Calculate the Vout vs Vin characteristics for a diode circui

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The discussion focuses on calculating the Vout versus Vin characteristics for a diode circuit with input voltages ranging from 5V to 10V peak. It addresses the behavior of diodes connected in anti-parallel, emphasizing that Vout will be zero when Vin is below 5.7V and will follow Vin above this threshold. The forward voltage drop of the diode (0.7V) is crucial in determining when the diode conducts, impacting the output voltage. The second diode's behavior is also analyzed, indicating it will conduct from 0V to -4.3V and be cut off at values less than -4.3V. Overall, the circuit limits Vout to a range defined by the diode characteristics and input voltage conditions.
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Homework Statement


Calculate the Vout vs Vin characteristics for input voltages of 5V to 10V peak.[/B]
upload_2017-1-29_0-53-5.png


Homework Equations

The Attempt at a Solution


Would Vout simply be nothing because the diodes are connected in anti-parallel?
Putting a volt probe at Vin and Vout gives something like: (Green is Vin, red is Vout)
Why is Vin not shown as a sine wave?
upload_2017-1-29_1-2-12.png
 
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Hi mveep, Welcome to Physics Forums!

mveep said:
Would Vout simply be nothing because the diodes are connected in anti-parallel?
What do you think might be effect of the 5 V sources in series with the diodes? Isolate one of the diodes branches just to see how it behaves:

upload_2017-1-29_6-11-59.png


What potential would V1 have to reach for D1 to conduct (be forward biased)?
 
gneill said:
Hi mveep, Welcome to Physics Forums!What do you think might be effect of the 5 V sources in series with the diodes? Isolate one of the diodes branches just to see how it behaves:

View attachment 112244

What potential would V1 have to reach for D1 to conduct (be forward biased)?
Something higher than 5V? If there's no potential difference (if V1 is smaller than 5), the diode is off, correct?
 
mveep said:
Something higher than 5V? If there's no potential difference (if V1 is smaller than 5), the diode is off, correct?
Remember that the diode is not ideal here (it's a 1N4001) so it will have a forward bias voltage contribution, too. What's the typical forward voltage of a silicon diode?
 
gneill said:
Remember that the diode is not ideal here (it's a 1N4001) so it will have a forward bias voltage contribution, too. What's the typical forward voltage of a silicon diode?
0.7V
 
Right. So add that to the 5 V you stated in your previous answer. That gives 5.7 V. So you can expect the diode to turn on when the potential V1 reaches 5.7 V.

Now let's extend our view of the circuit slightly to include the power source and resistor:

upload_2017-1-29_13-53-40.png


If the diode forward voltage remains essentially constant so long as the diode is conducting (a pretty good approximation for this situation), what do you expect to happen to V1 for different values of Vin?
 
gneill said:
Right. So add that to the 5 V you stated in your previous answer. That gives 5.7 V. So you can expect the diode to turn on when the potential V1 reaches 5.7 V.

Now let's extend our view of the circuit slightly to include the power source and resistor:

View attachment 112260

If the diode forward voltage remains essentially constant so long as the diode is conducting (a pretty good approximation for this situation), what do you expect to happen to V1 for different values of Vin?
V1 would be 0 if Vin is 5V to 5.6V, but the same as Vin if Vin is higher than 5.7V
 
mveep said:
V1 would be 0 if Vin is 5V to 5.6V,...
Make that 5.7 V rather than 5.6 V. You said the forward voltage of the diode is 0.7 V.

mveep said:
...but the same as Vin if Vin is higher than 5.7V
How? Which one of the two voltages in the diode branch can change while the diode is conducting? Will the 5 V source change value? Will the 0.7 V diode forward voltage change?
 
gneill said:
Make that 5.7 V rather than 5.6 V. You said the forward voltage of the diode is 0.7 V.
Ok, so V1 = 0 for Vin less than 5.7V

gneill said:
How? Which one of the two voltages in the diode branch can change while the diode is conducting? Will the 5 V source change value? Will the 0.7 V diode forward voltage change?
The 0.7 forward voltage will change
 
  • #10
mveep said:
Ok, so V1 = 0 for Vin less than 5.7V
No, if the diode is not conducting what's the current through the resistor?
The 0.7 forward voltage will change
No. It cannot. The forward voltage of a diode is (almost) constant.
 
  • #11
gneill said:
No, if the diode is not conducting what's the current through the resistor?

