enter image description here

Can anyone give me hint of this question. I know this is homework question. I know correct answer is 10 v.

But if I tried to solve it like this way:

(1) Z1 and Z2 limit output 5 V and Z3 and Z4 will give output 5 V. So total output is 10 v. And it will be subtracted from 15 v so output should be wrong.

  • \$\begingroup\$ The Zener diodes are in parallel, ok, not really but they are definitely not in series. So you therefore cannot add the zener voltages up like 5 V + 5 V = 10 V <= not OK here. \$\endgroup\$ – Bimpelrekkie Apr 25 '17 at 7:43
  • \$\begingroup\$ yes its my mistake \$\endgroup\$ – Beginner Apr 25 '17 at 8:08

Q1 is off so all the current goes through the Zeners. If they were not breaking down they would see 15V from VCC > 5V the breakdown voltage.

So the Zeners are breaking down and because their breakdown voltage is 5V, there is a 5V voltage drop across all of them.

So the voltage at the bottom of R2 & R3 is 5V, at the top is the common node VO - they are equivalent to a parallel resistor = 5k. They are in series with R1 = 5k.

With these hints you should be able to solve for VO.

| improve this answer | |
  • 1
    \$\begingroup\$ Downvote because you just gave the answer to the OP, instead of giving them clues so they could work it out for themselves. How is the OP supposed to learn?! \$\endgroup\$ – TonyM Apr 25 '17 at 7:40
  • \$\begingroup\$ @TonyM if I remove the last line it would be acceptable then. I'll edit my post. \$\endgroup\$ – vrleboss Apr 25 '17 at 7:56
  • \$\begingroup\$ No - OP has taken the answer and thanked you for it, moment's gone. Just delete the answer. RK has written what was needed below. \$\endgroup\$ – TonyM Apr 25 '17 at 8:00

Z1 and Z2 limit output 5 V and Z3 and Z4 will give output 5 V. So total output is 10 v. And it will be subcontracted from 15 v so output should be wrong.

Zeners don't limit output at 5V because they are not directly connected to the output. Output voltage will be any zener diode voltage (because they are all identical) plus voltage drop across R2 (or R3. They're identical, too).

So, since Q1 is off, Q1 and R4 can be deleted from the circuit. From the remaining circuit, you can easily see that \$V_O = V_Z+ I_{R2}\cdot R2\$, where \$I_{R2}\$ is the current flowing through R2 and \$V_Z\$ is the zener voltage, 5V in your case.

Tip: Zeners are identical, R2 and R3 are 10k and these strings are in parallel. And also R1 is 5k. This might tell you something :)

Now you can find the answer by yourself.

| improve this answer | |
  • \$\begingroup\$ Good answer but unfortunately, they can't find the answer for themselves - @HervéGrabas has already given it to them for nothing :-( \$\endgroup\$ – TonyM Apr 25 '17 at 7:55
  • \$\begingroup\$ @TonyM I think this answer fails to explain why there is 5V across the Zener which is the key of the problem set. \$\endgroup\$ – vrleboss Apr 25 '17 at 8:00
  • \$\begingroup\$ current through R1 is 3 mA. R2 is 1.5 mA. then how i get 10 v?? \$\endgroup\$ – Beginner Apr 25 '17 at 8:19
  • \$\begingroup\$ How did you calculate those values? Also, what's the current through R3, and where does it come from? \$\endgroup\$ – Finbarr Apr 25 '17 at 8:28
  • \$\begingroup\$ current through R1 is Vcc/R1= 3 mA. now R1 and parallel combination of r2 and r3 are in series so curren will be same. R2 and R3 have same value. so current will be divided equally. so current will be 1.5 ma \$\endgroup\$ – Beginner Apr 25 '17 at 8:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.