BJT AC Analysis

Outline: The re transistor model

Chapter 5. BJT AC Analysis

AC analysis through re modelCB, CE & CC

voltage-divider biascommon-emitter fixed-bias

emitter-bias & emitter-follower common-base configuration

BJT AC Analysis

A model is a combination of circuit elements, properly chosen, that best approximates the actual behavior of a semiconductor device under specific operating condition.

Transistor Modeling

There are three models: re model hybrid model hybrid equivalent model.

BJT AC Analysis

The superposition theorem is applicable and the the investigation of the dcconditions can be totally separated from the ac response.

Before discussing model, we must make preparation to the discussion.

An suitable Q-point has been chosen.Then the dc levels can be ignored in the ac analysis network.

BJT AC Analysis

Those important parameters, such as Zi , Zo , Vi , Vo , Ii and Io should be kept unchanged.

The coupling capacitor C1, C2 and bypass capacitor C3 are chosen to have a very small reactance. Therefore, each is replaced by a short circuit.

This also results in the “shorting out” of dc biasing resistor RE .

BJT AC Analysis

The input impedance Zi is defined from base to ground.

The input current Ii is defined as the base current of the transistor.

The input voltage Vi is defined between base and ground.

The output voltage Vo is defined between collector and ground.

BJT AC Analysis

The output current Io is defined as the current through the load resistor RC.

The output impedance Zo is defined from collector to ground.

By introducing a common ground, R1and R2 will be in parallel.

RC will appear from collector to emitter.

BJT AC Analysis

The transistor equivalent circuit also consists of resistors and independent controlled source.The circuit analysis techniques such as superposition, Thevenin theorem can be applied to determine the desired quantities.

There are important quantities indicating amplification effects:

BJT AC Analysis

Voltage gain: Vo / Vi

Current gain: Io / Ii

Briefly, steps to obtain ac equivalent circuit:

dc sources to zero

capacitors to short circuits

Removing elements bypassed by short circuits and rearranging network

BJT AC Analysis

Figure: Transistor circuit for ac analysis

BJT AC Analysis

The re Model for CBAs shown in the figure, it is the common-base BJT circuit.Now, re model is introduced.

On the input port, there is a resistor, re .

where re = 26mV / IE (Eq. 5.1)The subscript e of re indicates that it is the dc level of IE that determines the aclevel of the resistance.

BJT AC Analysis

On the output port, there is a controlled current source, denoted by a diamond shape.

The Ic is controlled by the level of Ie . The input impedance Zi is re .

The output impedance Zo .

In general, for common-base configuration, the Zi is relatively small and Zo quite high.

BJT AC Analysis

Voltage gain:Vo = - Io RL = - (- Ic) RL =Ie RL

Vi = Ii Zi = Ie Zi =Ie re

so thatAv = Vo / Vi = (Ie RL)/(Ie re) = ( RL)/ re RL / re

Current gain:Ai = Io / Ii = (-Ic)/(Ie) = (- Ie )/(Ie) -1

For common-base configuration, the Vo and Vi are in phase.

BJT AC Analysis

Figure: re model for common-base circuit

BJT AC Analysis

The re Model for CEAs shown in the figure, it is the common-emitter BJT circuit.

Now, replace the transistor with re model On the input side, there exists re .

Zi = Vi / Ii = [( +1)Ib re]/ Ib= (Ie re)/ Ib= Vbe /Ib

= ( +1) re re ( if is large enough)

Typically, Zi is at moderate level.

BJT AC Analysis

On the output side, the controlled-current source is connected between collector and emitter, Isource= Ib.

Or, Zo = ro , as shown in the figure. At this time, Ic Ib .

Ideally, the output impedance Zo .

Normally, voltage gain Av and current gain Ai are high.

For common-collector configuration, this model is also applicable.

BJT AC Analysis

Figure: re model for common-emitter circuit

BJT AC Analysis

CE Fixed Bias CircuitAs shown in the figure, it is the common-emitter fixed-bias configuration. The input signal Vi is applied to the base

and the output Vo is off the collector. The input current Ii is not the base current

and the Io is the collector current. For small-signal analysis, VCC is replaced

with ground.

BJT AC Analysis

Those dc blocking capacitors C1 and C2 are replaced with short circuits.

So the RB is in parallel with the input port and RC output port.

The parameters of Zi , Zo , Ii and Io should be in the same places as original network.

Finally, substitute the re model for the transistor.

BJT AC Analysis

In practice,

The re is determined from dc analysis.

In the small-signal analysis, we assume that , ro and re have been determined in advance.

The ro is obtained from the datasheet.

The is read from the datasheet or measured with testing instrument.

