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Physics · Ch 14 — Electronic Devices

Current Amplification Factors: $\alpha$ and $\beta$

14.21

Current Amplification Factors: $\alpha$ and $\beta$

Common-base current amplification factor, α\alpha. The common-base current amplification factor is defined as the ratio of the collector current to the emitter current:

α=ICIE\alpha = \frac{I_C}{I_E}

Since, from Section 9.19, ICI_C is always slightly LESS than IEI_E (a small fraction of the emitter current is inevitably lost to recombination in the base, forming IBI_B), α\alpha is always slightly less than 11 -- typically in the range 0.950.95 to 0.990.99 for a real transistor, the closer to 11 the thinner and more lightly doped the base.

Common-emitter current amplification factor, β\beta. The common-emitter current amplification factor -- by far the more commonly quoted and more practically useful of the two, since the common-emitter configuration (Sections 9.20, 9.22, 9.23) is the one most often used in practice -- is defined as the ratio of the collector current to the (much smaller) base current:

β=ICIB\beta = \frac{I_C}{I_B}

Because IBI_B is only a small fraction of IEI_E, β\beta is a comparatively LARGE number for a real transistor, typically ranging from about 2020 up to several hundred -- β\beta is, in effect, a direct measure of how much a transistor amplifies base current into collector current, and is the single figure most often used to characterise a given transistor's current-gain capability.

Deriving the relation between α\alpha and β\beta. Since α\alpha and β\beta are simply two different ratios of the SAME three currents, related by IE=IB+ICI_E = I_B + I_C (Section 9.19), either can always be derived from the other. Starting from β=IC/IB\beta = I_C/I_B, and substituting IB=IE−ICI_B = I_E - I_C (from the current relation):

β=ICIE−IC\beta = \frac{I_C}{I_E - I_C}

Dividing numerator and denominator by IEI_E, and using α=IC/IE\alpha = I_C/I_E:

β=IC/IE1−IC/IE=α1−α\beta = \frac{I_C/I_E}{1 - I_C/I_E} = \frac{\alpha}{1-\alpha}

Rearranging this same relation the other way gives α\alpha in terms of β\beta:

α=β1+β\alpha = \frac{\beta}{1+\beta} …