LED driver design


Joined Jun 5, 2013
You know the voltage the 10 LEDs in series will take.

You know the line voltage.

So you can calculate what voltage the capacitor must drop at 150 ma.

Do you know how to calculate the voltage across a capacitor at a specific frequency and current?



Joined Jan 29, 2010
hi q,

Capacitor impedance Zc =1/(2 * pi * F * C)

For 230Vac at 150mA , that is 230V/0.15A =1530 Ohms

So, 1530 = 1/( 2 * pi * 50 * Cap)

Work it out and post your answer


Thread Starter


Joined Nov 7, 2021
In the most simple version I do not have a filter cap but you can add one. In the plot the voltage across the 10 LEDs is about 33 volts. (green trace) The current is 120 or 100hz pulses averaging about 150mA. (blue trace)
What happen if i add 33v zener diode parallel to the led? Will the current change?


Joined May 15, 2009
Ask yourself what you want the Zener to do.
Why do you want it?
BTW I do not like capacitor limited power supplies. Reason is because of transients. Unlike an inductor used in fluorescent tubes capacitors let spikes through. A series resistor helps suppress these but an inductor opposes them (except of course for self capacitance between the wires). You really need to use, if you were to actually build one, continuous mains rated capacitors self-extinguishing. That's the safety aspect which is alarming - high voltages and no guidance on real components. I hope it is just a simulation exercise.

Thread Starter


Joined Nov 7, 2021
Ask yourself what you want the Zener to do.
Why do you want it?
I want the zener diode to act as a voltage regulator.
The circuit was designed by some one that does not understand the operation of the supply. The 2.2uF 400V cap limits the current into the load. So a constant current was changed to a constant voltage by a Zener
as ronsimpson explain before, adding zener diode creates a constant voltage circuit.

Thread Starter


Joined Nov 7, 2021
Screenshot_20211206-215133_P R O T O.jpg
I've already tried this circuit which is supposed to have 16 v output but when I measure it why do I only get ~2.2 volts? What is wrong with this circuit?


Joined Jun 17, 2014

Just to note, R2 is there mainly to keep the turn on surge through C1 low when the unit is first plugged in as well as for any impulses on the line later.

These circuits are very useful but for a 1 amp output i am not sure it is that good of an idea.
Assuming you have to drop 20 volts (just a rough idea for now) that means the cap has to have around 220vac across it all the while providing around 1 amp of average current. That would require a cap of something like 15uf but it has to be rated for constant AC use and it has to handle 2 times the peak voltage. That's possible in maybe a stand up metal cap but it might be on the expensive side.

I think i would feel better using a wall wart, either regulated or unregulated. If this is really homework though then you probably have to work through the theory that's all. You just have to know how to calculate current through a resistor/capacitor series circuit, and size R2 to limit the surge and R1 to bleed the cap in a reasonable time period.

Let us know what you come up with, but if you actually build this you better be very very careful as capacitors do not provide true galvanic isolation so there is always a shock hazard.