BJT AC Analysis

Input impedance Zi :From the figure, it is obvious that

Zi = RB|| re

For the majority of situation, RB is greater than re by more than a factor of 10.So we get

Zi re

BJT AC Analysis

Output impedance Zo:From the definition of Zo , we get

Zo = RC ||roIf ro 10RC , then

Zo RC

Voltage gain Av:From the figure, we get

Vo =( - Ib )(RC ||ro)

BJT AC Analysis

And

Finally,

Ib =Vi /( re)

Vo =( - Ib )(RC ||ro)So

=( - Vi /( re))(RC ||ro)

=( - Vi / re)(RC ||ro)

Av = Vo / Vi

= - (RC ||ro) / re

BJT AC Analysis

Phase relationship:The negative sign in Av ,reveals that a 180 phase shift occurs between the input and output signals.

This also means that a single-stage of amplifier of this type is not enough.The magnitude of output signal is larger than that of input signal. But the frequencies of them should be the same.

BJT AC Analysis

Figure: re model for CE fixed-bias circuit

BJT AC Analysis

Figure: phase shift of input & output

BJT AC Analysis

Example 5.4As shown in the figure, it is the common-emitter fixed-bias configuration. Determine:re , Zi , Zo, Av with ro = 50k .Solution: From dc analysis, we get

B

BECCB R

VVI

k

VV470

7.012 A04.24

BJT AC Analysis

Then, ac analysis

IE= (β+1) IB = (100+1) 24.04A= 2.428mA

re = (26mV)/IE = (26mV)/ 2.428mA = 10.71

re = (100)(10.71 ) = 1.071 kZi = RB|| re = (470 k ) || (1.071 k )

= 1.069 k

BJT AC Analysis

Zo = RC || ro = (3 k ) || (50 k )

= 2.83k

Av = - (RC ||ro) / re = - 2.83k / 10.71

= - 264

From the Av ,we can see that the output signal has been amplified but out of phase with the input signal.

BJT AC Analysis

Figure: Example 5.4

BJT AC Analysis

Voltage Divider BiasAs shown in the figure, it is the voltage divider bias configuration.Substituting re equivalent circuit, note that:

RE is absent due to the low impedance of the bypass capacitor CE .

When VCC is set to zero, one end of R1and RC are connected to ground.

R1 and R2 remain part of the input circuit while RC is part of output circuit.

BJT AC Analysis

Some parameter of the equivalent circuit: Input impedance Zi :

Zi = R1|| R2|| re

Output impedance Zo:

Zo = RC ||ro

If ro 10RC , thenZo RC

Voltage gain Av:

BJT AC Analysis

So,

Vo =( - Ib )(RC ||ro)=( - Vi / re)(RC ||ro)

Av = Vo / Vi = - (RC ||ro) / re

=[- Vi /( re)] (RC ||ro)

If ro 10RC , then

Av = Vo / Vi - RC / re

Phase relationship:180 phase shift occurs between the input and output signals.

BJT AC Analysis

Figure: Voltage divider bias & its equivalent circuit

BJT AC Analysis

Example 5.5As shown in the figure, it is the voltage divider bias configuration. Determine:re , Zi , Zo, Av with ro = 50k .

Solution: dc analysis, testing RE >10 R2 ,

901.5k > 108.2k

135k > 82k (satisfied)

BJT AC Analysis

CCB VRR

RV21

2

V81.2

Using the approximate approach (Sec. 4.5, p155), we obtain:

Vkk

k 222.856

2.8

VE = VB - VBE

= 2.11V/1.5kIE = VE / RE

= 2.81V-0.7V = 2.11V

= 1.41mA

re = 26mV / IE = 26mV / 1.41mA

= 18.44

BJT AC Analysis

ac analysis,

Zi = R1|| R2|| re

= 56k || 8.2k || (90)(18.44 )= 1.347k

Zo = RC ||ro

= 6.8k || 50k = 5.99k

Av= - (RC ||ro) / re = - 5.99k /18.44 = - 324

BJT AC Analysis

Figure: Example 5.5

BJT AC Analysis

CE Emitter Bias As shown in the figure, it is the emitter bias configuration, without CE .

Substituting re equivalent circuit, note that:

First, let us obtain Zb . The resistance ro is ignored for simplicity.

Vi =Ibre+ Ie RE =Ibre+(+1) Ib RE

So, Zb = Vi / Ib = re+(+1) RE

BJT AC Analysis

Vo = -Io RC

So, Av = Vo / Vi

At input port, Zi = RB ||Zb

At output port,Zo = RC

= - Ib RC = - (Vi / Zb) RC

= - RC / Zb

Phase relationship:180 phase shift occurs between the output and input signals.

BJT AC Analysis

Figure: Emitter bias & its equivalent circuit

BJT AC Analysis

Example 5.6As shown in the figure, it is the emitter bias configuration. Determine:re , Zi , Zo, Av.

Solution: dc analysis,

EB

BECCB RR

VVI)1(

kk

VV56.0)1120(470

7.020A89.35

BJT AC Analysis

IE= (β+1) IB = (120+1) 35.89A=4.34mA

re = (26mV)/IE = (26mV)/ 4.34mA= 5.99

Then, ac analysis

=1205.99 + 121560= 68.48 k

Zb = re+(+1) RE

BJT AC Analysis

Zi = RB|| Zb = (470 k ) || (68.48k )= 59.77 k

Av = - RC / Zb

Zo = RC = 2.2 k

= -120 2.2k / 68.48k

= -3.86

For emitter bias with CE , see Ex. 5.7.For emitter bias + voltage divider, see Ex. 5.8 & 5.9.

