Chapter 4 Bipolar Junction Transistors

Chapter 4 Bipolar Junction Transistors

Outline

Basic operation of the npn Bipolar Junction Transistor

Load-line analysis of a common-emitter amplifier

The pnp bipolar junction transistor

Small-signal equivalent circuits

The common-emitter amplifier

The emitter follower (common-collector amplifier)

The common-base amplifier

4.1 Basic operation of the npn transistor

Figure 4.1 The npn BJT

A npn BJT consists of a thin layer of p-type material between two layers of n-type material.

Two pn junctions are formed in the device.

The current flowing across one junction affects the current in the other junction. It is this interaction that makes the BJT useful as an amplifier.

Basic operation in the active region The common-emitter configuration Active region (normal operation)

Base-emitter junction is forward biased The base-collector junction is reverse

biased

Figure 4.2 An npn transistor with variable biasing sources (CE configuration)

4.1 Basic operation of the npn transistor

The emitter region is doped heavily.

The base region is very thin.

Amplification by the BJT

4.1 Basic operation of the npn transistor

Figure4.4 CE characteristics of a typical npn BJT

A small change in vbe can result in an appreciable change in ib, this causes a much larger change in ic, and in suitable circuits, it is converted into a much larger voltage change than the initial change in vbe.

The common-emitter current gain

Device equations

4.1 Basic operation of the npn transistor

B

C

i

i 10~1000

CBE iii

The common-base current gain

E

C

i

i

Exercise:

1. Find the relationship between and

2. Estimate and for Figure4.4

4.2 Load-line analysis of a CE amplifier

The power-supply voltages VBB and VCC

bias the device at an Q point for which the amplification of the input signal is possible.Exercise:

1. Write KVL for input loop.

2. Write KVL for output loop.

P144 Example 4.2

Figure4.6 Common emitter amplifier

4.2 Load-line analysis of a CE amplifier

Pay attention to Phase relationship between vin and vout

Figure 4.11 Amplification in the active region

4.2 Load-line analysis of a CE amplifier (ref 4.4)

Active region model — amplification (1) Saturation region model (2) Cutoff region model (3) Inverted (reverse) model (4)

Model B-E junction B-C junction

(1) forward reverse

(2) forward forward

(3) reverse reverse

(4) reverse forward

Figure Distortion illustration

4.2 Load-line analysis of a CE amplifier

Distortion occurs?

Example: Determine the diode states for the circuits shown below.

Assume ideal diodes.

Figure 4.12 The pnp BJT

4.3 The pnp bipolar junction transistor

Think about:

1. Circuit configuration

2. Active region — source polarity

4.4 Large-signal DC analysis of BJT circuit

Example: Find the Q-point (VBEQ, IBQ, VCEQ, ICQ) of the circuit (P151)

Think about:

1. Indicate the current flowing through RB and RC

2. Write KVL for input/output loop

=100

VBEQ is given (about 0.7V for Si, and 0.2V for Ge)

IBQ=(VCC-VBEQ)/RB

ICQ= IBQ

VCEQ=VCC-ICQRC

Figure 4.13 The BJT

Figure 3.11 Half-wave rectifier with resistive load

Example: Find the Q-point of the circuit (P174)

4.4 Large-signal DC analysis of BJT circuit

Think about:

1. Indicate the current flowing through RB ， RE and RC

2. Write KVL for input/output loop

Figure 4.14 BJT with two voltage supplies

IBQ=(VB-VBEQ)/[RB+(1+)RE]

ICQ= IBQ

VCEQ=VCC-IEQRE-ICQRC VCC-ICQ(RC+RE)

Analysis of the four-resistor bias circuit

4.4 Large-signal DC analysis of BJT circuit

Method 1 — Thevenin equivalent circuit 　

VBQR2VCC/(R1+R2)

Figure 4.15 The BJT with four resistors

Convert to Figure4.14

R1 and R2 are in series

Method 2 — Assume IR1>>IBQ,

IR2>>IBQ

IEQ=(VBQ-VBEQ)/RE

VBQ VCEQ

+

-

VCEQ=VCC-IEQRE-ICQRC VCC-ICQ(RC+RE)

4.5 Small-signal equivalent circuit

Figure 4.16 small signal equivalent circuits for the BJT

Input resistance rbe

rbe(r)=300+(1+)26(mV)/IEQ(mA)

Tansconductance gm

gm= /rbe

Output resistance rce

Usually large, negligiblerce

rce

DC supply voltage —— bias the device at a suitable operating point AC signal —— what we want to amplify The use of C1, C2 and Ce

Coupling — C1, C2

bypass — Ce

Circuit configuration What shall we do with this circuit?

DC bias circuit — determine Q point Small signal equivalent circuit

Find Av, Rin and Rout

4.6 The common-emitter amplifier (P164)

Figure 4.28 Common-emitter amplifier of Example 4.9.

C2

Ce

4.6 The common-emitter amplifier (P164)

Figure 4.28 Common-emitter amplifier of Example 4.9.

