Confused with simple diode circuits given I-V characteristics

WBahn

Joined Mar 31, 2012
29,979
Ok, so if Vs=13V and Is=0mA, I know Vout = Vs-IsR1=Vs
Now, for 2nd part if Vout is zero, Is=Vs/R1

And if I use condition for Q1 as Is=1ma, i would get a point of 12V
so my graph for Vout vs Is would just be essentially y=x, as for w/e conditions i will get pts that are (13V,13mA),(13.5v,13.5mA) etc.
Think about this. You first stated that if Vs=13V that if Is=0mA that Vout would be 13V, so one of the points would be (13V,0mA). You then stated that if Vout is 0 that Is=Vs/R1 which would make Is=13mA. So another point would be (0V, 13mA). Then you noted that if Is=1mA that Vout would be 12V, giving you the point (12V, 1mA). How do you then come up with your final conclusion?
 

Thread Starter

ModelIt

Joined Apr 12, 2014
12
Think about this. You first stated that if Vs=13V that if Is=0mA that Vout would be 13V, so one of the points would be (13V,0mA). You then stated that if Vout is 0 that Is=Vs/R1 which would make Is=13mA. So another point would be (0V, 13mA). Then you noted that if Is=1mA that Vout would be 12V, giving you the point (12V, 1mA). How do you then come up with your final conclusion?
Ohh so, I basically get a load line from 13mA diagonal to 13V, putting this on IV characteristic of diode, the Q point would be(0,13V), so the diode would be a short circuit and Vout then would be at 0V. Is that a correct line of thought?
 

WBahn

Joined Mar 31, 2012
29,979
Ohh so, I basically get a load line from 13mA diagonal to 13V, putting this on IV characteristic of diode, the Q point would be(0,13V), so the diode would be a short circuit and Vout then would be at 0V. Is that a correct line of thought?
You are starting to get the right idea. But you are ignoring polarities. Next time you jump start a car, think about the consequences if you hook up the cables backwards. Polarity counts. The voltage on the diode characteristic, Vd, is for the anode positive relative to the cathode. If Vout is positive, is Vd positive or negative? Similarly, the current in the diode, Id, is positive if it is flowing from anode to cathode. If Is is positive, is Id positive or negative?
 

Thread Starter

ModelIt

Joined Apr 12, 2014
12
You are starting to get the right idea. But you are ignoring polarities. Next time you jump start a car, think about the consequences if you hook up the cables backwards. Polarity counts. The voltage on the diode characteristic, Vd, is for the anode positive relative to the cathode. If Vout is positive, is Vd positive or negative? Similarly, the current in the diode, Id, is positive if it is flowing from anode to cathode. If Is is positive, is Id positive or negative?
I see that Vout and Vd have reversed polarity, so if Vout is +, Vd is -. And if Is is +, current flows from cathode to anode, so Id must be -.
But how does this affect the load line, does it move it into 3rd quadrant to get the Q point along the sloped line 1/R2?
 

daviddeakin

Joined Aug 6, 2009
207
Sometimes it can be very helpful to re-draw a circuit in a different orientation, which often makes it more intuitive. Does this help?

(You didn't add any sense of polarity to your voltage source, so I had to guess).
 

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WBahn

Joined Mar 31, 2012
29,979
I see that Vout and Vd have reversed polarity, so if Vout is +, Vd is -. And if Is is +, current flows from cathode to anode, so Id must be -.
But how does this affect the load line, does it move it into 3rd quadrant to get the Q point along the sloped line 1/R2?
Don't get hung up about quadrants and such. Just plot the points on the diode characteristic onto the load line. For instance, one point on the diode characteristic is Vd=-10V, Id=0mA, right?

We know that Vd = -Vout and Id = -Is, right?

So Vd=-10V means Vs = 10V and Id=0mA means Is=0mA, right?

Plot that point on the load line!

Since the diode characteristic consists of just straight lines, all you need to do is plot the endpoints of each segment (or, in the case of the two end segments, a second point somewhere convenient on the segment) and connect them with straight lines.

PLEASE, plot the load line and attach it here. Don't just talk about what it looks like. Sketch it!
 
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