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

Common emitter

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?

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  • \$\begingroup\$ 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) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 10 at 21:46
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Why is β ever used in this kind of analysis?

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.

My rule of thumb is that the quiescent current in R1 should be about ten times the expected base current at room temperature (and, thus, that the expected standing current in R2 is 90% of that in R1). My other rule of thumb is that if I'm doing this for real and not just some on-bench experimentation, I check the circuit operation over temperature.

...you should design them so the voltage across RE is 1 V. Why?

First, that's a rule of thumb, so it's yours to violate. Second, that gives a pretty good all-around stabilization of the emitter current over part to part and temperature variations. To be sure, again, you want to check your circuit (usually by Monte Carlo simulation) over variations in part parameters and temperature.

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  • \$\begingroup\$ This. It's a question I often use for junior design engineer interviews, and this answer would get 9/10 (because I never give 10 to anything). 1V is not magic, but less than that gets you close to temperature stability problems, and more than that wastefully eats into your voltage headroom. Beta only has to be 'big', so 1/big (the base current) is nearly zero, so don't try to bias a power transistor this way. \$\endgroup\$ – Neil_UK Aug 11 at 6:26
  • \$\begingroup\$ So the question may arise - for such a kind of emitter stage (well designed), has the value of beta (if in the region of 100 or even larger) any relevance? Its influence on the voltage gain is not too big. However, it determines the input resistance of the stage (rin=R1||R2||h11 with h11=beta/gm=h21/gm with gm=transconductance). The value of rin is of great importance for the resulting voltage gain in conjunction with the signal source resistance. \$\endgroup\$ – LvW Aug 11 at 8:06

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