Again, it may be that the author of the problem was wanting to give a problem that could be solved for a numerical value of R2. They didn't accomplish that, however.Just realized that the first sentenced of the of the first post by the TS stated:
"I was asked to find a value of R2 to make the diode current = .25 mA on a recent quiz."
That sounds definitive to me can can only be calculated as described above.
SG
Agreed.I think that the salient point was in post #16
"I have a feeling the value of R3 was accidentally cropped out, the rest of the quiz was incredibly simple"
It's not confusing to me- but apparently is to the TP. I'm just waiting to see if TP gets an answer on if R3 has a value or not. One way or another 250uA must appear on the R3 leg.What do you think is confusing about is?
Yes, it's a parallel circuit, but that doesn't help a lot since R2 and R3 are not in parallel -- and even if they were, that wouldn't allow for finding a numerical value for R2 independent of knowing what R3 was.
Hello,Just realized that the first sentenced of the of the first post by the TS stated:
"I was asked to find a value of R2 to make the diode current = .25 mA on a recent quiz."
That sounds definitive to me and can only be calculated as described above.
SG
MrAL we already verified that back in post 49,50So no matter how you look at it there is no single value for R2.
Yes you can but you cannot dismiss as you called it R3bNote that again there is no way to eliminate R3 from the calculation.
The VF of the diode was fixed at .7 volt by the TS in post #5. The current through the diode was fixed at .25ma in post #1. You don't need to make any assumptions if you have the facts.One thing I might consider would be the Vf of the diode. Another would be what would it take to limit the current through the diode to a safe value. There are a couple of very basic assumptions that could be made.
Available at your local Radio Shack!What's the RS part number?
You are correct the .7 volt was not given in #1 but it was in #5 which is what I said.Actually the 0.7Vf for the diode was not given in #1. And the 0.25mA? Hmmm... Kinda low but if that's the value...
Give us your answer.But given that it is 0.7Vf, the answer is easily found.
The 30 kΩ resistor almost certainly does that. I doubt there are many diodes (outside of an IC) that can't handle at least half a milliamp.One thing I might consider would be the Vf of the diode. Another would be what would it take to limit the current through the diode to a safe value. There are a couple of very basic assumptions that could be made.
In theory. But the two diodes would really have to be very well thermally coupled such as on the same die. Even then though the current will not be the same in the two diodes unless R3=R2.But shouldn't there be a diode in series with R2 that is thermally coupled to the other diode? Otherwise, when we have worked out the correct value for R2, it will only be correct at one temperature?
The value of 0.7v for the diode was given some time after the first post by the original poster.Actually the 0.7Vf for the diode was not given in #1. And the 0.25mA? Hmmm... Kinda low but if that's the value... But given that it is 0.7Vf, the answer is easily found.
What do you mean by "dismiss"?MrAL we already verified that back in post 49,50
Yes you can but you cannot dismiss as you called it R3b
Forget the diode! Remove it. You have three resistors. R1 in series with the parallel combination of R2 and R3. The voltage across the whole thing is 12 V and R1 is 30 kΩ. Given ONLY that information, find the value of R2 that results in the current in R3 being 0.25 mA.Actually the 0.7Vf for the diode was not given in #1. And the 0.25mA? Hmmm... Kinda low but if that's the value... But given that it is 0.7Vf, the answer is easily found.