I'm a beginner in circuit design and even if it applies to a real situation I'm doing this as an exercise.

After my thermostat broke some months ago I decided to replace it with an AC switch, which for my needs is fine since I really don't need automatic temperature control, manual control is ok.

Sometimes I've thought it would be nice adding an LED attached to it, and even if there's a low current in the switch it I think it's better to put the LED in a separate circuit. I've imagined this homemade solution using a ferrite core. I understand that the voltage/current on the secondary would be very weak but... Might it work? LED on a switch, as-is on the left and my proposed scheme on the right

On the left the as-is circuit, and on the right my modification. A weak or flickering light would be an acceptable result, considering it has no external power source. I'd prefer this to a battery-powered circuit. Could it put some impedance load on the AC line or disturb it somehow? Thanks in advance.

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    \$\begingroup\$ Basic current transformer so load dependant.LED needs reversed Diode protection or back to back LEDs \$\endgroup\$ Jan 4, 2021 at 14:32
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    \$\begingroup\$ Do you have the other line or neutral available? Switches with a neon onboard should be readily available and could be an easier and more code-compliant solution. \$\endgroup\$ Jan 4, 2021 at 15:07
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    \$\begingroup\$ Wouldn't a neon indicator lamp be easier? \$\endgroup\$
    – Simon B
    Jan 4, 2021 at 15:11
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    \$\begingroup\$ You could also use a double throw switch and on the output side use back to back LEDs with resistors in series. Just make sure to encapsulate well for high voltage \$\endgroup\$
    – user103776
    Jan 4, 2021 at 15:56
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    \$\begingroup\$ There is a great circuit on YouTube that does exactly what you want to, but it seems moderator Voltage Spike, on prompting of user ocrdu, both didn't think it was good enough to tell you about, as he/she has deleted my posts to you about it twice, then accused me of promoting my product and/or personal website (this seems retaliatory, as I don't have a personal website or product), so good luck. Check YouTube for circuits with BC547 transistors acting very similarly to what you're trying to do though. It's too bad SE has moderators that grade the amount of help people can give others. \$\endgroup\$
    – Dewey
    Jan 7, 2021 at 6:10

3 Answers 3


Could it put some impedance load on the AC line or disturb it somehow?

If you wire it as a current transformer (i.e. a single turn in the primary) then it should be OK but, you do need to know the full load current of the heater to calculate the number of secondary turns.

So, if the heater current is 10 amps and you wind a thousand turns on the secondary, it will drive 10 mA into an LED. But you need a reverse diode across the LED or maybe another LED wired in reverse to the first one. You don't need a capacitor or series resistor for this to work but, you do need to provide sufficient secondary turns so that the LED doesn't get fed too many mA: -

enter image description here

  • \$\begingroup\$ isn't a single turn making it a voltage transformer? shouldn't be primary only goes inside the ring? \$\endgroup\$ Jan 4, 2021 at 15:16
  • \$\begingroup\$ A single turn primary is totally synonymous with it being a current transformer. Voltage transformers have many multiple turns on their primary (usually). The single turn primary loops through the ring. \$\endgroup\$
    – Andy aka
    Jan 4, 2021 at 15:18
  • \$\begingroup\$ ah.. i get it. sorry.. i was totally using 2 turns on my mind. :))) \$\endgroup\$ Jan 4, 2021 at 15:19
  • \$\begingroup\$ Keep in mind that the "AC line" in my scheme is just a sensing line, with a very low current. I didn't measure it but I think the switch works at a maximum 0,1 ampere. \$\endgroup\$
    – ManuelJE
    Jan 4, 2021 at 15:25
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    \$\begingroup\$ @ManuelJE how many watts is your heater? The circuit you have drawn shows a switch that connects 220 VAC to a heater i.e. there is nothing about it that suggests it is a low power connection driving 100 mA to a heater (24 watts). If the heater current is genuinely 100 mA then a different approach might be suitable. \$\endgroup\$
    – Andy aka
    Jan 4, 2021 at 15:30

Might it work?

Yes, if you use components with appropriate values.

Could it put some impedance load on the AC line or disturb it somehow?

Yes, but unless you use an unnecessarily large transformer core, the load will be very small.

In your diagram, you show more turns in your primary winding than in your secondary. Almost certainly, you would want this the other way around. That is, you want the transformer to step-up the voltage. Also, I would put two LEDs (or one LED and one vanilla diode) in the circuit in anti-parallel configuration. In parallel, but with opposite polarities. That will help to keep the secondary conducting when the power is on. With only one LED, when the secondary stops conducting, there is typically a voltage spike. If your transformer has a sufficiently large core, that spike could damage the LED.

