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

enter image description here

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.

enter image description here

Figure 2: Diagram of AC equivalent circuit at high frequency

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

enter image description here

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,

enter image description here

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,

enter image description here

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 $Rtho = R_O || R_C$

$Cco = Cce + Cwo + Cmo$

Please log in to add an answer.

Continue reading...

The best way to discover useful content is by searching it.

Search