Aim of this chapter

EIE209 Basic Electronics

Basic Transistor Amplifiers

Contents

• Biasing

• Amplification principles

Aim of this chapter

To show how transistors can

be used to amplify a signal.

Basic idea

amplifier

Step 1: Set the transistor at a certain DC level

0.6V

7V

Step 2: Inject a small signal to the input and get a bigger output

— biasing

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_C

I_B

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 V_BE ≈ 0.7 V.

Then, I_B = (10 – V_BE )/ R_B = (10 – 0.7)/ R_B

I_C = bI_B = 100 (10 – 0.7)/ R_B = 10 mA

So,

R

_B

= 94kΩ

Also, V_CE = 10 – R_L I_C

Hence, 5 = 10 – 10R_L

b

dependent biasing — bad biasing

R_L R_B

V_BE +

–

V_CE +

– I_C

I_B

Transistor:

b = 100 V_CC=10V

This is a bad biasing

circuit!

because it relies on the accuracy of

b

,

but

b

can be ±50% different from what is

given in the databook.

Now, let’s go to the lab and try using R_B = 94kΩ and R_L = 0.5kΩ, and see if we get what we want.

A slightly better biasing method

Again, our objective is to find the resistors such that

I_C = 10mA and V_CE = 5V.

First, if I_B is small, we can approximately write

Suppose we get I_C = 10mA. Then R_L = 0.5kΩ. resistors will make sure I_B is much smaller than the current flowing down R_B1 and R_B2, which is consistent with the assumption.

A much better biasing method —

emitter degeneration

R_L

Again, our objective is to find the resistors such that

Stable (good) biasing

Summary of biasing with emitter degeneration:

Choose V_E , I_C and V_CE .

R

_E

R

_L

Use V_BE ≈ 0.6 to get V_B.

Then use

Terminology

The following are the same:

Biasing point

Quiescent point

Operating point (OP)

Alternative view of biasing

R_L

V_BE +

–

V_CE +

– I_C

V_CE +

–

V_R +

– +

–

V

_CC

V

_CC

I_C

V_CE I_C

V

_CC

R_{L Load line}

Slope=–1/R_L

What controls the operating point?

R_L

V_BE +

–

V_CE +

– I_C

V

_CC

V_CE I_C

Load line Slope=–1/R_L

V

_CC operating point

a bigger V_BE

a smaller R_L

What happens if

V

_BE

dances up and down?

Load line Slope=–1/R_L

V

_CC

a bigger V_BE = 0.65

a smaller V_BE = 0.6

The OP also dances up and down along the load line.

V_CE also moves up and down.

Typically, when

V

_BE

moves a little bit,

V

_CE

moves a lot!

Derivation of voltage gain

Clearly, Ohm’s law says that

V

_{CE =}

V

_{CC -}

I

_C

R

_{L fi D}

V

_{CE = -}

R

_L._D

I

_C

Last lecture: transconductance

Common-emitter amplifier

R_L

v_BE +

–

v_CE +

– I_C

V

_CC

R_B1

R_B2

The one we have just studied is called COMMON-EMITTER amplifier.

Small-signal voltage gain = –g_mR_L

That means we can increase the gain by increasing g_m and/or R_L.

Output waveform is anti-phase.

How do we inject signal into the amplifier?

R_L

V_BE +

–

v_CE +

– I_C

V

_CC

R_B1

R_B2

?

v

_in

±20mV

= V_CE + v~_CE

~

D

v_CE

or

Note on symbols

v_CE = V_CE + v~_CE

total signal

(large signal) operating point or

DC value or

quiescent point

small signal or

ac signal

=

+

Total signal

a

DC point

A

Small signal

a

or

D

a

Solution: Add the

same

biasing DC level

But, it is impossible to find a voltage source which is equal to the exact biasing voltage across B-E.

V_BE could actually be 0.621234V, which is

determined by the network RB1, RB2 and the transistor characteristic!!

