Electrical – Calculating base-emitter current of a BJT, with emitter connected to ground (no resistor)

For my opinion, it is somewhat confusing to use this term "re" (and your question can serve as a "good" example). This parameter re (called "intrinsic emitter resistance") is nothing else than the inverse of the transconductance re=1/gm. I think, it is not correct to call it "resistance" because it relates the output current and the input voltage (gm=dIc/dVBE). In contrast, the classical resistor definition relates the voltage across and the current through the two-pole element.

This transconductance appears in each gain formula, for example in common emitter configuration:

Gain A=-gmRc/(1+gmRe)=-Rc/(Re+1/gm)=-Rc/(Re+re) .

The value of gm=1/re can be found by differentiating the transfer function Ic=f(Vbe) at the selected quiescent current Ic and can be found as gm=Ic/Vt (Vt=temperature voltage). Hence, it is a parameter which is computed at the DC bias point.

Does this answer your question?

EDIT

From the gain formula (with emitter resistor RE) we can derive that the gain is reduced by the factor 1/(1+gmRE). From this we can again derive that the term (gmRE) is the loop gain of the feedback system. From feedback theory we know that the disrortions are reduced also by the same factor. Rewriting this factor we get

1/(1+gm*RE)=(1/gm)/(RE+1/gm)=re/(RE+re).

This prooves the result you have shown.

"... since the transistor acts like a resistor between collector and emitter ... "

No, not really. The collector of a bipolar transistor acts like a current source (or sink) whose value is determined by the base current and the h_FE of the device. However, the external circuit can limit the current to something less than this value, in which case the effective h_FE is lower.

"I see in transistor datasheets a maximum and minimum value of hfe."

Yes. The actual value varies considerably from device to device, even from the same manufacturing batch, and it also varies somewhat with the operating parameters (voltage, temperature, etc.) of the device. You really can't depend on having a particular (or even a constant) value, so you design your circuits so that they work over a range of values.

"... then what is the point of the adding a resistor to the collector end of the transistor?"

This is part of the circuit design. When you're creating a voltage amplifier, you use the collector current of the transistor to develop the desired voltage across the external resistor. This resistor is called the "load resistor", and it gives you a definite value of output impedance — the transistor by itself has a very high effective output impedance.

Example:

collector emitter voltage is 9 v ,Hfe = 100 , base emitter voltage is 9 , a resistor at the collector has resistance of 330 ohms and one at the base has resistance 10k ohms, tell me the current at the collector with steps.

OK, assuming you mean that 9V is applied to the base through a 10K resistor, 9V is applied to the collector through a 330Ω resistor, and that the emitter is grounded, the steps are as follows:

• The base current is \$I_B = \frac{V_{BB} - V_{BE}}{R_B} = \frac{9.00 V - 0.65 V}{10k \Omega} = 0.835 mA\$

• Assuming the transistor is not saturated, the collector current \$I_C = h_{FE} \cdot I_B = 100 \cdot 0.835 mA = 83.5 mA\$

• The voltage across the collecor resistor should be \$I_C \cdot R_C = 83.5 mA \cdot 330 \Omega = 27.5 V\$

• Since that value is higher than our supply voltage, the assumption made in the second step must be false — the transistor is saturated. Therefore, the collector current is determined entirely by the collector resistor and the collector supply voltage: \$I_C = \frac{V_{CC} - V_{CE(SAT)}}{R_C} = \frac{9.00 V - 0.3 V}{330 \Omega} = 26.4 mA\$