CHAPTER 1 Voltage Amplifier Amplifier Characteristics AV = Open circuit voltage gain AV vi = Open circuit voltage Ri = input resistance of the amplifier Ro = output resistance of the amplifier Voltage Amplifier Ai = short circuit current gain Ai ii = short circuit current Amplifier Characteristics Current Amplifier Passive and Active Components Analog and Digital Signals CHAPTER 1

Operational Amplifier Inverting Amplifier Op Amp Configurations Summing Amplifier Non - Inverting Amplifier If you don’t remember the formulas, remember these two characteristics and perform nodal analysis vo = Aod ( v2 – v1) Operational Amplifier Open loop mode v1 = v2 to create Aod =  Two main characteristics CHAPTER 8 No current going into the op-amp

CHAPTER 2 Forward Biased, DC Analysis AC Analysis Reverse Biased Model 1 V = 0 Model 2 V Model 3 V and rf Load Line  ID vs VD At 300K VT = 0.026 V Forward Biased, DC Analysis AC Analysis Thermal equilibrium, depletion region Reverse Biased Must perform DC Analysis first to get DC diode current, ID PN junction Group 5 N-type P-type Group 3 Calculate rd = VT / ID Insulator Conductor Semiconductor Extrinsic Semiconductor: Group 4 eg. Silicon and Germanium photodiode Intrinsic Other types of diode Solar cells Materials Bandgap Energy LED CHAPTER 2 Zener Diode

Vm is the peak value of the output voltage Half Wave V r = V m T P RC If the ripple is very small, we can approximate T’ = Tp where Tp is the period of the cycle VC = Vme – t / RC Full Wave V r = V m T P 2RC Vm is the peak value of the output voltage Capacitor Discharge V r = V m T ′ RC Ripple Voltage, Vr Vo = Vs - V Filter Vo = Vs - 2V PIV = 2Vspeak - V Center-tapped Duty Cycle Bridge PIV = Vspeak - V Full Wave Vo = Vs - V CHAPTER 3 Rectifier Half Wave PIV = Vspeak

Consider a silicon pn junction Consider a silicon pn junction. The n-region is doped to a value of Nd = 1016 cm-3. The built in potential barrier is 0.712 V. Determine the required p-doping You need to calculate ni  the required constants will be given 0.712 = 0.026 ln [(Na )(1016) / (1.5 x 1010)2] e (0.712/0.026) = [(Na )(1016) / (1.5 x 1010)2] Na = 1.76 x 1016 cm-3

A diode is biased at ID = 1mA A diode is biased at ID = 1mA. The peak to peak of its sinusoidal current is 0.05 ID. Find the value of the diode sinusoidal voltage, vd (peak-to-peak) AC analysis only since ID already calculated rd = VT / ID = 0.026 / 1mA = 26  vd = idrd = 0.05 mA (26) = 1.3 mV (peak to peak) rd id

Full wave rectifier with iDpeak = 0. 2 A and vOpeak = 12 V Full wave rectifier with iDpeak = 0.2 A and vOpeak = 12 V. Ripple voltage is to be limited to 0.25 V. The input signal is 120 V (rms) at 60 Hz. Assume V = 0.7 V Determine the required turns ratio Find the required value of the capacitor What is the PIV rating of the diode

Part (i) 0.7 + VO – VS = 0 VS peak = 12.7V Change to rms 8.98 V (rms) Turns ratio = 120 / 8.98 = 13.36 Part (ii) For full wave: V r = V m T P 2RC Vr is given as 0.25V R? vOpeak = 12 V R = 12 / 0.2 = 60  0.25 (2)(60)(C) = 12 (1/60) C = 6667 µF Part (iii) PIV for center-tapped: 2VSpeak - V = 2(12.7) – 0.7 = 24.7 V

Full wave battery charger circuit Full wave battery charger circuit. Given that VB = 9V and vs = 15 sin ( 2ᴨ60t) (V). Assume V = 0.7 V Determine R such that the peak current is 1.2 A Determine the fraction of time that each diode is conducting

Part (i) 0.7 + IDR + 9 – 15 = 0 R = (15 – 9 – 0.7) / 1.2 R = 4.42  Part (ii) VS must be at least 9.7 V 15 sin ( 2ᴨ60t) = 9.7 sin ( 2ᴨ60t) = 0.64667 2ᴨ60t = 40.29 and 139.7 Duty cycle for each diode = ( 139.7 – 40.29) / 360 ) = 27.6%