PNP transistor and argument with my teacher

[I like Serena]

I argued with my teacher in class for 20 minutes! And then after class for another 15 minutes!

Hmm, if an argument lasts that long there's usually a misunderstanding of some type.
Or someone just doesn't want to admit to being wrong, talking around the topic and still seem right.
Is your teacher the type that never wants to admit to being wrong?

 

[Ratch]

To the Ineffable All,

My goodness, we are sure getting wrapped around the axle on the polarity of the emitter-base junction. Let's just say that for a PNP, the emitter will be more positive than the base if the transistor is in its active region and conducting. So it doesn't matter what direction you do a KVL as long as you remember that.

Ratch

 

[I like Serena]

Please help me see the light!

Do you see the "light" now?

 

[Femme_physics]

Yes! Not only that but my classmate ran a simulation for me:

http://www.falstad.com/circuit/#$+1....5 R+240+112+240+96+0+0+40.0+6.0+0.0+0.0+0.5 Further confirming the result. He also ran it on another simulator and we get the same Re. :) In addition I printed those two simulators results and I am printing a short little summary of this discussion and replies here to further finalize my claims.

 

[Femme_physics]

May everyone who posted here tell me please what are their degrees (or relations to electronics), or is it too personal of a question? I know the degrees of those who has it posted on their profile

 

[Studiot]

I suggest that cooling things with your lecturer would be a good idea.
You will go on to greater things he will be stuck where he is.
You might win the battle but you will certainly loose the war.

 

[Femme_physics]

You're right, I know you are... you're absolutely right in fact, this issue got me worked up though, it's 1:15 A.M. here and I can't go to sleep :( , and I got to wake up at 5:30 to catch a morning bus for my shift. I hope it'll wear down soon so I could get a few hours of sleep. You'll have to excuse me for belaboring the point here. I do have all the evidence I need with the simulators printed and the replies here (also printed). That's enough. Don't worry, I'm only worried about getting justice across, and making sure I get a 100 on the test (Because I was correct about everything according to the class review he did today-- including this!). Many thanks, btw, to the people of this forum. I will do it gracefully. :) I'm a chick after all :P

 

[Studiot]

Sleep well

:zzz:

 

[Antiphon]

I have two degrees in EE, a BS and an MS. I have been designing and building circuits for decades. My circuits are on military ships at sea as we speak.

There is a very simple convention here.

When you are confronted with a voltage of the form Vab, you connect "the red" probe to "a" and the "black" probe to "b" of your hypothetical voltmeter.

For a properly biased NPN, Vbe is positive. For a PNP it's negative.

But don't believe me- believe the manufacturers data sheets. The 2N3904 and 2N3906 are the prototypical Phillips and flathead transistors of the world. Take a look at Vbe for each.

http://www.makershed.com/v/vspfiles/assets/images/2n3906.pdf

http://www.fairchildsemi.com/ds/2N/2N3904.pdf

 

[Femme_physics]

I believe you, Antiphon.

I have a followng question, just for better understanding. If this value Vbe = -0.7 volts was given to an NPN transistor in the same scenario, would then my teacher answer be correct?

And, also, does it even make sense to give Vbe value for an NPN transistor in minus?

 

[I like Serena]

Hmm, forget about a given VBE of -0.7 or +0.7 for a moment. We'll get back to that.
Just to improve understanding, suppose in your circuit the PNP transistor is replaced by an NPN one, leaving everything else the same.

What would VBE become?

 

[Femme_physics]

Let's see, VBE = -0.7

Vbe = Vb - Ve = 0 - 0.7 = -0.7 volts

Veb = Ve - Vb = 0.7 - 0 = +0.7 voltsSame result. Doesn't matter NPN or PNP. If Veb had been defined as -0.7 volts, THEN and only then my teacher answer would be correct

 

[I like Serena]

Uhh... if you replace just the PNP transistor by an NPN one, VBE becomes something entirely different...
Visual queues or something? Does anything *bump*?

 

[Femme_physics]

I'm drawing it in my notebook but I ain't seeing anything different...

Ratch said that of a PNP transistor

the emitter will be more positive than the base if the transistor is in its active region and conducting.

In NPN it appears to be opposite. Yea, you're right ILS, since the base is more positive than the emitter in NPN's, the emitter to the ground will be -0.7 and therefor the voltage that would fall on Re would have been 6.7 volts

 

[Jony130]

If Vbe is equal -0.7V for NPN transistor. This means that NPN transistor in is cut-off.
And we have a similarity situation for PNP. If Vbe = 0.7V for PNP then PNP BJT is in cut-off.

 

[I like Serena]

No... with NPN the transistor won't be in its active region and it won't be conducting... the current *bumps*...

 

[Femme_physics]

At cutoff the transistor does not allow current hence the premise that Ie = 1.2 mA is false to begin with, but the voltage on the resistor Re would still be 6.7 volts

 

[I like Serena]

Yes, the premise that Ie=1.2 mA would be false. It would be 0 mA. Good!

But the voltage across Re would not be 6.7 volts.
What would it be?

