I recently started reading Malvino Electronic Principles to further my understandings for hobby electronics. I've been stuck for a few hours on a problem out of chapter 3 of the 6th edition 3-29. "What value should R2 be to set up the diode current of 0.25 mA"
I'm not sure why I've been so stuck on this problem it seems easy enough on the surface. I know the answer is 23 k ohms but I can't get there.
The method that makes sense to me which doesn't give the correct answer is to calculate Vout needed by doing IR=V and that is .25mAx5kohm which gave me 1.25v for my Vout. from there I did R2 = R1/((Vin/Vout)-1) and that gives me 3.4k ohms which isn't the right answer.
I've gone around in circles but obviously I'm missing or misunderstanding a concept here and would like to be pointed in the right direction.
The 23k ohms was the answer given in the back of the book so I was attempting to follow the procedure to get an answer that matched. I can reasonably assume maybe there was an error in the answer given and it should be 2.3k that is close to the answer you gave you used 0.6v and the book consistently uses 0.7v drop across a diode so that could be the difference.
Is this how you would typically solve a problem like this? I'm not sure how practical this situation is or if it is a problem given as an exercise.
The trick is to calculate the resistance of the diode at that current. Then current supplied by the voltage supply can then be calculated in terms of R2. Next we use the current division theorm and equate it to 0.25ma. The resistance of the diode is 0.7/0.25E-3 = 2800Ω. The total current supplied by the voltage source is 12/[30E3+R2||(5E3+2K8)] . Then we multiply the above term by R2/[R2+7K8] and equate it to 0.25E-3 . Solving for R2 gives 22941Ω which is close enough to the answer you want. So it involves two equations and two unknowns (I_total,R2).
The method that makes sense to me which doesn't give the correct answer is to calculate Vout needed by doing IR=V and that is .25mAx5kohm which gave me 1.25v for my Vout. from there I did R2 = R1/((Vin/Vout)-1) and that gives me 3.4k ohms which isn't the right answer.