Question 155: Semiconductors and Logic Gates

Question

In a common emitter configuration with suitable bias, it is given that $${R_L}$$  is the load resistance and $${R_{BE}}$$ is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by : 
$$\beta $$ is curret gain, $${{\rm I}_B},{{\rm I}_C}$$  and $${{\rm I}_E}$$ are respectively base, collector and emitter currents.

Accepted Answer

Current gain ($$\beta $$) = $${{\Delta \,{I_C}} \over {\Delta {I_B}}}$$
Voltage gain  = $${{{V_{CE}}} \over {{V_{BE}}}} = \beta {{{R_L}} \over {{R_{BE}}}}$$
Power gain = voltage  gain x current gain = $${\beta ^2}{{{R_L}} \over {{R_{BE}}}}$$

Answer

$$\beta {{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_C}} \over {\Delta {{\rm I}_B}}},{\beta ^2}{{{R_L}} \over {{R_{BE}}}}$$

Answer

$$\beta {{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_E}} \over {\Delta {{\rm I}_B}}},{\beta ^2}{{{R_L}} \over {{R_{BE}}}}$$

Answer

$${\beta ^2}{{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_C}} \over {\Delta {{\rm I}_E}}},{\beta ^2}{{{R_L}} \over {{R_{BE}}}}$$

Answer

$${\beta ^2}{{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_C}} \over {\Delta {{\rm I}_B}}},\beta {{{R_L}} \over {{R_{BE}}}}$$

Semiconductors and Logic Gates

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