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 CIRCUIT


6 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 ohms


BUT 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 equation


But 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.

Attached Thumbnails

 

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?

Attached Thumbnails

 

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

Could you elaborate please?

I like Serena

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.

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 "+".


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 wanna get to the bottom of this so i will be more thorough and post the original exercise with my full solution. Im on the bus with 15% iphone battery so it might die any moment. Talk to u guys soon from my headquarters