M.B. Patil, IIT Bombay

BJT Differential Amplifier Common-mode and difference-mode voltages A typical sensor circuit produces an output voltage between nodes A and B (see Fig. 1) such that Vd Vd , Vo2 = Vc − , (1) Vo1 = Vc + 2 2 where Vc is called the “common-mode” voltage and Vd the “difference-mode” or “differential” voltage. The common-mode voltage is a result of the biasing arrangement used within the sensor Vo1 = Vc +

Vd 2

Vo2 = Vc −

Vd 2

Sensor circuit

Figure 1: Example of common-mode and difference-mode voltages. circuit, and it can be large (a few volts). The difference-mode voltage is the quantity of actual interest, and it is a measure of the quantity being sensed (such as temperature, pressure, or concentration of a species in a gas). The difference-mode voltage is typically much smaller (say, a few mV) as compared to the common-mode voltage. Of these, the common-mode voltage is obviously a necessary devil, something we cannot simply wish away1 . What we can do is to amplify only the differential quantity (Vd ) while “rejecting” the common-mode quantity (Vc ) by making use of a differential amplifier, as shown in Fig. 2. If Vi1 and Vi2 are expressed as Vi1 = Vic + Vid /2 and Vi2 = Vic − Vid /2, the output of Vi1

Vi1

Sensor circuit

Vo

Vi2

Vo Vi2

Differential Amplifier (a)

Vo1

Sensor circuit Vo2 Differential Amplifier

(b)

Figure 2: Sensor of Fig. 1 connected to a differential amplifier: (a) single-ended output, (b) differential output. 1

Another situation in which we want to reject the common-mode signal is a telephone line where we want to amplify the true signal – the difference between the voltages on the two wires – but reject the spurious common-mode electromagnetic interference riding on each wire.

2

M.B. Patil, IIT Bombay

the differential amplifier is given by, Vo = Ad Vid + Ac Vic ,

(2)

where Ad is the differential gain, and Ac is the common-mode gain. A good differential amplifier should reject Vic entirely, i.e., it should have Ac = 0. In reality, Ac for a differential amplifier is small but finite, and a figure of merit called the “Common-Mode Rejection Ratio” (CMRR) is used to indicate the effectiveness of the amplifier in rejecting common-mode inputs. The CMRR is defined as Ad (3) CMRR = . Ac In this experiment, our objective is to wire up a simple differential amplifier circuit and to see how its CMRR can be improved. The BJT differential pair The circuit shown in Fig. 3, known as the BJT differential pair, can be used to amplify only the differential input signal Vid = (Vi1 −Vi2 ) while rejecting the common-mode signal ViC = 12 (Vi1 + Vi2 ). The two resistors are assumed to be matched and so are the BJTs2 Q1 and Q2 . VCC RC

RC Vo

Vc1 Vi1

Vc2 Q2

Q1

Vi2

I0 −VEE

Figure 3: BJT differential pair. Let us first look at the large-signal behaviour of the circuit. The two emitter currents are given by Is Vi1 − Ve Is Vi2 − Ve Ie1 = exp , Ie2 = exp (4) α VT α VT 2

Note that Q1 and Q2 should not only be identical in construction (fabrication), they must also operate at the same temperature.

3

M.B. Patil, IIT Bombay

(with α = write

β ≈ 1), and they must satisfy the constraint Ie1 + Ie2 = I0 . From Eq. 4, we can β+1

I0 I0 , I = . (5) e2 1 + e−Vid /VT 1 + e+Vid /VT The collector currents are Ic1 = αIe1 ≈ Ie1 and Ic2 = αIe2 ≈ Ie2 . Fig. 4 shows Ic1 and Ic2 as a function of Vid = Vi1 − Vi2 . Note the “steering” effect of the input voltage on the collector currents. When Vid > 0 V, Ic1 is larger than Ic2 ; when Vid > 0 V, Ic2 is larger. When Vid is about 4 VT , almost the entire current I0 is conducted by Q1 ; when it is about −4 VT , it is conducted by Q2 . Ie1 =

linear region 1

Ic1 /I0 Ic2 /I0

0 −6

−4

−2

0

2

4

6

Vid /VT

Figure 4: Normalised collector currents Ic1 and Ic2 versus Vid . Our interest is in using the circuit as an amplifier, and we are therefore looking for the region of Vid where Ic1 and Ic2 vary linearly with Vid (see Fig. 4). In this linear region, the higher-order terms in Vid Vid 1 Vid 2 1 Vid 3 exp = + + + ··· (6) VT VT 2 VT 6 VT must be sufficiently small, i.e., Vid must be much smaller than VT , something like Vid < 0.2 VT , making the width of the linear region in Fig. 4 about 2×0.2×VT or 10 mV at room temperature. The output voltage Vo (see Fig. 3) is given by Vo = Vc1 − Vc2 = (VCC − RC Ic1 ) − (VCC − RC Ic2 ) = RC (Ic2 − Ic1 ) .

