Potential Difference Across Diode & Resistor

  • #1

Homework Statement



Assume an ideal diode with ##V_t=0.6##v, find the potential differences across the diode V_ab and across the resistor V_bc as the forward bias voltage is varied from 0 to 10v. Hint: equivalent circuits may be useful

Embedding the image wasn't working so: http://s28.postimg.org/l6b8yzqbw/image.jpg

Homework Equations




The Attempt at a Solution


I'm not really sure how to find the potential difference across the diode and resistor. From KVL I get ##V_0=V_{ab}+iR## although this doesn't seem useful since it's an equation with both of the quantities that I want. I'm pretty sure there's no potential difference across them until the voltage increases to 0.6V though.
 
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Answers and Replies

  • #2
berkeman
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Homework Statement



Assume an ideal diode with ##V_t=0.6##v, find the potential differences across the diode V_ab and across the resistor V_bc as the forward bias voltage is varied from 0 to 10v. Hint: equivalent circuits may be useful

Embedding the image wasn't working so: http://s28.postimg.org/l6b8yzqbw/image.jpg

Homework Equations




The Attempt at a Solution


I'm not really sure how to find the potential difference across the diode and resistor. From KVL I get ##V_0=V_{ab}+iR## although this doesn't seem useful since it's an equation with both of the quantities that I want. I'm pretty sure there's no potential difference across them until the voltage increases to 0.6V though.
It looks like the question just says the diode voltage is 0.6V, so the voltage across the resistor would just be the value of the voltage source minus the 0.6V diode drop,. Seems kind of too easy, though...
 
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  • #3
It looks like the question just says the diode voltage is 0.6V, so the voltage across the resistor would just be the value of the voltage source minus the 0.6V diode drop,. Seems kind of too easy, though...
Yea that's probably correct though since I'm supposed to find the ac voltage across the resistor now if the battery is replaced by an ac function generator. I'm supposed to let ##V_{ac}=V_0/sin wt## with V_0>V_t so I suppose this is also what the function generator is equal to. Now I have to graph V_{ac} and V_{bc} on the same scale versus t.
Using kirchoffs law would it end up looking like this?
http://s11.postimg.org/818tzi9f6/image.jpg

Sorry for terrible pictures I'm on my phone
 
  • #4
berkeman
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Vac=V0/sinwt
I think you mean Vac = Vo * sin(wt)

And your drawing is basically right for the ideal diode they are specifying.
 
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