Semiconductor Devices: PNP Biasing

In this video we explore biasing the PNP version of the BJT. We compare its operation to that of the NPN.
References: Semiconductor Devices: Theory and Application; Chapter 4, section 2; Chapter 5, sections 3 and 4.
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[@simonyoungglostog]

Thank you. That's the first video I've watched with NPN and PNP calculations on the same page.

[@khrystynastepura388]

Great video! You explain really well.

[@simonyoungglostog]

Still a great revision video. Thank you.

[@44mod]

I loved the video and could partly stay with you I am self taught. What video should I look at first in your playlist. I really like the way you explain everything

[@ElectronicswithProfessorFiore]

If you want to start at the beginning, I suggest the first DC Circuit Analysis video. Run through that playlist and then do the AC Circuits list. That will give you all of the background you'll need for any of the other lists.

[@44mod]

@@ElectronicswithProfessorFiore I have already started them. Thank you so much again. You are a blessing to me and I thank you again.

[@Chris-hi2hn]

Is there a rule of thumb for choosing RE and RC to compensate for a Beta that varies?

[@ElectronicswithProfessorFiore]

That depends on the bias used. For high bias stability, make Re around the same size (or larger) than the base resistor(s). The value of Rc will depend on the bias voltage and what you need for a gain.

[@WayneLambrightIII]

I think I found an error VCE which is 6.4 minus -0.7, should be 5.7. Could you confirm that your math was incorrect on the VCE with the NPN.

[@ElectronicswithProfessorFiore]

Ve is negative. Subtracting a negative is adding. 7.1 is correct.

[@willwill1738]

Thanks for your excellent kind work and a dedicated website with related books. I just came across and it is a treasure.

[@arkufahl]

I feel like something is off with the PNP calculation at 8:30. The 0.7v drop is "across" the emitter/base, not before it. So you can't use the 0.7v drop to calculate the drop across the 1K resistor. I can't simulate your circuit on a breadboard, because I'm a novice and don't know how to make negative voltages. However, when I setup a PNP circuit and measure... the voltage drop across 1K, plus the voltage drop across 3K, plus the 0.7v drop equals my source voltage... meaning, 1K voltage drop plus 0.7v can't equal the source voltage... there has to be some voltage left for the 3K resistor to drop, meaning 1K voltage drop has to be less than what you are showing. Right? What is it?... I'm not sure, but I'm pretty sure it isn't source voltage minus 0.7v. Help me understand.

[@ElectronicswithProfessorFiore]

First, notice in upper right it says "Approx: Vb = 0".
The issue here is that the drop across Rb is small enough to ignore, therefore we can approximate that the 5V supply drops across the 1k and Veb. We know Veb = 0.7 ("across", just as you said), leaving 4.3V for the 1k. Now, why is the drop across Rb small enough to ignore? That's because the drop is Rb times the current through it, which is Ib, and Ib is beta times smaller than the collector or emitter current. Seeing that Rb is around the same size as Re, it should be obvious that, with a beta of 100 or so, Vrb will be at least an order of magnitude smaller than the drop across Re.
So, yes, the supply is spread across three things (Vre, Veb, Vrb), but that last one is so tiny that we can ignore it and still get decent accuracy. A more precise analysis would show that the current is slightly smaller than this approximation, but that deviation would be smaller than the deviation you'd get from resistor tolerances in lab.

[@arkufahl]

Okay. I'm understanding the 0.7v drop placement now, and how the 4.3v drop across Re is derived. Thank you.
I also get why you are discarding the drop across Rb. However, for me, its only insignificant because of the value shown for Rb (3K). If that were to be, say 100K, its not insignificant (to me). So its a personal preference that I don't like seeing values ignored (as opposed to being calculated), because it "could" be significant depending on the circuit. Again, just my opinion. But feel free to correct me if I'm wrong.
Beta = Ic/Ib, so it can't be said that Ib is beta times smaller than Ic and Ie. Yes, Ib is "smaller" than Ic and Ie, but "beta smaller" than Ic only (I've seen it shown that Ie/Ib = alpha. So, Ib is "beta smaller" than Ic, and "alpha smaller" than Ie, but not "beta smaller" than both). Again, I think they are "close" in your example circuit, but its not the same calculation for both.
While I'm babbling about things... I think the approximations are what is throwing me off... For example, in reference to the NPN circuit... saying Ie and Ic are approximately the same is only "true" if Ib is a small current (in the microamp realm), which isn't always the case. For example, I am looking at one of my own circuits that has an Ib of 2.1mA, and an Ic of 2.83mA, resulting in an Ie of 4.93mA. So, Ic and Ie are not anywhere near approximately the same (Ie = Ic + Ib, Ic = Ib * beta). What I mean is that stating they are approximately the same is only "true" sometimes... Again, more of a personal preference to state the calculation (Ie = Ic + Ib) which is always true.
Thanks again for talking this through. I appreciate it. And again, feel free to correct anything I said that was incorrect... I'm just trying to learn. Take care.

[@ElectronicswithProfessorFiore]

@@arkufahl In earlier work it was shown that there are three regions of operation for a BJT: saturation, linear (constant current model), and breakdown. Circuits like this are operating in the linear region, and therefore we can expect there to be a decently large beta (or, if you prefer, an alpha close to unity). No amplifier bias circuit like this would be set up in saturation, so you'd never see a ratio of Ic/Ib of three, or similar. It will be the case that Ic and Ie are approximately the same. We use these and other approximations because, within these sorts of applications, we can calculate values with high accuracy in much less time than a more precise method. Bottom line, why spend three or four times the effort to produce an "improvement" that will be less than the variation caused by the tolerance of a single component?