Short answer
It will work but will have a lower \$\beta\$ (beta)
Why?
The BJT is formed by two p-n junction (either npn
or pnp
), so at a first glance it's symmetrical. But both the concentration of dopant and the size of the regions (and more important: the area of the junctions) is different for the three regions. So it simply won't work at the full potential. (like using a reversed lever)
Wiki about BJT: look especially the section Structure
and the reverse-active
operating mode
The lack of symmetry is primarily due to the doping ratios of the emitter and the collector. The emitter is heavily doped, while the collector is lightly doped, allowing a large reverse bias voltage to be applied before the collector–base junction breaks down. The collector–base junction is reverse biased in normal operation. The reason the emitter is heavily doped is to increase the emitter injection efficiency: the ratio of carriers injected by the emitter to those injected by the base. For high current gain, most of the carriers injected into the emitter–base junction must come from the emitter.
Another note: classical BJTs are created stacking the three regions in a linear way (see picture in the left), but modern bipolars, realized in surface (MOS) technology, will have also a different shape for collector and emitter (in the right):
In the left a traditional BJT, in the right a BJT in MOS technology (also called Bi-CMOS when both transistors are used in the same die)
So the behavior will be even more affected.
There are several gains associated with voltage amplifiers. Consider the following model
In this model, the gain \$A_{VO} \$ is the open circuit voltage gain of the amplifier which, in your circuit, is given by \$R_C/r_e\$
But note that the output resistance of the amplifier (which is about \$R_C\$ in your circuit) forms a voltage divider with the load \$R_L\$.
So, the loaded voltage gain is:
\$A_{V} = A_{VO} \dfrac{R_L}{R_{out}+ R_L}\$
But, note that the input voltage \$V\$ is less than the source voltage due to the voltage division between the source resistance and the input resistance of your amplifier. Thus, the loaded gain with respect to the source is:
\$A_{VS} = A_V \dfrac{R_{in}}{R_S + R_{in}} = A_{VO} \dfrac{R_L}{R_{out}+ R_L} \dfrac{R_{in}}{R_S + R_{in}}\$
So, you cannot expect to measure anything close to the open circuit gain \$R_C/r_e \$
Best Answer
This is the basic circuit you seem to be asking about. It's known as a common-emitter amplifier with emitter degeneration:
The two resistors have totally different functions.
RC is totally fundamental to the operation of a common-emitter amplifier. It determines the voltage gain of the amplifier.
If you removed it, you simply wouldn't have a working circuit, because there'd be no path for current to flow through the collector of the BJT. If you removed it and replaced it with a short circuit, the BJT would still produce current gain, but the output voltage would always be exactly equal to the V+ voltage, and the circuit just wouldn't be very useful.
Note: In some cases, RC is not present, but the load is connected from the collector to the positive supply, so that the load itself fills the role of RC.
RE, on the other hand, is a little more complicated. This resistor is why we call the circuit "emitter degenerate". Having RE means that an increase in collector current tends to reduce Vbe, which reduces the portion of the input voltage that contributes to gain. This is a form of negative feedback. The main benefits of this is that it increases the range of input bias where the circuit operates linearly, makes the circuit gain more stable if the BJT properties vary, and it increases the input resistance of the circuit.
If you removed RE and replaced it with a wire you'd just have a standard common-emitter amplifier.