Draw the ckt diagram of voltage divide bias circuit using CE configuration and explain how it stabilizes the operating point?
1 Answer

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  • The aim of any small ampr is to amplify all of the i/p s1 with minimum amount of distortion possible to the o/p signal in other words o/p must be an exact reproduction of the input signal. (Amplifies)

  • To obtain low distortion when used as ampr the operating point needs to be correctly selected.

  • The best possible position for Q point is as to the center position of the load line as reasonably possible, thereby producing a class A type ampr i.e. $VE = \frac{Vcc}{2}$

  • $R_1$ and $R_2$ forms a VH divider.

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  • This method of biasing greatly reduces the effect of varying beta ($\beta$) by holding the bare bias constant steady voltage level allowing for best stability.

Transistor bias voltage:

$VB = Vcc \times \frac{R_2}{R_1 + r_2}$

This makes the transistor ckt independent of changes in $’\beta’$ as the voltage at transistor bare emitter and collector.

The current flowing through $RB_2$ is generally set at 10 times the value of the required bare current. IB so that it has no effect on the voltage divider current or changes in $\beta$

  • Following equation shows how transistors Qpt is made stable by making parameter free from $\beta$ changes or temp changes.

Vc = Vcc = Ic Rc

= VE + CcE

$V_E = I_E R_E = V_B – VBE$

$VCE = Vc – VE = Vcc – (Ic Rc _ IE RE)$


$= VB_2 = R_2 \times \frac{Vcc}{R_1 + R_2}$

$IB_2 = \frac{VB}{R_2}$

$IB_1 = IB + IB_2 = \frac{Vcc – V_B}{R_1}$

$R_B = \frac{R_1 R_2}{R_1 + R_2}$

$Ic = \beta dc \times I_B$

$I_B = \frac{V_B – V_E}{R_B + ( 1 + \beta) R_R}$

Y = ABC + BCD + ABC and realize using basic logic gates.


= BC ( A + A ) + BCD

= BC + BCD

= B ( C + CD)

= B. (C + D)

Y = B.C + B.D

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