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

Discussion in 'The Projects Forum' started by veeresh540, Feb 9, 2015.

1. ### veeresh540 Thread Starter New Member

Dec 17, 2014
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Hi,

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2. ### bertus Administrator

Apr 5, 2008
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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

planeguy67 likes this.
3. ### bertus Administrator

Apr 5, 2008
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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

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4. ### bertus Administrator

Apr 5, 2008
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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

5. ### bertus Administrator

Apr 5, 2008
15,806
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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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6. ### bertus Administrator

Apr 5, 2008
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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

7. ### John P AAC Fanatic!

Oct 14, 2008
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Nobody else seems to be saying it, so I will--this thread is in the wrong section.

8. ### WBahn Moderator

Mar 31, 2012
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I tend to agree.

9. ### bertus Administrator

Apr 5, 2008
15,806
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Hello,

I moved the thread from the embedded forum to here.

Bertus

10. ### bertus Administrator

Apr 5, 2008
15,806
2,389
Hello,

Have a look at page 3 of the datasheet.
There is the CTR value given and how it is used.

Bertus

11. ### bertus Administrator

Apr 5, 2008
15,806
2,389
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

12. ### Dodgydave AAC Fanatic!

Jun 22, 2012
5,159
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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.

13. ### bertus Administrator

Apr 5, 2008
15,806
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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

Dec 17, 2014
13
0