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.
The Vout vs Vin characteristics for a diode circuit can be calculated by using Ohm's Law and the diode equation. First, determine the voltage drop across the diode by subtracting the forward voltage from the input voltage. Then, use Ohm's Law to calculate the output current. Finally, use the diode equation to calculate the output voltage. Repeat this process for different input voltages to create a Vout vs Vin graph.
The diode equation is a mathematical relationship that describes the behavior of a diode. It states that the current through a diode is equal to the reverse saturation current multiplied by the exponential of the forward voltage divided by the thermal voltage. This equation can be rearranged to solve for Vout, which is the voltage across the diode.
No, a linear approximation cannot be used to accurately calculate the Vout vs Vin characteristics for a diode circuit. Diodes have a non-linear relationship between voltage and current, so a linear approximation would not accurately represent the behavior of the diode.
The input voltage directly affects the Vout vs Vin characteristics for a diode circuit. As the input voltage increases, the voltage drop across the diode also increases, resulting in a higher output voltage. However, once the input voltage exceeds the forward voltage of the diode, the output voltage will remain constant.
The Vout vs Vin characteristics for a diode circuit provide important information about the behavior of the diode. They can be used to determine the voltage and current requirements for a specific application, as well as the maximum and minimum operating conditions for the diode. The characteristics also help in selecting the appropriate diode for a circuit and analyzing its performance.