![]() Where is the phase angle between V C and I L. Power Gain A p:įrom the circuit of Figure 2, the average power delivered to the load imprudence R L is given by: Then the overall current gain A Is is given by:įrom Equations (20) and Equation (23) we get, ….(24)Įquation (24) is true provided that the voltage source V s and the current source I s have the same source resistance R s. This current source drives the amplifier resulting in I b at the input terminals of the amplifier and current I L through the load impedance. We may replace the voltage source V s with series source resistance R s by what is known as Norton’s equivalent source shown in Figure 3(b), consisting of current source I s with source resistance R s in the shunt. Thus, A V forms a special care of A VS with R s = 0. This energy source then drives the amplifier represented by its input resistance R i. This form of an equivalent circuit for the energy source is known as Thevenin’s equivalent source. Then the overall voltage gain considering the source resistance is given byįigure 3(a) given the driven voltage source V s with source resistance R s in series. Source voltage V s applied at the input of an amplifier results in voltage V b between bae and emitter terminals (input terminals) of the transistor and voltage V c at the output. ….(17) Overall Voltage Gain Considering R s: If we include R L in parallel with R 0, we get the output terminal impedance Z t given by, In the calculation of Y 0, R L has been considered external to the amplifier. If the source impedance is purely resistive, then the output impedance Y 0 is real i.e. On substituting the value of I c from Equation (2) into Equation (12) we get,īut with V s = 0, Figure 2 gives (R s + h ie) I b + h re V c = 0Ĭombining Equation (13) and (14) we get, ……(15)Įquation (15) shows that the output admittance Y 0 is a function of source resistance R s. It is the ratio of the output current I c to the output voltage V c with V s = 0. Voltage Gain ….(11) Output Admittance Y 0: It is the ratio of the output voltage V c to the input voltage V b. Voltage Gain or Voltage Amplification | Analysis of Common Emitter Amplifier using H P arameter: Substituting the value of V c from Equation (7) into Equation (6) we get,įrom Equation (10) we find that the input impedance R i is also a function of load resistance R L. This is the impedance between the input terminals B and E looking into the amplifier as shown in Figure 2 and is, therefore, given by, Hence current gain ….(4) Input Impedance R i: Hence in the equivalent circuit of Figure 2, we have used the RMS value of voltages and currents namely I b, V b, I c, and V c.Ĭurrent gain is defined as the ratio of the load current I 1 to the input current I b. This results in the equivalent circuit of Figure 2. ![]() equivalent circuit of Figure 1(b).Īnalysis of Common Emitter Amplifier using H Parameter:įor analysis, we replace the transistor with its small-signal two generator h-parameter model. the operation, R 0 comes in parallel with R c, and effective load resistance R L = R C || R O. ![]() The value of capacitor C b is chosen so large that its reactance at the operating frequency is small and may be neglected. output voltage developed across R c is capacitively coupled to the next stage through the capacitor C b and R 0 is the effective impedance in the output circuit. Hence R e-C z combination is also excluded from the a.c. Similarly, the reactance of capacitor C z is so small at the lowest operating frequency that C z effectively bypasses all a.c. It is assumed that R b is large in comparison with the input resistance of the amplifier between base and ground and hence R b is neglected in the equivalent circuit. The R 1 – R 2 combination is equivalent to resistance R b (= R 1 || R 2) between base and ground. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |