1. 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

2. 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

3. 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 = 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

4. 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

5. 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 V. Determine the required p-doping
You need to calculate ni  the required constants will be given
0.712 = 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

6. 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 = / 1mA = 26 
vd = idrd = 0.05 mA (26) = 1.3 mV (peak to peak) rd id

7. 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

8. 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

9. 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

10. 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) =
2ᴨ60t = 40.29 and 139.7
Duty cycle for each diode = ( 139.7 – 40.29) / 360 ) = 27.6%