1. Recap in Unit 2
EE2301: Block B Unit 2

2. The Transistor
A transistor is a 3-terminal semiconductor device (cf Diode is a 2-terminal device)
Performs 2 main functions fundamental to electronic circuits:
Amplification – magnifying a signal
Switching – controlling a large current or voltage across 2 terminals (on/off)
2 major families of transistors:
Field Effect Transistors (FETs) – Unit 2
Bipolar Junction Transistors (BJTs) – Unit 3
Unit 8

3. Unit 8

4. Basic Structure of a MOSFET
A quick revision on
Basic Structure of a MOSFET
EE2301: Block B Unit 2

5. N-channel MOSFET n+ n+ pG D S D
Gate
Oxide
Substrate
IG=0 n+ n+ p G
IG=0
Bulk (substrate) S
Gate current sees a capacitor formed by the gate, oxide (which is insulating), and substrate (conducting)
EE2301: Block C Unit 2

6. NMOS is normally off
When VGS > VT
EE2301: Block C Unit 2

7. Operating Regions BREAKDOWN SATURATION CUTOFF When VGS - VT = VDS
ID = K VDS2
TRIODE/OHMIC
SATURATION
CUTOFF
EE2301: Block C Unit 2

8. Bipolar Junction Transistor
Let’s con’t Unit 3
Bipolar Junction Transistor
EE2301: Block B Unit 2

9. Block C Unit 3 Outline The bipolar transistor (BJT) operation
Construction of the BJT
Basic working principle of the BJT
Operation modes of the BJT
Transistor amplifier circuits
Biasing the transistor in a circuit
Small signal equivalent circuit (hybrid model)
G. Rizzoni, “Fundamentals of EE” Sections 10.1 – 10.4
EE2301: Block C Unit 3

10. Construction of a BJT Summary IE = IB + IC
1. A BJT has an emitter, collector and base, while a MOSFET has a gate, source and drain.
2. The operation of MOSFET depends on the voltage at the oxide-insulated gate electrode, while the operation of BJT is dependent on the current at the base.
3. BJTs are preferred for low current applications, while MOSFETs are for high power functions.
4. In digital and analog circuits, MOSFETs are considered to be more commonly used than BJTs these days.
Formed by joining 3 sections of semiconducting material of alternating doping nature
This can be either:
Thin n region sandwiched between p+ and p layers (called pnp-transistor)
Thin p region sandwiched between n+ and n layers (called npn-transistor)
Comprises three terminals (each one associated with one of the three sections) known as the emitter (E), base (B) and collector (C)
IE = IB + IC
EE2301: Block C Unit 3

11. Let’s have a quick revision on p-n junction first
How does it work?
Let’s have a quick revision on p-n junction first
EE2301: Block B Unit 2

12. p-n Junction
The pn junction forms the basis of the semiconductor diode
Within the depletion region, no free carriers exist since the holes and electrons at the interface between the p-type and n-type recombine.
EE2301: Block C Unit 1

13. Diode Physics iD + - + - Forward Biased: Diode conducts + + + + - - -
Forward Biased:
Diode conducts iD + + + +
- - -
+ + + - - - - + + +
- - -
+ + + - - - + + + +
- - -
+ + + - - - - + + + + - + - - - - + + + - + - - - + + + + - + - - - -
Semiconductor Electronics - Unit 1: Diodes

14. Diode Physics - + - + Reverse Biased: Little or no current + + + +
Reverse Biased:
Little or no current + + + +
- - -
+ + + - - - - + + +
- - -
+ + + - - - + + + +
- - -
+ + + - - - - + + + - - - + + + - - - + + + - - -
Website:
Semiconductor Electronics - Unit 1: Diodes

15. 3 different operation regions compared to MOSFET
EE2301: Block B Unit 2

16. Large Signal Model
Like the MOS, the BJT is also characterized by 3 operating regions, but differ in what each region means. The operation principle is also obviously different between the two.
For a BJT, it is easier and typical to think of the emitter current (IE) to be controlled by the base current (IB) and the voltage across the collector and emitter (VCE)
Cut-off region
BE junction is reverse-biased
Linear active region
BE junction is forward-biased
(VBE = Vγ typically around 0.7V)
BC junction is reverse-biased
Saturation region
BOTH BE and BC junctions are forward-biased (VBE = Vγ)
EE2301: Block C Unit 3

17. Cutoff mode
In the cutoff mode, the pn junctions at both the emitter-base and collector-base interfaces are reverse biased. This is similar to having two back to back reverse biased diodes, where it can be seen that no current can flow between the emitter and collector terminals. As such, the transistor is simply off. There is also no current flowing into the base in this mode.
To turn the transistor on, we must forward bias the base-emitter junction, VBE>0
EE2301: Block C Unit 3

