The Differential Amplifier

153

GAIN

DABJTCM

PHASE

DB

Temperature = 27

Case = 1

DEG

0.00

−180.0

−12.00

−234.0

−24.00

−288.0

−36.00

−342.0

−48.00

−396.0

−60.00

−450.0

100K

1M

10M

100M

10K

1G

Frequency in Hz

(A)

GAIN

DABJTCM

PHASE

DB

DEG

Temperature = 27

Case = 1

0.00

−180.0

−12.00 −234.0

−24.00 −288.0

−36.00 −342.0

−48.00 −396.0

−60.00	100K	1M	10M	100M	−450.0
10K					1G

Frequency in Hz

(B)

FIGURE 3.10

Common-mode gain frequency response of the BJT DA of Figure 3.6A with various values of parasitic emitter capacitance, Ce, showing how Ce can improve the DA’s CMRR by giving a low CM gain at high frequencies. (A) CM gain frequency response (FR) with Ce = 0. (B) CM gain FR with an optimum Ce = 4 pF. (C) CM gain FR with Ce = 4.2 pF. (D) CM gain FR with Ce = 3.8 pF.

3.5Input Resistance of Simple Transistor DAs

In general, simple, two-transistor DAs have much lower Rin for DM signals than for CM inputs; this is true over the entire frequency range of the DA. This effect is more pronounced for BJT DA stages than for JFET or MOSFET DAs. By way of illustration, consider the CM HIFSSM of the BJT amplifier

© 2004 by CRC Press LLC

154	Analysis and Application of Analog Electronic Circuits
GAIN			DABJTCM		PHASE
DB	Temperature = 27		Case = 1		DEG
0.00					−180.0
−12.00					−234.0
−24.00					−288.0
−36.00					−342.0
−48.00					−396.0
−60.00	100K	1M	10M	100M	−450.0
10K					1G
		Frequency in Hz
			(C)
GAIN			DABJTCM		PHASE
DB	Temperature = 27		Case = 1		DEG
0.00					−180.0
−12.00					−234.0
−24.00					−288.0
−36.00					−342.0
−48.00					−396.0
−60.00	100K	1M	10M	100M	−450.0
10K					1G

Frequency in Hz

(D)

FIGURE 3.10 (continued)

illustrated in Figure 3.6(A). When Rin( f ) DM = Vsc/Isc is plotted using MicroCap, at low frequencies Rin rπ + rx = 3 E3 ohms, and it begins to roll off at approximately 100 kHz. On the other hand, when the CM HIFSSM of Figure 3.6(B) is stimulated, the low frequency Rin( f ) CM 3 E8 ohms, or 2RE (1 + gm rπ), a sizeable increase over the DM Rin( f ) . The low-frequencyRin( f ) CM begins to roll off at only 200 Hz.

Ideally, the Rin( f ) for both CM and DM would be very large, to prevent loading of the sources driving the DA. High Rin( f ) DM can be achieved in several ways. One way is to insert a resistor Re′ between the emitter of each BJT and the Ve node (see Figure 3.5). It is easy to show from the midfrequency, common-emitter h-parameter MFSSM of the BJT DA for DM

inputs (see Figure 3.11(B)) that RinDM = Vsd/Ib = hie + Re′ (1 + β). The practical upper bound on Re′ is set by dc biasing considerations. The small-signal

input resistance, hie, can be shown to be approximated by:

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The Differential Amplifier

155

+Vcc

	RB	RC	RC	RB
		Vo
	Rs			Rs
+		R1 Ve	R1	+
Vs’				Vs
−	A			−

RE

−Vcc

	RC				RC
Rs					Rs
	vo				vo
	ib				ib
+	hie		+		hie
vsd	βib	R1		vs	βib
−	B		−		vec

ved	C

R1

2RE

FIGURE 3.11

(A) A BJT DA with extra emitter resistances, R1, which lower DM gain and increase DM input resistance. (B) Simplified MFSSM of the left side of the DA given DM inputs. (C) Simplified MFSSM of the left side of the DA given CM inputs.

hie VT β/ICQ

(3.20)

where VT = kT/q; β is the transistor’s current gain, hfe; and ICQ is the collector current at the BJT’s quiescent (Q) operating point.

Another approach is to use BJT Darlington amplifiers in the DA; a Darlington circuit and its simplified MFSSM are shown in Figure 3.12. Darlingtons can easily be shown to have an input resistance of:

© 2004 by CRC Press LLC

156	Analysis and Application of Analog Electronic Circuits

VCC

Rc

Q1

ib

Q2

vs

ve2

R1

A

	( oe = hre = 0
B1 hie1	C1
ib
vs
hfe	ib
	E1
	hie2
	B2
	ib = ib1(1 + hfe1)
	B
	ib

Rc

	C2
hfe ib	ve2
(1 + hfe )	R1

FIGURE 3.12

(A) A Darlington stage that can replace the left-hand BJT in the DA of Figure 3.11A. (B) MFSSM of the Darlington valid for DM excitation of the DA. The input resistance for the Darlington DA given DM excitation is derived in the text.

