What is the output voltage of this ideal diode circuit?

  • #1

Homework Statement


Given the following circuit:

circuit.png

what is the positive and negative peak values of Vo?
vi has a frequency of 1kHz, peak of 5V, and is a sine wave.

Homework Equations



x(t) = Vsin(2(pi)*f*t)
KCL, KVL, etc.

The Attempt at a Solution


[/B]
I made Vi to be 5sin(2(pi)1000t)
during its positive iteration it will have a peak voltage of 5V and the ideal diode D1 will be off ( Id1 = 0amps, Vd1 <0). The current across R2 (the resistor in series with the diode) will be 0 because it shares the current of D1. The current across R1 will also be 0 since it shares the current of D1:

Ir1 = Ir2 = 0amps.

Then the voltages of the resistors are as follow:

Vr1 = Ir1/r1 = 0 volts
Vr2 = Ir2/r2 = 0volts

I used KCL to find Vd1

Vi +Vd1 = Vr1 + Vr2
then Vd1 = -5 volts which agrees with Vd1 <0.

To find Vo I use KCL Vd1 + Vo = Vr2
then Vo = 5volts peak
but the answer is not 5 volts peak it is -2.5 volts peak. What have I done wrong? I also got a wrong answer for the negative iteration but maybe solving this issue will help me solve the other on my own. Notice, i treated Vo as an open and I connect both Vo and Vi to ground. Vo is supposed to be treated as an open right? Like a voltage meter? Thank you all.
 

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Answers and Replies

  • #2
BvU
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You describe this as if you are allowed to treat the diode as ideal -- fine if suggested in the wording of the exercise.
In that case the diode can be replaced by an open section during the positive half-cycle and by a short-circuit during the negative half.
So in the latter case the circuit you have is a voltage divider
 
  • #3
You describe this as if you are allowed to treat the diode as ideal -- fine if suggested in the wording of the exercise.
In that case the diode can be replaced by an open section during the positive half-cycle and by a short-circuit during the negative half.
So in the latter case the circuit you have is a voltage divider
Yes, the diode is ideal. I did analyze the positive half cycle by replacing the diode with an open but I did not get the correct answer.
 
  • #4
Delta2
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My original thought was +5V positive peak and -2.5V negative peak. Cant see where this goes wrong, what are the provided answer keys for the two cases?
 
  • #5
My original thought was +5V positive peak and -2.5V negative peak. Cant see where this goes wrong, what are the provided answer keys for the two cases?
Yes those are correct. For the negative cycle i replace the diode with a short but i end up with a negative current across the diode which is impossible for an ideal diode. I dont know how but i think it might be due to polarity. On the negative cycle I can keep every polarity of each component the same and every current direction the same as long as i just use a negative voltage value for V1 right or do i change the polarity of the V1 or the current? Well, I chose to keep everything the same except for V1 which i made negatice. Then the voltgae drop across each resistance was -2.5volts and the current across each component was -2.5/1000 because they share the same current but again this is an impossible value for a diode in forward bias. Am i doing something wrong?
 
  • #6
BvU
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negative current across the diode which is impossible
If 'negative' corresponds to 'upward' in your diagram, then that is quite possible. And 'down' or 'positive' the current can only be zero - as in the first part of your analysis (for the positive half-cycle).
 
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  • #7
If 'negative' corresponds to 'upward' in your diagram, then that is quite possible. And 'down' or 'positive' the current can only be zero - as in the first part of your analysis (for the positive half-cycle).
It turns out that I assumed wrong. D1 is on on the positive iteration and off on the negative. Thank you.
 
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  • #8
BvU
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D1 is on on the positive iteration
not in your picture -- unless you have a special definition of 'on' :rolleyes:
 
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  • #9
not in your picture -- unless you have a special definition of 'on' :rolleyes:
ah. THEN WHAT AM I DOING WRONG AHHHHH!!!!!
 
  • #10
Delta2
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When you say the diode is ON do you mean that
1) current flows through the diode?
or
2) current does not flow through the diode?

Because the way I understand, the diode is ON during the negative iteration, and the diode is OFF during the positive iteration.
 
  • #11
When you say the diode is ON do you mean that
1) current flows through the diode?
or
2) current does not flow through the diode?

Because the way I understand, the diode is ON during the negative iteration, and the diode is OFF during the positive iteration.
Yes that is what my instint tells me, the diode should be ON during the negative iteration but the diode current at the negative iteration was a negative value, which is impossible in an ideal diode.

My work for the negative cycle:
Vd1 = 0
Vi = Vr1 + Vr2 - Vd1
at this cycle vi = -5volts
since r1 = 1k and r2 = 1k then the voltage drops across both resistors are equal and since Vd1 = 0
then Vr1 = -2.5Volts and Vr2 = -2.5Volts and their currents will be -2.5/1000 = -2.5mA. The current across the resistors and diode are all the same since they are in series thus Id1 = -2.5mA BUT this is not legal for an ideal diode. The current across an ideal diode MUST be 0 or positive if Vd = 0. Right? Then I must assume that the diode is not ON in the negative cycle and it instead must be off.
 
  • #12
CWatters
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The Attempt at a Solution



I made Vi to be 5sin(2(pi)1000t)
during its positive iteration it will have a peak voltage of 5V and the ideal diode D1 will be off ( Id1 = 0amps, Vd1 <0). The current across R2 (the resistor in series with the diode) will be 0 because it shares the current of D1. The current across R1 will also be 0 since it shares the current of D1:

Ir1 = Ir2 = 0amps.

