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Given $\Delta V_{BE} = $ potential difference across base emitter, $\Delta I_C = $ change in collector current, $\Delta I_B = $ change in the base current and $\Delta R_{BE} = $ change in base emitter resistance, which of the following best represents transconductance $g_m$.

$(a)\;\frac{\Delta V_{BE}}{\Delta I_c} \\ (b)\;\frac{\Delta R_{BE}}{\Delta I_c} \\ (c)\;\frac{\Delta I_c}{\Delta I_B} \\ (d)\;\frac{\Delta I_c}{\Delta v_{BE}} $


$\Delta V_{BE} $ -Potential difference across base emitter

$\Delta I_C$- charge in collector current

$\Delta I_B$- change in base current

$\Delta R_{BE} $ change in base- emitter resistance     

1 Answer

The property to measure amplification in a transistor used an amplifier circuit is called transconductance.
To have large amplification a small change in $V_{BE}$ due to input signal, should result in large change in collector current $I_C$
Transconductane $g_m =\large\frac{\Delta I_c}{\Delta V_{BE}}$
Hence d is the correct answer.
answered Jan 28, 2014 by meena.p
edited Jul 14, 2014 by balaji.thirumalai

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