Rin = hie1 + hie2(1 + β1)

(3.21)

If an emitter resistor, R1, is used as in the first case, Rin can be shown to be:

Rin = hie1 + (1 + hfe1)[hie2 + R1(1 + hfe2)] hfe1 hfe2 R1

(3.22)

Thus, if hfe1 = hfe2 = 100 and R1 = 200 Ω, Rin > 2 MΩ.

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The Differential Amplifier

157

Another obvious way to increase Rin for DM excitation (and CM as well) is to design the DA headstage using JFETs or MOSFETs. Using these devices, the lowand mid-frequency Rin is on the order of 109 to 1012 Ω.

3.6How Signal Source Impedance Affects Low-Frequency CMRR

Figure 3.13 illustrates a generalized input circuit for an instrumentation DA. Note that the two Thevenin sources, vs and vs′ can be broken into DM and CM components. As defined earlier:

vsd ∫ (vs − vs′)/2	(3.23A)
vsc ∫ (vs + vs′)/2	(3.23B)

Superposition is used to compute the effects of vsc and vsd on the output of the DA. When vsc is considered, vs and vs′ are replaced with vsc in Figure 3.13; when vsd is considered, vs is replaced with vsd and vs′ with −vsd. Note that manufacturers usually specify a common-mode input impedance, Zic, measured from one input lead to ground under pure CM excitation, and a difference-mode input impedance, Zid, measured under pure DM excitation from either input to ground. The input Zs are generally given by manufacturer’s specs as a resistance in parallel with a small shunting capacitance, for example, Zic as 1011 Ω 5 pF. In the following development, the signal frequency is assumed to be sufficiently low so that the currents through the input capacitances are negligible compared to the parallel input resistance. Thus, only input resistances are used.

Is
Rs	V1
+
Vs	Rc
	R1

DA

Vo

Vs’	Rc
+
Rs + ∆R	V1’
Is’

FIGURE 3.13

A generalized input equivalent circuit for a DA.

© 2004 by CRC Press LLC

158	Analysis and Application of Analog Electronic Circuits

By Ohm’s law, the DM current into the noninverting input node is just:

id = 2vsd/R1 + vsd/Ric	(3.24)
From which,
id/vsd = 1/Rsd = 2/R1 + 1/Ric	(3.25)

can be written. Solving for the equivalent shunting resistance in Equation 3.25 yields:

R1 = 2 Rid Ric/(Ric − Rid)

(3.26)

In many differential amplifiers, Ric > Rid. In others, Ric Rid and, from Equation 3.26, R1 = •. Thus,

Zic = Zid Ric

(3.27)

Assume that Ric = Rid. Thus, R1 may be eliminated from Figure 3.13, which illustrates two Thevenin sources driving the DA through unequal source resistances, Rs and Rs + R.

Using superposition and the definitions in Equation 3.23A and Equation 3.23B, it is possible to show that a purely CM excitation, vsc, produces an

unwanted difference-mode component at the DA’s input terminals:
v	id	/v	sc	= R	ic	R/2(R	ic	+ R	)2	(3.28)
								s
Also, vsc produces a large CM component; the								R term is numerically negligible.
		vic/vsc = Ric/(Ric + Rs)								(3.29)
For purely DM excitation in vs, it can also be shown that
		vid/vsd = Ric/(Ric + Rs)								(3.30)
and
v	ic	/v	sd	= R	ic	R/2(R	ic	+ R	)2	(3.31)
								s

In order to find the CMRR of the circuit of Figure 3.13, Equation 3.2 will be used for vo, and the definition for CMRR, Relation 3.13. Thus:

v

= A

D

v

sc

R

ic

R/2(R

ic

+ R

)2

+ A

C

v

sc

R

/(R

+ R

)

(3.32)

oc

s

ic

ic

s

and

v

od

= A

v

sd

R

/(R

ic

+ R

) + A

C

v

sd

R

ic

R/2(R

+ R

)2

(3.33)

D

ic

s

ic

s

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The Differential Amplifier

159

∞

CMRRSYS

CMRRA

- ic /Rs +	0	∆R/Rs

CMRRA

FIGURE 3.14

The CMRR of a balanced input DA as a function of the incremental change in one input (Thevenin) resistance. Note that a critical value of R/Rs exists that theoretically gives infinite CMRR.

After some algebra, the circuit’s CMRR, CMRRsys, is given by

CMRRsys = [AD + AC R2(Ric + Rs )][AD R2(Ric + Rs ) + AC ] (3.34)

Equation 3.34 may be reduced to the hyperbolic relation:

CMRRsys = [(AD AC ) + R 2(Ric + Rs )] [(AD				AC ) R 2(Ric + Rs ) + 1] (3.35)
which can be approximated by:
CMRRsys			CMRRA			(3.36)
			CMRRA	R
					+ 1
			2(Ric + Rs )

in which the manufacturer-specified CMRR is CMRRA = AD/AC, and CMRRA R/2(Ric + R).

© 2004 by CRC Press LLC