# NPN Transistor voltage calculation

For the B) part, I don't get why is it that for calculating voltage between collector an the emitter (Vce) I take into account diode on base (0.7V), whereas for part A) it's ignored?

Why is Vce in part A and part B are calculated differently? Why is it that in part A voltage of 0.7V isn't taken into account? I get that if βf = infinite, and βf = ic/ib ib -> 0

• Your title mentions a PNP transistor, but your circuit only has an NPN one. Can you edit to clarify? Apr 2, 2017 at 4:32
• Vbe in part A is not ignored. It is simply computing Vce using a different mesh. With this model, when beta is finite, you can use ic=beta ib to find ic, and then use KVL at the output mesh to find Vce. When beta is infinite, you need another way to find Ic - so, you first find Vce and then deduce Ic. Apr 2, 2017 at 6:35
• @SredniVashtar funny, I didn't realize that Vce can be calculated either using left loop (base) or the right loop (collector). Turns out if you calculate Vce in part A using left loop (applying KVL): Vce = 103 * 10^(-6) * 10K + 0.7 = 1.73 V
– Jack
Apr 2, 2017 at 18:11
• Use more digits in the first calculation and you'll find 1.7295... instead of 1.728. Welcome to the world of approximate calculations :-) Apr 2, 2017 at 19:54

Your answers look great. I don't see a problem with them. They are even rounded, appropriately. (Except, perhaps, the part A collector voltage.)

In both cases, you are accounting for the diode at the base. In part B, it's simply that there is no additional voltage drop across the $10\:\textrm{k}\Omega$ resistor. So the collector is effectively tied to the same voltage as the base in part B. In part A, there is a voltage drop across the $10\:\textrm{k}\Omega$ resistor, so that voltage drop is added to the $700\:\textrm{mV}$ of the base-emitter diode. Part A doesn't ignore it. It simply adds to it a voltage drop needed for the resistor. The value of $V_{CE}$ is different in the two parts because it has to be different: in part B there is no voltage drop across the base resistor while in part A there is a voltage drop across the base resistor. The collector voltage has to be adjusted to account for this difference. But in neither case is the diode voltage ignored.

Check it out. You computed a base current of $103\:\mu\textrm{A}$ (rounded.) Multiply this base current by the value of the base resistor, $10\:\textrm{k}\Omega$, and you'll get about $1.03\:\textrm{V}$ (rounded.) That's the added voltage drop across the base resistor. Now, add the $700\:\textrm{mV}$ diode voltage to this value and you get your collector voltage of $1.73\:\textrm{V}$. Very close to the answer you gave in part A (which should have been rounded to the same number of places.)

So part A doesn't ignore it. Neither does part B.

The reality of a BJT is a little different. The next improvement to your approximate viewpoint is to realize that the collector current and the base voltage are related by something called the Shockley diode equation. So the "diode voltage" isn't a fixed value, as you assumed.

That equation uses some model parameters you weren't provided: an emission co-efficient (usually just 1) and a temperature-dependent saturation current. Plus, two physical constants (Boltzmann's constant and the constant representing the basic unit of charge) that, with temperature, provide a "thermal voltage" value resulting from large-number statistics applied to the energy distributed over the degrees of freedom at the atomic/molecular level.

There are still further improvements before you reach the very first plateau of BJT models. These improvements become the first level of the Ebers-Moll model. There are two further Ebers-Moll models before the Gummel-Poon model arrived. Then further improvements to that model, before the VBIC model arrived. Etc.

For most practical uses of discrete devices, you don't even need to get to the first level of the Ebers-Moll model. Just the Shockley diode equation is often enough for most practical applications for discrete devices. Not so, for IC design use. But at the level most of us normal mortals live in, it's enough. However, it's too often not enough to ignore the Shockley equation. So it would be good to take note of that much when you get the time to consider it.

I'd be remiss to fail to mention one more detail that you might not otherwise pick up from the above. The saturation current I mentioned is temperature dependent. So is the thermal voltage. The first has an opposite sign, so that increases in temperature result in a decrease of the saturation current. The second has the same sign, so increases in temperature result in increases in the thermal voltage. The Shockley diode equation almost never shows the temperature dependence of the saturation current, though. But it does often expose the thermal voltage, since it's a lot easier to remember and more fundamental, physics-wise. But the saturation current's sign overwhelms the sign of the thermal voltage in the Shockley equation, so that the resulting change in a BJT's diode voltage is negative -- not uncommonly resulting in values around $-2.2\:\frac{\textrm{mV}}{^\circ K}$. Just looking at the thermal voltage alone would give you nearly the opposite impression. (The equation for the saturation current involves a function of $T^3$, though even the power itself there is also a "parameter" that varies a little and can be as high as 4, or so.)

I add the above because the question comes from the physics stack exchange and it seems appropriate to give you some notice.

• "So the collector is effectively tied to the same voltage as the base in part B". But it's not, is it? Collector has 6V voltage source. How can be the node at collector be the same as base node?
– Jack
Apr 2, 2017 at 18:10
• @Jack In B, the base resistor doesn't have any current through it, so no voltage across it, so the base is at the same voltage as the collector. In A, there is base current through the base resistor, so there is a voltage across it and this has to be added to the base voltage itself to get the collector voltage. In neither case is the collector tied to the 6V source -- there is the $680\:\Omega$ resistor there. But it drops a different voltage in the two cases because there is a different current through it. Just look at your answers, to see that. It all works out.
– jonk
Apr 2, 2017 at 21:19
• @Jack If you are into physics (and I can't tell, all I know is that this question came from that group), then I recommend "Matter & Interactions" to you. You will gain a thorough understanding of electric fields, charges, and atomic level details of why current flows at all, etc. It's a very, very good book if you want to know how things work, in detail. No electronics book I've read comes close. You'll understand details few practical electronics engineers know. It's a good and very readable tour de force. Nothing like it.
– jonk
Apr 2, 2017 at 21:22

A) Vbe is used to calculate Ib and thus Vce = V+ - βIbRc is in the linear region

B) As β goes to ∞ and Ib goes to 0 then it no longer is linear and Vce is saturated where in this case Vce(sat) = Vbe or Vbc=0 so different rules apply.

## Side note

B) is purely hypothetical , because when Vce goes towards saturation β drops as much as 10% of the linear value and datasheets rate Vce(sat) for some rated current. Vce(sat)=0.1V is never reached with the negative feedback from Collector to base or ANY value of β.

• why different rules? So what if it's saturated? Why Vce in part B isn't 6v - voltage across 680 ohm R? I guess, why does Vce differ in part A and B? Just because Ib is 0 in part B?
– Jack
Apr 2, 2017 at 17:55
• Saturation is defined by Vbc=0 but β starts to reduce rapidly below Vce=2V. Why different rules? because with this config with negative feedback Vce cannot go below Vbe with infinite β and Vce rises as β specs drop. Vce(sat) is given here as =0.1 which is almost always rated at Ic/Ib=10 Note in my formula which is same as question but simplified, you cannot multiply ∞ *0 ( β * Ib ) in this hypothetical question. Apr 3, 2017 at 1:55

For the B) part

since beta = infinity, Ib = 0, and Vce = Vbe + 0 * 10K = Vbe.

the rest is simple.