The turns ratio, and the core size will depend upon the current flowing through your power line, the current you want to drive your LED, and whether you want to run your transformer in linear mode (avoiding core saturation) or saturated mode. For most applications, one chooses a transformer core that is big enough to avoid saturation. Saturation introduces non-linearity, and in the case of a transformer that is driven by a sinusoidal voltage source, it can lead to excessive currents.

However, in your circuit, core saturation may be your friend. In a transformer driven by a sinusoidal current source, saturation can serve to limit secondary voltage (and consequently current). Further, if your concern is simply detecting current in a power line, rather than measuring it, non-linearity can allow a secondary output that is relatively constant over a wide range of primary current levels.

First, how to determine whether a core will go into saturation. If you have a current transformer with a single turn primary, the magneto-motive force \$F_m\$ is just the current.


H is given by

\$H = \frac{\displaystyle F_m}{L_m}\$

where \$L_m\$ is the effective length of the magnetic circuit.

Then we will deny reality, and assume that there is a constant \$\mu\$ that relates \$B\$ and \$H\$ so:

\$B=\mu H\$

Putting it all together, the core will saturate if

\$\frac{\displaystyle \mu I_{max}}{\displaystyle L_m} \gt B_{sat}\$

Now, if you are operating in linear mode, (no core saturation), then the ratio between the current through your power line and the current on the secondary side is just the turns ratio.

If, for example, the current through your switch is 10A, and you want your LED to run on 20mA, then your turns ratio needs to around 1:500. (Of course unless you use a bridge rectifier, the LED will conduct only half the cycle so your average LED current will be 10mA). If the current through your switch is only 50mA, because, say it is only driving a relay, and you wanted to run your LED with 20mA, you would use a 2:5 ratio.

No suppose you want to saturate your transformer core. In that case, at each half cycle, \$B\$ swings between \$B_{sat}\$ and \$-B_{sat}\$, and the rest of the time, \$B\$ stays constant. When \$B\$ swings between its extreme values, there is an EMF (i.e. voltage) pulse generated in the windings which has a Volt-Second value of

\$ET = 2NB_{sat}A_{cross}\$

where \$A_{cross}\$ is the effective cross section of the core. There will be one positive going pulse and one negative going pulse per cycle. If the mains frequency is f (probably 50Hz or 60Hz) then the average voltage of, say the positive pulses only would be

\$V_{avg} = 2NfB_{sat}A_{cross}\$

Notice that the equation does not depend upon the current through the power line. Once core saturation is reached, the \$ET\$ (volt-seconds) of the pulse is independent of any further increase in current. This, may be an advantage in your application. However, the width of the pulse, will narrow as the power line current increases. That is, higher currents will generate shorter pulses of higher voltage. This may necessitate the use of an RC or LC filter to smooth the pulse.

So, with a given core effective cross section, core effective length, \$\mu\$, \$B_{sat}\$, and knowing that the core is saturated, we can pick a resistor, (possibly with smoothing capacitor and/or inductor) that will drive the LED with a smaller sized core, and without precise knowledge of the current in the power line.


What I've done previously - put a green LED in parallel - this tells you when the heat is Off, which is nearly as good.

Trusting that the valve will not close with the 2 mA in the LED. You should really put a diode antiparallel with the LED to protect it, ymmv.


simulate this circuit – Schematic created using CircuitLab

What I would do if I needed an On indicator:


simulate this circuit

Trusting / hoping that
a) the valve needs about 300 mA at 24 V
b) the valve works fine with ~ 21 V with the Zener in series
c) the zener can dissipate about 1 W (or try to find a lower voltage if they even exist, or use two or three 1N4001s in series, with another one the other way in parallel).

  • \$\begingroup\$ This does mean, however, that your LED is connected to the AC source with no isolation, which may be a safety issue for 220V, especially if the LED is somewhat user accessible (which seems relatively likely) \$\endgroup\$
    – BeB00
    Jan 5, 2021 at 20:04
  • \$\begingroup\$ @BeB00 Yes I wouldn't recommend this for direct 220 V - and you can't just use a resistor dropper there. I was thinking of a heating system's isolated 24 V AC, which is safer. Also the LED is no more or less accessible than the switch which OP admits wiring up. If it's mains it should all be in an earthed metal box. \$\endgroup\$
    – tomnexus
    Jan 5, 2021 at 20:41
  • \$\begingroup\$ I guess my main thought would be that mains switches are deliberately and very well designed to protect users from any dangerous voltages, with lots of certifications and standards, whereas the LED housing would be designed by a complete novice. The latter seems a lot more dangerous \$\endgroup\$
    – BeB00
    Jan 5, 2021 at 20:49

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