opto-coupler: help with finding collector current.

In some guides on opto-couplers reference is made to the detector continuous collector current (IC). In some data sheets this is listed and the maximum current shown. However, in the data sheet for the Fairchild 4N25 it is not. The only reference to IC is in a chart which I do not fully understand.


What is this showing?

How do I find the maximum allowed current on the output side?
Does the chart even show this?
Do I really need to know?

It represents VCE(SAT) vs Collector current, at 4 different If (led current) values

The datasheet clearly indicates Ic max in the “absolute maximum ratings” :

Ic = 50mA
and Ic =100mA for less than 1ms

https://www.google.com/url?q=http://www.vishay.com/docs/83725/4n25.pdf&sa=U&ei=NGhmUqHRLJGDhQecooC4Bw&ved=0CAcQFjAA&client=internal-uds-cse&usg=AFQjCNG7Vhx4XbTQyldmIqjIdRodvUSZzw

and yes you need to know : you don’t want to put a load that needs more than Ic max
you need to know the Vce(SAT) at this Ic, this will give you the power dissipated, which must not exceed 150mW
etc…

Hi,

thanks for the reply but the data sheet you reference is for the vishay chip. The Fairchild data sheet does not seem to have it, http://www.fairchildsemi.com/ds/4N/4N25M.pdf

While the data sheets are very similar they do not appear to be exactly the same.

sorry, you are right, there are little differences (maybe the M suffix ?? ). Anyway, both are rated 150mW maximum (output power) and in the fairchild datasheet, you can see that Vce(SAT) increases very quickly when Ic > 2mA , which increases the power dissipated.

Sorry, I still do not understand this.

Does this mean I should try to keep the current on the output side at 2mA or below? 2mA seems very little since the maximum is 30V

Or is this about how much current can pass across the output depending on the current into the LED? If so, I still don't understand how I calculate the maximum current allowed on the output side.

Thanks for your help and sorry I am being thick.

it depends on what you want to do with the optocoupler.
You don’t need to keep Ic <= 2mA, but if you want it to be safe, you need to look at the curve and make sure that Ic*Vce(SAT) < 150mW
The fig. shows how Vce(SAT) varies when Ic varies, given 4 different If (LED current) values.
If you want to drive the optocoupler with an arduino output, AMHA, you should keep If = 10mA (it is well within the abs. max ratings of arduino, and it gives a lower Vce(SAT) than lowers If do. (If=20mA gives better Vce(SAT) values, but it is the high side of max. ratings of arduino outputs)
All design choices, depend on your project and its needs (speed etc…) and without information about your project, I can’t tell much more…

BTW, I don’t understand your 2mA vs 30V comparison… one is a current, the other is the max voltage you can apply between collector and emitter

OK, I may be slowly getting there.

The higher the amp then the lower the voltage needs to be to reach saturation. At 20mA saturation is achieved at about 0.2V and at 10mA saturation is achieved at about 2V. Is this correct?

I am not sure I understand what the saturation is though. Is it, the closer you are to saturation the more current is allowed to flow on the output side? The brighter the LED is then the lower the resistance in the photo detector?

I still do not understand how I find the maximum current for the output side.

This is not for use on a project. I just want to understand more. When the max current is clearly listed in the data sheet ( such as in the Vishay document) i understand it. But when it is not clearly listed (as in the Fairchild document) I have no idea how to find it.

In normal junction transistor operation in a given circuit, as you increase the base current Ib (or in a phototransistor, increase the light intensity) the collector current Ic increases. At some point, continuing to increase the base current or light intensity does not produce a further increase in collector current. When this happens, the transistor is said to be saturated.

Vce(sat) is the voltage between the collector and emitter of a transistor, when the transistor has reached saturation. The power dissipated by the transistor is Vce(sat)*Ic and if you exceed the maximum power for a significant length of time, the transistor will overheat and be destroyed.

Vce(sat) and Ic depend on your circuit design, so as outlined by a previous poster, you need to apply various rules to figure out a safe operating condition.

not exactly, but..... not far even if they are in the same package, the optocoupler contains 2 devices : an LED and a phototransistor .

When used in switching applications, a transistor may be compared with a....switch :grin: . 2 states : open/closed What you want is : no current when in open state, and the less "resistance" when in closed state. Then the lower Vce(SAT) the best.

with a "normal" transistor, the saturation state is obtained by setting Ib >> Ic/hfe . When it is fully saturated, the voltage between collector and emitter is at its lowest value. The Ic and Ib values needed are given in the transistor datasheet.

To make it simple, in an optocoupler, Ib depends on the light it receives, the more light it receives, the higher Ib is . Thus the 4 different curves in the figure.

But when it is not clearly listed (as in the Fairchild document) I have no idea how to find it.

When the datasheet doesn't indicates it , and if you need to know it, you may calculate it with the maximum output power (Pmax) : let's say you know what Ic you need -> look at the curves and note the corresponding Vce(SAT) . If Vce(SAT) x Ic is less than Pmax, then you're OK - But ...... being within the limits only means you won't fry it , it doesn't mean you are using it well. Example : the vishay datasheet says IcMax = 50mA - If I look at the curves, I cannot even see the Vce(SAT) value for Ic=50mA . If I set Ic=10mA (which is far below 50mA) , I can see tha Vce(SAT) (with If=10mA too) is more than 2V ! OK 2V x 0,010 = 0,02 W , below 150mW, but 2V are lost in the transistor .

Edit : while I was writing this, Jremington posted a shorter an actually, maybe better explanation 8) - I post mine anyway, but... ;)

Finally making headway.

Thanks for the replies and help.

The current on the output side will depend on the current on the input side, and the voltage across the output side. For a given input current ( If ), that could be any point along the corresponding line on the chart.

The key fact to realise, is that the current and voltage on the output side, also depend on the impedance of the rest of the circuit on the output side. If you energise the output side with voltage V and resistance R, then the equation for the output side is something like V = IR + Vc where Vc is the voltage across the output side of the device.

Or Vc = V - IR.

This corresponds to a sloping diagonal line on your chart. Your operating point of the device is going to be where this sloping diagonal line intersect the curvy line on your chart for a particular value of If.