# Diode Analyses

Discussion in 'Homework Help' started by RidaElChami, Aug 17, 2015.

1. ### RidaElChami Thread Starter New Member

Aug 17, 2015
8
0
Hello,

I am new here. I need urgently to find how to do calculation and Plot Vout/Vin for the attached circuit.

Thank you,

Rida.

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2. ### #12 Expert

Nov 30, 2010
16,704
7,354
The symbol on the left is a pictograph of an AC voltage but the labels say DC. Which is it?

3. ### RidaElChami Thread Starter New Member

Aug 17, 2015
8
0
You are right ! it is an AC voltage.

4. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
If you are plotting Vout vs. Vin, then that is the DC transfer characteristic, which could then be used to determine Vout vs. time for an arbitrary (AC in this case) Vin.

So which are you interested in? Vout vs. Vin, or Vout vs. time?

Also, as this is Homework Help and note Homework Done For You, you need to show YOUR best attempt to solve YOUR homework problem. That gives us the necessary starting point to help guide you through your difficulties.

5. ### RidaElChami Thread Starter New Member

Aug 17, 2015
8
0
This is a homework problem ! i am interested in (Vout vs. Vin). I started with assuming diodes states ...

6. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
I realize that this is a homework problem, which is why you need to show your best attempt. It doesn't have to be correct, but it does have to be there. Again, that is our starting point.

For Vout vs. Vin, you need to identify the input voltages at which the diodes change state. So start from a large negative Vin and determine the state of the two diodes and what Vout is for that Vin, and then determine the first value of Vin for which one of the diodes changes state and determine Vout at that point. Then determine the next value of Vin at which the other diode changes state and the value of Vout at that point. Finally, determine Vout for a large positive Vin. Those are your corner points and your Vout vs. Vin characteristic is just those points connected by straight lines.

7. ### RidaElChami Thread Starter New Member

Aug 17, 2015
8
0
thank you. I Tried to do it, but i have problems !! the results are not very clear.

Note : Vdon : Constant Voltage Drop.

1 - Large negative Vin : D1 and D2 ON

Vout = -Vdon ;
V
in < -2*Vdon;

2- D1 ON and D2 OFF :

V
out = Vin + Vdon;
D1 ON : --> V
in > -2*Vdon;
D2 OFF : --> V
in > -((R1+R2)/R1)*Vdon;

3-
D1 and D2 OFF : Large positif Vin

V
out = (R2/(R1+R2))*Vin;
D1 ON : ---> V
in > -((R1+R2)/R2)
D2 ON : ---> V
in > -((R1+R2)/R1)

And how to know by calculation that we can not have D1 OFF and D2 ON at teh same time.

8. ### RidaElChami Thread Starter New Member

Aug 17, 2015
8
0
sorry :

3- D1 and D2 OFF : Large positif Vin

Vout = (R2/(R1+R2))*Vin;
D1 ON : ---> Vin > -((R1+R2)/R2)*Vdon
D2 ON : ---> Vin > -((R1+R2)/R1)*Vdon

9. ### MrAl Distinguished Member

Jun 17, 2014
2,562
518
Hi,

There seems to be a problem here. With an AC input, the two diodes would conduct heavily when the AC input went negative and that could blow them out. This means there should be some limit somewhere placed on something, like either the amplitude of the AC input or the input source resistance.

10. ### RidaElChami Thread Starter New Member

Aug 17, 2015
8
0
Hi,

My Problem is plot Vout Vs. Vin ! i don't mind if i will blow them or no .. And as you told there is a limit on datasheet ! but this is not my problem for now.

11. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Let's take your last question first:

The surest way is to assume that D1 is OFF and that D2 is ON and see what that would require in terms of the voltages in the circuit.

If D1 is OFF, that means that Vin > Vout (ignoring diode drops -- i.e., assuming Vdon = 0 V -- for this quickie analysis), which means that current is flowing from left to right (toward Vout) in R1. But for D2 to be ON, that means that Vout < 0V, which means that current is flowing upward (toward Vout) in R2. This violates KCL at Vout and is therefore not possible.

In general in a circuit like this, each diode will start off in a particular state for large negative voltage and then, as the voltage is increases, each diode will change state at most one time (it's possible it might never change state) and, once changed, will remain in that state from that point on.

Now, having said this, we have to recognize the limitation of assuming Vdon = 0 V. If Vdon is nonzero (say 0.7 V), then we CAN have current flowing away from Vout even if D1 is OFF because it can flow through R1. So it may be possible to have D1 OFF and D2 ON, but if that turns out to be the case, then the case where D1 is ON and D2 is OFF will not be possible. It will be one or the other. Which is the case depends on the relative sizes of the resistors.

So now let's look at your work:

For case 1, you have a contradiction that you aren't spotting. Let's say that Vin = -10 V and Vdon = 0.7 V. If I look at the effect of D1, then Vout would be -9.7 V, but if I look at the effect of D2, then Vout would be -0.7 V. It can't be both. The problem isn't with the analysis, per se, it's with the circuit (and it's directly related to the point made by MrAl). If we assume that most of the current flows through the diodes (i.e., that R1 and R2 are large enough so that little current flows through them) and if we assume that the diodes are matched, then it would follow that the voltage drop across each diode would be the same and Vout would be half of Vin. As Vin increases and the current drops, the resistors come into play and a general (parameterized) analysis becomes quite complicated. But for your analysis, the best thing is to note that if Vin < 2·Vdon that bad things happen and to start your analysis at that point.

As you go from Case 1 to the next case, the question of which diode turns off first depends on the resistors. Remember that the turn-off point happens when the diode current is reduced to zero but the voltage is still equal to Vdon. So what relationship must be true in order for D1 to turn off first and what must be true for D2 to turn off first?

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12. ### RidaElChami Thread Starter New Member

Aug 17, 2015
8
0
Thank you for your reply ! I have remarked the contradiction on the case 1 .. And when i did the simulation with spice i found a very high current through diodes "when ON" (1TA).

In other circuits I used a way to solve problems:

DIODE OFF : Vd < 0;
DIODE ON : Id>0;

in each case i verify these conditions ! And i can make the plots without any problem ... So I don't have a big idea to solve such circuit. I think the circuit is not true and not good to analyse it.

Please find attached the solution done by the prof. That I am certain it is all false !!

Thank you very much.

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13. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
I would recommend finding a different professor or, possibly, a different school.

That your prof's solution is bogus can be established by looking at just a single data point -- namely if Vin = 0 V, then how can Vout be anything other than 0 V given that there is no other power source in the system. Yet your prof claims that Vout = Vdon under those conditions, meaning that the circuit is dissipating energy even though no energy is being delivered to it. You might ask your prof what the current in each of the devices is when Vin = 0 V and to explain how KCL is satisfied at the Vout node.

Now, profs can and do make mistakes, even simple ones that they should have caught themselves by asking simple sanity check questions. Been there, done that, and have no doubt whatsoever that I will be there and do that more than once in the future. The real telling thing will be how the prof handles being asked to justify their obviously wrong solution.

DIODE OFF: Vd < Vdon & Id = 0
DIODE ON: Id > 0 & Vd = Vdon
TRANSITION POINT: Vd = Vdon & Id = 0

That last one is the one that is most useful in analyzing diode circuits of this type.

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14. ### RidaElChami Thread Starter New Member

Aug 17, 2015
8
0
Thank you ! your details was very useful for me.