Solve a Diode Problem: Voltage Sources, VDD, Part A

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In summary, the conversation discusses the use of voltage sources and diodes in a circuit. The conversation also explores different scenarios and calculations for the current in a specific diode.
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Homework Statement


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The Attempt at a Solution


I know this might seem like a stupid question but are the 10/-20V voltage sources like the 5.3V or are they VDD (which is something different?) I'm also stuck on part a. If D2 is off, that means its not conducting, so zero current, but would there still be a voltage drop across it? Then since its -20V on one side does that mean V(D2)=20V gain to the node on top?
 
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  • #2
asdf12312 said:

The Attempt at a Solution


I know this might seem like a stupid question but are the 10/-20V voltage sources like the 5.3V or are they VDD (which is something different?) I'm also stuck on part a. If D2 is off, that means its not conducting, so zero current, but would there still be a voltage drop across it? Then since its -20V on one side does that mean V(D2)=20V gain to the node on top?

Think of the 10/-20V sources as batteries which connect to the indicated terminals and then to ground.

If you are told to assume that a diode is off, simply remove it from the circuit (it supposedly conducts no current so is equivalent to an open circuit. If you are told to assume that a diode is on, for a first pass analysis replace it with a battery equivalent to the junction forward bias (0.7V in this case).
 
  • #3
OK, I guess I can treat the 10V as a voltage source that is grounded then. So it won't be part of the loop equation, then I do I(D1)=(5.3+V(D1))/6k where V(D1)=0.7, I(D1)=-1mA. Since there can't be negative current, D1 on/D2 off doesn't work. D1/D2 both off obviously doesn't work either. When I simulated the circuit for D1 off/D2 on I got I(D2)=34mA. is this the answer?
 
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  • #4
asdf12312 said:
OK, I guess I can treat the 10V as a voltage source that is grounded then. So it won't be part of the loop equation, then I do I(D1)=(5.3+V(D1))/6k where V(D1)=0.7, I(D1)=-1mA. Since there can't be negative current, D1 on/D2 off doesn't work. D1/D2 both off obviously doesn't work either. When I simulated the circuit for D1 off/D2 on I got I(D2)=34mA. is this the answer?

You can't ignore the 10V supply because it has a path through your loop to ground via the 1.93 kΩ resistor. In effect you have a second loop (stick a battery in there for the 10V supply and wire its negative lead to the ground symbol at the bottom to "see" the loop).

You might find that nodal analysis will treat you well here. Only one node to worry about!

Your simulation yields the correct current for ID2 (it's what I've just calculated).
 
  • #5


Without more context or a diagram, it is difficult to provide a specific answer. However, I can provide some general information about diodes and voltage sources that may help with solving the problem.

A diode is a semiconductor device that allows current to flow in one direction but not in the other. It typically has two terminals, the anode and the cathode. The anode is the positive terminal and the cathode is the negative terminal. When a diode is forward biased, meaning the anode is at a higher voltage than the cathode, it allows current to flow. When it is reverse biased, meaning the cathode is at a higher voltage than the anode, it does not allow current to flow.

A voltage source is a device that provides a constant voltage, such as a battery or a power supply. In this problem, the 10/-20V voltage sources are most likely referring to two different voltage sources - one with a positive voltage of 10V and one with a negative voltage of -20V.

VDD is a common abbreviation for the power supply voltage in a circuit. It is typically a positive voltage, such as 3.3V or 5V.

In part a of the problem, it is stated that D2 is off. This means that the diode is reverse biased and does not allow current to flow. In this case, there would still be a voltage drop across the diode, but it would be very small (typically less than 0.7V for a silicon diode). The voltage drop would depend on the characteristics of the diode and the current flowing through it. Since the voltage sources are not connected to the diode, they would not have any effect on the voltage drop across it.

I hope this information helps in solving the problem. Remember to always carefully read the given information and consider the properties of diodes and voltage sources in your solution.
 

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

A diode is a semiconductor device that allows current to flow in only one direction. It is made up of a P-N junction, where the P side has excess positive charge and the N side has excess negative charge. When a voltage is applied in the forward direction, the diode conducts current, but when the voltage is applied in the reverse direction, the diode blocks current flow.

2. What is the purpose of solving a diode problem?

Solving a diode problem involves analyzing the behavior of a circuit containing one or more diodes. This is important in order to understand the voltage and current characteristics of the circuit, and to determine the appropriate values for components such as resistors and capacitors.

3. What is a voltage source and how does it affect diode behavior?

A voltage source is a component in a circuit that provides a constant voltage. In the context of a diode problem, the voltage source determines the direction and magnitude of the voltage applied to the diode. This voltage affects the diode's ability to conduct current, as well as its breakdown voltage (the voltage at which it starts to conduct in the reverse direction).

4. What does VDD stand for and how is it related to diode problems?

VDD stands for "drain-to-drain voltage" and is commonly used in diode problems involving MOSFETs (metal-oxide-semiconductor field-effect transistors). It refers to the voltage difference between the drain and source terminals of the MOSFET, and is a key factor in determining the MOSFET's operating region (e.g. cutoff, saturation, etc.).

5. What is involved in solving a diode problem?

To solve a diode problem, one must analyze the circuit using Kirchhoff's laws and diode equations. This includes determining the voltage and current characteristics of the diode, as well as the overall behavior of the circuit. It may also involve finding the appropriate values for resistors or other components in order to achieve a desired output.

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