PNP transistor and argument with my teacher

[Femme_physics]

I just had arguments with my teacher about my test answer in class. I'm a little upset, but maybe I'm acting like a spoilt brat (I'll be the first to admit it) I went out and I'm using a hallway computer in my college so I can't post images. I still disagree with him-- But I'll try to explain it as best as I can.

Say I have a PNP transistor. I have a source of 6 volts. I have an unkown resistor RE, and a currect Ie = 1.2 mA. I am also told that VBE = -0.7 volts. VBE is connected to the ground.

THE FOLLOWIG IS A DRAWING OF ONLY THE LEFT SIDE OF THE CIRCUIT6 volts ----/\/\/\/\/\-------(EMITTER--->BASE)----GROUND

(the jagged lines describe the resistor RE)

Consider the right side of it (with the collector) irrelevent. Both me and my teacher did.



So, in the solution, can I just do:

Sum of all voltages= 6 - ReIe - 0.7 = 0
5.3 = 0.0012 x Re
Re = 4416.6 ohmsBUT my teacher say I should write VBE in plus in my equation! That since it's been defined as -0.7 volts, it's a mistake to write it in minus in the equationBut that doesn't make sense to me. Vbe is supposedly an electrical component, not a voltage source. There must be a voltage drop there, it can't act as a votlage source!
Please help me see the light!PS If anyone could tell me with explanation who is wrong and who is right (his explanation that it can act as a voltage source is weird to me!) he should still be in class in the next 2 hours...otherwise I'll talk to him next week

 

[I like Serena]

Hey Fp! :)

Looks to me like a misunderstanding.

What you write looks correct to me.
Indeed the transistor is not a voltage source and you'll have voltage drop from E to B.Apparently VBE is the voltage from E to B, which is negative.
If you write it down as a formula, you should have:

ƩV = +6 - ReIe + VBE​

Note the plus in front of VBE.
Is that perhaps what your teacher meant?
Did you perhaps write this down with a minus in front of VBE?If you fill in the number for VBE, you get:

ƩV = +6 - ReIe - 0.7​

which is what you wrote down and which is correct as far as I can tell from this.(Btw, VBE in a similar situation with an NPN transistor would be +0.7 V.)

 

[Studiot]

Do you mean diagram 1 or 2 in my sketch?

As you have described it the transistor would burn out.

 

[Femme_physics]

Let's stick to just PNP - how come you set it as plus, but plug it in minus? And how come in NPN it is plus?

 

[maimonides]

A pn junction will conduct for Vnp > 0
A np junction will conduct for Vpn < 0
Hint: You will also have to think about the polarity of your power supply and about sign conventions for currents.

 

[Femme_physics]

Diagram B

 

[I like Serena]

As I understand your description, you have the following diagram:

Is that what you intended?

 

[Femme_physics]

Correct ILS, plus the the fact that current Ie flows downwards from the resistor ( as it should be ) and equals 1.2 mA

 

[Femme_physics]

A pn junction will conduct for Vnp > 0
A np junction will conduct for Vpn < 0
Hint: You will also have to think about the polarity of your power supply and about sign conventions for currents.

Could you elaborate please?

 

[I like Serena]

Correct ILS, plus the the fact that current Ie flows downwards from the resistor ( as it should be ) and equals 1.2 mA

Okay, okay. I added the current. :)
What about my earlier post?

 

[Femme_physics]

I read your earlier post wholly through and replied to it. Its right under Studiot's first post

 

[Studiot]

OK, taking ILS diagram now makes sense

V_be = -O.7 means The V_b - V_e = -0.7

that is (0) - (+0.7) = -0.7

Which is as it should be.

 

[Femme_physics]

@ Studio - So in KVL you plug it counts as addition to the 6 volts like my teacher proposed?

 

[I like Serena]

Oops. I didn't recognize it as an answer to my post.

Let's stick to just PNP - how come u set it as plus, but plug it in minus?

This is about sign conventions.

V_BE could be either V_B - V_E or V_E - V_B.
From your context I deduce that the sign convention your teacher is using is that:

V_BE = V_B - V_E​

With V_B = 0 [V] and V_E = +0.7 [V], it follows that V_BE = -0.7 [V].

With this sign convention, since you already know that in your formula you need -0.7, you need to plug in V_BE as "+".

And how come in NPN it is plus?

Using the same sign convention, in a similar NPN circuit you have:

V_BE = V_B - V_E = 0.7 [V] - 0 [V] = +0.7 [V]​

That is, Vb must be higher than Ve (contrary to a PNP transistor).

 

[Studiot]

If you got that, good, because my next statement may make sense and may not. Tell me if it doesn't.

You are mixing up voltages measuresd across a component with voltages appearing at particular points in the circuit.

The voltage across the base emitter is -0.7 volts (because the transistor is pnp)

but the voltage at the emitter is +0.7 volts in relation to the circuit reference.

You have to decide if you are going to use voltage drops or voltages at circuit points and then use them in a consistent manner.

Obviously the voltage drop = difference between the voltages at two points.

However we normally count this positive going up the circuit and negative going down.

Are you still with me?

 

[maimonides]

I was referring to the fact that pnp circuits often have negative supply, so ground is the highest potential.
And, as the others have pointed out, Vbe = -Veb

 

[Femme_physics]

ILS - in ur first post u said I was correct, now from what I read out you say I'm wrong... Originally you said I write it in plus in KVL but plug as minus... I get your voltage explanations of VBE = vb - ve ... What i don't get is how I plug it in the formula for a KVL loop where i denote the voltage drops... Up till this day, all my voltage drops were minused UNLESS it was an opposite voltage source or the current was flowing the opposite direction. I'm still confused of whether I did right or not...or whether my teacher was correct. Ill get home in 15 mins and i could use my scanner... I want to get to the bottom of this so i will be more thorough and post the original exercise with my full solution. I am on the bus with 15% iphone battery so it might die any moment. Talk to u guys soon from my headquarters

 

[I like Serena]

What you wrote in your opening post is all correct.

I already said it might be a misunderstanding with your teacher.
What formula exactly did your teacher disagree on?

 

[Studiot]

This is about sign conventions.

VBE could be either VB - VE or VE - VB.

I would be interested, where have you seen the second 'convention'.

I have only ever seen 'first minus second' in electrical circuits.

This is then consistent with the zero referenced node voltages

thus V_b = V at b minus zero and is positive if V_b is more positive than zero.

I would imagine the other convention to be a nightmare to work with.

 

[I like Serena]

I would be interested, where have you seen the second 'convention'.

As you know, I'm not an electrical engineer.
So I can't say for sure if the second convention is used in electrical engineering.
I'd have to research it to find out.
(But I find it hard to believe that all electrical engineers world wide use the same convention.)

I do know that in different fields of science different conventions are used and also that different conventions are used in the same field.
I have learned to always check which convention is used and not assume anything.
Often the convention must be deduced from the context.

As for other scientific fields with a different convention, the first that springs to mind is vector algebra:
$\vec{AB} = \vec B - \vec A$.

 

[Femme_physics]

Finally home.

I can scan and write in more details.

Let me start with the replies:

I would be interested, where have you seen the second 'convention'.

I have only ever seen 'first minus second' in electrical circuits.

This is then consistent with the zero referenced node voltages

thus V_b = V at b minus zero and is positive if V_b is more positive than zero.

I would imagine the other convention to be a nightmare to work with.

I have also always only seen the "first minus second convention"

If you got that, good, because my next statement may make sense and may not. Tell me if it doesn't.

You are mixing up voltages measuresd across a component with voltages appearing at particular points in the circuit.

The voltage across the base emitter is -0.7 volts (because the transistor is pnp)

but the voltage at the emitter is +0.7 volts in relation to the circuit reference.

You have to decide if you are going to use voltage drops or voltages at circuit points and then use them in a consistent manner.

Obviously the voltage drop = difference between the voltages at two points.

However we normally count this positive going up the circuit and negative going down.

Are you still with me?

I believe so. If I get to choose though whether I'm going to use voltage drops or voltages at circuit points (I believe we did the latter when we just started studying electricity) I prefer to choose voltage drops-- seems faster. You said "You are mixing up voltages measuresd across a component with voltages appearing at particular points in the circuit."

Not really. I know for sure that the voltage drop FROM ve to the ground is 0.7. Whereas Ve must be 0.7 and Vground = 0 ... I know what every point equals to. I reckon.

I was referring to the fact that pnp circuits often have negative supply, so ground is the highest potential.
And, as the others have pointed out, Vbe = -Veb

Oh yes, duly noted that.

------------------------------

What you wrote in your opening post is all correct.

I already said it might be a misunderstanding with your teacher.
What formula exactly did your teacher disagree on?

I'm not sure which misunderstanding.

Let me be as clear as I can be, I have the scanner so let's start with the ENTIRE exercise:

http://img403.imageshack.us/img403/2739/theexercise.jpg

My full solution (and my teacher solution below it):

http://img502.imageshack.us/img502/5513/solutionstuff.jpg

 

[Studiot]

"You are mixing up voltages measuresd across a component with voltages appearing at particular points in the circuit."

Not really.

That's good

You need voltage drops for KVL so you have to perform subtractions if you calculate these drops from the circuit voltages at circuit nodes. It is important to keep a consistent direction for these.

I see you are studying the common base configuration.

 

[maimonides]

I hope i´m not heaping confusion on confusion, but:
Kirchhoff can be done in two ways: Voltage drops negative and sources positive
or the other way round (see: http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws).
And then U=RI and current direction and sign conventions come in.

 

[I like Serena]

Let me keep it simple: your solution is correct and what you wrote down as your teacher's proposal is wrong.

 

[Femme_physics]

http://img402.imageshack.us/img402/9233/thedifference.jpg

It's soundproof logic!

 

[I like Serena]

It's soundproof logic!

Yep!

And it matches with the convention that VBE = Vb - Ve.

 

[Femme_physics]

Studiot, maimonides, just to make it more than official-- do you also agree with ILS that I was right and my teacher was wrong? I argued with my teacher in class for 20 minutes! And then after class for another 15 minutes!

Yep!

And it matches with the convention that VBE = Vb - Ve.

As it should be!

 

[maimonides]

Your drawings in posts 21 and 25 are perfectly ok, as far as I can tell anything at all.

 

[Femme_physics]

Thanks, maim. Studiot?

 

[Studiot]

Your value for R_e and analysis are correct. There is a forward voltage drop across both the emitter resistor and the emitter base junction for Kirchoff purposes.
I would not use K myself, however. Just a simple voltage divider. Then the issue would not arise.

Well done.

 

[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)?