Determine Resistor Value in circuit with a diode

I'm ashamed to ask because I actually finished the entire course this problem is from (6 years ago), and because I'm probably missing the obvious, but how would I go about determining the value of R2 in this circuit in order to make 0.25mA flow through the diode?



I'm hoping this can be done without advanced analysis, because I have only refreshed my memory to include thevenin, norton, and ohm's law, but that's where I'm at in the book.

By The way, I know the anser is 24K Ohms, but I don't know how to figure that out without peeking at the answer.

 

Were you told to assume a forward voltage drop of a particular value for the diode?

 

Sorry, R2 is actually 23 K Ohms. That was a typo.

I think I figured out a way to solve it, but if anyone else has an alternate method to this, please post it...I am curious.

I knew that the current through the diode, and R3 was 0.25mA, so the Voltage across R3 had to be 1.25V.

Backing up across the diode puts us at 1.95V at the junction between R1 & R2

That means the drop across R1 had to be 10.05V. That puts the R3 current at 0.335mA. Subtracting the branch current of R3 (0.25mA) we have 0.085mA left for the R2 branch.

with a 1.95 V drop across R2, and a current of 0.085mA, the resistance is 22941.176 Ohms, or rounded to 23K Ohms.

I'm going to sleep now...this made my head hurt

 

Yes....Thanks...we must have been typing at the same time....the drop was for the 2nd apporoximation of 0.7V....

 

If the diode forward voltage drop is 0.7525V at 0.25mA then the answer of 24kΩ makes sense.

 

0.7V is too high forward drop for a forward current of only 0.25mA. The Fairchild data sheet for the 1N4001 shows a typical forward drop of about 0.65V at 25mA and less than 0.4V (extrapolated from the Figure 2 curve) at 0.25mA (a factor of 100 times smaller current).

 

I probably shouldn't have put the 1N4001 in there (drew it in multisim), please assume a 0.7 v drop. I'm mainly asking if I got the mechanics of the solution correct or if there was a different way.

 

The method that you use seems to be correct.