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I'm looking for an isolated power supply with the intention of driving the LED of an opto-coupler. The voltage regulation is not important and current output needed is not high. Cost is a major factor.

What kind of circuitry can be employed for this?

Off the shelf solutions from major semiconductors have higher current output of ~200 mA and are more expensive.

Update 1: Reason for isolated supply for an opto-coupler. It is for detecting a fuse blow in a circuitry, where I'm using the fuse as a path for powering the LED in a opto-coupler. enter image description here

EarthLord
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    Well, how high is not high? How expensive is not expensive? 1W or 3W LED? EL817c or similar? – tlfong01 Aug 26 '21 at 07:16
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    EL817c would be good. So something for 5 mA current output. – EarthLord Aug 26 '21 at 07:35
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    Isolated or not isolated? – winny Aug 26 '21 at 07:35
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    Isolated as mentioned – EarthLord Aug 26 '21 at 07:36
  • You might like to read Part C of my answer to the following EESE Q&A to see how to use 3V3 Rpi/Arduino GPIO pins to drive EL817C/EL354 with 5mA. How to properly use a relay module with JD-VCC from Arduino/Raspberry? Asked 1 year, 2 months ago Active 8 days ago Viewed 11k times https://electronics.stackexchange.com/questions/505318/how-to-properly-use-a-relay-module-with-jd-vcc-from-arduino-raspberry/508672#508672. – tlfong01 Aug 26 '21 at 08:19
  • There is some confusion here. EL817C input current is around 5mA, but output is not strong enough to drive a LED. Usually EL817C output drives a small transistor. eg. 2N2222, which in turn drives an up to 500mA load, or with a relay to 10A or higher. – tlfong01 Aug 26 '21 at 08:25
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    Another confusion is about isolation. If you use an optocoupler, the input and output side is of course optically isolated. So you don't need any "isolated" power supply. – tlfong01 Aug 26 '21 at 08:27
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    @EarthLord: Why do you need a power supply to drive the input to the opto-coupler? They are normally driven by a signal. You use them to electrically isolate the signal source from the signal consumer while still passing the signal. – JRE Aug 26 '21 at 08:28
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    Thanks for the comments. Added the reason in the question. – EarthLord Aug 26 '21 at 12:41
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    No reason for that LDO. Just adjust the resistor value accordingly to the get the current you need (5 mA?). Please note that you will be bleeding said current past the fuse when it's blown and charging whatever output capacitor you have there. Is this ok in your design? – winny Aug 26 '21 at 13:29

2 Answers2

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The cheapest option and least risk option is to get one of those trough hole DC-DC SIP packages.

If you can't fit that size, take a look at small push-pull transformers. Driven by your mcu or an TI SN6501 for example.

Jeroen3
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Your circuit isn't isolated so you don't need opto although it may help in some other regards. If you do use opto then just put the opto-LED and series resistor across the fuse.

This approach is common in industrial controls fuseholders.

enter image description here

Figure 1. A DIN rail-mounted fuseholder with integral LED blown fuse indicator. Image source: Weidmüller.

Transistor
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