18. Linear active mode Vγ VCB > 0 B n E C p n+
The active region is characterized by a small base current greatly amplified to give a much larger collector current, such that IB << IC
It is common to expression their relationship by a parameter symbolized by β: IC = βIB, where β is typically large in the active region (from 50 to above 400)
As a result of this large current gain, we find that IC ~ IE
Holes in the base are injected into the emitter Vγ B
VCB > 0 IB n IE E C p IC n+
Electron flow
Electrons in the emitter are injected into the base
Some of the emitter electrons recombine in the base but most it is swept into the collector
Large electric field across the junction sweeps electrons in the base into the collector with minimal recombination occurring in the thin base layer.
EE2301: Block C Unit 3

19. Saturation mode Vγ VCB < 0 B n+ n E C p
The saturation region is characterized by low current gain between the base and collector.
Base current is now larger due to hole injection into the collector from the base
Holes in the base are also injected into the collector now Vγ
VCB < 0 B IB n+ n IE E C IC p
Electron flow
Potential gradient between collector and emitter is greatly reduced since VCE < Vγ (but still positive), having an effect on the collector current
Accelerating electric field across the junction is now switched off. Electrons diffuse into collector.
EE2301: Block C Unit 3

20. BJT I-V Characteristic (1)
When Collector (C) is open
BE junction similar to diode
IBB injects current to forward-bias the BE junction
Unit 8

21. BJT I-V Characteristic (2)
Collector Characteristic: Family of curves (for each value of IB, an IC-VCE curve can be generated)
Unit 8

22. BJT I-V Characteristic (3)
BE forward-biased, CB reverse-biased (amplifier operation)
Both junctions are forward-biased
VCE is small
Breakdown region: Physical limit of operation of the device
Both junctions are reverse-biased
Unit 8

23. Bipolar Junction Transistor
Recap in last lecture
Bipolar Junction Transistor
EE2301: Block B Unit 2

24. Construction of a BJT IE = IB + IC
Formed by joining 3 sections of semiconducting material of alternating doping nature
This can be either:
Thin n region sandwiched between p+ and p layers (called pnp-transistor)
Thin p region sandwiched between n+ and n layers (called npn-transistor)
Comprises three terminals (each one associated with one of the three sections) known as the emitter (E), base (B) and collector (C)
IE = IB + IC
EE2301: Block C Unit 3

25. BJT I-V Characteristic (3)
BE forward-biased, CB reverse-biased (amplifier operation)
Both junctions are forward-biased
VCE is small
Breakdown region: Physical limit of operation of the device
Both junctions are reverse-biased
Unit 8

26. Large Signal Model
Like the MOS, the BJT is also characterized by 3 operating regions, but differ in what each region means. The operation principle is also obviously different between the two.
For a BJT, it is easier and typical to think of the emitter current (IE) to be controlled by the base current (IB) and the voltage across the collector and emitter (VCE)
Cut-off region
BE junction is reverse-biased
Linear active region
BE junction is forward-biased
(VBE = Vγ typically around 0.7V)
BC junction is reverse-biased
Saturation region
BOTH BE and BC junctions are forward-biased (VBE = Vγ)
EE2301: Block C Unit 3

27. Basic Techniques to determine operation region
EE2301: Block B Unit 2

28. Block C Unit 3 P.1 C B E Identify the terminals of the transistor
Determine whether it is PNP or NPN
PNP : VBE and VBC < 0  forward biased
NPN : VBE and VBC > 0  forward biased
Determine VBE and VBC to see if they are forward or reversed biased
Determine the operation region
Cutoff if both are reversed biased
Active linear if VBE is forward and VBC is reversed
Saturation if both are forward biased
VBC = (reversed)
VBE = VBC –VEC
= -0.7 – (-0.2)
= (reversed)
Both reversed biased  cutoff
EE2301: Block C Unit 3

29. Block C Unit 3 P.1 C B E VBE = 0.7 (forward) VBC = VBE –VCE
= 0.7 – (0.3)
= 0.4 (forward)
Both reversed biased  saturation
EE2301: Block C Unit 3

30. Block C Unit 3 P.1 C B E PNP type VBE = -0.6 (forward) VBC = VBE –VCE
= -0.6 – (-4.0)
= 3.4V (reversed)
VBE is forward and VBC is reversed  active linear
EE2301: Block C Unit 3

31. Finding the operating mode 1
Given following voltmeter readings, find the operating mode of the BJT:
V1 = 2V; V2 = 1.3V; V3 = 8V
The first question to ask is whether the BJT is on/off. This depends on VBE and if there is a base current. So we try to find VBE first:
VBE = V1 – V2 = 0.7V (this indicates the BJT is on)
Furthermore, IB = (4 – V1)/40 = 50µA (current flows into the base as if the BJT is on)
Now we need to check if it is in active and saturation mode. We know that if the BJT is in active mode, then β will be large. Therefore, we need to find the ratio of IC over IB.
VCC = ICRC + V3 12 = IC*1 + 8 IC = 4mA
β = 4/0.05 = 80 (BJT is in active mode)
To further verify this, we can check whether IE ≈ IC:
IE = V2/321 = 4.05mA (hence equal to IC + IB)
EE2301: Block C Unit 3