(7)

4

M.B. Patil, IIT Bombay

In the linear region, Ic1 =

αI0 αI0 αI0 = , ≈ −V /V T id 1+e 1 + 1 − Vid /VT 2 − Vid /VT

(8)

Ic2 =

αI0 αI0 αI0 = . ≈ +V /V T id 1+e 1 + 1 + Vid /VT 2 + Vid /VT

(9)

The output voltage is Vo = RC (Ic2 − Ic1 ) = For Vid VT , we have

αI0 [(2 − Vid /VT ) − (2 + Vid /VT )] . (2 + Vid /VT )(2 − Vid /VT ) Vo = −

αI0 2 Vid × . 4 VT

(10)

(11)

Since αI0 /2 = Ic1 = Ic2 when Vid ≈ 0, and with gm = IC /VT = (I0 /2)/VT , we get Ad '

Vo = −gm RC . Vid

(12)

Note that Vo does not involve the common-mode voltage Vic at all, exactly as we would like. In a real circuit, this is not quite true, as we shall see later. Small-signal analysis The expression in Eq. 12 can also be obtained using small-signal analysis of the BJT differential pair. The ideal current source in Fig. 3 is replaced with an open circuit in small-signal analysis. Using the T equivalent circuit for the BJTs, we then obtain the small-signal equivalent vid circuit shown in Fig. 5 (a). By symmetry, we have ie1 = −ie2 = . The output voltage is 2 re vo = vc1 − vc2 = −αie1 RC − αie2 RC = −2α

vid RC = −gm RC vid → Ad = −gm RC , 2re

(13)

using gm = α/re . The same relationship can be obtained using the equivalent circuit shown in Fig. 5 (b) where the hybrid-π equivalent circuit is used for the BJTs. Implementation using discrete transistors The BJT differential pair is an integral part of op amp integrated circuits. In this experiment, we will make up the circuit using discrete transistors. Since the transistors are supposed to be identical in all respects and also operating at the same temperature, it is best to use emittercoupled transistors in a chip providing an array of BJTs, such as CA3096 or LM3086. Fig. 6 shows a simple implementation of the circuit of Fig. 3 with a “crude” current source, viz., a simple resistor REE . With Vi1 = Vi2 = 0 V, the emitter voltage is about −0.7 V, and the

5

M.B. Patil, IIT Bombay

RC

RC vc1

vc2

vo

+

vid 2

ie1

re

re

ie2

−

RC

RC

α ie2

α ie1

vid 2

+

vid 2

vc1 ib1

rπ

vc2

vo βib2

βib1

(a)

rπ

ib2

−

vid 2

(b)

Figure 5: Small-signal equivalent circuit of the BJT differential pair: (a) using T equivalent circuit for the BJTs, (b) using the hybrid-π equivalent circuit for the BJTs (the output resistance of the BJTs ro is assumed to be large). VCC

VCC

Vi1

RC

RC

RC

RC

Vc1

Vc2

Vc1

Vc2

Q2

Q1

Vi2

Q2

Q1

REE

REE

(a)

−VEE

(b)

−VEE

Figure 6: Practical implementation of the BJT differential pair: (a) with general inputs Vi1 and Vi2 , (b) with Vi1 = Vi2 = 0 V. bias current I0 is therefore I0 =

−0.7 − (−VEE ) VEE − 0.7 = , REE REE

(14)

each transistor carrying I0 /2. Let us find the differential- and common-mode gains for this amplifier. Consider Vi1 = vid /2 and Vi2 = −vid /2. The small-signal equivalent circuit for this situation is shown in Fig. 7. Writing KCL at the common emitter, we get

6

M.B. Patil, IIT Bombay

RC

RC vc1

vc2

+

vid 2

ie1

vc1

α ie2

α ie1

re

re

ie2

RC

RC

−

vc2 α ie

α ie vid 2

vic

vic ie

re

re

ve

ie

ve 2 ie

REE

(a)

REE

(b)

Figure 7: Small-signal equivalent circuit for the differential amplifier of Fig. 6: (a) computation of Ad , (b) computation of Ac . 1 vid 1 −vid ve − ve + − ve − = 0, re 2 re 2 REE

(15)

giving ve = 0, i.e., the common emitter as AC ground. The circuit now reduces to that shown in Fig. 5 (a), and we have ie1 = −ie2 =

vid vid 1 , vc1 = −α ie1 RC = −α RC = − gm RC vid . 2 re 2 re 2

(16)

