文档库

最新最全的文档下载
当前位置:文档库 > Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

chapter5

Ampli?ers

Bipolar and MOSFET transistors are both capable of providing signal ampli?cation.There are three ampli?er types that can be obtained using a single transistor.These ampli?ers are described in the chart below.

Bipolar Technology

Con?guration Signal Applied To Output Taken From

Common-emitter Base Collector

Common-base Emitter Collector

Common-collector Base Emitter

MOS technology

Con?guration Signal Applied To Output Taken From

Common-source Gate Drain

Common-gate Source Drain

Common-drain Gate Source

The common-collector ampli?er and the common-drain ampli?er are often referred to as the emitter follower and the source follower,respec-tively.

There are several frequently used two-transistor ampli?ers to be con-sidered as well.These are the Darlington con?guration,the CMOS inverter,the cascode con?guration and the emitter-coupled(or source-coupled)pair.The cascode ampli?er and the coupled-pair ampli?er are available in both bipolar and MOS technologies.

Each of these ampli?er types will have its own characteristics:voltage and current gain,and input and output resistance.Analysis of com-plicated circuits can be simpli?ed by considering the large circuit as a combination of simpler blocks.

In this chapter,we will present the bipolar case?rst and then repeat our analyses for the MOS equivalent circuits.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.1Common-emitter ampli?er.

5.1The Common-Emitter Ampli?er

The schematic for the resistor-loaded common-emitter ampli?er is shown in Figure5.1.The circuit load is shown as resistor R C.Let us start by evaluating the ampli?er’s transfer function as the value of the input source V I is increased.

With V I=0,transistor Q1is cut o?.There is no current?ow in the base,so collector current is also zero.Without current in the col-lector,there is no voltage developed across R C,and V o=V CC.As V I increases,Q1enters the forward active region and begins to conduct current.Collector current can be calculated from the diode equation:

I C=I S exp

V I

V T

(5.1)

The large-signal equivalent circuit is provided below.As the value of V I increases,there is an exponential gain in collector current.As collector current increases,the voltage drop across R C also increases until Q1enters saturation.At this point,the collector to emitter voltage of Q1has reached its lower limit.Further increase of the input voltage will provide only very small changes in the output voltage.

The output voltage is equal to the supply voltage minus the drop across the collector resistor:

V o=V CC?I C R C=V CC?R C I S exp

V I

V T

(5.2)

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.2Common-emitter ampli?er large signal equivalent circuit.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.3Common-emitter ampli?er small signal equivalent circuit. Plotting the transfer function shows an important result.A small incremental change in V I causes a large change in V o while Q1operates in forward active mode.The circuit exhibits voltage gain.

We can use the small-signal equivalent circuit shown in Figure5.3to calculate the gain.In this analysis,we do not include high frequency model components.We also ignore the internal resistance of the source V I and the resistance of any load driven from V o.

The small signal analysis gives

V o=?g m V I(r o R C)(5.3) The unloaded voltage gain is then given as

A V=V o

V I

=?g m(r o R C)(5.4)

The input resistance is given as

R I=r b(5.5)

and the output resistance is

R o=r o R C(5.6) If the value of R C is allowed to approach in?nity,the gain equation for the common-emitter ampli?er reduces to

A V=?g m r o=?V A

V T

(5.7)

Finally,we can calculate the short circuit current gain.If we short the output,we obtain

I =

I o

I I

=

g m V I

V I

b

=g m r b=β(5.8)

Example

Use the circuit de?ned in Figure5.1with R C=10K?,I C=50μA, beta=100,and r o=∞.Find input resistance,output resistance,volt-age gain and current gain for the common-emitter ampli?er.

Solution

Input resistance:R I=r b=β/g m=100/(50μA/26mV)=52K?Output resistance:R o=r o R C≈R C=10K?

Voltage gain:A V=?g m R o=?(50μA/26mV)/10K?=?19.23 Current gain:A I=β=100

Our analysis to this point has ignored external loading e?ects.Let us add some base resistance and load resistance to our circuit and?nd the e?ects on voltage gain.

We will start our analysis by assuming that V I’s DC level is adjusted to maintain I C=50μA.Let R b=10K?and R L=10K?.The small signal equivalent circuit is shown in Figure5.5.

From this we have

V I=V S

r b

R b+r b

and

V o=?g m V I(R o R L)=?g m V S

r b

R b+r b

R o R L

R o+R L

so that

A V=

V o

S =?g m

r b

b b

R o R l

o L

A V=?

50μA

26mV

52K?

62K?

5K?=?8.065

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.4Resistor loaded common-emitter ampli?er.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.5Small signal equivalent circuit for the resistor loaded common-emitter ampli?er.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.6Common-emitter ampli?er with emitter degeneration resistor, R E.

