Couplers - optic, inductive, capacitive circuits - Page 1
Forum section: electronics / beginners

author: nForce
author_rank: Frequent Contributor
timestamp: December 19, 2018, 07:02:24 pm
content: I have an optoisolator circuit (in attachment).What does those 4 resistors do? Or is that a potentiometer at Vcc?Thank you.

author: RoGeorge
author_rank: Super Contributor
timestamp: December 19, 2018, 07:14:58 pm
content: 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).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 19, 2018, 07:46:05 pm
content: 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

author: David Hess
author_rank: Super Contributor
timestamp: December 20, 2018, 01:11:49 am
content: 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.

author: nForce
author_rank: Frequent Contributor
timestamp: December 20, 2018, 07:21:00 pm
content: Hmm, so those 2 transistors are actually something to prevent connecting Vcc and Gnd with a short circuit?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 20, 2018, 08:21:42 pm
content: What two transistors? What short?Tim

author: nForce
author_rank: Frequent Contributor
timestamp: December 20, 2018, 09:02:21 pm
content: Oh, sorry I meant resistors.

author: nForce
author_rank: Frequent Contributor
timestamp: December 22, 2018, 05:31:50 pm
content: I meant a resistor on the left side and a resistor on the right side. If a resistor weren't there we would connect Vcc through potentiometer to ground.Do you see it on schematics?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 22, 2018, 06:52:41 pm
content: The transistors aren't shorts, so I don't see it...Tim

author: nForce
author_rank: Frequent Contributor
timestamp: December 23, 2018, 10:31:49 am
content: Why do we have then one resistor on one side and the other on the other side? Someone has commented here, that it's a load resistor. But if it isn't short between Vcc through potentiometer to ground, then we don't need those two resistors?Thanks, I really appreciate the help for beginners.

author: RoGeorge
author_rank: Super Contributor
timestamp: December 23, 2018, 11:46:36 am
content: In theory, yes, the potentiometer might be enough. In practice, no.You MUST put those two resistors because:In the first place, safety reasons. One must never design a circuits that can be destroyed by simply turning a potentiometer. Somebody else, or even you by mistake, might turn that potentiometer to the minimum, or to the maximum, and damage something. That reason alone should be enough to justify the extra two resistors.For this particular schematic, there is one more reason that made the two resistors necessary: the adjustment range.Note there is a neat trick in that schematic, which includes a feedback loop:Quote from: David Hess on December 20, 2018, 01:11:49 amFor the schematic to work best, it would be necessary to have 2 optocouplers with the same gain. In real life, there are always small fabrication differences between parts of the same model. In our case, we use the potentiometer to slightly trim any possible differences in the gain of each optocoupler, so the potentiometer value will be probably 10 times less than the resistors. If not, at the slightest rotation we will introduce too much differences.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 23, 2018, 11:46:59 am
content: Still don't get it.These?That's just done to set a minimum value for each half of the potentiometer. So each load resistor is adjustable from R1 (the blue circled resistor) to R1 + R2 (R2 = pot end-to-end value).Again, there's no ground around here, I don't know what you're talking about shorting.Tim

author: nForce
author_rank: Frequent Contributor
timestamp: December 23, 2018, 12:53:05 pm
content: T3sl4co1l:Here:

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 23, 2018, 08:54:33 pm
content: Okay so the transistor?I said earlier they aren't shorts. Usually when a statement contradicts what someone thinks, they make note of it. But you didn't make note of it until now, which was confusing.So yeah, the transistor is only capable of drawing 10s of mA. It is not a hard switch. For resistances smaller than (+V) / (10 of mA), the collector voltage (in the fully-turned-on (saturated) condition) will be some volts. Only when the resistance is larger than this limit, will the collector voltage be able to pull down near ground (to the Vce(sat) spec of the device). In either case, there is never a short, because the transistor is not a hard switch and can only draw some current, or there is resistance, or both.HTH,Tim

author: nForce
author_rank: Frequent Contributor
timestamp: December 29, 2018, 01:05:05 pm
content: Quote from: T3sl4co1l on December 23, 2018, 08:54:33 pmWhy has to be resistance larger to pull down voltage? So we make short if the resistance is larger.

author: nForce
author_rank: Frequent Contributor
timestamp: December 30, 2018, 05:49:34 pm
content: I have another question:Can someone explain how does this circuit work:https://www.analog.com/en/products/ad202.html#product-overview?I haven't analyzed circuits much, so I need some help.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 31, 2018, 04:15:31 am
content: Externally, or internally?Externally, it's magic. You wire up power, and it transmits power and an analog signal.Internally, it looks like two transformers coupled with a synchronous de/modulator (also known as a chopper amp).Tim

author: David Hess
author_rank: Super Contributor
timestamp: December 31, 2018, 05:59:19 pm
content: Check out figure 12 on page 8 ofLinear Technology application note 9for a discrete implementation and description.