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I am considering using 50VAC supply (ELV safer ?) instead of 240V AC (50Hz) for the attached light schematic.

Diagram of ONE existing light with 12 LEDS.

There will be 20 of these lights = 240 LEDs.

Step LED Light 1.2W

5mm white LEDs.

Schematic is as read from the current circuit board.

The AC power supply will feed approx. 20 x 1.2W lights.

I am looking to use a toroidal transformer (240VAC - 50VAC) with appropriate ampere rating.

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    \$\begingroup\$ Did you have a specific question? \$\endgroup\$ – Chris Stratton Oct 12 at 5:19
  • \$\begingroup\$ Your first sentence is missing "I am" at the start as is your last sentence which, I think, starts with the word "Looking". Please fix your grammar capitalisation and punctuation. 'A' or 'amps', not 'AMPS'. All of these affect legibility and credibility. Welcome to EE.SE. \$\endgroup\$ – Transistor Oct 12 at 6:54
  • \$\begingroup\$ I would advise against 50V PSU as it is going outdoors, and if water gets into any wiring/touches positive and negative terminal, it is more likely to corrode (due to electrolysis) \$\endgroup\$ – F.Ahmed Oct 12 at 8:48
  • \$\begingroup\$ You can get 12V power adapters for dirt cheap anyways/ or just use one from an old internet router \$\endgroup\$ – F.Ahmed Oct 12 at 8:48
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    \$\begingroup\$ DANGER The original circuity is lethally dangerous and should NEVER be used for something like garden lights. ALL parts of the circuit should be considered as being at full mains potential as they at any time may be. This type of supply is only suitable (if then) for circuitry which is always well insulated from human contact. Garden lights don't qualify!. || For 50 VAC remove (short) R! C1, select R2 as required and place a large capacitor on the LED side of R2 to remove the flicker. If 50VAC used Vdc_vpk is about 70 Volts = 70/20 = 3.33V/LED. May be marginal depending on LEDs used. \$\endgroup\$ – Russell McMahon Oct 12 at 11:35
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UPDATED - MAJOR CHANGE due to new information:

WARNING The question has probably been misinterpreted by everyone (including me) due to a lack of complete information. I/We thought there were 20 series LEDs per string and that the OP's diagram showed the WHOLE LED setup and that each LED shown was a 1.2W unit.

With the added (clear and nicely lit and focused) photos it now seems likely that:

    1. Each "light: has 12 series LEDs.
    1. Each light is 1.2W so about 100 mW per LED so ABOUT 33 mA/LED and very roughly about 40V DC per light.
    1. A total of 20 lights = 20 x 12 = 240 LEDS is proposed.

This changes several major things.
Power per light is 1.2W, so a stabilising resistor can be relatively low power.
V_string_DC inside a light is now 12 x Vf_LED or maybe 36 - 40 VDC.
With added headroom voltage to keep LED current relatively constant a DC voltage of 45-50 Vdc would be adequate.


STILL RELEVANT. Lightly edited.

DANGER The original circuity is lethally dangerous and should NEVER be used for something like garden lights. ALL parts of the circuit should be considered as being at full mains potential as they at any time may be.

This type of supply is only suitable (if then) for circuitry which is always well insulated from human contact. Garden lights don't usually qualify!
IF the mains voltage is fed into a light by manufacturer approved wiring that meets relevant regulatory code and if the light also meets code then it MAY be OK - but is not a good idea.


For 50 VAC remove (short) R1 & C1, select R2 as required and place a large capacitor on the LED side of R2 to remove the flicker. If 50VAC used Vdc_vpk is about 70 Volts = 70/20 = 3.33V/LED. May be marginal depending on LEDs used.

With each light needing 45-50 VDC an AC 50V feed is rather higher than desirable.
See below for voltage sizing.

If using two strings in parallel as has been suggested would lead to much greater dissipation if 50 VAC is used.

IF you have a choice of VAC then testing the LEDS at desired brightness and current will allow a suitable value of VAC. R2 is set internal already and can be left unchanged.

  • Operate a light from a DC supply with R1 and C1 removed (shorted).

  • Vary supply to give about 1.2W dissipation. This will be at ABOUT 40VDC.

  • Determine Vac to give the above DC value.

  • Choose a per light filter cap OR a per whole 20 light feed filter cap (much larger).

Per light cap probably 100 uF plus.
Per whole string maybe 2200 - 3300 uF

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Without any reservoir capacitor on the output of the bridge rectifier, the LEDs will flicker strongly at 100Hz.

To convert it to 50V AC, you will need to re-calculate C1 and R2 to get the correct current through the LEDs.

Also white LEDs require about 3.0 to 3.3V to run. A string of 20 thus requires about 60 to 66V. The peak voltage of 50V AC (about 70V) will just about drive them, but it will be even more flickery. Better to split the LEDs into two strings of 10 LEDs.

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  • \$\begingroup\$ See my comment "WARNING ... on the ops question \$\endgroup\$ – Russell McMahon Oct 13 at 9:48
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For assured safety you need to use a DOUBLE INSULATED transformer, as you have no safety ground on the secondary circuit which possibly may be subject to the effect of weather (especially rain, damp) and human contact.

There is much that could be improved regarding the circuit, as noted in the other replies, notably regarding flicker at 2 X line frequency. Some people notice the effect more than others.

You won't need C1 (and R1) with 50V AC input to the bridge rectifier (50V AC may be a bit low for 20 LEDs in series)

Add a reservoir capacitor across the DC output of the bridge and set R2 for the desired LED current. Iled ~= 400mA from the LED chain Vf/Pd where Vf ~= 3V (typical for a white LED) and Pd = 1.2W.

The pk-pk ripple on the reservoir capacitor ~=(Iled.t(non-conducting))/Creservoir where t(non-conducting) is the rectifier non-conduction period (typically ~7ms for 50Hz AC, 6ms for 60Hz. As this period is longer for 50Hz, we'll use that.

Re-arrange and choosing Vripple(pk-pk) to be, say, 5V we get ...

Creservoir=(Iled.t(non-conducting))/Vripple = (0.4x7.10^-3)/5 = 560uF

Choosing the nearest larger E6 value gives 680uF

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    \$\begingroup\$ At 1.2W, VF ~= 3V say, I = P/V = 1.2/3 = 400 mA. \$\endgroup\$ – Russell McMahon Oct 13 at 9:42
  • \$\begingroup\$ My above comment is WRONG. It's 1.2W for 12 LEDS in series. So about 30 mA/string per light with 12 x 20 = 240 LEDs total \$\endgroup\$ – Russell McMahon Oct 13 at 9:58
  • \$\begingroup\$ It seems that the original question has changed. I've based my answer on the question as originally phrased. \$\endgroup\$ – Graham Stevenson Oct 13 at 10:10
  • \$\begingroup\$ Yes, I agree - the original question gave incomplete & hence ambiguous information. I interpreted it wrongly. I assumed that the 12 LEDs shown were for example and that he intended 20 LDs in series. He didn't. :-( \$\endgroup\$ – Russell McMahon Oct 13 at 10:20
  • \$\begingroup\$ Oh joy ! Well the general form of the answer holds true. You can stick whatever values you like into it. \$\endgroup\$ – Graham Stevenson Oct 13 at 10:22

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