Electronics > Beginners

Couplers - optic, inductive, capacitive circuits

(1/4) > >>

nForce:
I have an optoisolator circuit (in attachment).

What does those 4 resistors do? Or is that a potentiometer at Vcc?

Thank you.

RoGeorge:
The circled resistors are the loads for the two open collector transistors.  Probably a potentiometer and 2 resistor (judging by the arrow from Vcc to it, the two resistor on top seems to be the uninspired drawing of a potentiometer).

T3sl4co1l:
The assumption is that the gain of the two devices is equal down to a constant (gain error) term.  The output from the device is in terms of collector current, so the voltage gain is that current times the collector load resistor.  You could vary just one of them, but you won't know which one naturally has the higher gain so either one may be higher or lower.  Putting the trimpot between them (yes, it's a potentiometer, just drawn oddly) allows that to be set.

In practice, the LED emission, photodiode sensitivity, and transistor hFE, will all vary with manufacture, age and temperature, in inconsistent ways; I doubt that this circuit will do very well, maybe within 5 or 10% of input value.  If circuit characterization is acceptable, and possibly component matching as well, it might be possible to refine that limit to 1%, even 0.1%; but this is probably going to be more costly than a better method in the first place (the next best method being a dual-photodiode sensor like the IL300, and the next after that, probably an ADC+iso+DAC combo).

Tim

David Hess:
Just in case what T3sl4co1l wrote is not clear, the entire circuit is a linearized optocoupler.  The potentiometer adjusts the matching by trimming the gain.

nForce:
Hmm, so those 2 transistors are actually something to prevent connecting Vcc and Gnd with a short circuit?

Navigation

[0] Message Index

[#] Next page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod