2N3904 driving FOD814 - currents...

Thread Starter

daman

Joined Jun 14, 2012
15
Hi,

I have a small project and I need to use 2N3904 to drive FOD814 (which is then used to drive some load).
Here's the circuit:


Could somebody please help me calculate currents etc to see if I can drive 20mA load at OPTO_OUT?
Here's my math (and please be lenient/understanding...):

At 2N3904:
Vbe (base-emitter) = 0.7V
Ib (base) = (3.3-07)/10k = 0.26mA
Vce (collector-emitter) = 0.2V

As for FOD814 part:
Ir1 = (12V - Vd - Vce) / R1 = (12V - 2.1V - 0.2V) / 1000 = 9.7mA
Ir2 = (12V - Vfod - Vce) / R2 = (12V - 1.2V - 0.2V) / 1000 = 10.6mA

So Ic (collector current) = 20.3mA

Considering hFE = 60 I will need to increase Ib (eg. changing R3 to 3k4) to be able to achieve 20mA at the collector.

Now - at 20mA FODs CTR is about 110% so I should be able to drive 20mA load at the OPTO_OUT...?

Does that make _any_sense...?

Thanks,
 

ronv

Joined Nov 12, 2008
3,770
Hi,

I have a small project and I need to use 2N3904 to drive FOD814 (which is then used to drive some load).
Here's the circuit:


Could somebody please help me calculate currents etc to see if I can drive 20mA load at OPTO_OUT?
Here's my math (and please be lenient/understanding...):

At 2N3904:
Vbe (base-emitter) = 0.7V
Ib (base) = (3.3-07)/10k = 0.26mA
Vce (collector-emitter) = 0.2V

As for FOD814 part:
Ir1 = (12V - Vd - Vce) / R1 = (12V - 2.1V - 0.2V) / 1000 = 9.7mA
Ir2 = (12V - Vfod - Vce) / R2 = (12V - 1.2V - 0.2V) / 1000 = 10.6mA

So Ic (collector current) = 20.3mA

Considering hFE = 60 I will need to increase Ib (eg. changing R3 to 3k4) to be able to achieve 20mA at the collector.

Now - at 20mA FODs CTR is about 110% so I should be able to drive 20mA load at the OPTO_OUT...?

Does that make _any_sense...?

Thanks,
Hmm, could get a little complicated.
Let's work backwards.
The curve for the 814 CTR is a typical curve at 25C. It is also with the voltage from collector to emitter still at 5 volts - so not turned on hard. I see another curve for when it is saturated but they don't say which model it is for. Bottom line is I think I would drive it harder. Maybe 40 ma.
The only current that counts towards CTR is the current thru R2 and the opto, so you need to adjust for that then add the current thru the LED.
A similar thing for the 3904 as far a saturation voltage. HFE is calculated with a volt from collector to emitter. To get it down into saturation you need to drive the base harder. Usually a gain of 10 is used for this. So if you want 40ma and .2 volts drive it with 4 ma.
 

DickCappels

Joined Aug 21, 2008
6,381
It all makes sense, but it would be a good idea, as ronv suggest, to over-drive both Q1 and U6 by a considerable margin.

The gains of Q1 and the transistor inside U6 are temperature dependent and the CTR of U6 will also decrease as the LED ages.

upload_2016-5-26_13-13-34.png
 

Thread Starter

daman

Joined Jun 14, 2012
15
Great - thanks!
So based on what you're saying I need to change R3 to something close to 1k and R2 to around 270Ohm - right...?
 

ErnieM

Joined Apr 24, 2011
8,041
(ignore this comment please... Misread the schematic. Leaving this here so following comments make sense.)

Did anyone mention R1 is not doing anything useful here?
 
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