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10. DC (DIRECT-COUPLED) AMPLIFIERS

AC COUPLED SMALL SIGNAL AMPLIFIERS

ADVANTAGES:

1. Signal, load and the amplifier bias are separate.One can work on the bias calculations stage by stage w/o worrying about an interaction with the signal

source or the load.2. No dc current flows through the load or through the signal source.

(Magnetic saturation of the load or the signal source is prevented. Also, their dc resistances do not interferewith the bias.)3. Stages can easily be cascaded. Design of a stage involves only the ac loading effects of its neighbors, dcconditions of the adjacent stages are completely independent of each other.

DISADVANTAGES:

For dc isolation and stability AC coupled amplifiers depend on the isolating capability of capacitors whileexpecting the capacitors to be fully transparent (almost like a short circuit) at the signal frequency.

a. At low frequencies capacitors fail to act like short circuit.Therefore, the AC coupled amplifier behaves like a high pass filter and becomes useful only above a

certain cut-off frequency.b. 3 capacitors are needed for each amplifier stage. Capacitors are bulky and costly and cannot be inte-

grated on a silicon chip.c. "DC-like" slowly varying signals cannot be amplified because of impractically high values of capacitors

required.

DIFFERENTIAL AMPLIFIERS

10. 1 THE DIFFERENCE AMPLIFIER (THE DIFFERENTIAL AMPLIFIER)COMMON-MODE RESPONSEDIFFERENTIAL - INPUT RESPONSEA. The Tail Current SourceB. The BJT Current Source

10. 2 BJT DIFFERENCE AMPLIFIER with ZENER DIODE REGULATED CURRENT SOURCE10. 3 OUTPUT RESISTANCE OF A BJT CURRENT SOURCE

EXAMPLE

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1. v1 v2 but signs are opposite. 2. v1 v2 and signs are the same.

The outputs : 1. vOUT = vO1 "Single ended"

2. vOUT = vO2 "Single ended"

outputs are measured wrt the common (ground).

3. vOUT = vO1 - vO2 "Differential output"outputs are measured wrt the each other.

COMMON-MODE RESPONSE

Right Left Differential Output vO1 - vO2 0 always in common mode

Therefore, 1. Zero response for common-mode input--differential output operation

What if single ended output, vO1 or vO2 ?Since iE1 = iE2 = I O / 2 and fixed, iC1 = iC2 = fixed vO1 = vO2 = fixed,Therefore, 2. Zero response for common-mode input--single-ended output operation.

Note that, in the single-ended output case, the output voltage contains a bias voltage since it is measured wrtthe ground. However, the measured voltage does not change with (does not respond to) the common modeinput. In the differential output, the bias voltages are also canceled leaving an absolute zero output.

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Summary: The difference amplifier circuit shown above does not respond to common mode signals appliedto its input. It "rejects" common-mode signals

DIFFERENTIAL - INPUT RESPONSE

I will assume the signals are small so that I can do a small-signal (linear) analysis.

v1 = HvINDM L 2 v2 = -HvINDM L 2

v1 = v2 v1 - v2 = vINDM

SSAC Equivalent:

The pivot point behaves as if it is tied to ground. Its potential (voltage) cannot change.Therefore it behaves like a "Virtual Ground" .

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So, equivalently:

DvO1 = - R C h fe DiB D iB = ikjjj

V indm 2R + h ie

y{zzz

Single - Ended Output DifferentialGain Avdm =D ikjjj

DvO1V indm

y{zzz=

R C h fe2 HR + h ieL

If R issmallthen A vdm =R C h fe

2 h ie=

R C h fe2 h fe .

kT qICQ=

12

R C ICQ

kT q

Then :

1. Gain for Differential Input SingleEndedOutput :Avdm =

12

R C ICQkT q where I CQ

> IO 2

2. Input Impedance R indm =V indmI indm

where V indm

2= HR + h ieL.DiB and I indm = D iB

Therefore, R indm = 2 h ie since R 1 & R 2 are external.

3. Output Impedance HR out Lsingle - endedoutput = R CIn conclusion: The design of differential amplifier is very simple.

