Electronics > Beginners

REALLY basic question on circuit analysis.

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zorthgo:
This is kind of a simple question. But I can't put my finger on it...

I am studying for my electronics test next week, so I am going over stuff that we did in class. One of the examples, although very easy has gotten me stumped. On the attached file under case a and b, I need to find ID, which is the current that will go through the load resistance. My problem is that I don't recognize the equation that gives me the 3mA for the ideal case and the 2.94mA for the actual diode case. At first I thought it was  voltage/current division but for that I would need one of the two resistance on the numerator like:

ID=Vth(R1/(R1+R2))

But the formula only has

ID=Vth/(R1+R2)

It is missing resistance on top.

Can anyone tell me what formula that is? Is it a weird form of current/voltage division that I don't recognize?

Thanks! :)

zorthgo:
Also... If it is just the basic ohms law, doesn't the diode have any resistance?

IanB:
It's simple Ohm's law. The current ID through the diode is the current in the circuit, given by

ID = VTh / (RTh + RL)

The diode is "ideal" which means it passes current with no voltage drop. Therefore the diode behaves just like a piece of wire in this case.

In the second case, the diode is "real" and has an (assumed) voltage drop of 1.2 V. So the voltage driving current round the circuit is now reduced by the diode drop. So subtract 1.2 V from 60 V and repeat the above calculation.

zorthgo:
Thanks IanB!

I have another question if you wouldn't mind. I am trying to solve a similar exercise but this time I have to find the resistance of RL. I am positive that I need to use the formula ID=IseVf/.026. But the problem is that my number is coming out way too large. So large that my calculator can't even handle it. So, there must be something wrong. I would really appreciate if anyone could give a quick look at it.

IanB:
Did you miss a "-1"? That is to say,

ID = IS(eVf/0.026 - 1)

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