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

[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?
answer: 42.5

Homework Equations

The Attempt at a Solution

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

 

[gneill]

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?

 

[CoolDude420]

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

 

[gneill]

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?

 

[CoolDude420]

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:

 

[gneill]

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).

 

[CoolDude420]

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?

 

[CoolDude420]

Here is how I got the t values:

 

[gneill]

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!

 

[CoolDude420]

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

 

[CoolDude420]

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