EEs: Zener + full wave rectifier problem help

[pX]

Alright I hate to do this, but this problem is stumping me. I know I am probably making the problem a lot harder than it is, as I tend to do.
Can someone tell my how I would tackling this problem

HERE

this is for a lab I have tomorrow, he will give us the answer beforehand I think, but it bothers me to not know how work it before the lab.
thank you.

 

[Superdoopercooper]

You still need hlep on this... it was a yesterday questoin. ;-(
EDIT:

CRAPPP. :disgust: :| I tried for about 5 minutes to explain it... but words just don't work. I need to sit you down in front of a white board and work it out.

Do you still need help???

 

[pX]

i kinda made something up, something simple, and it was right.

in a nutshell, 20mA should go to RL, and 25mA thru R; you know the voltage drop over RL is 10v,,,, and supposedly the drop across R is 6V (which seems wrong to me because that is not the average voltage there, its really 16 - (VR/2) where VR is the ripple voltage). I still don't see why this works, I think the question could stand to be worded a lot better.

maybe using 16V is right because that is worst case, i.e. when the full wave rectified signal is 16V you have the most current going thru the R resistor, thus you use that value,, that kinda makes sense.

 

[Superdoopercooper]

Well... I guess when you assume RL is infinite.... that makes EVERYTHING so much easier. Plus, if he gave you a Zener current, you should be all set up. maybe I misunderstood what was sought in the problem. Oh well... if you want I can try to explain it to you.... or you can do like I did a lot in college... just let it fall into the never-ending abysses in my mind... never to be remembered or cared about. 😉 😀

 

[blahblah99]

well.... lets see,

you have 12Vrms at the transformer, which gives about 17Vp-p. Assuming the 1V drop across the two diodes, that'll leave you 16V at the capacitor. With a 10V zener, the voltage across R is 6V, and the voltage across RL is 10V.

If you want to limit the current going into the zener and RL to 25mA, 6/R = 25mA , R=240. If RL is infinity, you get 25mA through zener and still have 10V.

But if Iz = 5mA, you have 20mA through RL, and 10/RL = 20mA, RL = 500 Ohms.

 

[Superdoopercooper]

I knew it was easier than it sounded. 😀 shouldn't have had that beer when I got home from work. 😉