#### CBSE NOTES CLASS 12 PHYSICS

#### CHAPTER 14 SEMICONDUCTORS

**Transistor as a device**

**Transistor as an amplifier****Amplification of dc voltage:**In the active region transistor behaves like an amplifier. For V_{o}versus V_{i}curve, the slope of the linear part of the curve represents the rate of change of the output with the input. It is negative because the output is V_{CC}– I_{C}R_{C}and not I_{C}R_{C}. Hence, the output voltage of the CE amplifier decreases as input voltage increases. In this case the output is said to be out of phase with the input. If we consider ΔV_{o}and ΔV_{i}as small changes in the output and input voltages then $\frac{\mathrm{\Delta}{\mathrm{V}}_{\mathrm{o}}}{\mathrm{\Delta}{\mathrm{V}}_{\mathrm{i}}}$ is called the small signal voltage gain A_{V}of the amplifier.If the V

_{BB}voltage has a fixed value corresponding to the midpoint of the active region, the circuit will behave as a CE amplifier. The dc base current I_{B}would be constant and corresponding collector current I_{C}will also be constant.The voltage gain A

_{V}can be expressed in terms of the resistors in the circuit and the current gain of the transistor as follows.We have, V

_{o}= V_{CC}– I_{C}R_{C}Therefore, ΔV

_{o}= 0 – R_{C}Δ I_{C}Similarly, from V

_{i}= I_{B}R_{B}+ V_{BE}ΔV

_{i}= R_{B}ΔI_{B}+ ΔV_{BE}But ΔV

_{BE}is negligibly small in comparison to ΔI_{B}R_{B}in this circuit.$$\mathrm{O}\mathrm{r},\mathrm{}{\mathrm{A}}_{\mathrm{V}}\mathrm{}=\mathrm{}\u2013\mathrm{}\frac{{\mathrm{R}}_{\mathrm{C}}\mathrm{}\mathrm{\Delta}\mathrm{}{\mathrm{I}}_{\mathrm{C}}}{{\mathrm{R}}_{\mathrm{B}}\mathrm{}\mathrm{\Delta}{\mathrm{I}}_{\mathrm{B}}}=\mathrm{}\u2013{\mathrm{\beta}}_{\mathrm{a}\mathrm{c}}\left(\frac{{\mathrm{R}}_{\mathrm{C}}}{{\mathrm{R}}_{\mathrm{B}}}\right)$$

Also, the dc voltage V

_{CE}= V_{CC}- I_{C}R_{C}would remain constant. The operating values of V_{CE}and I_{B}determine the operating point, of the amplifier.**Amplification of ac signal**If a small sinusoidal voltage with amplitude v

_{s}is superposed on the dc base bias by connecting the source of that signal in series with the V_{BB}supply, then the base current will have sinusoidal variations superimposed on the value of I_{B}. As a consequence the collector current also will have sinusoidal variations superimposed on the value of I_{C}, producing in turn corresponding change in the value of V_{O}. We can measure the ac variations across the input and output terminals by blocking the dc voltages by large capacitors.Let us superimpose an ac input signal v

_{i}, to be amplified, on the bias V_{BB}(dc). The output is taken between the collector and the ground.To start with let us assume that v

_{i}= 0. Then applying Kirchhoff’s law to the output loop, we get,V

_{cc}= V_{CE }+ I_{C}R_{C}And from the input loop, we get,

V

_{BB}= V_{BE}+ I_{B}R_{B}When v

_{i}is not zero, we get,V

_{BE }+ v_{i }= V_{BE}+ I_{B}R_{B}+ ΔI_{B}(R_{B}+ r_{i})Or v

_{i}= ΔI_{B}(R_{B}+ r_{i}) = r ΔI_{B}$${}_{\mathrm{a}\mathrm{c}}\mathrm{}=\mathrm{}{\left(\frac{\mathrm{\Delta}{\mathrm{I}}_{\mathrm{C}}}{{\mathrm{\Delta}\mathrm{I}}_{\mathrm{B}}}\right)}_{{\mathrm{V}}_{\mathrm{C}\mathrm{E}}}=\frac{{\mathrm{i}}_{\mathrm{c}}}{{\mathrm{i}}_{\mathrm{b}}}$$

The power gain A

_{p}can be expressed as the product of the current gain and voltage gain. MathematicallyA

_{p}= β_{ac}× A_{V}- The transistor is not a power generating device. The energy for the higher ac power at the output is supplied by the battery.