Another circuit option for the common-emitter ampli?er is shown in Figure5.6.Here we see the addition of a series resistance between the emitter and ac ground.The presence of this resistance increases output resistance,increases input resistance and decreases transconductance. The resulting decrease in voltage gain leads us to call the presence of this resistance emitter degeneration.The equivalent circuit in Figure5.7 will be used to determine input resistance and transconductance while Figure5.8will help us calculate output resistance.

Let us look at input resistance assuming r o→∞and R b=0.From Figure5.7,we see that

V I=r b I b+(I b+I c)R E=r b I b+I b(β+1)R E=I b(r b+(β+1)R E)

R I=V I

I b

=r b+(β+1)R E

Ifβis large,we can say that R I≈r b+βR E,and sinceβ=g m r b,we have R I≈r b(1+g m R E).

Considering transconductance,Figure5.7again shows that

V I=r b I b+(I b+I c)R E=I c

β

r b+I c

1+

1

β

R E=I c

1

g m

+R E+

R E

β

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.7Small signal equivalent circuit for the common-emitter ampli?er with emitter degeneration.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.8The test current I x is used to calculate the output resistance. Ampli?er transconductance is then

G m=I c

V I

=

1

1

g m

+R E+R E

β

Ifβis large

G M≈

1

1

g m

+R E

=

g m

1+g m R E

Output resistance is determined by using a test current and calculating the resulting voltage.

We?rst assume that R C is very large and can be ignored.Next we note that the entire test current?ows in the parallel combination of r b and R E.This gives

V I=I X(r b R E)

We also note that current?owing through r o is given by

I(r o)=I x?g m V I=I x+I x g m(r b R E)

Using these results we?nd voltage V x

V x=?V I+I(r o)r o=I x[r b R E+r o(1+g m(r b R E))] Finally,we have R o=V x/I x such that

R o=r b R E+r o(1+g m(r b R E))

The second term is much larger than the?rst,so we can neglect the?rst to obtain

R o=r o(1+g m(r b R E))=r o

1+g m

r b R E

b E

If we divide both the numerator and denominator of the fractional term by r b and use the identity r b=βg m,we arrive at

R o=r o

1+

g m R E

1+g m R E

β

Ifβis much larger than g m R E,this reduces to

R o=r o(1+g m R E)

Finally,we can evaluate the voltage gain of the degenerated common-emitter ampli?er using these simplifying assumptions,but also assuming a?nite value of R C:

A V=?G m(R o R C)=?

g m

1+g m R E

r o(1+g m R E)R C

r o(1+g m R E)+R C

=?

g m R C

1+g m R E+R C

o

If R C/R o is small compared to(1+g m R E),the voltage gain reduces to

A V≈?R C R E

This is a very important result.If all our assumptions are valid,we can design ampli?ers whose gain is independent of g m andβvariations.

5.2The Common-Base Ampli?er

The common-base ampli?er has a signal applied to the emitter and the output is taken from the collector.The base is tied to ac ground.This circuit is frequently used in integrated circuits to increase collector re-sistance in current sources.This technique is called cascoding.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.9Common-base ampli?er and simpli?ed“T-model.”

The hybrid-pi model is an accurate tool,but it is di?cult to use for this circuit.Gray and Meyer suggest a simpli?ed“T-model”that is easy to use and understand,although it is limited to low frequency cases where R C is much smaller than r o of the transistor.The circuit schematic and simpli?ed“T-model”are shown in Figure5.9.Note that r o should be ignored unless R C≈r o,at which time r o should be included in the analysis.

The simpli?cation process results in the creation of a new circuit ele-ment r e.This resistance is the parallel combination of r b and a controlled current source modeled as a resistance of value1/g m.Thus,

r e=

1

g m+1

r b

=

1

g m(1+1

β

)

=

β

g m(β+1)

=

α

g m

Ifβis large,r e≈V T/I C.

By inspection,the input resistance is seen to be R I=r e.Output resistance is similarly R o=R C.Transconductance is G m=g m.From this,we?nd the voltage gain and current gain:

A V=G m R o=g m R C

A I=G m R I=g m R e=α

Note that the current gain of this ampli?er topology is always less than unity.This makes the cascading of several ampli?ers impractical without some type of gain stage between the common-base stages.

In addition to cascoding,common-base ampli?ers are not subject to high-frequency feedback from output to input through the collector-base capacitance as are common-emitter ampli?ers.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.10Emitter follower and small signal equivalent circuit.

5.3Common-Collector Ampli?ers(Emitter

Followers)

The common-collector ampli?er has its input signal applied to the base and the output is taken at the emitter.In this circuit,input resistance depends on the load resistance and output resistance depends on the source resistance.The circuit schematic and small-signal equivalent cir-cuit are shown in Figure5.10.

By summing currents at the output node,we can?nd the voltage gain V o/V s:

A V=V o

V s

=

1

1+R s+r b

L

If we replace the voltage source and R s with a test current source

R I=V x

I x

=r b+R E(β+1)

The input resistance is increased byβ+1times the emitter resistance. Replacing R L with a test voltage source allows us to determine the output resistance:

R o=V x

I x

≈1

g m

+

R s

β+1

The output resistance equals the base resistance divided byβ+1,plus 1/g m.

One of the major uses of emitter followers is as an“impedance matcher.”It has high input resistance,low output resistance,voltage gain near unity and can provide current gain.The emitter follower is often placed between an ampli?er output and a low impedance load.This can help reduce loading e?ects and keep ampli?er stage gain high.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.11Common-emitter and common-collector two-transistor

ampli?ers.

5.4Two-Transistor Ampli?ers

Typical one-transistor ampli?ers can provide voltage gain of several thousand depending on loading.Thus,most practical IC ampli?ers re-quire several stages of ampli?cation to provide the levels of performance needed for circuit applications of today.Analysis of cascaded stages can be completed by considering each transistor as a stage,but several widely used two-transistor“cells”exist.These can be considered single stages and analysis can be simpli?ed.Five common subcircuits will be discussed here:the common-collector,common-emitter ampli?er(CC-CE);the common-collector,common-collector ampli?er(CC-CC);the Darlington con?guration;the common-emitter,common-base ampli?er, also known as the cascode;and the common-collector,common-base ampli?er,also known as the emitter-coupled pair.

MOS equivalents to the cascode and emitter-coupled pair circuits ex-ist,but analogues for the CC-CE,CC-CC and Darlington con?gurations can be better implemented as physically larger single transistor designs.

5.5CC-CE andCC-CC Ampli?ers

CC-CE and CC-CC con?gurations are shown in Figure5.11.Note that some type of bias element is usually required to set the quiescent oper-ating point of transistor Q1.The element may be a current source,a resistor,or it may be absent.Q1is present for two main reasons.It in-creases the current gain of the stage,and it increases the input resistance of the stage.These circuits can be considered as a single“composite”transistor as long as Q1’s output resistance doesn’t a?ect the circuit

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.12Two-transistor ampli?ers can be represented by one composite transistor.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.13Small signal equivalent circuit for the composite transistor. performance.Let us determine the circuit values of r b,r o,g m,andβfor the composite transistors.We’ll denote the values for the composite transistor with a su?x(c)as shown in Figure5.12.

The small signal equivalent circuit for the composite transistor is shown in Figure5.13.We assume the e?ects of r01are negligible for this analysis.

We?rst look at input resistance with the composite emitter grounded. The input resistance to Q2is simply r b2.Thus,the input resistance to the composite transistor looks like the input resistance to the common-emitter ampli?er with emitter degeneration.In this case

r b(c)=r b1+(β+1)r b2

Next,we consider transconductance.Transconductance of the composite transistor is the change in collector current of Q2for a given change in

the e?ective v be of the composite.We need to know how v2changes for a change in the composite v be.This circuit looks like the emitter follower found in the previous section:

v2 v be =

1

1+r b1

(β+1)R b2

Since collector current of the composite is identical to the current of Q2,

we have

I C(c)=g m(c)v be(c)=g m2v2=g m2V be(c)

r b1 (β+1)r b2

and so the composite transconductance is

g m(c)=I C(c)

v be(c)

=

g m2

1+r b1

(β+1)r b2

In the case where I bias=0,emitter current of Q1is equal to base current of Q2.This results in r b1=βr b2,and transconductance of the

composite simpli?es to

g m(c)=g m2 2

Current gain of the composite is the ratio of I C(Q2)to I B(Q1).The base current of Q2is equal to the emitter current of Q1such that

I C(c)=I C(Q2)=βI E(Q1)=β(β+1)I B(Q1)

so that

β(c)=β(β+1)≈β2

Finally,by inspection we have r o=r o2.

