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Explain high frequency equivalent circuit of BJT.
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At increasing frequencies, the reactance xc will decrease in magnitude resulting in a shorting effect across the output and a decrease in gain.

For high frequency response, various parasite capacitances (cbe, cbc, cce) of the transistors are included along with the wiring capacitors ( cwi, cwo) for analysis. For high frequency response, cs, cc, ce are assumed to be in short circuit state. Input capacitance ci includes wiring capacitance cwi, the transistor capacitance. Cbe and miller capacitance cmi. The o/p capacitance co includes wiring cce and miller capacitance cmo

Cmi = cbc (1-av)

Cmo = cbc ($1-_1$/av)

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for the input network, the – 3db frequency is defined by

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rthi = $r_s$ // $r_1$ // $r_2$ // r$\pi$

ci = cwi + cbc + cmi

cmi = cbc (1-av)

at very high frequency, the effect of ci is to reduce the total impedance of the parallel combination of $r_1$, $r_2$ , $r_\pi$ & ci. The route is a reduced level of voltage across ci, a reduction in In and a gain for the system.

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Rtho = rc // $r_l$ // re

$c_0$ = cwot + cce + cmo

Cme = cbe ( 1 - $1^av)$

$\because$ 1 >> yav cmo = cbc

At very high frequency, XCO will decrease and consequently reduced the total impedance of o/p parallel branches. The net result is vo will also decline towards ‘o’ as the reactance xc becomes (zero) or smaller.

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