# find o/p voltage

Discussion in 'Electronics Homework Help' started by strameshkumar, Feb 23, 2012.

1. ### strameshkumar

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0
Feb 23, 2012
find the Vo of the circuit and explain it?

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2. ### davennModerator

13,991
2,018
Sep 5, 2009
ok so what have you so far considered / worked out?

break the circuit down into stages and go from there
initially work out the voltage at the common point of the 2 x 10k resistors.
That is a standard voltage divider.

the resistors are the same value so that should tell you something about the
voltage measured at that point

tell me what you think that voltage should be and why you think so

cheers
Dave

3. ### electr0Dave

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Feb 23, 2012
Hi

Beware, then there is another resistor that forces greater current flow.

Consider the two resistors as one. Make an equivalent resistor.

This'll make you have a value nearest to the real.

4. ### jackorocko

1,284
1
Apr 4, 2010
you sure about that? Besides it really against the policy to just give answers in the homework section.

Last edited: Feb 23, 2012
5. ### strameshkumar

25
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Feb 23, 2012
hi
when i simulate in Multisim its 1.44V, but divider ckt theroritically 1.66v across diode + load resistor.
if i minus 0.7v of diode 0.96 is Vo. there is too much of difference

6. ### jackorocko

1,284
1
Apr 4, 2010
I am with you, in that the output voltage is 1.67V minus the diode voltage drop, but proteus and multisim both seem to disagree that there is a 0.7 volt drop across the diode. Not sure why either to be honest. proteus has the output voltage at 1.3 with an ideal battery. *shrugs*

7. ### electr0Dave

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Feb 23, 2012
I only simulates this in proteus and this is the value it gave.

For me, this value is right. Of course, this value can be different of real.

8. ### davennModerator

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2,018
Sep 5, 2009

in the homework section we help people to work out the answers for then selves
instead of some one coming along and spilling the beans

The Op didnt even get a chance to respond to my original question

I wanted the OP to learn about resistors used as a voltage divider and move through the circuit from there

Dave

9. ### electr0Dave

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Feb 23, 2012
Oh ... OK, sorry.

But this question also made me think, because I can not solve this with analysis of simple circuits.

Now I just read this topic and wait for the right moment to give a clue.

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,505
2,852
Jan 21, 2010
But the question is, how would you calculate this value? (Or I presume that is the question) "Because I simulated it" seems to be a very unsatisfactory explanation of how you arrived at the figure.

edit: the "answers" you've got are a good way of determining if your calculations result in a correct value.

11. ### Laplace

1,252
185
Apr 4, 2010
As shown in the attached diagram there are several ways to look at the circuit for analysis. I prefer the one where the diode offsets some of the Thevenin equivalent voltage from the input source, and the remainder goes through a two-thirds voltage divider to the output.

If the diode had no forward voltage drop, then the current would be 167 uA. A typical diode with a forward current of 100 uA will have a voltage drop of 0.5 V which will rise to 0.7 V at 5,000 uA. So Vo will be somewhere between two-thirds of (2.5 - 0.5) and (2.5 - 0.7). Given the simulation result for Vo can you calculate what diode voltage drop the simulator used?

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