6.101 Problem Set 2 Solutions

Problem Set 2 Solutions 6.101 Analog Electronics Lab Spring 2007 Problem 1 (a) (6pts) VB = 0V ⇒ IB = 0 IC = β + 0 = 0 VC = 6 − IC · RC = 6.0 − (0 · 10KΩ) = 6.0V (b) (6pts) VB = 1V ⇒ VE = VB − 0.7V = 0.3V IE = VE 0.3V = = 0.3mA ≈ IC RE 1KΩ VC = 6 − IC · RC = 6.0 − (.3mA · 10KΩ) = 3.0V (c) (6pts) VB = 2V ⇒ VE = VB − 0.7V = 1.3V IE = VE 1.3V = = 1.3mA ≈ IC RE 1KΩ VC = 6 − IC · RC = 6.0 − (1.3mA · 10KΩ) = −7.0V < VCEsat ≈ 0.2V The transistor is on because of the large base voltage, but it must be in saturation and at the edge of the saturation region. Thus VCE = 0.2V . VC = VE + VCEsat ≈ VE + 0.2V = 1.3V + 0.2V = 1.5V Problem 2 (12pts) 50mV = 12.5mA 4Ω VT 26mV rd ≈ = = 2.1Ω ID 12.5mA � � 4Ω = 6 sin(100t)mV = 3.9 sin(100t)mV 2.1Ω + 4Ω ID = IR = vout vOUT = VOUT + vout = 50mV + 3.9sin(100t)mV 1 Problem 3 (a) (15pts) IE ≈ IC = 4mA VE = IE RE = 4mA ∗ 100Ω = 0.4V VB = VE + 0.6V = 0.4V + 0.6V = 1.0V Make the resistor divider voltage equal the desired base voltage: VB = 1KΩ 15V = 1.0V → R = 14KΩ 1KΩ + R Use standard values, either 13KΩ or 15KΩ. (b) (15pts) VB = 15 1000Ω − (1000Ω||R)IB 1000Ω + R Both 13KΩ and 15KΩ are much greater than 1000Ω so (1000Ω||R) ≈ 1KΩ. VB ≈ 1.0V − 1000IB IC = β VB − 0.6V β+1 100 IC 10 VB − 0.6V = 10 11 1000Ω 10 VB = 1 − (VB − 0.6V ) 11 VB ≈ 0.81V IB = IC ≈ 1.9mA Problem 4 (20pts) RT H = R1 ||R2 = 20KΩ||15KΩ = 8.57KΩ VT H = R1 15KΩ VCC = 10 = 4.29V R1 + R2 15KΩ + 20KΩ IEQ RT H + VT H 1+β � � 8.57KΩ 10 = IEQ 1KΩ + + 0.7V + 4.29V 101 VCC = IEQ RE + VEB + IEQ = VB = (10 − 0.7 − 4.29)V 5.01V = = 4.62mA 1KΩ + 8.57KΩ 1.085KΩ 101 IEQ 4.62mA RT H + VT H = 8.57KΩ + 4.29 = 4.68V 1+β 101 2 Problem 5 (a) (10pts) A large capacitor, often lovingly referred to as a Big Fat Capacitor (BFC), can be placed in parallel with the emitter resistor. This does not affect the DC bias stability of the amplifier, but does drastically increase the AC gain. (b) (10pts) The Q point is the operating point of circuit. It is determined by the bias currents and voltages that result from resistor values chosen in the design of a circuit. On a load line graph, the Q point illustrates the operating collector current, base current, and voltage difference between the collector and the emitter of a transistor. Choose a Q point that allows for sufficient output swing, reasonable current levels, and keeps the amplifier in the desired region of opera­ tion (usually the forward active region, VCE¿ 0.2V). The x intercept of a load line is the open circuit voltage (i.e., take out the transis­ tor) between the collector and emitter nodes of the transistor. The y intercept of a load line is the short circuit current between the col­ lector and the emitter of the transistor (i.e. the current through the collector if you placed a wire between the collector and the emitter). 3