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I found a CNA1011K photocoupler laying around and I wish to try it out. Could someone explain what the parameters from the datasheet mean and how to use them for particular circuit parameters?

  1. LED maximum current If: 50mA - that's understood. How about the voltage across it? Standard LEDs have nominal voltage and current ratings, but not this one.

  2. Reverse voltage and current: why should I care about them?

  3. Forward voltage: @ If= 50mA is 1.2V. How do I ensure a 50mA current? If I use a 50Ohm resistor, as in the circuit diagram, the current will be 24mA, right?

  4. Is it better to alter the light intensity by increasing the voltage or the current through the LED?

  5. How do I know what base current range I can generate with LED? How to choose the RL value based on an arbitrary Vcc voltage, what is the Vcc range?

Nazar
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  • RE: "Is it better to alter the light intensity by increasing the voltage or the current through the LED?". The LED has a defined I-V characteristic. You can't change one without changing the other. See my answer here for one way of finding how the drive circuit (source and resistor) interacts with the LED to determine the operating point. – The Photon Apr 09 '15 at 15:53

2 Answers2

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The forward current is the absolute maximum rating. Just about any current will suffice to power the LED. Minimize the current for your application, generally 5 to 10 mA is sufficient for firing the LED.

The reverse voltage should be limited to prevent damage to the device. Although I don't see any reverse protection in the symbol, just keep the maximum reverse voltage below what the manufacturer recommends.

You do not have to ensure 50mA forward current on the emitter, it's just an LED. The resistor is the same as in a normal LED, size it to provide your desired current flow with your desired supply voltage. The diagram on the data sheet doesn't say what the supply voltage is.

The only thing that can change the emitter intensity is current through it. The current through it depends on the supply voltage and the series resistor (or other current limiting means). Why would you want to change the intensity anyway? That device is a simple optocoupler. It is meant to switch on and off, not convey an analog signal. Actually from the packaging of that device, it is most likely meant to be used with an encoder wheel with gaps to alternately pass and interrupt the emitter light.

The maximum open collector voltage appears to be 30 VDC. Choose your load again, based on your collector supply to ensure reliable and clean switching of the output. Almost any value will work as long as you don't exceed the output transistors current rating of 20mA. The lower the load resistor however, the longer the output rise time will be.

R Drast
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  • But how do I know what is the minimum current through the LED that will turn on the transistor? Then, I should assume the transistor to be a switch and the voltage on the emitter would be Vcc-0.5V? Then, the the collector voltage range could be 0.5V<Vcc<30V? So, if I power LED with 3.3V and choose the LED resistor of 220 Ohm, the forward current should be 15mA. Will this be enough light to turn on the transistor? Then if Vcc=3.3V and want transistor current to be 5mA, I should use 660 Ohm resistor, right? Then, the Vout will be 3.3V-0.5V=2.8V? – Nazar Apr 09 '15 at 15:50
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    As soon as you hit the operating threshold of the LED, and there is a light path through the coupler, the transistor will be biased on. The resistor values you give are reasonable. If you want to see exactly when the LED turns on, supply it with a variable voltage source, and slowly ramp up the voltage until you get significant current flow... it will jump from 0 or a few microamps to milliamps quickly (still use a series resistor to protect it, maybe 50 ohms) – R Drast Apr 09 '15 at 16:27
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  1. The voltage across the LED will be less than 1.2V typically but it won't change much at any reasonable current. See this graph- at 5mA you have 1.1V and at 50mA 1.2V, typically. Different units will tend to have somewhat different forward voltages, and the voltage will change with temperature.

enter image description here

  1. You don't need to care about reverse voltage and leakage current unless you are going to apply a reverse voltage.

  2. Forward voltage is typically as the above graph. You should not attempt to get 50mA- that's the maximum. Something more like 10-15mA will give you longer life. At 10mA, the forward voltage is a bit less than 1.2V, so if you have a 5V source, the resistance could be 380\$\Omega\$ (maybe a standard value of 360 or 390 ohms could be used).

  3. You need to limit the current through the LED with a resistor or a current source- feeding an LED with a 'stiff' voltage source will likely lead to tears.

  4. If you look at the coupled characteristics here:

enter image description here

You can see that at 10mA you'll typically see maybe 250uA of collector current. You should add quite a bit of margin to that because of variations with time and LED aging, so maybe you can count on 100uA of collector current. Your comments indicate that you want 5mA collector current- that's not really feasible reliably with this opto.

Spehro Pefhany
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  • Oh, I see, thank you. On #5 little confused. Is Ic=Ie? Why do they mention Vce=10V? Say my Vcc= 3.3V, and with conditions you mentioned with 250uA (100uA) if I put 100 Ohm resistor for RL, the output voltage will be 25mV (10mV) at the emitter, and Vce = 3.275V (3.29V)? – Nazar Apr 09 '15 at 21:52
  • Yes, Ic == Ie. They mention Vce because at low Vce the current transfer ratio Ic/If decreases. It will be close to the same at 3V and at 10V, but much less at a fraction of a volt. – Spehro Pefhany Apr 09 '15 at 23:52
  • Let's assume that I use 380 Ohm to create a 13mA forward current. Then, according to Ic-If plot that will give me about Ic=250uA. How do I use this information to calculate the output voltage (Ve) given arbitrary Vcc and RL? – Nazar Apr 13 '15 at 14:05
  • @Naz If R >> Vcc/250uA then the voltage will likely be less than 1 volt. If R << Vcc/250uA, then Vce = "Ve" ~= Vcc - 250uA * R. – Spehro Pefhany Apr 13 '15 at 14:29
  • Just to confirm, if Vcc = 3.3V and I want my High Ve ~ 3.3V, I can choose RL=100 Ohm. So, Ve would be 3.3V - 250uA*100Ohm ~ 3.275V. But why is there an absolute maximum rating for collector current Ic= 20mA, if you say that even 5mA collector current is not feasible? – Nazar Apr 13 '15 at 17:24
  • You presumably want the voltage to be < Vil, so R > 20K or so. The reason for the abs max is that the maximum and minimum are quite different- the coupling specification is very loose. So perhaps you can't get 5mA reliably long-term over temperature, doesn't mean that you can't get >20mA under some conditions, and that could damage the unit. – Spehro Pefhany Apr 13 '15 at 17:32