1. CHAPTER 3 Range of power supply Voltage Regulator VRi = VPS - VZ
Assume: IZmin = 0.1 IZ(max)
Range of power supply
Voltage Regulator
The load resistor sees a constant voltage regardless of the current
VL = VZ
VRi = VPS - VZ
II = IZ + IL
Zener Diode
CHAPTER 3

2. 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 = Vm e – 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
Multiple Diode Circuit
Ripple Voltage, Vr
Vo = Vs - V
Filter
Vo = Vs - 2V
PIV = 2Vspeak - V
Peak and RMS
V rms = V peak 2
Center-tapped
Duty Cycle
Bridge
PIV = Vspeak - V
Full Wave
Vo = Vs - V
CHAPTER 3
Rectifier
Half Wave
PIV = Vspeak

3. Chapter 4 Bipolar Junction Transistor

4. REMEMBER THIS
Current flow in the opposite direction of the electrons flow;
same direction as holes e e e I h h h

5. Transistor Structures
The bipolar junction transistor (BJT) has three separately doped regions and contains two pn junctions.
Bipolar transistor is a 3-terminal device.
Emitter (E)
Base (B)
Collector (C)
The basic transistor principle is that the voltage between two terminals controls the current through the third terminal.
Current in the transistor is due to the flow of both electrons and holes, hence the name bipolar.

6. 3 Regions of Operation Active Operating range of the amplifier.
Base-Emitter Junction forward biased.
Collector-Base Junction reverse biased
Cutoff
The amplifier is basically off. There is voltage but little current.
Both junctions reverse biased
Saturation
The amplifier is full on. There is little voltage but lots of current.
Both junctions forward biased

7. OPERATIONS - npn VBE = V
FORWARD ACTIVE MODE
VBE = V +
VBE
The base-emitter (B-E) junction is forward biased and the base-collector (C-B) junction is reverse-biased,. -
Since the B-E junction is forward biased, electrons from the emitter are injected across the B-E junction into the base  IE
Once in the base region, the electrons are quickly accelerated through the base due to the reverse-biased C-B region  IC iB
Some electrons, in passing through the base region, recombine with majority carrier holes in the base. This produces the current IB

8. TO ILLUSTRATE E B C - VBE + Imagine the marbles as electrons
A flat base region with gaps where the marbles may fall/trapped – recombine
A sloping collector region represents high electric field in the C-B region
Hence, when enough energy is given to the marbles, they will be accelerated towards to base region with enough momentum to pass the base and straight ‘fly’ to the collector

9. MATHEMATICAL EXPRESSIONSIB + IE
VBE -
IE = IS [ e VBE / VT -1 ] = IS e VBE / VT
Based on KCL: IE = IC + IB
No. of electrons crossing the base region and then directly into the collector region is a constant factor  of the no. of electrons exiting the base region
IC =  IB
No. of electrons reaching the collector region is directly proportional to the no. of electrons injected or crossing the base region.
Ideally  = 1, but in reality it is between 0.9 and
IC =  IE

10. Based on KCL: IE = IC + IB IC =  IB IC =  IE
IE =  IB + IB = IB(  + 1)
IE = IB(  + 1)
Now
With IC =  IB  IB = IC / 
Hence,
IE = [ IC / ] (  + 1)
IC = IE [  /  + 1 ]
Comparing with IC =  IE
 = [  /  + 1 ]

11. OPERATIONS - pnp IE = IS [ e VEB / VT -1 ] = IS e VEB / VTC E
OPERATIONS - pnp IB IC
FORWARD ACTIVE MODE -
The emitter – base (E- B) junction is forward biased and the base-collector (B- C) junction is reverse-biased,.
VEB IE +
VEB = V
IE = IS [ e VEB / VT -1 ] = IS e VEB / VT
**Notice that it is VEB
Based on KCL: IE = IC + IB

12. SUMMARY: Circuit Symbols and Conventions
Based on KCL: IE = IC + IB
npn bipolar transistor simple block diagram and circuit symbol.
Arrow on the emitter terminal indicates the direction of emitter current
pnp bipolar transistor simple block diagram and circuit symbol.

13. IC =  IB IC =  IE IE = IB(  + 1) Based on KCL: IE = IC + IB
NPN
PNP
IE = IS [ e VBE / VT ]
IE = IS [ e VEB / VT]
IC =  IB
IC =  IE
IE = IB(  + 1)
𝜶= 𝜷 𝟏+𝜷
𝜷= 𝜶 𝟏−𝜶
Based on KCL: IE = IC + IB

14. EXAMPLE
Calculate the collector and emitter currents, given the base current and current gain. Assume a common-base current gain,  = 0.97 and a base current of iB = 25 µA . Also assume that the transistor is biased forward in the forward active mode.
Solution: The common-emitter current gain is
The collector current is
And the emitter current is

15. Examples
EXAMPLE 1
Given IB = 6.0A and IC = 510 A. Determine ,  and IE
Answers:
 = 85 =
IE = 516 A
EXAMPLE 2
NPN Transistor
Reverse saturation current Is = 10-13A with current gain,  = 90.
Based on VBE = 0.685V, determine IC , IB and IE
Answers:
IE = (e 0.685/0.026) = A
IC = (90/91)(0.0277) = A
IB = IE – IC = 0.3 mA

16. BJT: Current-Voltage Characteristic IC versus VCE

17. Common-Emitter Configuration - npn
The Emitter is common to both input (base-emitter) and output (collector-emitter).
Since Emitter is grounded, VC = VCE
With decreasing VC (VCE), the junction B-C will become forward biased too.
The current IC quickly drops to zero because electrons are no longer collected by the collector
Node B 0V

18. Characteristics of Common-Emitter - npn