Hello people, i'm new to electronic circuits.

I have a trouble to calculate the resistors for a optocoupler type "CNY74-4" (quadruple optocoupler IC)

I need to isolate two parts of a electronic circuit, the first work at +12/+20V DC, the second at +5V DC.

There is a simple formula to calculate the two resistor for each optocoupler?

Thanks to all.

Roberto

 

Have you done any calculations on LEDs and transistors before? It is much the same The input is a LED and the output is a transistor. I used this datasheet http://www.vishay.com/docs/83526/83526.pdf
Take a look at Fig. 4 - Forward Voltage vs. Forward Current. This will give you the LED voltage drop for a given current. From the datasheet I think a LED current in range 10-20mA should be enough to turn the transistor into saturation. Go for a LED current equal to 15mA

 

Have you done any calculations on LEDs and transistors before? It is much the same The input is a LED and the output is a transistor. I used this datasheet http://www.vishay.com/docs/83526/83526.pdf
Take a look at Fig. 4 - Forward Voltage vs. Forward Current. This will give you the LED voltage drop for a given current. From the datasheet I think a LED current in range 10-20mA should be enough to turn the transistor into saturation. Go for a LED current equal to 15mA

Ok, that means a 1.3KΩ resistor, because R=V/I => R=20Vdc/0.015mA => R=1.333Ω

If you see the CNY74-4 datashet, there is a simple schema with a resistor between the emitter and ground, what is the correct formula to calculate this resistor?

Many thanks,

Roberto

 

Kind of wrong. The LED will also have a voltage drop. Take a look once more on figure 5. This will show the voltage drop versus current. Find the voltage drop at 15 mA and take this into your calculations please

 

Ok, looking the figure 4, at 15mA there is about 1.05V, whereby (20-1.05) / 0.015 => 18.95 / 0.015 = 1263ohm

Is correct?

 

Ok, looking the figure 4, at 15mA there is about 1.05V, whereby (20-1.05) / 0.015 => 18.95 / 0.015 = 1263ohm

Is correct?

Yes that is correct. I do not think 1263 ohm is any standard resistor. But it should have much to say if you use 1200 or 1300 ohm that is both standard resistor value in the E24 series. At least in this case. Just use what you can find in your lab

 

I will use a trimmer instead a resistor.

As regards the resistance of the transistor, which logic do i use?

 

I would not have used a trimmer. You do not need a value that is 1263 ohm. If you feel for it use two resistors in series to get close to the value. You can use figure 2. If the opto is connected to some 5 volt logic gate or controller. Use anything from 1Kohm to to 10K. I guess a 4.7Kohm resistor will work just fine. From the datasheet at page two. We see that the Saturation voltage, collector emitter will be at typical .3 volt at LED current equal to 16mA and IC = 2 mA.

 

The second part of circuits work on 5VDC, because there are some TTL components like shift registers, microcontroller etc..
Loking table at page 2, "Collector emitter leakage current" for the Vce = 5V is 100 nA, so the resistore value is 5/0.001 = 5000ohm.
Use the 4.7Khom commercial value is correct.

Is sufficient to use 1/4W power resistor on not?

Thanks,

Roberto

 

As you know P=U*I. Assume using figure 2 in the datasheet. The voltage drop in the transistor will be typical .3 volt at IC=2mA. So the current in a 4.7K resistor will be using typical values 4.7/4700 A. It should be easy now to get a picture of the resistor wattage

 

In summary, hereafter you can see the formulas for resistors wattage:

Diode Resistor
Voltage => 20 Vdc - 1.05 => 18.95
Current => 0.015
Wattage => P=V*I => 18.95 * 0.015 => 0.28W

Transistor Resistor
Vce = 5V - 0.3 => 4.7
Current = 100nA
Wattage => 4.7 * 0.001 => 0.047W

these formulas are right?

Thank you

Roberto

 

Kind of correct. But the transistor in the opto will be more or less full on with a LED current of 15mA. Hence the emitter resistor will control the current. I do not know were you got the 100nA from. Replace the 100nA with the current determined by the transistor Vce(which you set correct to 0.3 volt typical value) and the emitter resistor

 

Kind of correct. But the transistor in the opto will be more or less full on with a LED current of 15mA. Hence the emitter resistor will control the current. I do not know were you got the 100nA from. Replace the 100nA with the current determined by the transistor Vce(which you set correct to 0.3 volt typical value) and the emitter resistor

Hello t06afre, thanks for the attention.
If I understand correctly, the resistance of the transistor depends on what you connect to it.

In this case, connected to the transistor there is an Arduino port (see http://www.arduino.cc/)

So, for example, if the maximum current supported is 20mA:
R => (5 - 0.3) / 0.020 => 4.7 / 0.020 => 235Ω
P => 4.7 * 0.020 => 0.094W

 

No need to have the transistor conducting max current. And the higher current. The higher Vce voltage drop. By using a 4.7Kohm emitter resistor. You will have values well inside the spec of the opto. And that is also important. So stick with figure 2 in the datasheet. You have got how to calculate the LED series resistor correct. But use a emitter resistor in the range of 4.7Kohm. The output to the Arduino input port will be at the point named Channel II. in figure 2. And you will no problem at all.