# Efficient way to drop AC voltage in LED constant-current driver

I am trying to power a string of 24 LEDs which are connected in series (re-used from an LED bulb) from 120 V mains AC. Each LED has a Vf of 3.2 V, and a 150 mA maximum rating.

To limit the LED current to 100 mA, I was planning on using the LC5202D. It's a constant-current driver IC that comes in a DIP package (which I prefer over SMD.)

My plan was to use a buck topology as shown in the datasheet:

This leaves a problem. If my AC input is 170 V0 after rectification (120 * 1.414), but my LEDs are only consuming 76.8 V (24 * 3.2), what am I going to do with the rest of the voltage?

Considering the IC is only rated at 1.78 W power dissipation, it won't be able to drop 5 W of power (50 V * 100 mA).

Can I simply put a Zener diode after the rectifier to drop the 50 V, assuming I can find one with the proper power dissipation?

Another solution might be to use a flyback circuit with a transformer, but that's going to involve more parts and complexity than I want.

Maybe this is much easier and I'm just missing something.

• Current is measured in milliamperes (mA.) Capacity of batteries is measured in milliampere hours (mAh.) Which did you mean?
– JRE
Sep 16, 2022 at 16:45
• Oh my bad, I'm so used to typing in mAh all the time. Thanks for the callout, I'll fix it Sep 16, 2022 at 16:47
• Have a look at youtube.com/watch?v=vwJYIorz_Aw&t=241s for some good insights into how switching converters work. You should find it helps your understanding. Sep 16, 2022 at 17:34

Short answer: you don't need to do anything, the 5202 takes care of that for you. You limit the voltage to LEDs to limit their current, the 5202 limits the current directly. The power dissipation isn't equivalent to a series current limiting resistor because it's operating in PWM mode with an inductor to smooth the pulses.

• Ah, I see. I think the problem is that I'm thinking of this as more of a Linear Regulator. 120V go in, 80V gets used, the other 40V gets dissipated as heat somewhere. But that's not how these switch regulated (?)IC drivers works. Sep 16, 2022 at 17:34
• The datasheet has a good description of the operating cycle. The 5202 first turns on its internal MOSFET, providing a current path between OUT and SEN. Current flows from the source, through the LED string, the inductor (which limits the rate of current change), the 5202's internal MOSFET, and Rs. Once the voltage across Rs rises past a threshold value, the MOSFET is turned off, and the inductor discharges through D1, maintaining current through the LED string. After a fixed period of time, the MOSFET is turned back on and the cycle repeats. The timing can be altered to adjust brightness.
– vir
Sep 16, 2022 at 17:54

The circuit is essentially a Switched Mode power supply (SMPS), using the resistor Rs to sample the current circulating.
LC5200 switches fully On or Off depending of some parameters, for instance to avoid undevoltage operation, but in normal operation it switches the current On-Off passing through the inductor L1.
Dissipation in normal operation: the voltage drop when the internal MOSFET is On is about 2 Ohms, the voltage drop on the MOSFET and dissipation will be negligible. This is the main advantage of the SMPS.

About the current-set value, it could be seen that, when the current goes above the V_Sen (=V_ref +/-30mV), the cycle is Off; that +/-30mV makes the hysteresis for the On-Off. As the current on inductors does not vary abruptly, it gives time for that On-Off done by the LC5200. The current (on the LEDs) is then kept approximately constant:
I_LED = V_ref / R_shunt.

Can I simply put a Zenor diode after the rectifier to drop the 50 volts?

No, you wouldn’t need (explained above) it and shouldn’t do that, as it would be dissipating power (V*i) as heat; invalidating the usefulness of a SMPS.