# Necessity of beta in common emitter design

I'm thinking about a common emitter circuit like this one:

Initial assumption: If the base current is very small compared to the emitter current, then the collector and emitter currents are practically equal. This is an equivalent assumption to saying $$\\alpha\$$ is so close to 1 it may as well be 1.

If you know all the resistor values, you can find the base voltage, take off 0.7 V, and have the emitter voltage. With that, you can find the current through $$\R_E\$$. Since with the above approximation the collector current is equal to this, you can find the voltage at the collector by multiplying that current by $$\R_C\$$. Am I correct in calling this the quiescent collector voltage?

The voltage gain can also be obtained, by using $$\r_{ej}\$$, which is 26 mV (thermal voltage) divided by the emitter current found above. $$\R_C/r_{ej}\$$ should give the voltage gain.

From what I can see, if you allow the initial assumption then both quiescent collector voltage and voltage gain can be found without the transistor's $$\\beta\$$. The assumption seems reasonable to me, so my main question is: Why is $$\\beta\$$ ever used in this kind of analysis? It appears to be, so either my initial assumption is dodgy or my working following it is.

A smaller question about the circuit above:
I have read various guides on designing such a circuit, and a few of them say you should design them so the voltage across $$\R_E\$$ is 1 V. Why? Why not 2 V, or 0.5 V, or anything else?

• The irrelevance of H bias is actual bottom line in preference to negative feedback CE design due to massive non-linearity of Vbe controlled collector current. But vis-a-vis Ve, even 0.3V is adequate relative to Vbe / T{'C} ratio sensitivity but depends on overall Vbe(ac/dc ratio) – Tony Stewart Sunnyskyguy EE75 Aug 10 at 21:46

In this particular bit of analysis, for normal small-signal transistors (i.e., $$\\beta \simeq 100\$$ or so), the value of $$\\beta\$$ really only enters into things in the choice of values of R1 and R2. Assuming zero base current, all you need to do is get the ratio of R1 and R2 correct. But a real base current will pull $$\V_{be}\$$ down. So you want to, first, take that into account, and second, don't make R1 and R2 too large.