1 Similarly, vc2 = + gm RC vid . If the output is taken in a single-ended fashion (say, vo ' vc1 , we 2 1 have Ad = − gm RC . If it is taken in a differential fashion (vo ' vc1 − vc2 ), we have Ad = gm RC . 2 To find the common-mode gain, we apply a small signal vi1 = vi2 = vic , as shown in Fig. 7 (b). Because of symmetry, ie1 = ie2 ' ie , vic = ie (re + 2 REE ), and we get vc1 = vc2 = − α ie RC = − α

vic RC . re + 2 REE

(17)

If the output is taken in a single-ended manner (vo = vc1 or vc2 ), the common-mode gain is3 Ac =

vc1 α RC RC =− ≈− . vic re + 2 REE 2 REE

(18)

If the output is taken in a differential manner, vo = vc1 − vc2 = 0 since vc1 and vc2 are the 3

Note that Ac is independent of bias, and we can expect the expression to hold even for large values of the common-mode voltage.

7

M.B. Patil, IIT Bombay

same4 , giving Vc = 0. For this reason, the output of the first stage of an op amp is taken in a differential manner and fed to the second stage, thus resulting in a large value of CMRR. Improved current source The CMRR of the circuit in Fig. 6 is CMRR =

Ad

=

Ac

1 1 gm RC × = gm REE , 2 RC /2REE

(19)

when the output is taken in a single-ended manner. By increasing REE , the CMRR can be improved; however, this will reduce the bias current I0 (see Eq. 14) and therefore the differential 1 gain Ad = gm RC . Is there a way to provide the desired bias current and simultaneously achieve 2 a high CMRR? The key is to use a current source instead of the resistor REE . Fig. 8 shows a simple current mirror which can provide a current I0 which is nearly independent of the voltage VC4 . The

Iref Vc3 Q3

R

I0 Vc4 Q4 −VEE

Figure 8: A simple current mirror. operation of the current mirror is straightforward: The voltage VC3 is equal to VB3 which is about −VEE + 0.7 V, and the current Iref is therefore Iref =

0 − (−VEE + 0.7) VEE − 0.7 = . R R

(20)

If transistors Q3 and Q4 are identical and are operating at the same temperature, the two collector currents given by IC3 = Is eVBE3 /VT and IC4 = Is eVBE4 /VT

(21)

are identical. Furthermore, if β is sufficiently large, we can ignore the base currents and get I0 = IC4 = IC3 ≈ Iref ≈ 4

VEE − 0.7 . R

(22)

In practice, it is still possible for the output voltage to have a common-mode component (much smaller than the single-ended output case) due to mismatch in the resistance values and BJT parameters [1].

8

M.B. Patil, IIT Bombay

This is indeed what we are looking for – a constant current I0 which is independent of VC4 . This is the basic idea behind a current source. In practice, IC4 would show a small variation with VC4 due to the Early effect (see Fig. 9), ∂IC4 with the slope equal to I0 /VA , where VA is the Early voltage (typically greater than 50 V) ∂VC4 of the BJT5 . Consequently, the small-signal equivalent circuit of this current source would simply be a resistance ro = VA /I0 , the output resistance of Q4 . 1.5

IC4

Iref

R 10 k

IC3 Q3

IC4 VC4

IC3 , IC4 (mA)

IC3 1.0

0.5

Q4 −VEE −15 V

0 −15

−10

VC4 (volts)

−5

0

Figure 9: IC3 and IC4 versus VC4 for the simple current mirror of Fig. 8 (representative plot). Fig. 10 (a) shows an improved differential amplifier circuit using the simple current mirror as the current source. The corresponding small-signal equivalent circuit is shown in Fig. 10 (b). Since the small-signal circuit is the similar to that of Fig. 7, we can use our earlier expressions for Ad and Ac by replacing REE with ro . If single-ended output is considered, we get |Ad | =

1 gm RC , 2

|Ac | =

RC , 2 ro

(23)

where ro is the output resistance of Q4 . Since ro is typically much larger than REE , a significant reduction in Ac is obtained. References 1. A.S. Sedra and K.C. Smith and A.N. Chandorkar, Microelectronic Circuits Theory and Applications. New Delhi: Oxford University Press, 2009. 2. P.R. Gray and R.G Meyer, Analysis and Design of Analog Integrated Circuits. Singapore: John Wiley and Sons, 1995. 5

Note that near VCE4 = 0 V (i.e., VC4 = −VEE ), the currents drop since Q4 is not in the active region any more. This region should of course be avoided.

9

M.B. Patil, IIT Bombay

VCC RC

RC

Vc1

Vc2

RC

RC vc1

vc2 α ie2

α ie1 Vi1

Q2

Q1

Vi2

vi2

vi1 ie1

I0

R

re

re

ie2

ve

Vc4 Q3

ro

Q4

(a)

−VEE

(b)

Figure 10: (a) Improved differential amplifier using a simple current mirror, (b) small-signal equivalent circuit.