Question

can you do 4.83

Ar- Q Sea 100 V, what does the gain become? age at the collector. (b) Replacing the transistor by its T model, da the small-signal equivalent circuit of the a plifier. Analyze the resulting circuit to dete mine the voltage gain t/ 04.81 Consider the CE amplifier circuit of Fig. 4.43(a). It is required to design the circuit (i.e., find values for I and Rc) to meet the following specifications: (a) R,5kn (b) the voltage gain from base to collector is the maximum possible, consistent with the re- quirement that the collector voltage never falls below the base voltage with the signal between base and emitter being as high as 5 mV. 4.86 Refer to the voltage-gain expression (in terms a the transistor B) given in Eq. (4.75) for the CE amplifier with a resistance R, in the emitter. Let Assume that , is a sinusoidal source with zero de component, the available supply Vcc 5 V and that the transistor has B 100 and a very large Early voltage. What base-to-collector volt- age gain does your design provide? If R, 10 kn, what is the overall voltage gain? +15 V Rc D482 In the circuit of Fig. P4.82, is a small sine- wave signal with zero average. The transistor B is 100 (a) Find the value of Rg to establish a dc emitter current of about 1 mA. (b) Find Re to establish a de collector voltage of about +5 V. (c) For R-5kn and the transistor r, 100 kn, draw the small-signal equivalent cir- cuit of the amplifier and determine its overall voltage gain. R,-25 kn R RE V 4.83 The amplifier of Fig. P4.83 consists of two iden- tical common-emitter amplifiers connected in cascade. Observe that the input resistance of the -15 V Flg. P4.82 gee nert page for dorme P483 112 5 FS F7 FB % 5 & 0 7 8 9 + II

Answer 1

(a) the base voltage is determined from V_CC - V_B/R_1 = V_B/R_2 + V_B - V_BE/(beta + 1) R_E V_B (1/R_1 + 1/R_2 + 1/(beta + 1) R_E) = V_CC/R_1 + V_BE/(beta + 1) R_E V_B = (V_CC/R_1 + V_BE/(beta + 1) R_E) R_1 Vertical-bar Vertical-bar R_2 Vertical-bar Vertical-bar (beta + 1) R_E V_B = (15/100 + 0.7/101 times 3.9) times 100 Vertical-bar Vertical-bar 47 Vertical-bar Vertical-bar (101 times 3.9) V_B = 4.5 V then, V_E = V_B - V_BE = 4.5 - 0.7 = 3.8 v I_E = V_E/R_E = 0.97 m A I_C = alpha I_E = 100/101 times 0.97 = 0.96 m A v_c = V_CC - I_C R_C = 15 - 0.96 times 6.8 = 8.5 v (b) small - signal model - \r\n g_m = I_C/V_T = 0.96/25 times 01^-3 = 0.038 Ohm^-1 r_e = alpha/g_m = beta/beta + 1/g_m = 100/101/0.038 = 26 Ohm (c) input resistance is, R_in1 = R_1 Vertical-bar Vertical-bar R_2 Vertical-bar Vertical-bar (beta + 1) (r_e + R_E) = 100 Vertical-bar Vertical-bar 47 Vertical-bar Vertical-bar (101) (26 times 10^-3 + 3.9) = 3.50 k Ohm V_v1/V_sig = R_in1/R_sig + R_in1 = 3.5/5 + 3.5 = 0.41 (d) the input resistance on the second stage is the same as the input