BJT AC Analysis

Figure: Example 5.6

BJT AC Analysis

Emitter-Follower

As shown in the figure, it is the emitter-follower configuration.Actually, it is a common-collector network.

The output is always slightly less than the input, but this is good for practical use.

Also the Vo is in phase with Vi and this accounts for the name of “emitter-follower”.

BJT AC Analysis

The emitter-follower configuration has a high input impedance and a low output impedance.

This is the reason why it is used for impedance matching purpose.

This can give a weak load to the previous stage and a strong output ability to the next stage.

BJT AC Analysis

Substituting re equivalent circuit, note that:

First, the same as before, Zb is obtained:

The resistance ro is ignored because for most applications a good approximation for the actual results can still be obtained.

Zb = re+(+1) RE

Zi = RB ||Zb

Then,

BJT AC Analysis

Ie = (+1) Ib

So,

At output port,

= (+1) (Vi /Zb )

Ee

ie Rr

VI)1(

)1(

If is sufficiently large, we get

Ee

ie Rr

VI

This means that is Ie generated by Vi .

BJT AC Analysis

Also, Vo is the potential drop across RE .So we construct a network from the viewpoint of output port.

Furthermore, from this network it is obvious that

So, by setting the Vi to zero, we getZo = RE ||re

ieE

Eo V

rRRV

BJT AC Analysis

This leads to

Phase relationship:Vo and Vi are in phase.

eE

E

i

ov rR

RVVA

Since, RE is usually much greater than re ,

Av 1

BJT AC Analysis

Figure: Emitter-follower & its equivalent circuit

BJT AC Analysis

Example 5.10As shown in the figure, it is the emitter-follower configuration. Determine:re , Zi , Zo, Av.Solution: dc analysis,

EB

BECCB RR

VVI)1(

kk

VV3.3)1100(220

7.020A43.20

BJT AC Analysis

IE= (β+1) IB = (100+1) 20.43A=2.063mA

re = (26mV)/IE = (26mV)/ 2.063mA= 12.6

Then, ac analysis

=10012.6 + 1013.3k= 334.56 k

Zb = re+(+1) RE

BJT AC Analysis

Zi = RB|| Zb = (220 k ) || (334.56k )= 132.72 k

Zo = RE ||re = 3.3k ||12.6 = 12.55

eE

E

i

ov rR

RVVA

6.123.3

3.3k

k9962.0

For some variations of emitter follower configuration, see Fig. 5.54 & 5.55

BJT AC Analysis

Figure: Example 5.10

BJT AC Analysis

Common-base Configuration

As shown in the figure, it is the common-base configuration.

It has a relatively low Zi and high Zo and a current gain less than 1.However, the Av can be quite large.

Substituting re equivalent circuit into the network, note that:

BJT AC Analysis

First,

The resistance ro is typically in the Mand can be ignored in parallel with RC.

Zi = RE || re

Then, Zo= RC

And Vo = - Io RC = Ic RC = Ie RC

With Vi = Ie re ,

= RC / re So that Av = Vo / Vi

BJT AC Analysis

At last, assuming RE >> re , we get

Ii = Ie

AndIo = - Ie

= -So that Ai = Io / Ii

= - Ii

-1

Phase relationship:Vo and Vi are in phase in common-base configuration.

BJT AC Analysis

Figure: Common-base & its equivalent circuit

BJT AC Analysis

Example 5.11As shown in the figure, it is the common-base configuration. Determine:re , Zi , Zo, Av and Ai.

Solution: dc analysis,

E

BEEEE R

VVI

k

VV1

7.02 mA3.1

re = (26mV)/IE = (26mV)/ 1.3mA= 20

BJT AC Analysis

Then, ac analysis

Zi = RE || re

Zo= RC

= 1 k || 20 = 19.61

= 5 k= 245Av= RC / re =0.985 k / 20

Ai = - -1= -0.98

BJT AC Analysis

Figure: Example 5.11

BJT AC Analysis

Content not discussed but still important

• The hybrid equivalent mode (Sec. 5.5)

• The hybrid model (Sec. 5.6)• Collector Feedback (Sec. 5.13)

• Darlington Connection (Sec. 5.20)

• Cascaded Systems (Sec. 5.19)

• Current Source (Sec. 5.22, 5.23)

BJT AC Analysis

Summary of Chapter 5The re model is applied to the analysis of transistor configurations: Common-emitter:

Fixed bias; Voltage divider; Emitter bias Common-collector: emitter-follower Common-base.

The calculation of those parameters, like Vi , Vo , Ii , Io, Zi , Zo, Av and Ai ,in the circuits.