Question

Solve for the following using:

1) exact method

2) approximate method

compare if approximate method is sufficient for analysis. which solution is better?

Example For the voltage divider bias circuit, RB1-13K2, R 2.2k2 , Rc-68K(2. RE-1.5K(2.Vcc-15 V. VCEsat -0.3 V and β of the Si transistor is 80. Determine the following O" (a) DC base voltage and DC emitter voltage of the transistor amplifier (b) DC operating voltages and currents of the transistor amplifier (lBo, Ico, IEo and VcEO) (c) Collector saturation level of the transistor (d) Maximum DC collector-emitter voltage of the transistor (e) Region of operation of the transistor

Answer 1

Applying Kirchhoff's voltage rule V_cc - 1 times 13 k ohm - V_BE - I_E times 1.5 k ohm = 0 V_BE = 0.7 volts for s_i Therefore V_cc - 13 k (I) - 2.2 k (7 - I_B) = 0 also I_E/I_B = 1 + beta = 81 I_E = 8 + I_B Because V_cc - (13 k) I - 0.7 - 81 times 1.5 k I_B = 0 ...(1) & V_cc - (13 k) I - 2.2 k (I - I_B) = 0 ..(ii) from (1) & (II) I = 965 mu A I_B = 10 mu A a) V_B = V_cc - I times 13 k = 15 - 965 times 10^-6 times 13 times 10^3 [V_B = 2.455 volts] [V_E = 1.755 V] b) [I_B Q = 10 mu A,] [I_CQ = beta]_BQ = 800 mu A]] I_E Q = 81 times I_BQ = 810 mu A] V_CE Q, V_CC - I_CQ times 6.8 k - V_CE Q - I_E Q times 1.5 k = 0 Therefore [V_CE Q V_cc - I_CQ times 6.8 k - I_EQ times 1.5 k] \r\n [V_C EQ = 8.345 volts] c) for saturation V_CE = 0.3 V Therefore V_cc - I_c
The approximate method is only suitable to instances when the input resistance at the base terminals is very large as compared to R2 of potential divider