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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Physics
Chapter
Semiconductors

BITSAT 2014 · Physics · Semiconductors

For a CE transistor amplifier, the audio signal voltage across the...

For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2.0kΩ2.0 \mathrm{k} \Omega is 2.0V2.0 \mathrm{ V}. Suppose the current amplification factor of the transistor is 100 , What should be the value of RB\mathrm{R}_{\mathrm{B}} in series with VBB\mathrm{V}_{\mathrm{BB}} supply of 2.0V2.0 \mathrm{ V} if the dc base current has to be 10 times the signal current?

Options

  1. A14kΩ14 \mathrm{k} \Omega
  2. B18kΩ18 \mathrm{k} \Omega
  3. C10kΩ10 \mathrm{k} \Omega
  4. D5kΩ5 \mathrm{k} \Omega

Correct answer

A. 14kΩ14 \mathrm{k} \Omega

Step-by-step solution

The output ac voltage is 2.0V2.0 \mathrm{ V}. So, the ac collector current iC=2.0/2000=1.0mA\mathrm{i}_{\mathrm{C}}=2.0 / 2000=1.0 \mathrm{ mA}. The signal current through the base is, therefore given by iB=iC/β=1.0mA/100=0.010mA\mathrm{i}_{\mathrm{B}}=\mathrm{i}_{\mathrm{C}} / \beta=1.0 \mathrm{ mA} / 100=0.010 \mathrm{ mA} The dc base current has to be 10×0.010=0.10mARB=(VBB−VBE)/10 \times 0.010=0.10 \mathrm{ mA}\mathrm{R}_{\mathrm{B}}= \left(\mathrm{V}_{\mathrm{BB}}-\mathrm{V}_{\mathrm{BE}}\right)/\mathrm{I}BAssumingVBE=0.6V,RB=(2.0.0.6)/0.10=14kΩ_{\mathrm{B}}Assuming \mathrm{V}_{\mathrm{BE}}=0.6 \mathrm{ V}, \mathrm{R}_{\mathrm{B}}=(2.0 .0 .6) / 0.10=14 \mathrm{k} \Omega
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