I'm going to post here some problems I'm solving.

I'm going to have exam next 27/06/2014

My first problem is about diodes and it says:

Draw the characteristic curve with Is=2nA (T=293ºK). The curve should be drawed using 2 different temps: 20ºC and 25ºC.

 

So you have the characteristic equation of a diode. Looks like a plug and chug problem -- right?

 

So you have the characteristic equation of a diode. Looks like a plug and chug problem -- right?

I don't know Vd or η...

 

hi Psy,
Helpful data on this link.
http://www.allaboutcircuits.com/vol_3/chpt_3/1.html
E

 

hi Psy,
Helpful data on this link.
http://www.allaboutcircuits.com/vol_3/chpt_3/1.html
E

Hi Sir Eric...

As I said before, I don't have Vd neither η.

The equation we have been using is:

 

Sure you do! In order to make a plot you have to have an independent variable. This is "V sub D" or the voltage across the diode which you vary from 0 to some positive value.

η is the ideality constant which is 1 for an ideal diode and between 1 an 2 for a real diode. "V sub T" is the Thermal voltage kT/q or about 26 mV (25.851997 mV)

You have everything you need to make a plot of "I sub D" versus "V sub D".

 

Sure you do! In order to make a plot you have to have an independent variable. This is "V sub D" or the voltage across the diode which you vary from 0 to some positive value.

η is the ideality constant which is 1 for an ideal diode and between 1 an 2 for a real diode. "V sub T" is the Thermal voltage kT/q or about 26 mV (25.851997 mV)

You have everything you need to make a plot of "I sub D" versus "V sub D".

Yes, I know about that. But the problem doesn't says anything about Vd and also says to consider it for 2 temps. But I'm going to try!

 

Of course it doesn't say anything about "V sub D". You are supposed to know that to draw a characteristic curve you have to have something that changes over some range of values and your job is to see what happens to "I sub D" when you do that. You do it once for 20° C = 293° K, and then you do it again for 298° K

If I were doing it I'd use an Open Office spreadsheet to make a table of values for each temperature and "V sub D" is varied from say -1 Volt up to say +1 Volt. what you see in the table and the graph is highly counter intuitive because exponential functions get REALLY REALLY BIG VERY VERY QUICKLY. They also get REALLY REALLY SMALL VERY QUICKLY.

You really can't mess this up as long as you state your assumptions.

 

I'm not being capable of plotting a clear line in excel... Don't know why!

See attached image, please!


Edit;

I think I got something now but I still wonder why there isn't a "negative part" of the curve as we see in books or around the web!

 

You have the axes switched. "V sub D" should be on the horizontal axis (like in the second picture), and you clearly have it on the vertical axis. "I sub D" is a function of "V sub" with parameter T (Temperature).

As you should see from the table values, "I sub D" should be negative for values of "V sub D" that are less than 0. The exponential will be less than 1. Subtracting 1 from a number less than 1 will give you a negative number.

 

Yes, that's why i have switched them. Now it's closer but still not perfect!

 

So, is that curve correct?

I need to move on!
Can you guys help?

 

h Psy,
It compares reasonably well with the plot from the link I posted.

I didn't check your maths.

E

 

h Psy,
It compares reasonably well with the plot from the link I posted.

I didn't check your maths.

E

Thanks for helping...

Now, I'm struggling to understand the attached one. It's another problem and teacher asks to plot Vout (manually, not with LTSpice).

If the diode is "OFF", the circuit is "open" and Vout = V2


To try to understand the circuit when the diode is "ON" I have simulated the circuit at LTSpice but I can't understand why voltage at Va is 9.32951V

 

There is 1K of resistance between +12V and +6V. If the diode was not there (shorted), the voltage at the point between the resistors would be 9V and the current would be 6 mA. Does that help?

 

There is 1K of resistance between +12V and +6V. If the diode was not there (shorted), the voltage at the point between the resistors would be 9V and the current would be 6 mA. Does that help?

How do you get 9V with diode shorted?? Aren't the 2 voltage sources subtracting each other?? 12-6=6. Using a voltage divider we would have (500/500+500)*6 = 3.

 

From II KVL we find Va voltage

Va = V1 - Id * R1

And

Id = (V1 - V2 - Vd)/(R1 + R2) = (12V - 6V - 0.65V)/1K = 5.35V/1K = 5.35mA

So

Va = 12V - 5.35mA*500R = 12V - 2.675V = 9.325V

 

How do you get 9V with diode shorted?? Aren't the 2 voltage sources subtracting each other?? 12-6=6. Using a voltage divider we would have (500/500+500)*6 = 3.

With respect to each of the voltage sources, what effect does that 3V differential have?

 

From II KVL we find Va voltage

Va = V1 - Id * R1

And

Id = (V1 - V2 - Vd)/(R1 + R2) = (12V - 6V - 0.65V)/1K = 5.35V/1K = 5.35mA

So

Va = 12V - 5.35mA*500R = 12V - 2.675V = 9.325V

I just can't understand why Va = V1 - Id * R1

For the current, if I try to do it using "mesh" theorem, I would write

-Vin + R1*I1 + 0.7 +R2*I1 + 6 = 0

With respect to each of the voltage sources, what effect does that 3V differential have?

I think I didn't understood your question!

 

I just can't understand why Va = V1 - Id * R1

For the current, if I try to do it using "mesh" theorem, I would write

-Vin + R1*I1 + 0.7 +R2*I1 + 6 = 0

Your "mesh" looks good. But we don't need mash here.
All we need here is II Kirchhoff's law.
http://forum.allaboutcircuits.com/showpost.php?p=463363&postcount=19

And your circuit look like this

So we can write KVL

V1 = VR1 + VD1 + VR2 + V2 and from this we can find Va.

Va = V2 + VR2 + VD1

or

Va = V1 - VR1