% Nodal analysis - single stage transistor amp % Created by: Jon Erickson 20 Sep 2017 % This solves the single stage amplifier example from class. % Define circuit parameters using those given in PS2, prob 3 Vin = 2; %Volts <--always a good idea to specify units in comments Rin = 50; R1 = 2800; %Ohms R2 = 35e3; R3 = 2e3; R4 = 250; R5 = 2500; B = 200; %beta value %Define forcing vector [I] I = [Vin; 0; 0; 0]; % Define Conductance matrix [G] (4x4 in this case, since we have 4 nodes to solve for). G = [1, 0, 0, 0; 1/Rin, -(1/Rin + 1/R1 + 1/R2 + 1/R3), 1/R3, 0; 0, (B+1)/R3, -((B+1)/R3 + 1/R4), 0; 0, B/R3, -B/R3, 1/R5 ] % OK, now let's see how to solve matrix equations. We want the nodal % voltages vis: [G^-1]*([G][V] = [I]) % Let's compute the nodal voltages: V = inv(G)*I % The variable V a 4x1 column vector, that is V = [Va; Vb; Vc; Vd]; Va = V(1); % <--returns the 1st element of V, ie what we prev defined to be Va Vb = V(2); % <-- returns the 2nd element of column vector V, and so on Vc = V(3); % <-- the 3rd element of V Vd = V(4); Vout = Vd; %output voltage Av = Vout/Vin %voltage gain i3 = (Vb-Vc)/R3 i4 = Vc/R4 currentGain = i4/i3 %Note: The voltage gain is negative, which means the output signal is %INVERTED relative to the input. A positive swing on a sine wave at the %input appears at the output as a negative swing, and vice-versa

G = 1.0000 0 0 0 0.0200 -0.0209 0.0005 0 0 0.1005 -0.1045 0 0 0.1000 -0.1000 0.0004 V = 2.0000 1.9603 1.8853 -18.7590 Av = -9.3795 i3 = 3.7518e-05 i4 = 0.0075 currentGain = 201.0000