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Common-Emitter Amplifier

In this amplifier circuit we are using a voltage divider bias. Why can we consider that voltage produced by R2 is actually providing voltage to the base. What is R1 doing?

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  • \$\begingroup\$ I'd go back to your original question and tag this additional question to it. It makes more sense to do that rather than open a new question \$\endgroup\$
    – Andy aka
    Mar 19 '13 at 11:57
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    \$\begingroup\$ @Andyaka Adding additional questions to a question is not the idea. You want each question to be relatively independent of the others, they might relate, but if the question is different and will cause different answers, editing it will muddy the waters for the answers that are written. \$\endgroup\$
    – Kortuk
    Mar 19 '13 at 13:34
  • \$\begingroup\$ @Kortuk ok dude \$\endgroup\$
    – Andy aka
    Mar 19 '13 at 13:45
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My answer was not correct as @MikeJ-UK pointed out.

The voltage at the base is Vcc * R2 / (R1 + R2) - Ib( R1R2/(R1 + R2) ) o

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ If that were true, Ib would be zero. Vb = Vcc X R2/( R1 + R2 ) - Ib * ( R1//R2) \$\endgroup\$
    – MikeJ-UK
    Mar 19 '13 at 12:21
  • \$\begingroup\$ Ugh, pressing enter posts the message, and then you get locked out from editing. But yes you were correct and my original post has been edited. \$\endgroup\$
    – efox29
    Mar 19 '13 at 12:52
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R1 and R2 form a voltage divider between the supply and ground. The output of this divider provides the voltage and current to bias the transistor base.

This is a good place to do a Thevenin substitution. You can model the voltage divider as a fixed voltage source in series with a single resistance. The impedance of the Thevenin source, which is the resistance in series with the perfect voltage source, is the parallel combination of R1 and R2. To get the voltage of the fixed voltage source, look at the open circuit output of the divider, which is Vcc * R2(R1 + R2).

You haven't shown any numbers, so I'll leave the equations in symbolic form.

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What is R1 doing?

For a voltage to to exist across R2, there must be a current through R2 (Ohm's Law).

If R1 is removed, where would the current \$I_B\$ and the current required to produce the desired \$V_B\$ across R2 come from?

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  • \$\begingroup\$ I mean can we not supply the desired voltage to base through R1. What is the advantage of voltage divider bias? \$\endgroup\$
    – Ali Khan
    Mar 20 '13 at 2:05
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    \$\begingroup\$ That's a good question. In fact, the base bias current can be supplied through R1 alone. The problem is that the operating point with such a bias circuit is remarkably unstable against temperature changes as well as typical variations in beta from unit to unit. By adding R2 and designing its current to be roughly 10 times the base bias current, variations in temperature and beta have much less effect on the operating point. \$\endgroup\$ Mar 20 '13 at 2:23

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