# Help with operating an optoisolator

• Dextrine
In summary, rbelli1's circuit calculates Vo to be 2.624V using the formula I_f*I_emitter*432, but when he builds it, he gets Vo = 4.9. He thinks his understanding of how optocouplers work is off, but he is not sure. He plans to try changing the resistor value to 430 to see if that will correct the issue.
Dextrine
Hello everyone,

I am in need of some help operating the following component:
http://www.vishay.com/docs/83615/il211at.pdf

The 360V and 5V inputs are fixed, as well as the need for 2.5V at the indicated location.

The trouble I'm having is really understanding how the chip works as well as how to have it operating in, I believe it's called, linear mode?

Essentially, I want the 2.5V to change as a function of the 360V changing as well.My current understanding is that the current going through the diode is halved (because of the CTR) and then that current is represented by the transistor behaving as a sort of resistor, which causes a sort of voltage divider behavior. Thanks for any help!

I don't really understand what you are trying to do .

Please post a circuit diagram and explain your problem a bit more clearly .

Nidum said:
I don't really understand what you are trying to do .

Oh, I'm sorry. I thought i had posted it:

That's not really the proper way to operate that type of device. The CTR is specified as a minimum and typical value with an unbounded maximum. CTR decreases over time under load.

You can fix this by adding a second isolator as feedback. How much does the input voltage change? What is the impedance of the source?

BoB

Dextrine said:
Essentially, I want the 2.5V to change as a function of the 360V changing as well.
You want a circuit where the presence or absence of the 360V on one side causes a voltage on the other side to jump from 2.5V to ...what...?

First you stated that the 360V input was fixed, then you stated that it varies. What is the problem you are trying to solve with this circuit?

I should wait for your clarification rather than guess, but does the 360V represent the peak of a 240V ± 6% ac source that you are trying to read safely with an ADC, with the zero point at 2.5V?

Thanks for all the responses guys, I'll try to be clearer:

I am using the UCC28950 pretty much exactly how it is outlined in their application note. The only differences are that the output will be 360V DC and the regulation mechanism I plan to employ has optoisolation between the control stage and the power stage. So the signal going into the EA- pin will come from the optocoupler in a common-emitter configuration. (basically the last picture i posted will be how I isolate the power stage and control stage from one another)

Here is the datasheet for the UCC28950. Luckily, there is an excel sheet that they link to which I am using to select components (along with double checking to make sure it follows with their math on the datasheet).

http://www.ti.com/lit/ds/symlink/ucc28950.pdfSo, in short, my question could most clearly be:

How can I best operate the optocoupler in linear mode so as to allow optically isolated regulation.

I have attached the screenshot of the powerstage schematic as presented in the datasheet.

#### Attachments

• pwrstage.png
42.8 KB · Views: 554
https://www.eeweb.com/company-blog/power_integrations/20-w-standby-power-supply-reference-design

This is the common way. U2 sets the voltage. The switching controller can sense if the voltage is at the correct voltage. There is a small linear region around the set voltage. This allows the design to be relatively immune to difference in the CTR of the opto. U2 has a gain stage included to further stabilize the setpoint. This design can be modified to allow use of a plain zener diode in place of U2 but with less accuracy.

BoB

Dextrine
Thanks for the reply rbelli1, I guess what I'm really getting it is, how do you choose resistor values so that the optocoupler operates in the linear region, perhaps some document that contains a crash course on operating optocouplers in the linear region or an explanation of it? The topology makes sense, but it's just the specifics of ensuring that the device will stay operating in linear mode.

See the datasheet for the Tinyswitch-4. An example included is the zener only design.

BoB

Dextrine
Hmmm, so I've read through the application notes for the optocoupler and I came up with the following circuit which according to my math, should give me Vo=2.624V for the following reasons:
CTR = 1.1 @ 5mA
Vce (sat) = 0.4V

So in my circuit:
I_f=5mA
I_emitter = 5.5mA
I_emitter*432 = 2.376V
V_o = V_CE = 5-2.376=2.624

However, when I actually build this, I get Vo = 4.9, basically 5 almost as if there is no current transferred at all
.

Can someone let me know if my understanding of how optocouplers work is off or my analysis on this circuit?

Also, even taking into consideration the 1.3 forward voltage and assuming CTR=.5, I remade the circuit with the input resistor now 430 so that If = 0.01157, I_emitter = .005785 and I still get Vo=5 instead of the calculated 2.49.

Also, the device is actually this one: http://datasheet.octopart.com/MOC212R1M-Fairchild-datasheet-1287.pdf

I tried hooking up the input side to a constant current source but no matter how high i put it (up to about 100mA), no current was transferred to the other side. I even got a new optocoupler and still have the same problem... I'll keep trying stuff and posting anything here in case anyone else ever has the same problem.

I figured out the issue, I had tied the base and emitter to ground (as per an older schematic), however, upon more reading, I discovered that there should be some resistance (or capacitance) between base and ground. This fixed my problem. Thank you to everyone who helped.

The transistor in there is essentially just the same as any other BJT. It is just optimized to be light sensitive. You could ignore the LED part and use it like any other BJT.

You could saw the top off of most other BJT's and aim a light at them to make an opto-isolator. The usable wavelengths are unknown and the CTR is going to be poor but it will work.

BoB

Dextrine

## 1. What is an optoisolator and how does it work?

An optoisolator is an electronic component used to isolate circuits from one another. It is made up of a light-emitting diode (LED) and a phototransistor or photodiode. The LED emits light, which is then detected by the phototransistor/diode and converted into an electrical signal. This signal is then used to control the circuit connected to the optoisolator, while keeping it electrically isolated from the other circuit.

## 2. How do I choose the right optoisolator for my application?

The first step in choosing an optoisolator is to determine the requirements of your application, such as voltage, current, and speed. Then, look for an optoisolator that meets those specifications. Additionally, consider the type of isolation required (e.g. optocoupler, solid-state relay) and the package size and mounting options.

## 3. Can I use an optoisolator to protect my circuit from high voltage or noise?

Yes, optoisolators are commonly used for circuit protection. They provide electrical isolation between circuits, which can help protect against high voltage spikes or noise. However, it is important to choose an optoisolator with a high enough breakdown voltage and proper isolation ratings for your specific application.

## 4. How do I test if an optoisolator is working properly?

To test an optoisolator, you can use a multimeter to measure the LED and phototransistor/diode separately. The LED should have a forward voltage and the phototransistor/diode should have a resistance or voltage drop when activated by light. You can also use an oscilloscope to check the output signal and make sure it is switching properly.

## 5. Are there any safety precautions I should take when using optoisolators?

Yes, always follow the manufacturer's guidelines and datasheet when using optoisolators. They should be operated within their specified ratings and never be exposed to voltages or currents higher than what they can handle. Additionally, take proper precautions when dealing with high voltages and always use proper insulation and grounding techniques.

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