# Calculating Diode Current in R-V Circuit w/ Sinusoidal Input

• Engineering
• CoolDude420
In summary, the diode can conduct during the negative half cycle of the AC waveform when the source voltage is 3 V.
CoolDude420

## Homework Statement

R = 1 kohm and Vs(t) is sinusoidal of (peak) amplitude 3 V. The diode is modeled by the series combination of an ideal diode and 0.7 V voltage source.
For what percentage of time will the diode conduct?

## The Attempt at a Solution

I'm getting 76%, the answer is meant to be 42.5%

Why don't you start with a sketch of the source voltage for one cycle. Then indicate the part or parts of the cycle where the diode can conduct.

Hint: If you're getting a conduction percentage over 50% then you're saying that the diode can conduct during during at least part the negative half cycle of the AC waveform...does that seem reasonable?

gneill said:
Why don't you start with a sketch of the source voltage for one cycle. Then indicate the part or parts of the cycle where the diode can conduct.

Hint: If you're getting a conduction percentage over 50% then you're saying that the diode can conduct during during at least part the negative half cycle of the AC waveform...does that seem reasonable?

Shaded part is where it can conduct

First, you've labelled both the top and bottom of the voltage axis with 3 V. How is that possible? Shouldn't one of them be negative?

Second, you're implying with the two shaded areas that the diode can conduct both forwards and backwards, but not for some band of voltages of either polarity near zero volts. Is that really what a diode does? If the diode were ideal with no forward voltage drop (so 0 V instead of 0.7 V), would the sine curve be entirely shaded and the diode conducting continuously for the whole cycle? What would be the difference between that diode and a piece of wire?

gneill said:
First, you've labelled both the top and bottom of the voltage axis with 3 V. How is that possible? Shouldn't one of them be negative?

Second, you're implying with the two shaded areas that the diode can conduct both forwards and backwards, but not for some band of voltages of either polarity near zero volts. Is that really what a diode does? If the diode were ideal with no forward voltage drop (so 0 V instead of 0.7 V), would the sine curve be entirely shaded and the diode conducting continuously for the whole cycle? What would be the difference between that diode and a piece of wire?

I don't think my edited picture is showing up. Here it is:

CoolDude420 said:
I don't think my edited picture is showing up. Here it is:
It's better to add new material to the end of a post rather than overwriting or replacing previously published material, particularly if the previous material has already been commented on in later posts. Otherwise anyone who comes along later will have no idea what's happened and why the thread conversation refers to nonexistent things. You WILL get hit with infraction points if a moderator has to step into fix things.

So, your new diagram looks much better. How will you determine the portion of the cycle where the diode conducts? Can you determine the phase angle where it first turns on? (Note that you can use either time or angle to determine the fraction of the cycle. Angle is probably more straightforward, the angle going from 0 to 2 π over a whole cycle).

Alright. I tried it with using t as the variable. Put this into my calculator. I am getting 48.62% which is wrong. Ho w do I do it using phase angle and ignoring the t?

Here is how I got the t values:

Just solve for the angle when source voltage turns on the diode. So:

##3 sin(θ_1) = 0.7##

Find ##θ_1##. Then find the corresponding angle ##θ_2## when it turns off. Hint: Symmetry is your friend here!

gneill said:
Just solve for the angle when source voltage turns on the diode. So:

##3 sin(θ_1) = 0.7##

Find ##θ_1##. Then find the corresponding angle ##θ_2## when it turns off. Hint: Symmetry is your friend here!

View attachment 109933

gneill said:
Just solve for the angle when source voltage turns on the diode. So:

##3 sin(θ_1) = 0.7##

Find ##θ_1##. Then find the corresponding angle ##θ_2## when it turns off. Hint: Symmetry is your friend here!

View attachment 109933

omg. got the right answer. just realized my calculator was in degrees. derp

## 1. How do I calculate the current through a diode in an R-V circuit with a sinusoidal input?

To calculate the current through a diode in an R-V circuit with a sinusoidal input, you can use the formula I = V/R, where I is the current, V is the voltage, and R is the resistance. This formula assumes that the diode is an ideal diode with no voltage drop. If you are using a non-ideal diode, you will need to use a more complex formula that takes into account the diode's forward voltage drop.

## 2. What is the difference between calculating diode current in an R-V circuit with a sinusoidal input and a DC input?

The main difference between calculating diode current in an R-V circuit with a sinusoidal input and a DC input is that the sinusoidal input varies over time, while the DC input remains constant. This means that the current through the diode will also vary over time in the case of a sinusoidal input, while it will remain constant in the case of a DC input.

## 3. Can I use Ohm's Law to calculate diode current in an R-V circuit with a sinusoidal input?

Yes, you can use Ohm's Law to calculate diode current in an R-V circuit with a sinusoidal input. However, as mentioned earlier, this assumes that the diode is an ideal diode with no voltage drop. If you are using a non-ideal diode, you will need to use a more complex formula that takes into account the diode's forward voltage drop.

## 4. What other factors should I consider when calculating diode current in an R-V circuit with a sinusoidal input?

In addition to the diode's forward voltage drop, you should also consider the frequency and amplitude of the sinusoidal input, as well as the resistance in the circuit. These factors can affect the amount of current flowing through the diode and should be taken into account when making calculations.

## 5. Are there any online tools or calculators available for calculating diode current in an R-V circuit with a sinusoidal input?

Yes, there are several online tools and calculators available for calculating diode current in an R-V circuit with a sinusoidal input. Some popular options include CircuitLab, Falstad Circuit Simulator, and EasyEDA. These tools allow you to input your circuit parameters and will calculate the diode current for you.

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