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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:enter image description here

\$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!

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  • \$\begingroup\$ Start by solving for the base current, treating the BE junction as a diode in series with Rb. The collector current doesn't really effect the base current. Then use beta to get the collector current and calculate Vout. If your calculated Vout is less than Vce(sat), then you must be in saturation mode, so you know Vout is Vce(sat), and Ic = (Vcc-Vce(sat))/Rc. \$\endgroup\$ – mkeith Feb 11 '17 at 6:48
  • \$\begingroup\$ Instead of a numerical answer, give an equation as the answer. Then plug in some reasonable numbers (your room temperature, for one). \$\endgroup\$ – Whit3rd Feb 11 '17 at 7:06
  • \$\begingroup\$ Like many beginners you try to start the whole calculation with Vbe. Take it from me: calculations regarding Vbe are practically pointless and you don't need to know Vbe, just assume it's 700 mV. Now use beta and Rb to calculate what Ic can flow. \$\endgroup\$ – Bimpelrekkie Feb 11 '17 at 12:32
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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).

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  • \$\begingroup\$ This looks good, except that (Vcc-Vbe)/RB is not the base current. It only contributes part of the base current right? because there's the current from Vin. \$\endgroup\$ – John Smith Feb 11 '17 at 7:40
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    \$\begingroup\$ John Smith - OK, I know what you mean - however, in accordance with normal/classical usage of such a circuit I have assumed that the circit has a DC operational point which is determined by external fixed DC supply voltages only (Vcc in this case). That means: Any signal voltage Vin must be, of course, coupled to the amplifier stage using a capacitor - unless the signal voltage has a (large) internal source resistor with a value that must be known for solving the task. But it is not known. \$\endgroup\$ – LvW Feb 11 '17 at 7:48
  • \$\begingroup\$ I see! Yes, yes, I think you're right. So V_in is just V_CC minus the voltage drop across the resistor R_B right? \$\endgroup\$ – John Smith Feb 11 '17 at 7:53
  • \$\begingroup\$ Yes - otherwise (without a coupling capacitor) the voltage source Vin would provide a short for the current IB provided by Vcc. \$\endgroup\$ – LvW Feb 11 '17 at 7:55
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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).

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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.

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  • \$\begingroup\$ But the problem is supposedly asking for the real value of \$V_{BE}\$. What happens after I assume that \$V_{BE}\$ is 0.7 V? \$\endgroup\$ – John Smith Feb 11 '17 at 6:38
  • \$\begingroup\$ Because in active region, the base-emitter junction is modelled as a diode, which usually has a conduction voltage of 0.7V. Look up google for "Bjt pi model" \$\endgroup\$ – C K Feb 11 '17 at 6:41
  • \$\begingroup\$ Sorry sorry, search "bjt active mode dc model" instead \$\endgroup\$ – C K Feb 11 '17 at 6:42
  • \$\begingroup\$ @JohnSmith you assume an operation mode, put the variables in, then check if your assumption is correct. If not, you try another operation mode. You can find lots of material on that topic. What you need to search is "BJT DC analysis" \$\endgroup\$ – C K Feb 11 '17 at 6:48
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    \$\begingroup\$ @ÇetinKöktürk, this does not seem like the kind of problem where you are supposed to just assume Vbe is 0.6 or 0.7. I believe the OP is supposed to calculate Vbe based on the given information and the diode equation. Otherwise, why would the problem give Is? You can use a load-line approach or just guess and test iteratively (using a spreadsheet) until you find a current and voltage which satisfies all constraints. \$\endgroup\$ – mkeith Feb 11 '17 at 6:56
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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.

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  • \$\begingroup\$ dannyf, what do you mean with "...no basis..."? Have you any objections against combining the two equations: Ic=beta*IB and IB=(Vcc-VBE)/RB ? \$\endgroup\$ – LvW Feb 11 '17 at 11:45
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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.

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