Given ( Avdm & Rindm ) or ( Avdm & Rout ) or ( Rindm & Rout )use appropriate pairs of equations (two equations) to determine the two unknowns, I CQ & RC . I O > 2 I CQ

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DifferentialGain Hfor Single Ended Outputoperation LAvdm =12

R C ICQkT q

Avdm =12

HDCbias onR CL

kT q =12

2.59 V

25.9 mV

= 50

R C .I CQ H VLCC - VCBQ where V CBQ Zero or reverse biasingPotential Max. Differential Gain achievable

HAvdm Lmaxpossible 12

VCC

kT q ~ 250 for V CC = 12 VDC.

THE TAIL CURRENT SOURCE

A. SIMPLE V EE R E CIRCUIT

Assuming v1 and v2 do have a common DC component. Then,

Hv1LDC - R.I BQ1 - VBEQ1 = VEQ

IO = -- VEE - VEQ

R E=

VEE + VEQ

R Eand 2 IEQ1 = IO

Therefore, I O & I BQ & I CQ depend on what Hv1LDC is. This creates a disadvantage for this simple implementa-tion of the tail current source that has to be kept in mind.

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a. Differential Mode Gain:SSACC:

Because of virtual ground behaviour of point "X" (the tail) for pure differential input,2 R E 's get shorted to ground with no effect on the differential response. Therefore, the equations derivedearlier with an ideal current source biasing the tail should be applicable to this circuit.

1. Gain for Differential Input SingleEndedOutput :Avdm =

12

R C ICQkT q where I CQ

> IO 2

2. Input Impedance R indm = 2 h ie since R 1 & R 2 are external.

3. Output Impedance : HR out Lsingle - endedoutput = R C b. Common Mode (Response) Gain:

SSAC:

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SSAC analysis on the left half:

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V incm = R DiB + h ie .DiB + 2 R E H1 + h feLDiBAmplitute of small signalof "v o1 " HVo1Lcm = D vo1 = - R C .h fe .DiB

Common Mode

HSSAC

L Gain A vcm =

HVo1Lcm

V incm=

- R C .h fe .DiB

HR + h ie + Hh fe + 1L 2 R E DiBAvcm =

- R C h fe R + h ie + Hh fe + 1L 2 R E

If R E is large enough, A vcm R C

2 R E

Avcm R C

2 R E

ICQICQ

I 1kt q M

I 1kT q M

, I O > 2 ICQ

Avcm =Avdm

R E Io2 I

1kT q M

=Avdm

DCvoltagedrop onR E2 kT q

Ultimate low A vcm will be attained with values approaching Avdm > Avdm

VEE 2 kT q

Make Pick V EE much larger than kT q

Common - Mode - Rejection - Ratio = C.M.R.R =D Avdm

Avcm >

VEE2 kT q

Conclusions :

1. Differential Response is same as thedifferential amplifier with an ideal current source.

2. Common mode response is not zero. CMMR is not infinite.

3. Common mode rejection can be

improved by increasing V EE but it will always be inferior.

B. BJT CURRENT SOURCES

1. Simple BJT Current Source

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If a unbypassed REX

is present the equivalent small signal resisitance of the resulting current source becomesmuch higher than hoe

- . Therefore, do not bypass REX of the current source circuit. (See next section for proof.)

OUTPUT RESISTANCE OF A BJT CURRENT SOURCE WITH EMITTER DEGENERATION

It's small signal equivalent circuit to calculate r o using the I test , V test method.

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J0

- VTest N=ikjjj

R BX + h ie + R EX - R EX

- R EX + h oe - 1 .h fe R EX + h oe - 1y{zzz J

I1I2N

LinearSolve Aikjjj

R BX + h ie + R EX - R EX

- R EX + h oe- 1 .h fe R EX + h oe

- 1y{zzz, J 0- VTest NE

99-R EX VTest

I 1hoe .h fe - R EX MR EX + I

1hoe + R EX M Hhie + R BX + R EX L=,

9- Hhie + R BX + R EX LVTest

I 1hoe .hfe - R EX MR EX + I

1hoe + R EX M Hhie + R BX + R EX L==

I2 = -Hh ie + R BX + R EX LVTest

I 1hoe .h fe - R EX MR EX + I 1h oe + R EX M Hh ie + R BX + R EX L

r o ==VTestITest

=VTest- I2

I 1hoe .hfe - R EX MR EX + I1hoe + R EX M Hhie + R BX + R EX L

hie + R BX + R EX

r 0 = + ikjjj

1h oe

+ R EX y{zzz+

R EXh ie + R BX + R EX

ikjjj

1h oe

.h fe - R EX y{zzz

a. For small REX (approaching Common-Emitter)