32. Finding the operating mode 2
Part II: Find the state of the BJT once again if VBB is now short-circuited. Note that all the voltmeter readings will change as a result.
Once again we start with the base to check if the BJT is on or off first. We can check for VBE or the direction of the base current.
If VBB is short circuited, what then are the options we can choose from?
Option 1: If we assume (or guess) the base is also grounded as a result, then IB = 0
Option 2: If we assume the base is at a positive voltage, then IB flows out of the base
Which ever option you choose, the conclusion is still the same  the BJT is in cutoff
EE2301: Block C Unit 3

33. Finding the Q-point (by graph)
Just like the MOS, the approach (and motivation) for finding the Q-point is the same for the BJT. Once again, the Q-point is found at the intersection of the load line and the I-V curves.
IBB = 0.15mA; VCC = 15V; RC = 375Ω
Load line: Apply KVL around the right side mesh,
VCC = VCE + ICRC
Since we want to draw this on the I-V curve, we should express this as IC against VCE:
IC = VCC/RC – VCE/RC
EE2301: Block C Unit 3

34. Finding Q-point (by calculation)
Given: RB = 62.7kΩ; RC = 375Ω; VBB = 10V; VCC = 15V; Vγ = 0.6V; β = 145
Find the Q-point , assuming that the transistor is operated in active region
We start by assuming the BJT is on VBE = Vγ.
IB = (VBB – Vγ)/RB = 0.15mA
IC = βIB = 21.75mA
VCE = VCC – ICRC = 15 – 21.75*0.375 = 6.84V
How do we know the transistor is operated in active linear region?
We must test whether VBE is forward biased and VBC is reversed biased
VBE = VBB- IBRB = 10 – (62.7)(0.15) = 0.595V>0 (forward baised);
VBC=VBE – VCE = = <0 (reversed biased)
EE2301: Block C Unit 3

35. Some More Examples
EE2301: Block B Unit 2

36. Block C Unit 3 Q.3
Given the circuit of Figure P10.3, determine the operation point of the silicon device with β = 100. Assume a 0.6-V offset voltage In what region is the transistor?
Answer:
IC=1.25mA; VCE=8.101V; active region
EE2301: Block C Unit 3

37. Block C Unit 3 Q.6
Given the circuit of Figure below, determine the operating point of the transistor. Assume a 0.6-V offset voltage and β = 150. In what region is the transistor?
Answers:
IC=2.25mA; VCE=7.857V; active region
EE2301: Block C Unit 3

38. Hybrid model equivalent circuit
The BJT is inherently a non-linear device (basic circuit theory is based on linear problems)
We can instead define the linear characteristics of the BJT for a given Q-point for small variations about the Q-point
We can do so using a small signal equivalent circuit that will now allow us to linearize the behavior of the BJT for a given operating point
Now we can use linear circuit theory analysis applicable for the conditions are the parameters of the model are valid
For the BJT, one popular model is the h-model (h for “hybrid” as in h-parameters)
EE2301: Block C Unit 3

39. Mathematics will not covered in this course
h parameter model
Linearizes the characteristics of a BJTT (which is otherwise nonlinear) for a given operating point
Only valid for a particular operating point
Only valid for small varying signals
h11 (input resistance): hi; h12 (reverse transfer voltage ratio): hr
h21 (forward transfer current ratio): hf h22 (output conductance): ho
Mathematics will not covered in this course
±50uA
±2.5V
Vi=VBE; i1=IB;V0=VCE; i0=IC
EE2301: Block C Unit 3

40. End of Block C Unit 3
EE2301: Block B Unit 2

41. Finding the operating region 3
Given following voltmeter readings, find the operating mode of the BJT:
V1 = 2.7V; V2 = 2V; V3 = 2.3V
Once again, we start with the base of the BJT.
VBE = V1 – V2 = 0.7V (hence BJT is on)
This time VBB is unknown, so we cannot find IB directly. But we follow the same path as previous, it looks like we can still find IB through IC and IE.
IE = V2/RE and IC = (VCC-V3)/RC
But RE and RC are both unknown, so it seems we cannot find any of the currents.
However, we can find VBC or VCE:
VBC = V1 – V3 = 0.4V (This is sufficient to forward-bias the CE junction BJT is in saturation)
VCE = V3 – V2 = 0.3V (low value indicative of saturation)
EE2301: Block C Unit 3