Opto-coupler choice for DC  - low forward current - Page 1
Forum section: electronics / beginners

author: ssashton
author_rank: Frequent Contributor
timestamp: April 12, 2023, 10:24:51 am
content: Hi,Just a quick question. I want to take a 24v logic signal (max about 1KHz) and isolate it from a 3v3 GPIO.In order not to burn too much power in the series dropper resistor, can I get some suggestions for an opto-coupler that does not need much forward current, please?Thanks!EDIT: Seems like 4N35 will work okay if 5mA is enough to switch it on.

author: Terry Bites
author_rank: Super Contributor
timestamp: April 12, 2023, 11:01:18 am
content: For a 3V3 low cost device such as H11L1M would do.It needs a couple of mA to turn on and the output side will operate down to 3V.

author: spostma
author_rank: Regular Contributor
timestamp: April 12, 2023, 03:40:24 pm
content: Maybe a CPC1001N with a parallel reverse diode, and a 22K series resistor for a 1mA LED current?If isolation is not required, just use a 47K : 6K8 input divider, and a 3V3 transzorb on the processor power line.The input protection diodes easily can swallow the worst-case 0.5mA input current.

author: floobydust
author_rank: Super Contributor
timestamp: April 12, 2023, 07:23:57 pm
content: I use dual channel SMTMOCD217Mfor low input current like 1mA.

author: Zero999
author_rank: Super Contributor
timestamp: April 12, 2023, 09:52:17 pm
content: The minimum required LED forward current depends on the current through the phototransistor and the current transfer ratio of the opto-isolator.I would just use a high value resistors. A bog standard PC817 will do. It has a minimum CTR of about 50%, so go with 33k of the 24V side and 10k on the 3.3V side.

author: ssashton
author_rank: Frequent Contributor
timestamp: April 13, 2023, 10:09:21 am
content: Thanks for everbody's suggestions.Wouldn't 33K on the 24V input side of a PC817 (forward voltage about 1 - 1.2V) give less than 1mA LED current? Is that enough to enable it?To calculate the required 33K on the 24V input side and the 10K on the 3.3V side did you do it like this?Assume a 10K pull-up on 3.3V.3.3V though 10K is 0.33mA.PC817 CTR is about 50% at low levels (graph only goes down to 1mA and 70% CTR), so we need minimum 2 * 0.33mA = 0.7mA of LED forward current.1.2V LED forward voltage and 0.7mA forward current gives us 33K series resistance with 24V.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 13, 2023, 10:52:38 am
content: Just use a weaker pull-up on the isolated side. CTR will be poorer at low currents. Most any phototransistor type will do. Beware of noise, especially at the high impedances: a shunt resistor should be used across the LED to prevent turn-on at marginal voltages (drawing say If/5 at nominal voltage), and some filter capacitance may be welcome on the receiver side (as much as you dare, up to your bandwidth limit). Somewhat higher analog bandwidth can also be combined with a digital filter (whether linear, or hysteretic as you might use for switch debounce). You might also add a Schmitt trigger if the logic input doesn't otherwise integrate one.Incidentally, speed doesn't need to be low; by cascoding the phototransistor, bandwidth can be greatly extended. Or a photodiode type, used with a transimpedance amplifier. Not needed here, but can be handy elsewhere.Tim