Aim of this chapter

  EIE209 Basic Electronics Basic Transistor Amplifiers

Contents

• Biasing • Amplification principles
• Small-signal model development for BJT

Aim of this chapter To show how transistors can be used to amplify a signal. amplifier

  Basic idea — biasing

  Step 1: Set the transistor at a certain DC level Step 2: Inject a small signal to the input and get a bigger output

  — coupling

  7V

  0.6V

  amplifier

• –

  R L = 0.5kΩ

  V BE

  ) / R

  B

  = (10 –

  0.7 )

  / R

  B

  I C

  = b

  I B

  = 100 (10 –

  0.7 )

  / R

  B = 10 mA

  So, R B

  = 94kΩ

  Also,

  V CE

  = 10 – R

  L

  I C

  Hence, 5 = 10 – 10 R

  L So,

  = (10 –

  I B

  Then,

  I C

  Biasing the transistor

  To set the transistor to a certain DC level = To set

  V CE and I

  C

  R

  L

  R

  B

  V BE

  V CE

  I B

  V BE ≈ 0.7 V.

  Transistor: b = 100

  V CC =10V

  Suppose we want the following biasing condition:

  I C = 10 mA and

  V CE = 5 V

  Find R

  B

  and R

  L

  Start with

• –

  

b dependent biasing — bad biasing

  =10V Now, let’s go to the lab and try using = 94kΩ and

  V R R

  CC B L = 0.5kΩ, and see if we get what we want.

  I C

  R

  L …totally wrong! We don’t get I = 10mA and V = 5V

  R

  C CE B

• I

  B

  V

• CE

  

This is a bad biasing circuit!

• – V

  BE because it relies on the accuracy of , b

• – but can be ±50% different from what is

  b

  Transistor: given in the databook. b = 100

  A slightly better biasing method

  Again, our objective is to find the resistors such that = 10mA and = 5V .

  I V C CE

  =10V

  V CC First, if

  I is small, we can approximately write

  B

  I C

  R R R

  94 L ^{B 2 B1}

  R

  B1 0.6 = 10 ¥

  = fi

  R R R

  6

• +

  + ^{B1 B2 B 2}

  I B

  V CE

• Suppose we get = 10mA. Then R = 0.5kΩ.

  

I

C L

• – V

  BE

  R

  B2

  We can start with = 940Ω and = 60Ω. Such R R

• – resistors will make sure

  I is much smaller than B

  the current flowing down R and R , which is

  B1 B2 consistent with the assumption.

  What we need in practice is to fine tune R or R

  B1 B2 such that is exactly 5V.

  V CE A much better biasing method — emitter degeneration

  Again, our objective is to find the resistors such that = 10mA and = 5V .

  I V C CE

  =10V

  V CC Set V = 2V, say. Then, R = 2V/10mA = 0.2kΩ.

  

E E

  I C

  R

  L

  R

  B1

  Surely, R = 0.5kΩ in order to get V = 5V.

  L CE

• I

  B

  V CE

  Finally, we have = + 0.6. Therefore, if is

  V V

  I B E B

• – small compared to and , we have

  I I

• RB1 RB2

  R

  B2

  R R

  74 V E ^B1 E =

• – R
_{B 2}

  26 Hence, R = 740Ω and R = 260Ω.

  NOTE: b is never used in calculation!! Stable (good) biasing

  Summary of biasing with emitter degeneration: =10V

  V CC Choose V , I and V .

  E C CE

  I C

  R

  L

  R

  B1

• I

  B R R

  V E L

  B

  V CE

• – + Use ≈ 0.6 to get .

  V V R

  

BE B

B2

  R

  V E E

  Then use

• –

  R

  _{B B}
₁ 10 - V =

R

_{B 2 B}

  V

  to choose R and R such that I is much smaller the

  B1 B2 B current flowing in and .

  R R

Terminology

  The following are the same: Biasing point Quiescent point Operating point (OP) DC point Alternative view of biasing

  I C

  V CC

• I

  C

  R R

  V L L R

  Load line

• – Slope=–1/

  R

• L
• V

  CC

• – V

  CE

• +

• – V

  I V

  operating point

  BE C CE

• –

  –

  V CE

  V CC What controls the operating point?