Exactly the same biasing V_BE

The wonderful voltage source: capacitor

R_L

V_BE +

–

V_CE +

– I_C

V

_CC

R_B1

R_B2

V_C +

– The capacitor voltage is exactly equal to V_BE because DC current must be zero

Solution — insert coupling capacitor

R_L

v_BE +

–

v_CE +

– I_C

V

_CC

R_B1

R_B2

v

_in

±20mV

– + DC voltage equal to exactly the same biasing V_BE

This is called a

coupling capacitor

Complete common emitter amplifier

R_L

v_BE +

–

v_CE +

– I_C

V

_CC

R_B1

R_B2

v

_in

v

_o

+

–

+

–

– + _{+ –}

coupling capacitors

~

Can we simplify the analysis?

v

_in

v

_o

+

–

+

–

common emitter

amplifier

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?

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:

Small-signal model of BJT: objectives

r

_in

R

o

To find:

r

in

R

_o

G

_m

r

_in

R

_o

r

_in

R

_o

A

_m

+ –

G

_m

v

_in

_A

m

v

in

or

Derivation of the small-signal model

r

_p

Input side:

i

_B

v

_BE +

–

r_in = vB E

i_B =

v_{B E} i_C / b

For small-signal,

r_in = DvBE Di_B =

Dv_BE

Di_C / b

= b

(Di_C / Dv_BE ) =

b

g_m

where g_m is the BJT’s transconductance

Derivation of the small-signal model

Output side:

Dv_CE = -Di_C ¥ R_L

= -g_mDv_BE ¥R_L

where g_m is the BJT’s transconductance R_L

v_CE +

– I_C

R

_L

v

_CE

=

V

_CC

– I

_C

R

_L

V

_CC

For small-signal,

g

_m

v

~_BE

v

_CE

~ +

Derivation of the small-signal model

Output side:

where r_o is the Early resistor of the BJT. R_L Including BJT’s Early effect

Initial small-signal model for BJT

“MUST” REMEMBER

Small-signal

BJT parameters:

Initial small-signal model for FET

g

_m

v

~_GS

v

~_DS

+

–

r

_o

G

S

D

v

_GS

~ +

–

Similar to BJT, but input resistance is ∞.

Small-signal

FET parameters:

g_m = 2 K I_D

r_o = 1

l

l is the channel length modulation parameter

K is a semiconductor parameter

FET model

Example: common-emitter amplifier

R_L

v_BE +

–

V

_CC

R_B1

R_B2 +

v_in –

+ v_o –

Assume the coupling caps are large

enough to be considered as short-circuit at signal frequency

B C

E E

g_m~v_BE

r_p ro

R_L

V_CC is ac 0V.

Complete model for common-emitter amplifier

Complete model:

g_m~v_BE

Simplified model:

Total input resistance

R_in = R_B1 ||R_B1 || r_p

Total output resistance

R_o = R_L||r_o

Voltage gain

v_o

Alternative model for common-emitter amplifier

g_m(R_L||r_o) v_BE R_L||r_o

R_B1 ||R_B1 || r_p

+ v_in –

+ v_o – +

v_BE –

~

Total input resistance

R_in = R_B1 ||R_B1 || r_p

Total output resistance

R_o = R_L||r_o

Voltage gain

v_o

v_in = -gm(RL || ro)

Output in Thévenin form:

– +

More about common-emitter amplifier

g_m(R_L||r_o) v_BE R_L||r_o

R_B1 ||R_B1 || r_p

+ v_in –

+ v_o – +

v_BE –

~ –

+

~

Because the output resistance is quite large (equal to R_L||r_o ≈ R_L), the 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_L. This makes it not suitable to deliver current to load.

1kΩ, for example

10Ω

Bad idea — wrong use of common-emitter amplifier

But the output circuit is:

– + 200v_in

1kΩ

10Ω

+ v_o –

The effective gain drops to

Proper use of common-emitter amplifier

Now the output circuit is:

– + 200v_in

1kΩ

10MΩ

+ v_o –

The effective gain is

200_¥ 10

7

ª 200

The load must be much larger than R_L.

How can we use the amplifier in practice?

1kΩ

v_BE +

–

+10V

R_B1

R_B2 +

v_in –

5mA

How to connect the output to load?

