Trying to understand an optocoupler section of a schematic that I am designing - Page 1
Forum section: electronics / projects

author: n3rdx
author_rank: Contributor
timestamp: August 02, 2018, 10:04:35 pm
content: Hello, I previously developed functional prototypes of a MCU based controller/power switch. I am trying to breadboard a new version of the original schematic (A) which was experimentally improved to (B) in the attached schematic fragment. I don't have access to the actual hardware prototypes anymore, and am breadboarding another version from scratch. Both boards A, B were put into use, and they worked as intended. What I noticed in reviewing the two schematics is that only B had resemblance to the reference diagram of a CEL PS2801 / PS2501 type optocoupler which shows a non-inverted output. Then, could I have a bad schematic of item A, or is that a valid way to take an inverted output as I have seen it being discussed on various web forums.What I found in breadboarding the PS2801 optocoupler with a 3V input, 560R resistor to ground (left side of schematic) and 18V supply to Vcc and 1K2 ohms value for RL, I do get a non-inverted output but at much lower voltages than I am expecting. I see using my oscilloscope, only a "high" of ~8V. Why?So, I guess I am confused at this moment of the use of the series resistor on the right hand side, as I have seen a series resistor between Vcc and the photo-transistor collector, and in yet another version of my optocoupler sections (there were three variations! A, B, C) I experimentally used a series resistor before the collector, a series resistor after the emitter to ground and tapped the output from the emitter pin. Is this then an effective potential divider?I'm looking to learn about the best practices used for the output side of an optocoupler.

author: iMo
author_rank: Super Contributor
timestamp: August 02, 2018, 10:39:06 pm
content: Keep it simple.When you connect the 3V signal to the LED's Anode (Cathode via 560ohm to ground) the LED will LIT.The LED's forward current will be If= (3-1.1)/560=3.4mA.There is the CTR (current transfer ratio Ic/If ) defined, min is 80%, thus the Ic current of the photo-transistor will be min Ic = 0.8 * 3.4 = 2.7mA.When the LED lits, the transistor will be opened (conducting), the Ic current will be min 2.7mA.With 1.2k resistor in the emitter (the collector at +18V) the voltage at the emitter (against gnd) will be min Ve = 0.0027*1200=3.2V.In case you want to see "full" 18V at the emitter (Ve)a) the CTR must be higher, and/or,b) the If current into the LED must be higher (the LED resistor smaller than 560 ohm), and/or,c) the emitter resistor must be of higher value, such the given Ic current creates a higher voltage drop at the emitter resistor (your dynamic/pulse_shape characteristic may suffer due to higher RC, where C is a parasitic capacitance).In case you want "inverting" behavior, it means when there is the 3V signal at the LED side there will be 0volt at transistor side, you have to connect the emitter to gnd, and you wire a resistor from collector against +18V.Collector voltage will be the output. With 3Volt at LED side (LED lits) the transistor will be opened (conducting) and the voltage at Vc will go to 0V (almost).When 0 Volt at the LED side the LED does not lit, the transistor is closed (not conducting) and there will be full 18V at the connector.Mind the calcs above apply as well.Edit: added point c).

author: n3rdx
author_rank: Contributor
timestamp: August 03, 2018, 01:48:58 am
content: Quote from: imo on August 02, 2018, 10:39:06 pmFor any others who are considering developing optocoupler circuits and who need to perform such calculations: Since imo mentioned CTR (Current Transfer Ratio), I would recommend CEL's Application NoteAN3004How to simulate Current Transfer Ratios (CTR) and long-term CTR degradation in transistor optocouplerswhich discuss degradation of CTR over the operational lifetime with respect to operating temperature. Available here:http://www.cel.com/appnotes.do?command=showByType&group=2#

author: n3rdx
author_rank: Contributor
timestamp: August 04, 2018, 03:46:14 am
content: Quote from: imo on August 02, 2018, 10:39:06 pmFrom a data sheet I see that optocouplers are often provided with 'ranks'Notes: *1. CTR rank (PS2801-1 only)K : 300 to 600 (%)P : 150 to 300 (%)L : 100 to 300 (%)N : 80 to 600 (%)I also see that the CTR degrades over time. I could (as a designer) choose the minimum CTR and select component values for Rf, Rl, but wouldn't that change from device to device, or between batches of devices? How would one account for that for the long term? Also, what if we design for 80%, and populate the values, but the device installed in the PCB performs at a high CTR, say, 600% ?

author: iMo
author_rank: Super Contributor
timestamp: August 04, 2018, 01:09:59 pm
content: When you stick with recommended LED If current then when designed for 80% and having part with 600% could mean you just burn more electricity than required.It could be the "more light" may cause some "saturation effects" in the transistor (it should be equivalent to increasing Ibe in a transistor) such the dynamic params might be affected (longer time to remove excessive charges), experts from semiconductor physics area may discuss..PS: you may try: look at the transistor's output signal with an o'scope, change the If, and observe the rising/falling edges..