So there is this nightlight (a little flower that you plug in directly into a mains socket and switching it on gives you some light to light up your bedroom at night) which I dismantled and tried to know how its wired up. This is what I got:


simulate this circuit – Schematic created using CircuitLab

The mains is 230V AC @ 50 Hz and the capacitor C1 is of the shown type: enter image description here except its marking is 224K 200V which I presume is 22E4 pF (ie 220 nF), 200V maximum voltage rating and K = 10% tolerance.

My questions are :

  1. Is the R1-C1 combination acting as a filter here? Is this common in this kind of rectifier circuit?
  2. I plan to make the same circuit, can the diodes be replaced by others from the 1N400X series?
  3. The markings on C1 shows it's a 220nF cap, but measuring capacitance across the cap gave a reading of 580nF. Is this the result of the other parts of the circuit?
  4. I plan to make the same circuit, is it a safe bet to use a 2.20 uF capacitor as C1 ?

EDIT: I had inadvertently swapped the values for R1 and R2. Now the circuit appears to be correct. I made the following measurements across the components during operation:

  • \$\approx\$ 14mA through each LED
  • \$\approx\$ 2.95 V drop across each LED
  • \$\approx\$ across each diode
  • \$\approx\$ 1.37 Vdc across R2
  • \$\approx\$ 217.5 Vac drop across R1

  • 3
    \$\begingroup\$ electronics.stackexchange.com/questions/5572/… for an explanation of your R1-C1. BE VERY CAREFULL!!! THIS IS NOT A "SAFE" CIRCUIT as it is not isolated from the mains supply. \$\endgroup\$ – brhans Mar 29 '16 at 18:02
  • \$\begingroup\$ I'm puzzled. In a typical supply like that, C1 acts as a series current source, R1 is in series with it to protect from the switch-on surge, R2 is very small as the LEDs are current driven, and the 10uF will have only a low voltage rating. As drawn, R1 and C1 do nothing, R2 acts as the current source very inefficiently. Is R2 value correct? Is R1/C1 circuit correct? What's the C2 voltage rating, as drawn it needs to be 400v? \$\endgroup\$ – Neil_UK Mar 29 '16 at 18:07
  • \$\begingroup\$ Can you double-check the value of R2. At 680 kΩ it will pass about 0.6 mA to the LEDs (allowing for smoothing of peak voltage by C2) and this seems very low unless they're super efficiency LEDs. The diodes have to be rated for peak inverse mains voltage, 350 to 400 V. \$\endgroup\$ – Transistor Mar 29 '16 at 18:08
  • \$\begingroup\$ @brhans I did away with the switch in the schematic, but your statement still holds as the nightlight switch is only a single-pole one and hence there is still 230Vrms across the switch when the light is flicked off. \$\endgroup\$ – K. Rmth Mar 29 '16 at 18:08
  • 1
    \$\begingroup\$ You don't need to measure the current through the LEDs it will be determined by R2. The 6 to 8 V dropped across the LEDs won't make much difference to your calculations. A voltage reading across C2 might be useful but be careful. You could solder on some temporary test leads, connect to your multimeter switched to (high-voltage) DC and switch on remotely (plugging the night-light into an extension lead and switching the extension lead on at the wall socket). \$\endgroup\$ – Transistor Mar 29 '16 at 18:20

1) C1 is a power-dropping element to reduce the high mains voltage down to something more appropriate for a string of LEDs. R1 is typically there to discharge C1 when the power is removed.

2) The 1N4007 is rated for 1000V. Since diodes are so remarkably inexpensive, it is false economy to use a lower-voltage rated diode in such a potentially dangerous circuit to save literally a few pennies. NOT RECOMMENDED to substitute a lower-voltage rated diode in this circuit. It is already living on the edge. No future in tempting fate.

3) You can't really accurately measure the value of most components IN CIRCUIT. Because of the influence of the other connected components. If you want an accurate measurement, you must measure/test it OUT-OF CIRCUIT.

3a) Furthermore, note that C1 should be a special "Y type" capacitor which is rated, designed, manufactured and tested for use in direct mains connection. Use of an ordinary capacitor for C1 is a serious fire hazard and NOT RECOMMENDED.

4) This is a critical and potentially dangerous circuit. Unless you have significant experience with direct mains-connected circuits like this it is NOT RECOMMENDED that you make ANY changes to the circuit. Furthermore, since you can't know what many of the component ratings are from observations, you are putting yourself at considerable risk attempting to reproduce this circuit.

Recommend finding "BigClive" on YouTube. He has done a tear-down and circuit analysis of this very "flower LED night-light" and many similar cheap Chinese LED products. Highly recommended.

Ref: https://www.youtube.com/channel/UCtM5z2gkrGRuWd0JQMx76qA

  • \$\begingroup\$ Cheers for the BigClive channel; looks very interesting. The nighlight links are here and here \$\endgroup\$ – K. Rmth Mar 29 '16 at 18:58

First, I think you've got some of the part values wrong. R1 is probably 100K or 1M. R2 is probably more like 6.8K (or 680 ohms), however the circuit looks correct.

  1. C1 is a dropping capacitor. R1 is to discharge the capacitor (a bleeder resistor) so you don't get a shock by touching the prongs of the plug after it has been unplugged.

  2. Don't do this.. this circuit is not safe for tinkerers. You can use other types but it should be a rectifier diode capable of a fair amount of current. The voltage rating is not too critical - 50V is probably okay.

  3. You can't accurately measure the value of parts in-circuit in many cases. This is probably one of those cases.

  4. The safest bet is not to touch this circuit with a 10-foot (3m) pole.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.