Electronics > Beginners
Couplers - optic, inductive, capacitive circuits
RoGeorge:
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 am ---the entire circuit is a linearized optocoupler. The potentiometer adjusts the matching by trimming the gain.
--- End quote ---
For 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.
T3sl4co1l:
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
nForce:
T3sl4co1l:
Here:
T3sl4co1l:
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
nForce:
--- Quote from: T3sl4co1l on December 23, 2018, 08:54:33 pm ---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).
HTH,
Tim
--- End quote ---
Why has to be resistance larger to pull down voltage? So we make short if the resistance is larger.
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