**1 Answer**

written 22 months ago by |

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.

- 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 = VBE + VE

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

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

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

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

$Ic = \beta dc \times I_B$

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

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

Y = ABC + ABC + BCD

= BC ( A + A ) + BCD

= BC + BCD

= B ( C + CD)

= B. (C + D)

Y = B.C + B.D