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From http://www.electronics-tutorials.ws/amplifier/amp_2.html, i understand that the voltage gain of a common emitter BJT is given by below equation; where RL is the load resistance, RE is the resistance between emitter and ground and Re is an inherent BJT emitter resistance.

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This voltage gain is applicable for both AC/DC input voltage. How about the biased DC voltage? Does it get amplified as well?

Basically i would like to know whether a BJT (common-emitter example) also amplifies the biased voltage (applied at the base). In the absence of any input voltage, can i say that the output voltage (at the emitter) is equal to biased voltage (0.7V for example) multiplied by the voltage gain [-RL/(RE + Re)]?

Thank you.

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    \$\begingroup\$ Please note that the quantity "Re" is not an ohmic resistor but a dynamic value and should be labelled "re". More than that, for a proper understanding it is good to realize that re is NOT an "inherent" emitter resistance (some authors use such a term for a simplified explanation). Instead, it is the inverse transconductance re=1/gm. The transconductance gm establishes the connection between input (voltage at the base) and output (collector current): gm=dIc/dVbe. \$\endgroup\$ – LvW Nov 9 '17 at 10:30
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    \$\begingroup\$ Suppose that "biased voltage" is 0.7 V and the voltage gain of the circuit is 20, the supply voltage is 10 V, which are very reasonable values. Then how can that 0.7 V be amplified to 20 x 0.7 = 14 V? So the conclusion is: no, the DC biasing voltage is not amplified. Only variations superimposed on that DC bias voltage are amplified. So then it is possible to Vbe = 0.7 V DC, + 10 mV AC and amplify that to 5 V DC (DC bias at collector) + 200 mV AC (input variation amplified 20 times). \$\endgroup\$ – Bimpelrekkie Nov 9 '17 at 10:34
  • \$\begingroup\$ How does the bjt differentiate between bias voltage and dc variation voltage? How example if the DC variation voltage is 2V, what would be the output voltage at the collector? \$\endgroup\$ – happynomad Nov 9 '17 at 10:45
  • \$\begingroup\$ happynomad, as an answer to your question it may help to realize that the BJT works as a current source. Hence, it conveys the input voltage (DC as well as DC variations) into an output current (resp. current variations). This current (and its variations) is changed into voltages due to RL according to ohms law. \$\endgroup\$ – LvW Nov 9 '17 at 11:05
  • \$\begingroup\$ But it still doesn't really answer my question:( \$\endgroup\$ – happynomad Nov 9 '17 at 11:31
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Amplification, by definition, is associated with change in the voltage (or current or whatever parameter involved). So when you say that the voltage gain of the amplifier is say 10, you mean that a change in the input voltage by a certain amount would result in 10 times change in the output voltage. So, in fact, the formula you have quoted in the question is the ratio of the change in the output with change in the input voltage: $$\frac{dV_o}{dV_i} = -\frac{R_L}{R_E + R_e}$$ So actually amplification of the bias voltage itself does not make sense because there are no changes involved in that voltage.
So, suppose you are getting a gain of 10 with input bias of 0.7V and output at 4V. What your equation means is that if you increment the input bias (at 0.7V) by 0.1V (just as example) the output bias will go down by 10*0.1 = 1V from 4V.
But, be aware, that it is just the small signal gain of your circuit. So if you swing the input bias too much then you will not get the 10x change at the output as before (but a little less).

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