# Optocoupler to drive LEDs (maybe with transistor)

I wonder if there is some simple solution to drive LEDs powered from single rectified voltage.

Assume AC is 220 V, each LEDs is 3.0 - 3.2 V 150 mA. I reckon series connection made it simpler but not sure if the optocoupler is able to provide the current.

Adding a BJT/MOSFET at the opto output should be OK but I am not sure how to connect it. Protective part can be advised but I wish to have the basic ON/OFF function working first.

EDIT on 2019-01-28 12:34

Actually, my design consists of two 30xLEDs paths and wish to end them with an optocoupler to Neutral.

Thanks to JonRB and Transistor, they discovered several fatal issues, so i retry and hope this would make more sense.

Refer to this active current limiter, i wonder if the optocoupler can attach to it. Now, with 60 x LEDs (3.0 - 3.2 V 150 mA):

100k Resistor to turn IRF740 FET On.
8 Ohm Resistor with BC817 BJT to limit the current to ~150mA.
Optocoupler PC817 would shutdown both transistors.


I think IRF740((400Vds) should be capable of AC peak voltage and current limiter can save the LEDs from destory as well.

When in doubt, check the datasheet:

https://www.futurlec.com/Datasheet/LED/PC817.pdf

The datasheet has two key pieces of information pertinent to your use-case

1. Collector current - 50mA
2. Collector-Emitter voltage - 35V

Your use case is targeting 150mA and thus this part is not suitable. Likewise you are stating this is powered from 220V (thus a peak of ~310V) yet this part has a Vce of 35V

Basically this part is not suitable for direct blocking.

A suitably rated FET would be. However... It isn't that simple. I have included an example circuit WITHOUT values as working with main potential isn't something you should do, unless you know what you are doing

The FET needs to have a blocking potential greater than the maximum DCLink voltage. It must be able to sink your target LED current.

The Zener is acting as a crude voltage regulator to limit the voltage applied to the gate. The resistive chain is dropping the vast majority of the DCLink potential.

simulate this circuit – Schematic created using CircuitLab

• I understand PC817 won't work. I'd first remove the optocoupler label. I also try to avoid another track from the main. maybe I should add this to my question as well. btw, do you think it is impossible? although PS2533 is not popular, do you think it could do the job? – Simon Jan 27 '19 at 10:40
• Please note that your circuit will destroy the LEDs - you have not included a current-limiting component. While I understand that you were focussing on the LED switching, the OP clearly does not understand the implications of your simplification. – WhatRoughBeast Jan 27 '19 at 19:17
• As it stands yes (unless you say you had 200odd in series and the RDS on is sufficient). – JonRB Jan 27 '19 at 19:19

The PC817 datasheet lists the Absolute Maximum Ratings as

Collector-emitter voltage    80 V
Emitter-collector voltage     6 V
Collector current            50 mA
Collector power dissipation 150 mW


With a 220 V AC supply and a diode rectifier you will have $$\ 200 \sqrt 2 = 350 \ \text {V DC}\$$ on the opto-terminals. It will die instantly.

Meanwhile, you have 30 LEDs in series with a minimum forward voltage drop of 3 V giving 90 V across the LEDs. Therefore the worst case voltage drop across the two 120 Ω parallel resistors is 350 - 90 = 260 V. The resistors will limit the current to $$\ I = \frac {V}{R} = \frac {260}{60} = 4.3 \ \text A \$$. The LEDs will be spectacularly brightly for a very short while.

Note that the power dissipated in each resistor is given by $$\ P = I^2R = 2.15 \times 120 = 555 \ \text W \$$.

Back to the drawing board, I'm afraid.

• ok. thanks. seem I have to vanish this idea....maybe try a relay at rectifier input instead. – Simon Jan 27 '19 at 10:50
• @Simon Your idea isn't impossible. You've just not considered the parts to use properly; you need to think about ratings when selecting parts. – Hearth Jan 27 '19 at 12:08
• @Simon - You seem to have concluded that you only need to use a relay rather than an optoisolator. This is not true. Please pay attention to the calculations involving LED current and resistor power. If you use a relay or a MOSFET (perJonRBs answer) you will destroy your LEDS. If, somehow, you don't, you will instead destroy your 180 ohm resistors. – WhatRoughBeast Jan 27 '19 at 19:16
• ok, I should not give up so easily since I started up this question. please check if this current limiter works? – Simon Jan 28 '19 at 8:26
• Have you calculated the power dissipation in your current limiter yet? (I haven't but I suspect that it's going to be very high.) – Transistor Jan 28 '19 at 13:58