Why is the voltage output in diode analysis circuits sometimes confusing?

[CoolDude420]

Homework Statement

Hi,

In some of the diode analysis examples that we did in class, we often end up with circuits like the ones shown here. I'm quite confused because sometimes I understand them and other times I just don't know how these voltages are coming from. It really makes me question what I really know or don't know.

Here for example, in circuit 1, why isn't the output voltage Vo = (1000)(id)+5V?

In circuit 2, why isn't the output voltage Vo = (1000)(id)+5V again?

The correct answers are shown below the circuits(from lecture slides)

Homework Equations

The Attempt at a Solution

 

[gneill]

First, look at the direction of $i_D$ indicated in each circuit. What will be the polarity of the potential change across the resistor in each case? Revise your suggested solutions accordingly.

Second, a sum of known fixed potential changes will be a constant regardless of the current. In both cases you can do a "KVL walk" from one terminal of Vo to the other that only passes through fixed potential changes. So that must fix the value of Vo no matter what.

There's no reason why both solutions cannot be true (once you fix your suggested solutions).

 

[CoolDude420]

First, look at the direction of $i_D$ indicated in each circuit. What will be the polarity of the potential change across the resistor in each case? Revise your suggested solutions accordingly.

Second, a sum of known fixed potential changes will be a constant regardless of the current. In both cases you can do a "KVL walk" from one terminal of Vo to the other that only passes through fixed potential changes. So that must fix the value of Vo no matter what.

There's no reason why both solutions cannot be true (once you fix your suggested solutions).

I'm not exactly what you're asking. Should it be -id? So in circuit 1, Vo = (1000)(-id) +5? And applying KVL to circuit 1, we get Vs = (1000)(-id) +5, so Vo=Vs?

 

[gneill]

I'm not exactly what you're asking. Should it be -id? So in circuit 1, Vo = (1000)(-id) +5? And applying KVL to circuit 1, we get Vs = (1000)(-id) +5, so Vo=Vs?

Both drawings specify a direction for $i_D$. Your (original) proposed solutions contradicted this definition.

 

[NascentOxygen]

Here for example, in circuit 1, why isn't the output voltage Vo = (1000)(id)+5V?

You need to draw the "voltage arrow" across each component before trying to add the individual voltages. Sometimes the summation of two voltages involves adding the negative of one you have already drawn, or even both.

Think about what you do when adding vectors.

 

[CoolDude420]

Both drawings specify a direction for $i_D$. Your proposed solutions contradict this definition.

Okay, for circuit 1,
Vo =(1000)(-id) +5
and for circuit 2,
Vo=(1000)(-id) + 5

 

[gneill]

Okay, for circuit 1,
Vo =(1000)(-id) +5
and for circuit 2,
Vo=(1000)(-id) + 5

Yes. But it is also true that in circuit 1 Vo = Vs, and in circuit 2 Vo = Vs + 0.5V.

 

[CoolDude420]

Yes. But it is also true that in circuit 1 Vo = Vs, and in circuit 2 Vo = Vs + 0.5V.

I understand the circuit 1 now, but with circuit 2, the Vo symbol is the voltage across the 5v voltage source and the resistor, so why are we including the voltage across the 0.5V terminals of the battery?

 

[gneill]

I understand the circuit 1 now, but with circuit 2, the Vo symbol is the voltage across the 5v voltage source and the resistor, so why are we including the voltage across the 0.5V terminals of the battery?

There's more than one path you can take between the two nodes that define Vo. Your solution takes one of them. Follow the other.

 

[CoolDude420]

There's more than one path you can take between the two nodes that define Vo. Your solution takes one of them. Follow the other.

ohhhhhhhh. I see now. So both answers are correct then?

 

[gneill]

ohhhhhhhh. I see now. So both answers are correct then?

Yes.