  I C

  V CC

  Load line

  I C

  R Slope=–1/

  L

  R

  L a bigger

  V BE

• V

  CE

• –

  a smaller

  V R

  operating point

  BE L

• – V

  CE

  V CC V or I controls the OP

  BE B CONCLUSION:

  R also controls the OP L V What happens if dances up and down?

BE

  I C The OP also

  V CC

  dances up and Load line down along the

  I Slope=–1/ C R

  R

  L L a bigger = 0.65

  V BE load line.

• a smaller = 0.6

  V BE

  V also moves up

  V CE CE

• and down.
• – V

  BE

• – Typically, when

  V BE moves a little bit,

  V V CE CE

  V CC moves a lot! Animation to show amplifier action

  Derivation of voltage gain

  DV DV

  o CE

  =

  Question: what is ?

  DV DV

  in BE

  V Clearly, Ohm’s law says that CC

  V R = V - I fi DV = -R .DI

  

I CE CC C L CE L C

C

  R

  L ? Then, what relates DI and DV +

  C BE D

  I V

  C CE

  g =

• m

  Last lecture: transconductance

• – D

  V BE

  V BE

• – Hence,

  D

  V CE

= -

g R m L

  D

  V BE Common-emitter amplifier The one we have just studied is called COMMON-EMITTER amplifier.

  V CC SUMMARY:

  R

  L

  I C

  R

  Small-signal voltage gain B1

  = – g R m L

• v

  CE That means we can

• – increase the gain by

  R v

  B2 BE increasing g and/or R . m L

• – Output waveform is anti- phase.
D v in

  or

  R

  CE

  D v

  ~

  CE ~

  V CE

  ±20mV =

  B2 ? v in

  B1

  R

  

V

CC

  I C

  v CE

• v
• –
Note on symbols Total signal

  V BE

  L

  R

  How do we inject signal into the amplifier?

• –

  = + a DC point

  ~

  V v

• = v

A

  total signal operating point small signal Small signal

  (large signal) or or

  DC value ~ ac signal

  a or Da

  or quiescent point Solution: Add the same biasing DC level

  But, it is impossible to find a

  V CC

  voltage source which is equal

  Exactly the

  to the exact biasing voltage

  same biasing

  R

  L

  R

  I B1 C across B-E.

  V BE

• could actually be

  V +

• – BE

  v

  0.621234V, which is

  CE

• –

  determined by the network ~

  R v

  RB1, RB2 and the transistor

  v

  in

• – characteristic!!

  ±20mV How to apply the exact ?

  V BE The wonderful voltage source: capacitor

  V CC

  R

  L The capacitor voltage is

  R

  I B1 C exactly equal to

  V BE

  0A

• because DC current must be zero

  V CE

• – R

  V B2 BE

  V C

• – –
Solution — insert coupling capacitor DC voltage equal

  V CC to exactly the same biasing

  V BE

  R

  L

  R

  I B1 C

This is called a

• – +
• coupling capacitor

  v CE

• –

  ~

  R v

  v in

• – ±20mV
Complete common emitter amplifier

  V CC

  R

  L

  R

  I B1 C

• – +
• – ~ v

  CE v

  o

• – –

  R v

  ~ v

• – in
• – coupling capacitors (large enough so that they become short-circuit at signal frequencies)
Can we simplify the analysis? We are mainly interested in the ac signals.

  The DC bias does not matter!

Can we create a simple circuit just to look at ac signals?

  common emitter ~

  ~ v v amplifier in o

• – –
Small-signal model v in v o
• –
• –

  common emitter amplifier ?

  ?

  What is the loading (resistance) seen here?

  What is the Thévenin or Norton equivalent circuit seen here?

  1

  2 Two basic questions: ~

  ~ Thévenin form

  or Norton form

  in

  in

  v

  m

  A

  in

  v

  m

  G

  m

  A

  o

  R

  r

  Small-signal model of BJT: objectives r

  o

  R

  in

  r

  m

  G

  o

  R

  in

  To find: r

• –

  o

  R

  in Derivation of the small-signal model Input side:

  v v

  B E B E

  r = =

  in

  i i

  i

  / b

  B C

  B

  r

  For small-signal, p

• v

  Dv Dv BE

  r = =

• – in

  Di Di / b

  B C

  b b = =

  r = b/g

  (Di / Dv ) g

  C BE m p m

  where is the BJT’s transconductance g

  m Derivation of the small-signal model Output side:

