# How to begin calculation for common emitter using PNP

Discussion in 'Homework Help' started by Watershadow, Jul 6, 2011.

Jul 6, 2011
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I am not sure how to begin the calculations as I have not done pnp transistors before. Also I believe I have to start with the flow of current. Is VR2 a good place to start?

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2. ### blah2222 Well-Known Member

May 3, 2010
573
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You do it the same way as for the NPN, except current directions and voltages change.

For the case of the NPN, electrons from the Emitter are thrown into the base and sucked into the collector, causing electron flow to be from emitter to collector and conventional current to be collector to emitter.

The PNP is now the opposite. "Holes" are injected into the base from the emitter and sucked into the base. "Holes" show the direction of conventional current.

In the active mode, the BE junction must be ON, so for "holes" to flow from emitter to base, there must be a 0.7 V drop (silicon) from emitter to base.

Thus, Vbe = -0.7 V, unlike the case with NPNs that have a Vbe = 0.7 V. The CB junction must be OFF and have an electric field that is strong enough to pull "holes" to the collector from the base. So, Vbc > 0.7 V, unlike Vbc < 0.7 V for the NPN case.

So, the DC analysis changes a little bit, in terms of directions and signs, but the ac analysis is the exact same. Fluctuations in currents in an NPN or PNP result in the same net direction of change so you use the standard Hybrid-PI or T model for continuing the analysis.

Also remember that capacitors are open circuits in DC, so this simplifies the circuit to a four resistor bias network. I would start by looking at what's going on at the base. You know that in the active mode this PNP will have a Vbe = -0.7 V and you can make use of KCL at Vb.

Hope that helped.

3. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,906
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That's as good a place as any to start.

Other then the voltages having a different sign there isn't any difference between this and the NPN version.

Jul 6, 2011
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Thanks for the tips, but I am still slightly confused. Never done pnp before.

I tried beginning with VR2 which is VR2 = [R2 / (R1+R2) ]Ecc = 2,609V
since the voltage of VB is "reversed" to find VR4 instead of saying 2,609V " -" 0,7V I would be adding 0,7V to get 3,309V.
So ICQ = VR4/ R4 = 3,309/ 10 k = which is 0,33mA which seems to be incorrect (quite small)
Unless there is another starting point. Since all resistor values are given, I am told to start with the flow of current? So where does the flow of current start?
Help would be appreciated.

Dec 26, 2010
2,147
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You should have started by finding VR1, since the base bias voltage is related to the positive rail for PNP. This would be Ecc*R1/(R1+R2) - curiously you have got the right number, so you must have actually done the calculation that way.

VBE is not reversed relative to the base bias divider voltage. ALL the voltages are "reversed" together, so VBE still subtracts from the bias supply.

R4 is not the emitter resistor. Read the diagram carefully, and you will see that R4 is actually the collector load. The emitter feed is made up of R3 and R5 in series.

6. ### hobbyist Distinguished Member

Aug 10, 2008
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Since all the voltages are referenced to the neg. term.
Shouldn't it still be analyzed as [VB = ECC x(R2 / (R1 + R2)] ?

The only rerason I interject this is, what if a schematic has both NPN and PNP transistors in it, then they need to learn to analyze according to where the ground symbol is placed.

Am I correct in assuming this ?

Thankyou.

Dec 26, 2010
2,147
301
You can do that, but then you have to subtract Vcc to get the voltage driving the PNP, which has its common reference at Vcc.

To get to the answer with least effort, I would start by going straight for V(R1) which is (VB-Vcc) as you define it, because it is the voltage between the PNP and its common reference. Maybe it is safer for a beginner always to work with voltages related to ground, but I'm not too sure about that.

The OP seems to have thought that R4 was functioning like the emitter load, so perhaps the most important thing for him to grasp is that a PNP (in common emitter) has a positive reference.

8. ### hobbyist Distinguished Member

Aug 10, 2008
789
73
Yeh, I see what your saying..
I agree..

Thanks...

9. ### Duane P Wetick Senior Member

Apr 23, 2009
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A rule of thumb that I have used is this: Going more positive on the NPN transistor base turns it on...conversely, going more negative on the PNP base turns it on. I hope this helps.

Cheers, DPW [Everyone's knowledge is in-complete...Albert Einstein]