 

[Femme_physics]

I still see 6.7 volts...

Soundproof logic #2!

http://img140.imageshack.us/img140/4461/harto.jpg

 

[I like Serena]

Do you agree that Ie=0 mA?

If so, can you apply KVL (and don't use that VBE=0.7 V because it isn't)?

 

[Femme_physics]

Yes, I agree that Ie = 0 mA

If so, can you apply KVL (and don't use that VBE=0.7 V because it isn't)?

No, VBE is -0.7, we agreed.

I'd just do

Sigma V = -0; 6 - 0 x Re - ...oh

Ok, nevermind, I see what you're talking about.

We got 0 voltage drop at Re! Thanks :)

 

[I like Serena]

Yep!

So what will VBE be then?

 

[Femme_physics]

VBE is the rest of it...6.7 Volts!

 

[I like Serena]

Almost.
You haven't applied KVL quite correctly yet...

 

[Femme_physics]

Sigma V = -0; 6 - 0 x Re - Vbe = 0

Vbe = 6 volts

 

[I like Serena]

Again... almost.
One thing left: the sign of Vbe.
Note (from the note master ) that Vbe = Vb - Ve.
What is Vb and what is Ve?
And what is therefore Vbe?

 

[jim hardy]

I am with Femme on this. Her teacher made a mistake.

My high school teacher (John B Davis) taught us boys " When they give you voltage between two leads, as in " Vbe" , your voltmeter red lead (positive) goes to the first letter and the black one (negative) to second letter".

I note that Fairchild datasheet for 2N3906, a common pnp, gives Vbe in operating range as ~ -0.65 volts, note minus sign.
That means a meter connected as above would find base negative wrt emitter, which is forward biased(emitter positive wrt base).

VBE(sat) Base-Emitter Saturation Voltage IC = -10mA, IB = -1.0mA
IC = -50mA, IB = -5.0mA -0.65(min) -0.85(typ) -0.95(max) V

Now here's a confusion factor - Same datasheet gives emitter to base reverse breakdown voltage as also negative,

V(BR)EBO Emitter-Base Breakdown Voltage IE = -10μA, IC = 0 -5.0 V

NOTE letters reversed for this parameter VEB not VBE . The O means is collector open circuited.
So this time a meter connected as Mr Davis told us would find emitter negative wrt base, reverse biased, which is what that parameter specifies - reverse withstand capability of EB junction.

It's one of those little points of technique that come from playing with real parts instead of simulated ones.

The conventions usually work if one is rigorous.
Only way her teacher is right is if that PNP has suffered an internal Collector to Emitter short . IMHO.

datasheet at http://www.google.com/url?sa=t&rct=...K0hKshhDb60zBONXQ&sig2=cupCh8ZbZ7737zChEnpyeQ

 

[Jony130]

Now here's a confusion factor - Same datasheet gives emitter to base reverse breakdown voltage as also negative, NOTE letters reversed for this parameter VEB not VBE . The O means is collector open circuited.
So this time a meter connected as Mr Davis told us would find emitter negative wrt base, reverse biased, which is what that parameter specifies - reverse withstand capability of EB junction.

It's one of those little points of technique that come from playing with real parts instead of simulated ones.

But I think that Vbe would be even more confusing then Veb.

And this is why manufactures use Veb instead of Vbe.

 

[I like Serena]

I have a followng question, just for better understanding. If this value Vbe = -0.7 volts was given to an NPN transistor in the same scenario, would then my teacher answer be correct?

And, also, does it even make sense to give Vbe value for an NPN transistor in minus?

To get back to your question.A transistor is never a voltage source, and it can't make 6.7 volts out of 6 volts.

However, VBE can have the "other" sign, but then it's in blocking mode as you just saw (and the current is zero).

If the transistor is in conducting mode, you will have a voltage drop of 0.7 volts in the direction of the arrow.

 

[jim hardy]

But I think that Vbe would be even more confusing then Veb. ]

I don't know why they do it, seems to me they could have just reversed the sign in datasheet instead of swapping the letters around..

And this is why manufactures use Veb instead of Vbe.

If you look at Fairchild datasheet for 2N3906 at
http://www.fairchildsemi.com/ds/2N/2N3906.pdf
you'll see they use V_EB in the "ABSOLUTE MAXIMUM RATINGS" and "OFF CHARACTERISTRICS" sections

but they use V_BE in the "ON CHARACTERISTRICS" section !
Both are negative.


The test question showed VBE = -0.7V
which by the convention i was taught would mean emitter positive wrt base, hence positive wrt circuit common , forward biased
so teacher should have written KVL as:
0 = 6 - 0.7 - ReIe ; traversing CCW around loop
and not not:
0 = 6 - ReIe + 0.7



Hmmm for sanity check i looked at OnSemi's datasheet -
they use same notations, V_BE and V_EB but both are positive !
http://www.onsemi.com/pub/Collateral/2N3906-D.PDF
Femme - that should REALLY confuse things !
You might print those datasheets and show to teacher - it's something students ought to be made aware of.

old jim