What is the current 'Ic' in the embedded circuit diagram? Thanks in advance..

bertus

Joined Apr 5, 2008
22,277
Hello,

The collector can not be connected to the 5V directly.
You will need a current limiting resistor.
The current in the collector is also dependent on the current through the led and the transfer characteristics of the optocoupler.

Bertus
 

bertus

Joined Apr 5, 2008
22,277
Hello,

When you use a 10K resistor on the collector, the collector current can be 5 Volts / 10K Ohms = 0.5 mA max.
The real value will depend on the saturation voltage of the transistor.
Look in the datasheet of the optocoupler to find this saturation voltage.

Bertus
 

bertus

Joined Apr 5, 2008
22,277
Hello,

If the powersupply is 5 Volts and the Vce(sat) is 0.2 Volts, the voltage accross the resistor will be 5 - 0.2 = 4.8 Volts.
The current will be 4.8 Volts / 10K Ohms = 0.48 mA.

Bertus
 

bertus

Joined Apr 5, 2008
22,277
Hello,

I said that the Ic is dependent of the led current, but you should look-up the transfer characteristics of the optocoupler in the datasheet.
When there is more current through the led of Ic times the transfer factor (most times given in a percentage), the transistor can be considered to be in saturation.

Here is the transfer characteristics graph from the datasheet:



You have a led current of about 1.6 mA.
Lower the 4k7 to 1k5 to have a led current of about 5 mA, to ensure saturation.

Bertus
 

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bertus

Joined Apr 5, 2008
22,277
Hello,

In your schematic is given a powersupply of 15 volts at the led side.
There is a 6.2 Volts zener in series.
The led forward voltage is about 1.2 Volts.
The voltage accross the 4K7 resistor is 15 - 6.2 - 1.2 = 7.6 Volts.
The current through the 4K7 is 7.6 Volts / 4K7 Ohm = 1.617 mA.

If you can not change the resistor, can you remove the zener?
In that case the current would be (15 - 1.2) / 4K7 = 2.936 mA.

Bertus
 

bertus

Joined Apr 5, 2008
22,277
Hello,

Using the formula from page 3 of the datasheet ( CTR = (Ic/If) X 100% ) will give you ( 0.48 / 1.1617 ) X 100 = 29.68
This would give you the A or B version of the optocoupler.

Bertus
 

Dodgydave

Joined Jun 22, 2012
11,303
Hi,
The optocoupler is KB844A only.We don't have provision to change it. We need to calculate the collector current(Ic) by using formula either (Ic=Il+IIl) or (Ic=Il-IIl).Please do let us know which formula is valid and the corresponding Ic calculation results.
This sounds like homework to me,

why dont you use an ammeter then you will know the current, if you cant understand the datasheet why are are you bothering.
 

bertus

Joined Apr 5, 2008
22,277
Hello,

I moved your post to your existing thread about the same issue (see post#14).

The 844A version is a selection of the 844 version.

Bertus
 
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