5.6The Darlington Con?guration

The Darlington con?guration is shown in Figure5.14.It is characterized by having the collectors tied together,and the emitter of one transistor drives the base of the other.This composite can be used in common-emitter,common-collector or common-base con?gurations.It is usually good design practice to have some type of bias element present for Q1, but that element is not required.When used as a common-emitter ampli?er,the collector of Q1connects to the output instead of to an ac ground,and it provides feedback that reduces r o(c).Input capaci-tance is also increased because the collector-base capacitance of Q1is connected between the input and output,resulting in Miller-e?ect mul-tiplication.These drawbacks often make it preferable to use the CC-CE con?guration.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.14Darlington con?guration.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.15Cascode ampli?er.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.16Small signal equivalent circuit for the cascode ampli?er. 5.7The CE-CB Ampli?er,or Cascode

The cascode ampli?er is shown in Figure5.15.We have seen how cas-coding increases output resistance in current mirrors.This is again an advantage in ampli?cation stages,since voltage gain typically resembles g m r o.Another advantage is that there is no high-frequency feedback path through the collector-base capacitance as occurs in the common-emitter topology.

The small signal equivalent circuit for the cascode is developed by taking the equivalent circuits for both CE and CB stages and putting them together.This circuit is shown in Figure5.16.We will use the equivalent circuit to determine input resistance,output resistance and transconductance.

We see by inspection that the input resistance of the cascode is simply r b of Q1.We also see that the current gain from emitter to collector of Q2is approximately one,so the transconductance of the cascode is approximately the transconductance of Q1,or g m1.

To calculate output resistance,we?rst simplify our equivalent circuit. Shorting v1to ground has the result of removing r b1and the controlled current source for Q1.We also note that r o1and r e2are now in parallel to ground.Since r o1is much larger than r e2,we can neglect r o1as well. We can redraw our equivalent circuit and connect a test voltage to our output as shown in Figure5.17.

Current I x1is given by

I x1=

V x

r o2+r e2

≈V x

ro2

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.17Redrawn small signal equivalent circuit for determining the output impedance.

This current is also equal to?I e.Current I x2?ows in the controlled current source,but this current is de?ned as

I x2=g m2v2=αI e=?αV x r o2

Total current I x=I x1+I x2such that

I x=V x

r o2

(1?α)≈

V x

βr o2

and

R o=βr o2

5.8Emitter-CoupledPairs

Emitter-coupled pairs,also known as di?erential ampli?ers,are probably the most often used type of ampli?er in integrated circuit design.The emitter-coupled pair provides di?erential input characteristics required for all operational ampli?ers.Cascading of sequential stages can be ac-complished without need for impedance matching,and relatively high gains can be realized in a small area of circuitry,especially when com-bined with current mirror active loads.The MOS di?erential ampli?er is called a source-coupled pair.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.18Emitter-coupled pair.

The schematic representation of the emitter-coupled pair is shown in Figure5.18.Note that the biasing current source can be a transistor source(current mirror)or a simple resistor.If a resistor is used,the current source I EE becomes zero.If a transistor is used,the transistor equivalent circuit replaces I EE and the resistor.

Let us?rst consider the large signal transfer characteristic of the emitter-coupled pair shown in Figure5.18.For simplicity,we will assume the bias current source output resistance and the output resistances of Q1and Q2are all in?nite.This assumption is valid for the large signal analysis,but not for the small signal analysis.We can also assume that Q1and Q2are identical transistors with the same saturation current I S. If we use Kircho?’s Voltage Law on the loop containing V I1,V I2and the base-emitter junctions of Q1and Q2,we obtain

V I1?V be1+V be2?V I2=0

Recalling that V be=V T ln[I C/I S],we can rewrite the equation above and solve for the ratio of collector currents I C1and I C2:

I C1 I C2=exp

V I1?V I2

V T

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure 5.19Collector currents as a function of the input voltage.

Next we sum currents at the node where the emitters of Q1and Q2are connected:?(I E 1+I E 2)=I C 1α+I C 2α

=I EE We now solve for the collector currents

I C 1=

αI EE 1+exp ?V diff V T and

I C 2=αI EE 1+exp V diff V T

These currents are plotted as a function of V diff in Figure 5.19.Note that the currents become independent of V diff for values greater than 3V T ,or about 75mV .At this point,all of the current I EE is ?owing in only one of the transistors.The current change is linear for a region slightly less than about ±2V T .

The output voltages are given as

V o 1=V CC ?I C 1R C

and

V o 2=V CC ?I C 2R C

However,it is often the case that the di?erential output voltage,V odiff =V o 1?V o 2,is of most interest.V odiff is plotted against V diff in Figure

5.20.

It is possible to extend the range of linear operation by the addition of emitter degeneration resistors as shown in Figure 5.21.The linear region

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.20Di?erential pair output voltage as a function of the di?erential input voltage.

Analog BiCMOS Design Practices and Pitfalls_CHAPER5

Figure5.21The emitter degeneration resistors,R E,extend the linear range of the emitter-coupled pair.