peter taylor
- Joined Apr 1, 2013
- 106
Does it? It controls the LED current, but that doesn't mean the LED intensity is tightly controlled. The intensity vs current curve is not flat. It varies with a number of parameters, including temperature and process variation. Plus, it does nothing about variation in the coupling gain between the LED and the phototransistor on the other side.Here's the feed back circuit with optocoupler.
View attachment 92380
Showing how critical it is to precisely control the LED intensity.
Abandon all hope, Ye who enter.Anyone who sees the title of this thread and expects it to be useful with "understanding a constant current circuit" will be very disappointed.
It doesn't, for obvious reasons: once the b-e junction starts to conduct, any increase in supply voltage would cause the base current, thus the collector current, to increase. The feedback introduced by R3 in your circuit helps but doesn't eliminate that problem.The common-emitter circuit I showed earlier does this.
I basically just picked wordpress.com and never tried any other services. Simply typing it away. Very easy - if I can do it, anyone can do it,Did you write the HTML yourself, or use a web page builder.
It all comes down to how "constant" is constant, which is another way of asking when is "good enough", good enough?Perhaps I'm not asking the right questions ?
Why do I need this circuit to provide constant current, when a BJT transistor is inherently already a constant current source.
The common-emitter circuit I showed earlier does this.
It also has a resister Re in the emitter that acts to cancel variations in Ic due to changes in beta due to temperature.
That is, without it, Ic = β.Ib. With it, Ic = (Vb - Vbe) / Re (Vb = base bias voltage).
Without confusing myself, or anybody else, how does adding the extra element improve this.
Obviously it acts to draw current away from the base as Ic increases, that is, negative feedback.
But Vbe is sitting right in its knee, or threshold region, and must change slightly.
This brings a large change in Ib. Have I just answered my own question ?
A stupid question is one not asked.Please bear with me. I'm a total newb to electronics. I apologize if this question is stupid.
As several others have already commented, the use in the posted references with a KA431 shunt regulator are not using the optoisolator in a linear mode.An opto-coupler used in a critical feedback path of a switch mode power supply, for instance, would have to have its intensity precisely controlled.
This is (currently) throwing a 404 Error.Almost every real world off-line SMPS used optocoupler and TL431 in the feedback loop
https://www.fairchildsemi.com/application-notes/an/an-4137.pdf
If you're responding to my post, the feedback loop only depends on the opto emitter being turned on or off; that's digital, not linearBut when it is switched on.
Does it have a constant current ? [vulgar language snipped]
I downloaded the file before the link became broken...This is (currently) throwing a 404 Error.