1)Identify the transistor operating point (Q pt.) Vceq = ? Icq = ? and Ibq = ? 2

Question

1)Identify the transistor operating point (Q pt.)   Vceq = ? Icq = ? and Ibq = ?

2)Calculate the transistor small signal input resistance r assuming the operating temp is 27 degrees celsius.

3)Find the small signal voltage amplifier circuit parameters ri, ro, and avo (avo = Vo / Vin with RL = ) and draw the equivalent circuit include on the equivalent circuit diagram the voltage source (Vs and Rs) the load resistance (RL) as well as ri , ro , and avo and assume XC1 = XC2 = 0.

A 2N222A BJT is shown, graphical analysis of the transistor shows that ?DC = ?AC = 150 and rOC = 50k?. Tests also show that VBE(on) = .7 Volts and VceSAT = .2 volts. The amplifoer design characteristics are Vcc = 10 volts, RB = 330k?, RC = 1.6k?, RL = 3k?, and C1 = C2 = 2.5 uF, find: 1)Identify the transistor operating point (Q pt.) Vceq = ? Icq = ? and Ibq = ? 2)Calculate the transistor small signal input resistance r? assuming the operating temp is 27 degrees celsius. 3)Find the small signal voltage amplifier circuit parameters ri, ro, and avo (avo = Vo / Vin with RL = ?) and draw the equivalent circuit include on the equivalent circuit diagram the voltage source (Vs and Rs) the load resistance (RL) as well as ri , ro , and avo and assume XC1 = XC2 = 0.

Explanation / Answer

    a) To find the operating point Apply KVL to the input section Vcc = RBIB + VBE IB = (VCC-VBE)/RB = (10-0.7)/330000 = 28A IC = IB = (1500(28) = 4.2mA Apply KVL to the output setion Vcc = ICRC + VCE VCE = Vcc - ICRC = 10 - (0.0042)(1600) = 10 - 6.72=3.28V VCEQ = 3.28V, ICQ = 4.2mA and IBQ = 28A b) Transconductance gm = IC / VT VT at 27o C is ˜ 26mV gm = 0.0042 / 0.026 = 0.16 mhos r = /gm = 150 / 0.16 = 931.5 c) Input resistance is ri = (RB || r ) IB = (VCC-VBE)/RB = (10-0.7)/330000 = 28A IC = IB = (1500(28) = 4.2mA Apply KVL to the output setion Vcc = ICRC + VCE VCE = Vcc - ICRC = 10 - (0.0042)(1600) = 10 - 6.72=3.28V VCEQ = 3.28V, ICQ = 4.2mA and IBQ = 28A b) Transconductance gm = IC / VT VT at 27o C is ˜ 26mV gm = 0.0042 / 0.026 = 0.16 mhos r = /gm = 150 / 0.16 = 931.5 c) Input resistance is ri = (RB || r ) ri = (330k||931.5) ˜ 931.5 Output resistance ro = roC || RC = 50k || 1.6k = 1.55k Voltage gain Avo = - gm ro = - (0.16)(1550) = 248

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