+ v_o –

Emitter follower

+10V

R_B1

R_B2 +

v_in –

Biasing conditions:

R_E

+ v_o –

I_C biased to 10mA

V_CE biased to 5V

Base voltage ≈ 5.6V Emitter voltage ≈ 5V Collector current ≈ 10mA R_E = 500Ω

R_B1:R_B2 ≈ 44:56 Say, R_B1 = 440kΩ

R_B2 = 560kΩ

Thus, for small signal, DV_E = DV_B

or v_o = v_in

Small-signal model of emitter follower

+10V

R_B1

R_B2 +

v_in –

R_E

+ v_o –

B C

E E

g_m~v_BE

r_p ro

R_E

R_B1 || R_B2

Small-signal model of emitter follower

Small-signal model of emitter follower

which is quite small (good)!! Output resistance is

Small-signal model of emitter follower

1 v_in R_E||(1/g_m)

r_p+(1+b)R_E

+ v_in –

+ v_o –

+ –

very large

very small

Large input resistance

Small output resistance

Voltage gain = 1

Good for any load

Draw no current from previous stage

A better “emitter follower”

+10V

R_B1

R_{B2 +}

v_o –

I

_E

+ v_in –

Input resistance is very LARGE because R_E = ∞.

Output resistance is 1/g_m.

Gain = 1.

This circuit is also called CLASS A output stage. Details to be studied in second year EC2.

∞

Common-emitter amplifier

with emitter follower as buffer

+10V

v_BE +

– R_B1

R_B2 +

v_in –

R_L

+ v_o – speaker 10Ω

common-emitter amplifier (high gain)

emitter follower (unit gain)

∞

1 g_m

FET amplifiers (similar to BJT amplifiers)

+10V

v_GS +

– R_G1

R_G2 +

v_in –

R_L

+ v_o – speaker 10Ω

common-source amplifier (high gain = –g_mR_L)

source follower (unit gain)

∞

1 g_m

Further thoughts

Will the biasing resistors affect the gain?

R_L R_bias

+ v_in –

+ v_o –

Seems not, because

Gain = –g_mR_L

which does not depend on R_bias .

Input source with finite resistance

The input has a voltage divider network.

Therefore, the gain decreases to

Example

By how much does the gain drop?

r_p

94k||600 = 596Ω 5mA

g_m = 5mA/25mV = 0.2A/V

r_p = b/g_m = 100/0.2 = 500Ω

Further thoughts

1kΩ 84kΩ

+ v_in –

+ v_o – 16kΩ

10V

Recall that the best biasing scheme should be b independent.

5mA

200Ω

One good scheme is emitter degeneration, i.e., using R_E to fix biasing current directly. Here, since V_B is about 1.6V, as fixed by the base resistor divider, V_E is about 1V.

Therefore, I_C ≈ V_E/R_E = 5mA (no b needed!)

Question:

Common-emitter amplifier with emitter

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

R_E

Answer:

The gain is MUCH smaller.

Common-emitter amplifier with emitter by-pass

R_L R_B1

+ v_in –

+ v_o – R_B2

V_CC

R_E C_E

Add C_E such that the effective emitter resistance becomes zero at signal frequency.

So, this circuit has good biasing, and the gain is still very high!

Gain = – g_mR_L

which is unaffected by R_E because effectively R_E is shorted at signal frequency.

Summary

Basic BJT model

(small-signal ac model):

g_mv_BE r_o

r_p

B

E

C

E +

v_BE –

g_m = IC

(kT /q) = I_C V_T

r_p = b

g_m

r_o =VA

I_C

V_T is thermal voltage ª 25mV

Summary

Basic FET model

(small-signal ac model):

Similar to the BJT model, but with infinite input resistance.

Therefore, the FET can be used in the same way as amplifiers.

g_mv_GS r_o

G

S

D

S +

v_GS –

Summary

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

g_mv_BE r_o

r_p

+ v_BE – +

v_in –

R_bias R_L

+ v_o –

Gain = –g_mR_L

Input resistance = R_bias || r_p (quite large — desirable)

Output resistance = R_L ||r_o ≈ R_L (large — undesirable) R_L

R_bias

+ v_in –

Summary

Emitter follower (EF) small-signal ac model:

g_mv_BE

r_p

+ v_BE – +

v_in

– Rbias

+ v_o –

Gain = 1

Input resistance = R_bias || [r_p+(1+b)R_E ] (quite large — desirable) R_E

R_bias

R_E +

v_in –