• v =

  V – I R CE CC C L

  V CC

  ~

  R v

  For small-signal, L CE

  ~ R

  L

  I D = -D ¥ C

  v i R

• – CE C L

  g v

  m BE = - D ¥ g v R

  m BE L

• v

  CE

• – where is the BJT’s transconductance g

  m Derivation of the small-signal model

• Including BJT’s Early effect

  Output side:

  ~

  r R v

  o L CE

  V Dv Dv CE CE CC

  ~ = -Di +

• – C

  R r g v

  

L o m BE

  R

  L

  I C

  D = -D v i ( R r )

  CE C L o

• g D v R r

  = - ( )

  

m BE L o

v

  CE

• – where is the Early resistor of the BJT.

  r

  o Recall: r = V /I , where V is typically about 100V. o A C A A very rough approx. is = ∞. r o Initial small-signal model for BJT

  Small-signal

  “MUST” REMEMBER

  BJT parameters:

  I I C C g = = m kT / q)

  V ( B

  C T

  b r = p

  ~ ~

  g

  ~

  m r v g v

  o CE

  r v

  m BE p

  BE

  V A r =

• – o
• – I

  C E V is thermal voltage

  T ª 25mV

BJT model

  V is Early voltage A typically ~ 100V

  Small-signal FET parameters: g

• –
• – Similar to BJT, but input resistance is ∞.

  is a semiconductor parameter FET model

  K

  parameter

  l l is the channel length modulation

  1

  o =

  r

  2 K I D

  m =

  ~

  Initial small-signal model for FET g

  GS

  G S D v

  o

  r

  ~

  v

DS

  GS ~

  v

  m

  All amplifier configurations using BJT can be likewise constructed using FET.

  B C E E g

  V CC is ac 0V.

  L

  R

  o

  r

  p

  r

  BE ~

  v

  m

• –
• v
• – Assume the coupling caps are large enough to be considered as short-circuit at signal frequency

  o

  in

  B2

  R

  B1

  R

  V CC

  BE

  L v

  R

  Example: common-emitter amplifier

  R _B1

|| R

_B1
• – + v
• v
• – ~
• – + v
• – + v

  || r

  R

  in

  = R

  B1

  || R

  B1

  || r

  p

  Total output resistance R

  o

  = R

  L

  o

• v

  Voltage gain v

  o

  v

  in

  = - g

  m

  ( R

  L

  || r

  o

  ) ª - g

  m

  R

  Simplified model: Total input resistance

  BE

  o

  in

  Complete model for common-emitter amplifier

  Complete model: g

  m

  v

  BE ~

  r

  p

  r

  o

  R

  L R _B1 || R _B1

  o

  in

  BE

  g

  m

  v

  BE ~

  R

  L

  || r

  

o

R _B1

  || R _B1 ||

  r

  p

• – ~
• – + v
• – + v

  L Alternative model for common-emitter amplifier Output in Thévenin form:

  Total input resistance ||

  R r

  L o

  R = R || R || r

  in B1 B1 p

• – ~

  v v v

  R || R ||

  r

• in B1 B1 BE o p

  Total output resistance

• – – –

  ||

  ~ R = R r o L o

  ( || ) g R r v

  L m o BE

  Voltage gain v

  o

  = - g ( R || r )

  m L o

  v

  in

  ª - g R

  m L More about common-emitter amplifier

  Because the output resistance is quite large (equal to R || r ≈ R ), the

  L o L

  common-emitter amplifier is a POOR voltage driver. That means, it is not a good idea to use such an amplifier for loads which are smaller than R . This

  L makes it not suitable to deliver current to load.

  1kΩ, for example ||

  R r

  L o practically no output!!

• – ~

  v v v

  R || R || 10Ω

  r

• in B1 B1 BE o p
• – – –

  ~

  ( || ) g R r v

  L

m o BE

  Transconductance g _m =

• –
• 10V
• v
• – +

• –

  speaker

  10Ω
• – The effective gain drops to 200 ¥
• – + v

  o

  1kΩ 10Ω

  in

  200 v

  But the output circuit is:

  = (0.2)(1k) = 200 or 46dB

  I _C / (25mV) = 5/25 = 0.2 A/V Expected gain = g _m R _L

• v

  in

  5mA

  o

  B2

  R

  B1

  R

  v BE

  1kΩ

  

Bad idea — wrong use of common-emitter amplifier

  10 1000 + 10 = 1.98 Proper use of common-emitter amplifier The load must be much larger than .

