"... 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 hFE of the device. However, the external circuit can limit the current to something less than this value, in which case the effective hFE 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\$
They apparently set the Rc/Re ratio to 3 because they wanted a voltage gain of about 3. Presumably that was desirable for the role of this circuit. Basically the Rc/Re ratio is the voltage gain assuming the transistor's gain is "large". Large means that you can approximate the collector and emitter currents as being equal, which also means the base current is 0.
You are right in that the Rc/Re ratio also sets the fraction of the supply voltage that the output can swing. As a simplification, consider that the transistor can vary from open to short C to E. When open, the collector (output) voltage is Vcc. When short, it is Vcc out of the voltage divider formed by Rc and Re. When Rc/Re is large, the output when the transistor is fully on approaches 0, and the output can swing the whole 0 to Vcc range. When Re is a significant fraction of Re+Rc, then it eats up some of the output voltage when the transistor is on, and the output can't swing as low to ground.
For good linear operation, you want to leave about a volt or so across C-E. The lowest output voltage swing is therefore Vcc-1V divided by the resistor divider, plus the 1 V across the transistor:
Vomin = [(Vcc - 1V) Re / (Re + Rc)] + 1V
At high Rc/Re ratios, this approaches the 1 V you leave across the transistor. For lower ratios, the output voltage swing "lost" to Re is significant. All this is to say, yes, you're right, the output voltage swing depends on Rc/Re.
There is no rule of thumb for setting Rc/Re. This is basically the voltage gain of the amplifier. You set this to what it needs to be for other reasons.
However, you can't just make the gain infinite since then other factors that we can reasonably ignore at modest gains get in the way. These other factors are often hard to know. I'd say, try not to exceed a voltage gain of about 1/5 the transistor current gain. That's of course a tradeoff I picked out of the air, but with a reasonably good transistor gain, like 50 or more, a voltage gain of 10 is doable. Beyond that, the approximation that the transistor gain is "large" and you can mostly ignore the base current becomes less valid. So does the approximation that the B-E voltage is fixed.
As you can see at high gains the exact parameters of the transistor matter more and more. Since transistor gain varies widely, we generally want circuits to work with some minimum gain, but not rely on any maximum gain. Put another way, we want circuits to work with transistor gain from some minimum to infinity. The higher the gain, the more sensitive the circuit is to the transistor's gain and other parameters.
As a exercise, see what happens in your circuit when Rc/Re = 3 and the transistor gain is 50 (a quite reasonable minimum guaranteed value for a small signal transistor). Then analyze it again with infinite gain. You'll see only a rather small difference. Now do the same with a gain of 30, and you'll see much more sensitivity to the transistor gain.
Best Answer
Usually, the external load is coupled to the collector via a dc blocking capacitor so, this means that the nominal dc bias point at the collector is unaffected by that external load. That is noteworthy for most designs of CE amplifiers.
Now, ignoring dc conditions, a fairly accurate assumption is that the ac gain is Rc/Re. If an ac impedance is across Rc then the gain effectively lowers. So, you choose a value for Rc that swamps the external impedance and the rule of ten is fine but can easy be a rule of 1 to 1 in many designs when impedance matching is important. If you study the common emitter amp, a decent approximation to the output impedance is the collector resistor, Rc.