TL;DR: For a 230 V indicator, use a 0.22 µF X2 (≥275 VAC) dropper with a bridge; “0.22uF … just right for a typical LED.” [Elektroda, jta, post #16713649]

Why it matters: This FAQ helps DIYers wire a safe, clear power‑on LED for mains or 12 V lamps without dimming bulbs or wasting heat.

Quick-Facts

Quick Facts

Does my original 230 V LED indicator scheme make sense?

No. Your drawing ties both LED leads to essentially the same point, so no voltage drop appears to drive it. A small shunt resistor in the lamp line could create a drop, but it wastes power and heats badly as lamp wattage rises. “That’s all theory” here; it’s not a good practical path. [Elektroda, wniedzie, post #16713599]

How do I wire an LED so it lights only when the 230 V bulb is on?

Place the indicator across the switched live and neutral, in parallel with the bulb. Note an edge case: the lamp terminals still carry mains potential even if the bulb is off or blown, so the LED can still see 230 V. This method indicates supply presence at the lamp, not filament health. [Elektroda, wniedzie, post #16713610]

What parts make a safe capacitor‑dropper LED on 230 V?

Use a bridge rectifier, a 0.22 µF X2 capacitor rated ≥275 VAC, a series resistor of sensible power, and a high‑value bleed resistor (≥1 MΩ) across the capacitor to discharge after unplugging. “Otherwise you have a chance to discharge it through your hand and it’s not pleasant.” [Elektroda, jta, post #16713649]

Can I run a single LED without a bridge and still get full brightness?

Yes, pair the LED with an antiparallel diode and increase the dropper to about 0.47 µF for similar brightness. Keep the X2 rating on the capacitor and include a bleed resistor for safety. Add a modest series resistor to tame inrush at switch‑on. [Elektroda, jta, post #16713649]

Is a series shunt resistor in the lamp line a good current sensor?

It works in theory, but it is poor practice. Even a 1 Ω shunt must dissipate significant power as bulb wattage rises, running hot and wasting energy. If chosen, the LED can sit across the shunt, but thermal sizing becomes the challenge. [Elektroda, wniedzie, post #16713599]

Can I drop voltage with series silicon diodes and power the LED?

Yes. Three or four rectifier diodes in series with the lamp create enough forward drop to feed an LED indicator. This approach avoids a large resistor but still needs proper mains insulation and wiring discipline. [Elektroda, jarek_lnx, post #16713773]

Do I need a capacitor if I already use a large series resistor?

No. With a high series resistance, the capacitor is unnecessary. If you deliberately choose a capacitive reactance dropper, reduce the series resistor to about 1 kΩ to limit surge and shape current. [Elektroda, jdubowski, post #16713615]

Will a 1 W, 12 V supply run a 3 W bulb plus an LED indicator?

No. A 1 W supply cannot deliver a 3 W lamp’s load. Use a 12 V bulb and a supply with power equal to or above lamp wattage; a 6 W unit suits a 3 W bulb plus overhead. [Elektroda, jta, post #16714462]

Should I add an LED in the wall switch to find it when lights are off?

Yes, a locator LED in the switch is useful. Wire it to the switch contacts using proper limiting components rather than directly across mains. It indicates the switch’s position when the room is dark. [Elektroda, Marian B, post #16714765]

What single-resistor value is typical for a simple 230 V LED indicator?

Approx. 470 kΩ in series with a rectifying diode and the LED works for a basic mains indicator. Connect as in the forum’s Diagram 2 guidance for 230 V setups, observing polarity and insulation distances. [Elektroda, kortyleski, post #16714321]

Why won’t the thread’s System #1 work at all?

Because it lacks a valid voltage drop across the LED; both sides sit at essentially the same potential. Without a drop or proper current limiting, the LED cannot conduct and light. [Elektroda, Freddy, post #16714366]

What capacitor rating is mandatory across mains?

Use X2‑class capacitors at ≥275 VAC for line‑to‑neutral droppers. These are designed for across‑the‑line duty and fail safely. Do not substitute general‑purpose film or electrolytic capacitors here. [Elektroda, jta, post #16713649]

Quick how‑to: build a safe 230 V LED lamp indicator

  1. Rectify LED current with a bridge and use a 0.22 µF X2 capacitor (≥275 VAC) as the dropper.
  2. Add a modest series resistor and a ≥1 MΩ bleed across the capacitor to discharge on power‑down.
  3. Enclose parts, maintain creepage/clearance, and test with the lamp on a fused lead. [Elektroda, jta, post #16713649]

Will the LED still see voltage if the bulb is blown or switched off?

Yes. The lamp holder remains at mains potential in that wiring, so the LED may still energize. This indicates supply presence at the fixture, not filament continuity. [Elektroda, wniedzie, post #16713610]
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