  R

• 10V

  L

  5mA 1kΩ R

  B1

  Now the output circuit is:

  nearly open circuit

• 1kΩ
• –
• v

  10MΩ

  v

• o

  10MΩ + o

  200 v R v

  in

• – –

  v

  in

• – –

  The effective gain is ₇

  10 ª 200 200 ¥ ₇ 1000 + 10

  How can we use the amplifier in practice?

  1kΩ

• v
• –
• 10V
• v

  v BE

  R

  B1

  R

  B2

  How to connect the output to load?

  o

  ?

• – speaker 10Ω

  in

• – 5mA

  Emitter follower

  Biasing conditions:

• 10V

  Base voltage ≈ 5.6V Emitter voltage ≈ 5V Collector current ≈ 10mA

  R

  I biased to 10mA B1 C

  R = 500Ω

  E

  : ≈ 44:56 R R

  

V biased to 5V B1 B2

CE

  Say, = 440kΩ R

  B1

  = 560kΩ R

  B2

• R

  B2

  v V = V – 0.6

  E B

• in
• – v

  Thus, for small signal, R

  o E

• – DV = DV

  E B

  = v or v

  o in v / v

  Gain = = 1 o in

• 10V

  v

  B2

  || R

  B1

  R

  E

  R

  o

  r

  p

  r

  BE ~

  m

  g

• v

  o

  E

  in

  B2

  R

  B1

  R

  Small-signal model of emitter follower

• v
• v

  o

• – R
• –
• – B C E E
• v

  B = v

  in i

  B r in

  = v i n i

  B = v

  BE

  E i

• v
• v
• v

  E which is quite large (good)!! Input resistance is

  B E i

  v

  E i

  B = r p

  E i

  B = r p

  E i

  E /(1 + b ) = r p

  B

  o

  in

  p

  Small-signal model of emitter follower

  B C E E g

  m

  v

  BE ~

  r

  r

  B2

  o

  R

  E

  R

  B1

  || R

• – r
• v
• (1 + b )R
• v

• v
• i
• g

  E

  r

  p

  m

  v

  E

  =

  1

  1 R

  E

  p

  m

  =

  1

  1 R

  m

  E

  E

  i

  m

  v

  m

  1 g

  ||

  = R

  b

  m

  E

  1 R

  1

  ª

  m

  E

  i

  m

  E

  Small-signal model of emitter follower

  B C E E g

  m

  v

  BE ~

  r

  p

  r

  o

  R

  E r out which is quite small (good)!! Output resistance is

  r

  out

  = v

  m

  i

  E

  = v

  BE

  v

  m

  B

  i

  m

  BE

  =

  m

  i

  BE

  =

• g
• g
• v
• 1 r
• –
• g
• g
• g
Thevenin form:

  Good for any load Draw no current from previous stage

  ) r

  Large input resistance Small output resistance Voltage gain = 1

  very small

  o

  in

  E

  p

  m

  ||(1/ g

  

E

  R

• – very large
• – + v
• – +
A better “emitter follower”

  in

  1 v

  Small-signal model of emitter follower

• v
• (1+b)R

  Input resistance is very LARGE

• 10V

  because R = ∞.

  E

  R

  B1 Output resistance is 1/ g . m Gain = 1.

• This circuit is also called CLASS A R
• B2

  v

  output stage. Details to be studied ∞ in

  v

• –

  o

  in second year EC2.

  1/ g

  I

  E

• – m
• 10V

  Common-emitter amplifier with emitter follower as buffer

• –
• v

  v BE

  R

  B1

  R

  B2

  in

  L

• v
• – R

  o

• – speaker 10Ω
• 10V

  common-emitter amplifier (high gain) emitter follower (unit gain)

  ∞

  1 g m

  I E

  FET amplifiers (similar to BJT amplifiers)

• –
• v

  v GS

  R

  G1

  R

  G2

  in

  L

• v
• – R

  o

• – speaker 10Ω
Further thoughts

  common-source amplifier (high gain = – g m R L ) source follower (unit gain)

  ∞

  1 g m

  I S

  Will the biasing resistors affect the gain? Seems not, because

  R

  L

  R

  bias

  Gain =

• –g R

  m L

• which does not depend on .