1. Draw the DC bias circuit to find the Q point

Ac signal — short circuit

Capacitor — open circuit

VBQVCEQ

+

-

Rule of thumb: VCEQ 0.3~0.7VCC

2. Draw the AC circuit with BJT

DC signal — Connect to the GND

Capacitor — short circuit

4.6 The common-emitter amplifier (P164)

Figure 4.28 Common-emitter amplifier of Example 4.9.

rce

3. Replace with the equivalent circuit

Find Av=Vo/Vin, Rin=Vin/Iin and Rout=Vo/Io|Vs=0

4.6 The common-emitter amplifier (P164)

rce

Think about: 1. the relationship between R1, R2 and rce

2. the relationship between rce, RC and RL

3. Write two equations for Vin and Vo

Find the voltage gain Av=Vo/Vin

4.6 The common-emitter amplifier (P164)

Find the open circuit voltage gain Avo=Vo/Vin|RL=0

Find the source voltage gain Avs=Vo/Vs

Find the input resistance Rin=Vin/Iin

　Find the output resistance Rout=Vo/Io|Vs=0

rce

Rout

Analysis of the CE amplifier without CE

4.6 The common-emitter amplifier (P164)

C2

Ce

DC operating point

Remains unchanged

AC performance

Need more detailed analysis

Exercise: 1. Draw the small-signal equivalent circuit

2. Find Av=Vo/Vin, Rin=Vin/Iin and Rout=Vo/Io|Vs=0 assume rce

3. Compare with the results for circuit with CE

4.6 The common-emitter amplifier (P164)

Think about: 1. the currents flowing through rbe, RE and RL’

2. Write two equations for Vin and Vo

Figure 4.27 Common-emitter amplifier.

4.6 The common-emitter amplifier (P162)

Think about: 1. Effects of RE1 on Q-point, Av,Rin and Rout

2. Effects of RE2 on Q-point, Av,Rin and Rout

4.7 The emitter follower (P166)

1. Draw the DC bias circuit to find the Q point

Given R1=R2=100k, =200, VBEQ=0.7V

find the Q point.

4.7 The emitter follower (P166)

R1

RL

(b) AC circuit

4.7 The emitter follower (P166)

Find Av=Vo/Vin, Rin=Vin/Iin and Rout=Vo/Io|Vs=0

　Think about: Write two equations for Vin and Vo

RS’ Rout

’

Rout

4.8 The common-base amplifier (P299)

1. Draw the DC bias circuit to find the Q point

4.8 The common-base amplifier (P299)

Find Av=Vo/Vin, Rin=Vin/Iin and Rout=Vo/Io|Vs=0

　Think about: Write two equations for Vin and Vo

RC

Rin Rout

’

Rout

Summary

4.8 The common-base amplifier (P299)

C-E amplifier C-C amplifier C-B amplifier

Av

Ri

Ro

inverting noninverting noninverting

amplification follower amplification

The overall voltage gain of cascaded amplifier stages is the product of the voltage gains of the individual stages.

Think over: If Avo1=100, Avo2=200, what is the overall open circuit voltage gain of cascaded amplifier?

Avo=Avo1 Avo2 ? (refer to p17)

Review 1.5 Cascaded amplifiers

Applications calling for high or low input impedance Applications calling for high or low output impedance Application calling for a particular impedance

Refer to examples shown on page 26-27

Review 1.5 Cascaded amplifiers

4.9 Cascode amplifier (P300)

4.9 Cascode amplifier (P300)

共集－共射组合放大电路，不仅具有共集电极电路输入电阻大的特点，而且具有共射电路电压放大倍数大的特点；

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共射－共基组合放大电路，共基极电路本身就有较好的高频特性，同时将输入电阻很小的共基极电路接在共射极电路之后，减小了共射极电路的电压放大倍数，使共射极接法的管子集电结电容效应减小，改善了放大电路的频率特性。因此，共射－共基组合放大电路在高频电路中获得了广泛的应用。该组合电路的电压放大倍数近似等于一般共射电路的电压放大倍数。

4.9 Cascode amplifier (P300)

AC coupling versus direct coupling (P32)

Figure 1.37 Capacitive coupling illustration

AC coupling — The DC voltages of the amplifier circuits do not affect the signal source, adjacent stages, or the load.

Amplifiers that are realized as integrated circuits are almost always DC coupled because the capacitors or transformers needed for ac coupling cannot be fabricated in integrated form.

BACK

Example

Frequency response (P30) The complex gain: The ratio of the phasor for the output signal to

the input signal

Bode plot (P271) How circuit functions can be quickly and easily plotted against fr

equency? (straight line approximation & smart scale)

4.10 Frequency response for an amplifier

Review

Logarithmic Frequency Scale

A decade is a range of frequencies for which the ratio of the highest frequency to the lowest is 10.

An octave is a two-to-one change in frequency.

Review: Passive low-pass filter

Figure 8.1 Low-pass RC filter.

=

Review 2.8 Active Filter-High pass filter

4.10 Frequency response for an amplifier

Question:1. What is the voltage

gain of the circuit?2. Can you draw the

frequency response of amplifier?

3. What kind of filter is it?

4. Do you remember the small signal equivalent circuit of the NPN transistor?

5. Any changes to (4) if the frequency is high?

4.10 Frequency response for an amplifier

The small signal equivalent circuit for the NPN transistor

Mid-frequency

Cbe

Cbc

Usually, Cbe is about 10PF ~ 1000PF

Cbc <10PF

At high frequencies

Miller effect

Question: (P296)

If Zf=-j/C for a CE amplifier, how do you find Zin,miller

and Zout,miller

Cbc

ibrbe

Cbe

(1+gmRL’)Cbc

Find the Thevenin equivalent resistance Rs’?

Figure 8.34 Simplified equivalent circuit for the common-emitter amplifier.

1. What kind of filter is it?

2. What is the break (cut-off) frequency?

3. What factors affect the value of the break frequency?

4. What do we desire about an amplifier?

4.10 Frequency response for an amplifier

Figure 8.36 High-frequency behavior of the common-emitter amplifier.

Conclusion: The common-base amplifier achieves wide bandwidth, but its input impedance is very low. Consequently, the mid-band gain can be quite small due to loading of source. CE-CC cascade amplifier combines the advantages of both.

4.10 Frequency response for an amplifier

Figure 8.47 Amplifier with coupling capacitors.

4.10 Frequency response for an amplifier

Figure 8.47 Amplifier with coupling capacitors.