Then the voltages of the resistors are as follow:

Vr1 = Ir1/r1 = 0 volts
Vr2 = Ir2/r2 = 0volts

I used KCL to find Vd1

Vi +Vd1 = Vr1 + Vr2

[/B]


You refer to KCL but the equation you give contains voltages not currents. Did you mean KVL?

Assuming you mean KVL...First you need to draw arrows on your circuit to define what a positive value means. For example when you refer to VR2 you are making an assumption about which end of R2 is positive. You need to mark that using an arrow on the circuit or you risk making a sign error. I'm not saying you got the sign of VR2 wrong, just that you need to be more rigorous in your application of KVL.

KVL also says the voltages add up to zero. I find it safer if you write your equation so it does just that. Eg A+B+C+D=0 is better than A+B=-(C+D).




 
  • #13
CWatters
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If you mean KCL then you need to define which node you are applying it to and if into or out of the node is positive. Also try writing the equation so it sums to zero as per the definition of KCL.
 
  • #14
Delta2
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Yes that is what my instint tells me, the diode should be ON during the negative iteration but the diode current at the negative iteration was a negative value, which is impossible in an ideal diode.

. The current across an ideal diode MUST be 0 or positive if Vd = 0. Right? Then I must assume that the diode is not ON in the negative cycle and it instead must be off.
I think you are misinterpreting this fact about the ideal diode. The current through an ideal diode when Vd=0 can be negative or positive. The sign of the current tell us nothing because it depends on what direction we consider as positive and what direction is positive is purely arbitratry. Just forget about the sign of the current and just look at the direction of the current.. The direction of the current does not violate the diode operation, the diode allows current as it should be allowing current ,during the negative portion of the sine wave.
 
  • #15

You refer to KCL but the equation you give contains voltages not currents. Did you mean KVL?

Assuming you mean KVL...First you need to draw arrows on your circuit to define what a positive value means. For example when you refer to VR2 you are making an assumption about which end of R2 is positive. You need to mark that using an arrow on the circuit or you risk making a sign error. I'm not saying you got the sign of VR2 wrong, just that you need to be more rigorous in your application of KVL.

KVL also says the voltages add up to zero. I find it safer if you write your equation so it does just that. Eg A+B+C+D=0 is better than A+B=-(C+D).




Yes sorry I skipped many steps. I did use KVL using the mesh current connecting V1, R1, R2, and the diode. I assumed the current to run from the Vi to R1, R2 and then the diode. I set up the polarities of the resistors to where the current enters through the positive terminal (resistors dont have positive terminal but its to show that resistors use power instead of produce). The diode's polarities are not really up for assumption; anode is positive, cathode is negative. Then my KVL equation is Vi -Vr1 -Vr2 + Vd1 = 0 which i transform to Vi = Vr1 + Vr2 -Vd1
 
  • #16
I think you are misinterpreting this fact about the ideal diode. The current through an ideal diode when Vd=0 can be negative or positive. The sign of the current tell us nothing because it depends on what direction we consider as positive and what direction is positive is purely arbitratry. Just forget about the sign of the current and just look at the direction of the current.. The direction of the current does not violate the diode operation, the diode allows current as it should be allowing current ,during the negative portion of the sine wave.
Hmm well my studies say that if a current is negative across a diode then I have assumed wrong. Question: when I analyze the positive cycle I use Vi and a current going clockwise(assumed it is going clockwise), when I analyze the negative cycle I use -Vi but keep the current running clockwise. Is this correct?
 
  • #17
CWatters
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Yes sorry I skipped many steps. I did use KVL using the mesh current connecting V1, R1, R2, and the diode. I assumed the current to run from the Vi to R1, R2 and then the diode. I set up the polarities of the resistors to where the current enters through the positive terminal (resistors dont have positive terminal but its to show that resistors use power instead of produce). The diode's polarities are not really up for assumption; anode is positive, cathode is negative. Then my KVL equation is Vi -Vr1 -Vr2 + Vd1 = 0 which i transform to Vi = Vr1 + Vr2 -Vd1
You don't have to define Vd1 that way around (arrow pointing down) but its OK.

So then I would solve for Vd1...

Vd1 = -Vi + Vr2 + Vr1
But Vr2 and Vr1 are zero so
Vd1=-Vi,.....................................(1)

Then apply KVL to the output loop...

-Vd1+Vr2-Vo=0...........................(2)

Substitute (1) into (2)...

Vi+Vr2-Vo=0
Vr2=0
So
Vo=Vi

Vi=5v so Vo=5v

With experience this result comes by inspection.
 
  • #18
Delta2
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Hmm well my studies say that if a current is negative across a diode then I have assumed wrong. Question: when I analyze the positive cycle I use Vi and a current going clockwise(assumed it is going clockwise), when I analyze the negative cycle I use -Vi but keep the current running clockwise. Is this correct?
yes it is correct, but notice that the positive direction is clockwise, which is opposite to the direction in which the diode allows current (which is counter clockwise). So negative current means counter clockwise which is the same as the direction that the diode allows current.
 
  • #19
CWatters
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Can also apply KVL to the outer loop....

Define Vo with the arrow point up....

Vi-Vr2-Vo=0
Vr2=0 so
Vi-Vo=0
Vo=Vi
 
  • #20
I am still having issues understanding, I might go to my professor for further guidance but I do appreciate the help guys. Thanks.
 
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