  R

  bias

  v

• o
• – v

  in

  However, a realistic voltage source has finite internal

• – resistance. This will affect the gain.

  Input source with finite resistance The input has a voltage divider network.

  R r

  || bias

  R

  p L

  v = v

  BE in

  R

  bias

  R || r R +

  bias s p

  R

  s

• Therefore, the gain decreases to
• v

  o

  v R || r

  v

• – o bias p

  in = (- )

  g R

  m L

• – ||
• bias

  v R r R

  in p s assuming r very large. o

  R

  s

  R g v r R v v v

  bias r m BE o L in BE o p

• – –
• –

  = 0.845 or - 1.463dB

  g

  R _bias || r

_p

50 + R _bias || r _p = 596|| 500 50 + 596||500

  ) = 169

  L

  R

  m

  = 100/0.2 = 500Ω Voltage divider attenuation = Hence, the gain is reduced to 0.845( g

  m

  = b/g

  p

  = 5mA/25mV = 0.2A/V r

  m

  5mA

  50Ω 94k||600 = 596Ω

• v

  bias

  

in

  BE

  p

  10V By how much does the gain drop? r

  600Ω

  o

  in

  1kΩ 94kΩ

  Example

• v
• – R
• – + v
• – 50Ω
• – + v

  Further thoughts Recall that the best biasing scheme should be b independent.

  10V One good scheme is emitter degeneration, i.e.,

  5mA

  using to fix biasing current directly. Here, R

  E

  1kΩ since V is about 1.6V, as fixed by the base

  84kΩ

  B

  resistor divider, V is about 1V.

  E

• v

  o

  Therefore, ≈

  I V /R = 5mA (no b needed!) +

  C E E

• – v

  in

  16kΩ

• – 200Ω

  Question:

  Will this biasing scheme affect the gain? Common-emitter amplifier with emitter degeneration

  Exercise: Find the small-signal gain of this amplifier.

  V CC R

  L Answer:

  R

  B1

• v g R
• o m L

  = Ê ˆ

  v v

  1 in

• o

  Á ˜

  1

• 1 + g R
• – m E

  Á ˜

  v

  in

  b

  Ë ¯

  R

• – B2

  R E

• g - R R

  m L L ª ª

• 1 g R R

  m E E The gain is MUCH smaller.

  

We have a good biasing, but a poor gain! Can we improve the gain?

  Common-emitter amplifier with emitter by-pass

  Add C such that the effective emitter

  E

  resistance becomes zero at signal frequency.

  V CC R

  L

  So, this circuit has good biasing, and the R

  B1

  gain is still very high!

• v
• o

  Gain = – g R

  m L

• – v

  in

  R

• – B2

  which is unaffected by because R

  R C E E E

  effectively is shorted at signal R

  E frequency. is called bypass capacitor. C E is Early voltage typically ~ 100V

  V A

  /

  is thermal voltage ª 25mV

  V T

  I C

  V A

  o =

  r

  m

  b g

  p =

  V T r

  I C

  ) =

  q

  kT

  Summary

  (

  I C

  m =

  BE

  B E C E

  p

  r

  o

  r

  BE

  v

  m

  Basic BJT model (small-signal ac model): g

• v
• – g
Summary

  Basic FET model (small-signal ac model):

  D G Similar to the BJT model,

• but with infinite input

  ∞ g v r v

  m GS o

  resistance. GS

• – Therefore, the FET can be

  S S used in the same way as amplifiers. Summary

  Common-emitter (CE) amplifier small-signal ac model:

• R g v r R v v

  v

  

bias r

m BE o L in BE o p

  R

  L

  R

• – – –

  bias

• v
• o
• – Gain =
• –g R v

  m L in

• – Input resistance = R || (quite large — desirable)

  r

  bias p

  Output resistance = || ≈ (large — undesirable) R r R

  L o L

• v
• – + v

• – Gain = 1 Input resistance = R
• –
• (1+b)R

  E

  in

  E

  R

  bias

  R

• – R
• v
• v
• – + v

  E

  ) (small — desirable) R

  m

  || (1/g

  E

  R

  ] (quite large — desirable) Output resistance =

  || [ r

  p

  bias

  o

  bias

  in

  BE

  p

  r

  BE

  v

  m

  Emitter follower (EF) small-signal ac model: g

  Summary

  o