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!
Best Answer
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).