Simple Diode In Parallel Circuit

Thread Starter

Student01

Joined Apr 15, 2009
35
Completely lost with this one guys, how do I know which diode is on and which is off?

I've applied KCL around the left-hand loop to find Current from source (Is), but I'm stuck after that.

 
For starters, you need to know the PWL model of the diode. Not having this, I'll assume they're 3 identical real diodes, and we have access to the V-I graph on the datasheet.

Strictly speaking the precise answer has to be found by iteration, but a close-enough answer can be found quite simply. You can assume that Vout is set by the voltage drop across D3, so Vout=0.7 V, and from this I2 can be calculated. This would mean that the voltage drops across D1 and D2 are 0.35 V each, so to find I1 you need to refer to the PWL model (or the datasheet) to find the current at which Vd=0.35 V.

You should find that I2>>I1, but strictly speaking all 3 diodes are "on". Warning - this may not be the answer your teacher is looking for - it depends upon the definition of a diode being "on".
 
I think when you use the piecewise linear diode model, in introductory electronics classes, diodes with voltages below V(d,on) are usually considered OFF. Otherwise, the calculations for I1 and I2 become a little complicated.
 

PRS

Joined Aug 24, 2008
989
I wouldn't swear to D1 and D2 being off, but if this is true then D3 gets all the current from (2.5-.7)/500=3.6mA and vout is .7 volts. I(1)=0
 
Tch, what nonsense are they filling the kids' heads with these days? A diode has either forward, reverse or zero bias - what possible use are the concepts of on or off? I'm glad I was never taught that, it may have crippled my understanding of diodes for years.

The Shockley equation isn't a bad model at lower currents, though it does start to drift off a bit at higher currents.

For the purposes of the OP question, it looks like teacher expects the answer "only D3 is on".

Back in the world of real diodes, all 3 will be forward biased. If the diodes were all identical 1N4148s (Fairchild datasheet) then D3 will drop 0.7 V at 5 mA bias (I2). That means D1 and D2 will drop 0.35 V each, and looking that up on the V-I graph gives the current I1 as 5 μA. That's a workable current, which is certainly not "off", but neither is it "on" by the definition given. May I suggest a new term, "standby"...;)
 
I would agree with you Darren that it is a model with very little use. I also remember going through the classes where that was the first model that was given. The homework answers must follow according to the stated model, right or wrong.
 

Thread Starter

Student01

Joined Apr 15, 2009
35
To be fair, the course work does emphasize that the models supplied are approximations and that real diodes deviate from the assumptions of the model. Personally, I find the concept of a diode being on or off to be confusing when supplied with a circuit and told to solve. But I suppose simplifications are necessary when learning the basics.
 

JoeJester

Joined Apr 26, 2005
4,390
To be fair, the course work does emphasize that the models supplied are approximations and that real diodes deviate from the assumptions of the model.
Real diodes can deviate from other real diodes of the same device number. Real diodes can deviate from real diodes of the same manufacturer of the same device number. You do have a minimum specified in the data sheet.

For instance, the 1N4148 that was mentioned has a minimum 620 mV Vf and a maximum of 720 mV with a 5 mA input. Change the voltage source in your circuit to a 5 mA constant current source and see if the model falls within those parameters.

Your professor is looking for a specific answer given your knowledge base at the time of the course. Conducting (on) or not conducting (off) are the two extreme cases they are seeking. My simulation software showed about 2.63 uA of current in the "off" diode path with the voltage drop split across the two diodes. Your purely academic exercise given the parameters you were given and your knowledge base, based on the chapter you were studying the diodes were either on or off. Given those choices, D1 and D2 were off. If you stated there were a few microamps flowing in the D1/D2 path, you would not be incorrect, because you could have argued that position from a point of knowledge. Others in your class might not have had your "advantage". Reading too much into a question, especially in the learning stage you presently reside, would add to the confusion of your classmates.

They will see the "discrepancies" when they do the labs ... and probably have to explain those discrepancies when in the labs.

Good luck.
 

Thread Starter

Student01

Joined Apr 15, 2009
35
I strongly dislike having to explain discrepancies between lab results and the learned models. It always feels like guesswork to me as the only thing we've really been taught is the approximated models and that "this is close, but not really what happens."
 

JoeJester

Joined Apr 26, 2005
4,390
"this is close, but not really what happens" ... nice. How much are you and your classmates paying for such an answer?

That's much like to an answer I hated ... It's in the software.

Of course if you inquired ... the response would be ... That's beyond the scope of this course.

Keep your grain of salt handy. :D
 

Thread Starter

Student01

Joined Apr 15, 2009
35
"That's beyond the scope of this course."
I think I've heard that particular phrase mentioned up-tenth times in this unit. It depends who you ask. Usually lab instructors are willing to spend a little one-on-one time to satisfy a student's curiousity. Lecturers are not.
 
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