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r 0 =1

h oe

+R EX

R BX

.1

h oe

.h fe r 0 =1

h oe

b. For large REX and small RBX (i.e. Common-Base)

r 0 =1

h oe

+R EX

R EX

.1

h oe

.h fe r 0 =h feh oe

Implement V BBX . RB1X RB2X with as small as RBX as possible.Solution: Use a zener diode with a small r Z to subsitute for RB2X . Then, RBX > r Z // R1 RBX > r Z

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10.2 BJT DIFFERENCE AMPLIFIER with ZENER DIODE REGULATED CURRENTSOURCE

Note that ( % S b L I O requires RBX

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VREX = VZ - VBEX and I EX > ICX > IO

Pick V Z , R EX such that R EX >VZ - VBEX

IO

Pick R 1 X such that D Z avoids operating in its soft breakdown.

IR1 - IBX = ID > IZMIN VR1 = VEE - VZ

Pick R 1 X =VEE - VZ

ID + IBX

h fex . h oex- 1

I-V chs. of BJT Current Source With Zener where: Slope = 1r o =hoexhfex

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Simple BJT Tail Current SourceSlope = 1r o = hoex

Example :

Given thespecs of a BJT DifferenceAmplifier, design its current source,calculate its common - mode and differentialmodegains,input impedances, CMRR and CMR range.

VZ = 6 V V CC = + 15 V V EE = 15 V VA = 100R C = 7 K b > 100 I O = 2 mA I ZMIN = 1.5 mA

1. Design of Zener Diode biased BJT Current Source :

R EX =

6 - 0.7

2 mA =

5.3

2 mA = 2.65 K W

R 1 X =15 - 6

IBX + IDMIN

=9

I2 mA100 M+ 3 mA

> 3 K W HIDMIN IZMIN L

2. Find A vd , A vcm , R indm

Upper common - mode range Hv1Lmax = Hv2LmaxIf R 1 and R 2 > 0 A vd =

12

R C .I CQ

kt q =12

H7 K L I2 mA2 M

0.0259=

3.50.0259

= 135

Avcm > R C

2 r o

= R C

2 Hh oex L- 1 .h fex

> R C2 I IcxVAX + VCEX M

- 1.h fex

where r o is the small signal resistance of the current sourceand is equivalent toR E of simple current sourcein theequation above.

Calculating V CEX ,

VCEX = VX - H- VEE + VZ - VBEQ L

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For HVCEX LHv1LDC = 0, Hv2LDC = 0 VX = - VBEQ , therefore,

HVCEX LHv1LDC = 0, Hv2LDC = 0 =- 0.7 - H- 15 + 6 - 0.7L> 9 V > VCESAT O.K.

Since I EX > IO ,

Hh oex L- 1 = ikjjjIcx

VAX + VCEXy{zzz

- 1

= ikjjj2 mA100 V

+ 9 V y{zzz- 1

> 55 K

Avcm =7 K W

2 * 55 K W * 100 I1 + 9100 M

> 6 * 10 - 4

CMRR =AvdmAvcm

=135

6.10 - 4

> 225000 = 107 dB of rejection HVery Good !LThiswill be trueas longas Hv1 CM LMIN < Hv1LCM =

Hv2LCM < Hv1 CM LMAX , the common - mode range of the amplifier.

Calculating the common - mode range minimum,

Since the current sourcebehavior collapses for V X - VEE + VREX + VCESAT

Thenthe common - mode voltage appliedto bases of Q 1 and Q 2 can not go below,

Hv

1 MIN= v

2 MIN L V

BEQ1 Hor 2 L+

HV

XLMIN V

BEQ1- V

EE+ V

REX+ V

CESAT

HVXLMIN = - 15 V + HVZ - VBEX L+ VCESAT = - 15 V + H6 - 0.7L+ 1 V = - 8.7 V

Hv1 CM LMIN = VXMIN + VBEQ1 = - 8.7 + 0.7 > - 8 V

Calculating the common - mode range maximum,

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Condition for proper operation of Q1 and Q2 : HV CC - RC I CQ L> V X + V CESAT

Substitute VX = v1 MAX - VBEQ

v1 MAX VCC - R C ICQ + VBEQ - VCESATv1 MAX = 15 V - 7 * 1 mA + 0.7 - 1 > 7.7

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