Understanding Diode Analysis: Exploring Voltage Output in Circuit Theory

In summary: The answer to both questions is that Vo is the sum of the voltages across the resistor and the voltage across the battery.Yes. The answer to both questions is that Vo is the sum of the voltages across the resistor and the voltage across the battery.
  • #1
CoolDude420
201
9

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.

02c303a362.jpg


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

 
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  • #2
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).
 
  • #3
gneill said:
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?
 
  • #4
CoolDude420 said:
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.
 
  • #5
CoolDude420 said:
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.
 
  • #6
gneill said:
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
 
  • #7
CoolDude420 said:
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.
 
  • #8
gneill said:
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?
 
  • #9
CoolDude420 said:
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.
 
  • #10
gneill said:
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?
 
  • #11
CoolDude420 said:
ohhhhhhhh. I see now. So both answers are correct then?
Yes.
 

1. What is a diode and how does it work?

A diode is an electronic component that allows current to flow in only one direction. It is made up of two layers of semiconductor material, typically silicon, with an impurity added to one layer to create a junction. When a voltage is applied in the forward direction, the diode allows current to flow through. However, when the voltage is applied in the reverse direction, the diode blocks the current.

2. What is the purpose of diodes in circuit theory?

Diodes are used in circuit theory to control the flow of current and to convert AC (alternating current) to DC (direct current). They can also be used to protect circuits from reverse current and voltage spikes, and to regulate voltage levels.

3. How do you analyze the voltage output of a diode in a circuit?

To analyze the voltage output of a diode in a circuit, you can use Kirchhoff's circuit laws and Ohm's law. Kirchhoff's laws state that the sum of the voltages around a closed loop in a circuit is equal to zero, and the sum of the currents entering and leaving a node is also equal to zero. Ohm's law states that the voltage across a resistor is equal to the current through the resistor multiplied by the resistance.

4. What factors affect the voltage output of a diode?

The voltage output of a diode is affected by the current flowing through it, the temperature of the diode, and the type of material used to make the diode. The voltage drop across a diode will also vary depending on the amount of current flowing through it.

5. How do you calculate the voltage drop across a diode?

The voltage drop across a diode can be calculated using Ohm's law. First, measure the current flowing through the diode. Then, using the voltage and current values, use the formula V = IR, where V is the voltage drop, I is the current, and R is the resistance of the diode. The resistance of a diode is typically very small, so the voltage drop will also be small.

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