The discussion focuses on calculating the Vout vs Vin characteristics for a diode circuit with input voltages ranging from 5V to 10V peak. The diodes, specifically the 1N4001 model, are connected in anti-parallel, affecting the output voltage based on the input voltage. When Vin is below 5.7V, Vout remains at 0V, while for Vin above 5.7V, Vout follows Vin, capped at 5.7V. The second diode behaves symmetrically, conducting from 0V to -4.3V when Vin is negative, and is cut off for values less than -4.3V.
PREREQUISITESElectrical engineering students, circuit designers, and anyone interested in understanding diode behavior in voltage regulation and clipping applications.
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:mveep said:
Something higher than 5V? If there's no potential difference (if V1 is smaller than 5), the diode is off, correct?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?mveep said:
0.7Vgneill said:
V1 would be 0 if Vin is 5V to 5.6V, but the same as Vin if Vin is higher than 5.7Vgneill said:
Make that 5.7 V rather than 5.6 V. You said the forward voltage of the diode is 0.7 V.mveep 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?mveep said:
Ok, so V1 = 0 for Vin less than 5.7Vgneill said:
The 0.7 forward voltage will changegneill said:
No, if the diode is not conducting what's the current through the resistor?mveep said:
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?gneill said:
Correct. So what is the potential drop across it?mveep said:
If there's no current, there's no potential drop right?gneill said:
Correct againmveep said:
So V1 would be Vin?gneill said:Correct again
So what does that make V1 (so long as the diode is not conducting)?
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.mveep said:
The voltage at V2 would always be Vin because the diode is in reverse saturation?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?mveep said:
reverse bias and reverse saturation were the terms I learned in classgneill 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.mveep said:
The 2nd diode would be cut off when Vin is positive.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.mveep said:
So this too (V2) would be Vin until -5.7V, and then capped at -5.7V?gneill said:
Right.mveep said:
What would Vout look like? Do you happen to have any idea why my PSpice generated graph is showing 0V for Vin?gneill said:
That would depend on what Vin looks like and how much of Vin lies between the +/- 5.7 V clipping limits of the circuit.mveep said:
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.