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What is the impedance between the base of the BJT and ground of the common emitter amplifier circuit? The answer should be in terms of RE and hfe (the small current gain of the transistor). Is this a high or a low impedance (relative to the output impedance)?

Hint, this isn’t that bad remember the rules and the definition of hfe. Write down ΔIE in terms of ΔVB and RE then write down ΔIE in terms of ΔIB and hfe. Now the impedance looking into the base can be calculated from Ohms Law and by answering the questions what is ΔIB for a change ΔVB.

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    \$\begingroup\$ How far have you got with this homework question so far? \$\endgroup\$ – RedGrittyBrick May 31 '13 at 10:15
  • \$\begingroup\$ I have an answer, for the first part is Ie=(0.6Re)/Vb ? then hfe= delta Ic/ delta Ib. So delta Ic= Vb/(0.6Re) \$\endgroup\$ – user24624 May 31 '13 at 10:18
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I don't like to give full answers for homework questions but I always think a circuit diagram helps. The voltage at the emitter is 0.6V less than the base. The voltage across the emitter resistor can be calculated using ohms law (V/I). Now consider what the value of a resistor would be if Ib was to produce the same voltage as the emitter. (Ve / Ib). Does the 0.6V offset matter if you are looking for a change ? Is the value of Ie significantly different from Ic? You have all the relationships - do your homework.

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