Electronics problems - again

MrAl

Joined Jun 17, 2014
11,389
Hi,

Sure no problem. I was going to post an exact response curve too but was holding off because that might reveal too much right now and i assumed you wanted to figure this out for yourself.
Once you're done i can post that if you like.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Hi,

Sure no problem. I was going to post an exact response curve too but was holding off because that might reveal too much right now and i assumed you wanted to figure this out for yourself.
Once you're done i can post that if you like.
Just wat to alert you that the current source has it's "arrow" upwards but the plus signal is at the bottom side of the current source symbol... I don't know if this has any influence in maths behind the resolution!

And also, le tme tell you that we want Vout (Is), meaning that we do not need to work with exact values for Is but with "Is" instead so that we can say that Vout is "x V" for a given "Is".
 
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Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
I have tried to solve the problema combining your method and our teachers method.

My work is in the attached file.

Can you check it?

Also attached a full version of the image in case the other is not viewable enough!
 

Attachments

MrAl

Joined Jun 17, 2014
11,389
Hi,

Yes i see what you mean about the two signs inside the current source on the left. The arrow points up, but the tip is marked (-) and the bottom marked (+), which makes no sense.
Normally the current arrow is drawn in the direction of current flow for conventional current flow, not electron flow.

You are going to have to ask your instructor if he/she meant to show the reversed sign for the reason of reversed flow, or it just shows that the starting current was negative (-5ma).
There is no way we can interpret something off the all like that.
Note that in Eric's circuit he assumed the same thing as i did.

So find this information out first. Note that if the current is reversed it changes the whole problem because then we really would be starting with a +5ma current which means the circuit would start out with BOTH diodes conducting, then gradually they turn off one by one. This is completely different than before.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Hi,

Yes i see what you mean about the two signs inside the current source on the left. The arrow points up, but the tip is marked (-) and the bottom marked (+), which makes no sense.
Normally the current arrow is drawn in the direction of current flow for conventional current flow, not electron flow.

You are going to have to ask your instructor if he/she meant to show the reversed sign for the reason of reversed flow, or it just shows that the starting current was negative (-5ma).
There is no way we can interpret something off the all like that.
Note that in Eric's circuit he assumed the same thing as i did.

So find this information out first. Note that if the current is reversed it changes the whole problem because then we really would be starting with a +5ma current which means the circuit would start out with BOTH diodes conducting, then gradually they turn off one by one. This is completely different than before.
Hello MrAl...

The "-" and "+" sings are to ignore indeed...

But the way you solved the problem is not quite what we are used to do.
Today I went to school to ask our teacher for some help and he has enlightened me about the solving process. Later today, If I have the time, I'll post here the process because we are asked to do more than just draw the transfer curve. We are also asked to find the Vout expressions for all 3 situations.
 

MrAl

Joined Jun 17, 2014
11,389
Hi,

Yes the method i gave solves for Vout for all three cases.

That's good to know about the current source, so i'll check out your work. But it already appears that you assumed the current source was negative, is that true?

With a quick look however it looks like you are doing it right.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Ok...

This is how we was supposed to solve the diodes problem correctly.
See attached pics!

One of the other 2 version was also corrected but I should convert Vin into Is.
 

Attachments

MrAl

Joined Jun 17, 2014
11,389
Ok...

This is how we was supposed to solve the diodes problem correctly.
See attached pics!

One of the other 2 version was also corrected but I should convert Vin into Is.

Hello again,

Yes that's another way of doing it. That uses the known voltages to calculate the total current and therefore some up with the required points needed. It does not yet show the final solutions, but that's just a matter of using the slope intercept form again for each interval.
 
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