0
8.8kviews
Explain high frequency equivalent circuit of BJT
5
75views

Figure 1: Diagram of high freq. circuit

Analysis of amplifier at high frequency means to find $fH_i$ and $fH_o$ (higher cut-off frequency on input side and output side).

At high frequency Ce, Cb, Cc act as short circuit. But junction and wiring capacitors are effective at high frequency.

Figure 2: Diagram of AC equivalent circuit at high frequency

Cci = Total capacitance on input side amplifier due to junction and wiring capacitor.

Figure 3: Diagram of Cci equivalent

Cwi = Wiring capacitor of input side (Two wires on input side act as 2 plate and air as dielectric)

Cmi = Miller input capacitance.

Any Capacitor connected between output and input of amplifier then its effect can be shown separately on input as well as on output by using miller theorem.

$Cmi = Cbc ( 1 - A_V )$ ….(i/p side)

$Cmo = Cbc ( 1 - \frac{1}{A_V})$ …(o/p side)

Capacitance at input side are parallel i.e $Cbe, Cwi$ and $Cmi$ which are combined to form single equivalent capacitance $Cci$,

$Cci = Cbe + Cwi + Cmi$

Similarly, capacitance at the output side,

$Cco = Cce + Cwo + Cmo$

$fHi$ can be obtained from input equivalent circuit by using thevenin's thorem,

Figure 4: Thevenin's equivalent circuit diagram

$Vthi = \frac{V_S R_i || R_B}{(R_i || R_B) + R_S}$

But $Rthi = R_S || R_B || R_i$

Connect Vthi and Rthi to Cci,

Figure 5: Equivalent circuit diagram

At mid frequency Cci act as open $V1 = Vthi$ .....at mid frequency .........(1)

But at high frequency $V1 = \frac{Vthi \times Xcci}{\sqrt{XCci^2 + Rthi ^2}}$

Put $XCci = Rthi$,

$V1 = \frac{Vthi}{\sqrt {2} } = 70.7 \% of Vthi$.....at high frequency...................(2)

From (1) and (2)

$V1$ at high frequency = 70.7 % of $V1$ at mid frequency only if $XCci = Rthi$

$\frac{1}{2 \pi fHi Cci} = Rthi$

$fHi = \frac{1}{2 \pi Rthi Cci}$

where, $Rthi = Rs || R1 || R2 || Ri$...( if $Ri$ is not given $\approx \beta r_e$)

$Cci = Cbe + Cwi + Cmi$

Similarly, on output side, we can get,

$XCco = Rtho$

$\frac{1}{2 \pi fHo Cco} = Rtho$

$fHo = \frac{1}{2 \pi Rtho Cco}$

where $Rtho = R_O || R_C$

$Cco = Cce + Cwo + Cmo$