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I'm designing a standard common emitter amplifier. My textbook states that it's standard to make \$V_C = 2/3 V_{CC}\$ and \$V_E = 1/3V_{cc}\$. I understand that it has to with maximizing the voltage swing, but it's affect isn't clear to me. Why is doing this a standard practice?

Say \$ V_{CC}=12V\$, then \$V_C=8V\$ and \$V_E=4V\$. Since the output of the amplifier is taken at the collector, the output is at \$ 8V \$ with no input signal. I'm told that the swing is +/- \$ 4V\$, so the range of the amplifier is \$ 4V \$ to \$ 12V\$. I guess I don't understand what is meant by swing and how one arrives at these numbers. I sort of get that it can go up to \$12V\$ until it clips the signal since the transistor can't go higher than \$ 12V \$, but why can't it pull down lower than \$4V\$?

Another thing; I've read that if \$ \beta \$ is large, say 200, then \$ I_E \approx I_C\$. That makes sense to me, since the collector current will contribute to most of the emitter current for large beta. However, my prof said for this reason we are allowed to make the collector and emitter resistors the same value. Why?

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  • \$\begingroup\$ Hmmm - Sounds wrong to me. The only time I would do that is if I required signals from both the collector and emitter at the same time (eg a phase splitter). Otherwise I would usually set Ve between 10 - 20% of the supply with Vc at about 60% of supply. The emitter resistor would then be decoupled with a suitable capacitor. If you did the 2/3, 1/3rd thing then the collector and emitter resistors would be made equal. Ib is less than 1% of the current and given that resistors would be 5% tolerance it wouldn't make any significant difference in production. \$\endgroup\$ – JIm Dearden Sep 4 '15 at 15:29
  • \$\begingroup\$ @JImDearden, what about the voltage swing? That seems like the number one reason this is a "standard" from my understanding. \$\endgroup\$ – Lefty Sep 6 '15 at 23:46
  • \$\begingroup\$ The reason you want some voltage at the emitter (say 20%) is for bias stabilisation. The bypass capacitor effectively 'smooths' this voltage keeping the emitter at this dc level even though the (ac) current through the transistor varies. Setting the emitter voltage higher (33%) means you have less swing at the collector. Its always a compromise. see circuitstoday.com/transistor-amplifier for an example design \$\endgroup\$ – JIm Dearden Sep 7 '15 at 15:07
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Of course, you are allowed to make Rc=Re. However, the question remains if this make much sense! Hence, I think the first rule you have mentioned (Vc=Vcc*2/3 and Ve=Vcc/3) is a good trade-off between allowable swing and good stabilization of the operational point. As you probably know, the resistor Re provides negative feedback for DC and stabilizes the bias point.

Regarding your example (Vcc=12V). In this case the possible (theoretical) swing is 12-8=+4V and 4-8=-4V with Vce=4V.

Of course, the resistor Re provides also signal feedback, thereby reducing the gain to a value of app. Rc/Re. For larger gain you can bypass the resistor Re with a suitable capacitor.

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  • \$\begingroup\$ Okay, so I think I understand the voltage swing aspect better after reading these responses. Can you please elaborate on how this configuration provides "good stabilization" of the operational point? Or, could you tell me what would be an example of poor stabilization? \$\endgroup\$ – Lefty Sep 4 '15 at 18:45
  • \$\begingroup\$ OK - do you know something about negative feedback? In this case it is simple: Ic rises because of temperatur increase. Hence, more voltage drop across Re. That means: Ve is larger and Vbe goes down. Reduction in Vbe brings Ic back (nearly) to its former value. This works because the BJT is a voltage-controlled device: Ic=f(Vbe). \$\endgroup\$ – LvW Sep 4 '15 at 18:52
  • \$\begingroup\$ I don't know a lot about feedback in amplifier circuits. (I've only dealt with feedback in regard to op amps). Also, where does temperature come into play here? \$\endgroup\$ – Lefty Sep 6 '15 at 23:44
  • \$\begingroup\$ It is a property of bipolar transistors that the collector current goes up if the temperature increases (for a constant bias voltage Vbe). The reason is an increased mobility of the charged carriers for higher temperatures. \$\endgroup\$ – LvW Sep 7 '15 at 7:46
  • \$\begingroup\$ But temperature isn't mentioned here, so I don't see where it's coming into play. \$\endgroup\$ – Lefty Sep 7 '15 at 13:32
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If the current through the resistors is approximately the same \$(I_C \approx I_E)\$, and you want the voltage drop across them to be the same \$(\frac {V_{CC}} 3)\$, then by Ohm's Law the resistors should be the same value.

$$V_E = V_{CC} - V_C$$ $$I_E \approx I_C$$ $$R_E = \frac {V_E} {I_E} \approx \frac {V_{CC} - V_C} {I_C} = R_C$$

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  • \$\begingroup\$ How obvious! Can't believe that hadn't occurred to me \$\endgroup\$ – Lefty Sep 4 '15 at 18:36
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LvW covered most of it. As far as why the collector can't swing below 4V, consider the bias voltages on the transistor. The emitter is at 4V, which means the base is at about 4.7V. If the collector goes below about 4.7V the base-collector junction will be forward biased and the transistor won't be in forward-active mode any more.

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