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

[asdf12312]

Homework Statement

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?

 

[gneill]

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).

 

[asdf12312]

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?

 

[gneill]

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 I_D2 (it's what I've just calculated).

 

[rezeew]

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.