Finding the Base-Emitter Voltage in an NPN Transistor

Question

The problem asks to find the the real value of base-emitter voltage, \$V_{BE}\$ and the corresponding \$I_C\$, and \$V_{OUT}\$ in the figure below:

\$V_{CE(sat)}\$ is given but the transistor is not necessarily in saturation mode; in fact, it's likely operating in the forward active mode so I need to solve for the actual value of the junction voltages first to determine its mode of operation. Only \$V_{CC}\$ is given and I have too many unknowns. I've tried using circuit analysis, KVL, and KCL techniques in addition to the transistor equation but there are more unknowns than there are equations that I can think of. Any idea? I just need enough equations. Thanks!

Answer 1

We have two equations for two unknown quantities (Vbe and Ic):

IC=beta(Vcc-Vbe)/RB and Ic=Is*exp[(VBE/VT)-1].

(For normal operation in the active mode the exponential expression is much larger than "1" - hence, we can neglect the "1").

• An exact solution is possible (graphical solution) if we plot both functions Ic=f(VBE). The point where both curves meet is the actual operating point (Ic and VBE).

• An exact numerical solution in one step is not possible (because of the exponential function). However, the two following alternatives exist:

(a) Iterative solution: Start with VBE=0.65 volts and verify - using both functions - if this value was too large or too small (and try a second run).

(b) Replace the exponential function by the first (two) part(s) of the corresponding power series: exp(x)=1 + x + x²/2! + ......This approximation allows a direct but approximate numerical solution (mathematical combination of both functions).

Answer 2

Before the equations, its useful to know this:

Vbe will be near 0.7 volts if Ie is near 1mA

Vbe will be near 0.52 volts if Ie is near 1uA.

Vbe will be near 0.34 volts if Ie is near 1nanoAmp.

Vbe will be near 0.16 volts if Ie is near 1picoAmp.

Note every factor of 1,000 less Ie results in (3*0.06volt) less Vbe.

Every factor of 2.718... results in 0.026 volts change in Vbe; the 0.026v comes from [ Q/K*T ]

Every factor of 2X results in 0.018volts; you will see 0.018v or 0.036v or 0.054v buried insider bandgap reference designs.

All these numbers are valid at room temperature; these voltages change by approximately -2milliVolts/degreeCentrigrade; that is, higher temperature results in lower Vbe (and Vdiode, if you measure a simple diode).

Answer 3

To solve this problem, first assume that the transistor is in active mode. For this Vbe should be 0.7 V.

So you can calculate the current flowing through Rb/base now.

You know beta already, so you can calculate the collector current, therefore the collector emitter voltage.

Based on our initial assumption, the transistor is in active mode. Therefore, Vc should be greater than Vb if it was correct. If not, start from beginning assuming it is in saturation. (Vce = 0.2V). Good luck.

Answer 4

Any idea? I just need enough equations.

The base side is essentially Rb in serial with a diode. Assuming reasonable Ic / Rc, you can solve it either graphically or numerically, by iterating assumed Vbe -> Ib -> Ie -> new Vbe -> new Ib .... you can also start with Ib as well.

This approach assumes a couple things that can be problematic in some cases:

1. constant beta: in reality, beta goes down as Ic goes down and up (at two extremes).

2. Vce remains reasonable, typically 10 - 20mv but hopefully 100mv or more. otherwise, the transistor is in saturation.

a more interesting case is one where the upper end of Rb is tied to the collector, introducing an element of negative feedback to stablize the dc working point.

the approach proposed by LvW is incorrect - there is no basis for the first equation.

Answer 5

When connected to ideal voltage sources, then assume that they can sink and provide enough current to maintain their voltages. That means \$V_{B}\$ is \$V_{IN}\$ and \$V_{E}\$ is GND. So \$V_{BE}\$ is simply \$V_{IN}\$.

Now for \$V_{OUT}\$, which \$V_{CE}\$, which is of course not the same as \$V_{CESAT}\$, which is typical \$V_{CE}\$ at saturation.

$$V_{CESAT}$$ $$if \quad B_{F}(V_{IN} - V_{BESAT})/((B_{F} +1)(R_{PB} + R_{PE})) > (V_{IN} - V_{BESAT} + V_{CESAT})/(R_{C} + R_{PC});$$

$$V_{DD} - (R_{C}+R_{PC})B_{F}(V_{IN} - V_{BESAT})/((B_{F} +1)(R_{PB} + R_{PE}))$$ $$if \quad otherwise$$

\$R_{PB}\$ is parasitic base resistance, \$R_{PE}\$ is parasitic emitter resistance and \$R_{PC}\$ parasitice collector resistance. These are usually 0.5-1.5 ohms, so only useful to take into account when BJT is wrking more in current mode. Such as now that base and emitter are bot connected directly to sources.

For even more accurate calculation, the \$V_{CESAT}\$ and \$V_{BESAT}\$ are just the minimum and maximum boundaries, taken as a single entity, that actual \$V_{CE}\$ and \$V_{BE}\$ can take at saturation given the current formulas from Ebers-Moll or Gummel-Poon models. This may not be as accurate as starting with said 2 previous models and using Convergence, Numerical Methods or Symbolic Engines, but this is good enough to determine the regions and modes the transistors are in.

EDIT:

By Ebers-Moll model, I do not mean the regular this: https://wikimedia.org/api/rest_v1/media/math/render/svg/e95565b2aa76041d3124461d12091effe1afe96a

I meant the complete: https://wikimedia.org/api/rest_v1/media/math/render/svg/4062fe7275c023cf696f4be157c3725d95299b07

Couldn't attach the images as they are SVG. I won't go to the trouble of typing it, either.