No. It cannot. The forward voltage of a diode is (almost) constant.
If the diode isn't conducting, there would be no current through the resistor correct?
 
  • #12
mveep said:
If the diode isn't conducting, there would be no current through the resistor correct?
Correct. So what is the potential drop across it?
 
  • #13
gneill said:
Correct. So what is the potential drop across it?
If there's no current, there's no potential drop right?
 
  • #14
mveep said:
If there's no current, there's no potential drop right?
Correct again :smile:

So what does that make V1 (so long as the diode is not conducting)?
 
  • #15
gneill said:
Correct again :smile:

So what does that make V1 (so long as the diode is not conducting)?
So V1 would be Vin?
 
  • #16
mveep said:
So V1 would be Vin?
Yes! So V1 "follows" Vin until the diode conducts. Then it is held at 5 V + VD no matter how high Vin goes. That is, this branch limits the maximum voltage of V1 to 5.7 V.

Do a similar analysis for the other diode branch alone.
 
  • #17
gneill said:
Yes! So V1 "follows" Vin until the diode conducts. Then it is held at 5 V + VD no matter how high Vin goes. That is, this branch limits the maximum voltage of V1 to 5.7 V.

Do a similar analysis for the other diode branch alone.
The voltage at V2 would always be Vin because the diode is in reverse saturation?
upload_2017-1-29_12-18-6.png
 
  • #18
mveep said:
The voltage at V2 would always be Vin because the diode is in reverse saturation?
I don't think that "reverse saturation" is a valid term. Perhaps you mean cuttoff (not conducting)? That would be true if Vin could only take on positive values. What happens as Vin goes negative?
 
  • #19
gneill said:
I don't think that "reverse saturation" is a valid term. Perhaps you mean cuttoff (not conducting)? That would be true if Vin could only take on positive values. What happens as Vin goes negative?
reverse bias and reverse saturation were the terms I learned in class

When Vin goes negative, it will add with the +5 source?
 
  • #20
mveep said:
reverse bias and reverse saturation were the terms I learned in class

When Vin goes negative, it will add with the +5 source?
You can think of it that way if you're doing KVL around the loop. You should be able to find the conditions for Vin where the diode is cut off and when it is conducting, just as for the other branch.
 
  • #21
gneill said:
You can think of it that way if you're doing KVL around the loop. You should be able to find the conditions for Vin where the diode is cut off and when it is conducting, just as for the other branch.
The 2nd diode would be cut off when Vin is positive.
When Vin is negative:
The 2nd diode will be conducting from 0->-4.3V
It will be cut off at any value of Vin is less than -4.3
 
  • #22
mveep said:
The 2nd diode would be cut off when Vin is positive.
When Vin is negative:
The 2nd diode will be conducting from 0->-4.3V
It will be cut off at any value of Vin is less than -4.3
No. Think about it, the two circuits are essentially the same (Vin, 5 V, resistor, diode), the only difference being the diode branch is inverted. So the two should behave symmetrically.

With the diode reversed, it's forward voltage drop's direction is reversed in the branch, too. So it still "adds" to the 5 V (it's potential drop has the same direction as that of the 5 V source). Note the potential drop for the diode in the image below:
upload_2017-1-29_16-12-19.png
 
  • #23
gneill said:
No. Think about it, the two circuits are essentially the same (Vin, 5 V, resistor, diode), the only difference being the diode branch is inverted. So the two should behave symmetrically.

With the diode reversed, it's forward voltage drop's direction is reversed in the branch, too. So it still "adds" to the 5 V (it's potential drop has the same direction as that of the 5 V source). Note the potential drop for the diode in the image below:
View attachment 112267
So this too (V2) would be Vin until -5.7V, and then capped at -5.7V?
 
  • #24
mveep said:
So this too (V2) would be Vin until -5.7V, and then capped at -5.7V?
Right.
 
  • #25
gneill said:
Right.
What would Vout look like? Do you happen to have any idea why my PSpice generated graph is showing 0V for Vin?
 
  • #26
mveep said:
What would Vout look like?
That would depend on what Vin looks like and how much of Vin lies between the +/- 5.7 V clipping limits of the circuit.
Do you happen to have any idea why my PSpice generated graph is showing 0V for Vin?
Nope. Often that sort of issue is due to forgetting a connection or making an incorrect parameter setting. The scales on the graph in your first post look suspicious, being on the order of 10-180.
 

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