Determine VPS, quiescent value, and resistor value

  • #1
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Homework Statement


1) What is the VPS voltage source value in Volts

a) 1 b) 2 c)2.5 d)4 e) 5

2) Determine the Resistor value

a)1.75k b) 2 k e) 2.5k d)2.285 k e) 5k

3) What is the ID quiescent value approximately in mA’s?

a) 2 mA b) 1.7 mA c) 1.5 mA d) 1 mA e) 0.6 mA



Homework Equations


V=IR
Id=(Vps/R)-(Vd/R)


The Attempt at a Solution


I get Id=1.7 mA for part 3. I just found that by using the graph it looked liked 1.7 so thats what I assume is the answer. I think R can be found using the graph and ohms law Vd=5 V and I= 2 mA which gives R=2.5kΩ. What im wondering is why do I use the farthest values here? Why dont I use the Vd and Id values are where the lines cross? Vps=RId+Vd=2.5(1.7)+.75=5v for part 1 this can be found using KVL because I had already found R, Id, and Vd.

I uploaded the entire quiz and a picture so that you could see the graph
 

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Answers and Replies

  • #2
gneill
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The black line is called a Load Line. It depends only upon the source voltage and its series resistance. It is essentially independent of the load attached. In this case the "load" is the diode, which has its own characteristic curve (blue). Where the two intersect is the operating point of the circuit. It's a graphic way to "solve" for the operating conditions of a circuit containing nonlinear elements (like a diode).
 
  • #3
CWatters
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What im wondering is why do I use the farthest values here?
You don't have to use the furthest values.

Ohms law V=IR looks like the standard equation for a straight line y=mx+c where c=0 and the slope of the line m = R.

So you could calculate the slope of the line using any two points to get R.

This has real life applications because you cannot always measure the short circuit current. Imagine you wanted to measure the internal resistance of a 12V car battery. There is no way you could measure the short circuit current to get a data point at V=0. Trying to short circuit a 12V car battery is dangerous, the current could be 100's of amps and the battery terminals might melt and throw hot lead at you or worse. However you could load up the battery with some known safe currents and measure the corresponding voltage. Then plot the graph, calculate the slope, and get the internal resistance that way.
 
  • #4
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Ok so then that would make my logic correct by the way thanks to both of you for conceptual info?
 
  • #5
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R=V/I if I put in .75/1.75=.44 k ohms hmmmm what did I do wrong
 
  • #6
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Hey on question 4 what equations should I use I can't find an equation the equates Reverse bias voltage and reverse bias current
 
  • #7
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Ah ha so if the load line depends on the source and series resistance then Vd=Vps right?
 
  • #8
gneill
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R=V/I if I put in .75/1.75=.44 k ohms hmmmm what did I do wrong
That would give the resistance of the diode at the operating point: the potential across the diode divided by the diode current.

To find R you want the slope of the load line at that operating point. But since the load line is a straight line, you can take the rise/run for the line using any convenient pair of points that lie on it. Its end points are very convenient indeed...
 
  • #9
gneill
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Ah ha so if the load line depends on the source and series resistance then Vd=Vps right?
No, Vd is potential across the diode, which is Vp minus the drop across R. It's a series circuit, so you need to count all the potential drops.
 
  • #11
gneill
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Vps=2.5Id+Vd ?
Is that "2.5" meant to be the 2.5kΩ resistance of R? If so, the equation looks fine.
 
  • #12
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Well by the graph I know Id=1.7 then Vps=(2.5)(1.7) + Vd
 
  • #14
gneill
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Well by the graph I know Id=1.7 then Vps=(2.5)(1.7) + Vd
Sure. But I don't understand why you're agonizing over Vps. Didn't you already decide that Vps is 5V from the load line endpoint? It will not change.
 
  • #15
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Oh so that's all I have to do on a problem like this is just say that 5V is Vps
 
  • #16
gneill
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Oh so that's all I have to do on a problem like this is just say that 5V is Vps
Yes, since you can read it directly off the given load line.
 
  • #17
CWatters
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Oh so that's all I have to do on a problem like this is just say that 5V is Vps
Yes.

In most real life problems you would know Vps and use that to construct the load line.

This problem is slightly different in that they are asking you to work out Vps from a load line someone has already constructed.
 

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