I was going through voltage divider bias configuration of BJT and then i encountered two problems:- The above two problems are nearly same except that the emitter terminal is grounded in case of (b) and baised in (a) .when i solved and found out the thevenin's voltage(between emitter and base terminals) then it was positive for (b) and negative for (a) .Negative value of the thevenin's voltage would suggest that the emitter base junction is reversed baised so The transistor must be operating in the cut-off region.

In some videos online i found that even when thevenin's voltage is negative the BJT was operating in the active region .Is this really possible ?If yes then how?

Edit:The calculation of Vth for (a) is given in this video.

Thank You for giving your time.

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    \$\begingroup\$ The base voltage is also different in each, so both are forward biassed. However both have ridiculous values for RE so what are you really trying to do? It's also entirely unclear what you mean by the "thevenin's voltage" (between which points on the circuit) here. \$\endgroup\$ – Brian Drummond Jan 6 '18 at 14:12
  • \$\begingroup\$ transistors in your circuits are saturated if we assume C,B and E terminals to be C=collector, B=base, E=emitter and transistors are NPN. Check the connections and resistor values. R1 and R2 can be swapped or RE can be radically too small. \$\endgroup\$ – user287001 Jan 6 '18 at 14:13
  • \$\begingroup\$ @BrianDrummond I am finding thevenin's voltage between base and emitter terminals. Also can you give me a circuit daigram with thevenin's equivalent to shiw that it is forward baised? \$\endgroup\$ – Pink Jan 6 '18 at 14:18
  • \$\begingroup\$ The voltage on the base relative to ground is not relevant. What matters is the base voltage relative to the emitter. \$\endgroup\$ – Spehro Pefhany Jan 6 '18 at 14:21
  • \$\begingroup\$ Then you will find it positive in both cases, unless you make a mistake somewhere. If you show your working, someone may be able to see the mistake. \$\endgroup\$ – Brian Drummond Jan 6 '18 at 14:22

In both cases, (a) and (b), your circuit is essentially the one on the left, below. The equivalent circuit, after the biasing pair is converted to its Thevenin equivalent is on the right, below:


simulate this circuit – Schematic created using CircuitLab

For the left side

  • Case (a): You have \$V_{CC}=+20\:\text{V}\$ and \$V_{EE}=-20\:\text{V}\$.
  • Case (b): You have \$V_{CC}=+20\:\text{V}\$ and \$V_{EE}=0\:\text{V}\$.

\$V_{EE}\$ is the only difference.

For the right side

  • Case (a): You have \$V_{TH}\approx -11.538\:\text{V}\$, \$V_{EE}=-20\:\text{V}\$, and \$R_{TH}\approx 1.735\:\text{k}\Omega\$.
  • Case (b): You have \$V_{TH}\approx 4.231\:\text{V}\$, \$V_{EE}=0\:\text{V}\$, and \$R_{TH}\approx 1.735\:\text{k}\Omega\$.

In both cases, an NPN transistor would have its base-emitter junction forward biased and not reverse-biased. You should be able to easily see this fact. Note that in both cases, \$V_{TH} \gt V_{EE}\$.

Also, looking at the right side schematic, you can trivially find that:

$$V_{TH}-I_B\cdot R_{TH}-V_{BE}-I_E\cdot R_E=V_{EE}$$

This can be solved for \$I_B\$, knowing that \$I_E=\left(\beta+1\right)\cdot I_B\$:

$$I_B=\frac{V_{TH}-V_{BE}-V_{EE}}{R_{TH}+\left(\beta+1\right)\cdot R_E}$$

You should be able to go from there to the rest of any analysis needed. (Assuming you can treat \$V_{BE}\$ as a constant. If not, things get more complex.)

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You have written R2=22 kOhm. In the video R2=2200 Ohm. The latter is ok for active region.

The linked lecture is not exactly crap, but uses a not so common convention. It does not in the beginnig show which are the nodes from which the Thevenin's equivalent is seen. T's equivalent in his calculations have earth as the other terminal. It becomes clear later. Accepting this the result is ok. Vth is negative but causes forward voltage from B to E. Why forward? because Vth is less below zero than -Vee.

If we want to calculate T's equivalent for the voltage divider formed by R1 and R2 and power supplies, all seen between base and -Vee (a) or between base and GND (b), the resulted Vth have plus in the base of the transistor. This is common in all textbooks I have seen. I recommend to allways show in the beginning which are the nodes of the T's equivalent.

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  • \$\begingroup\$ Sory the resistor was 2.2 kohm. \$\endgroup\$ – Pink Jan 6 '18 at 15:08

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