Differential_amplifier_15538443713351126285c9dc8935d11b.pdf

DIFFERENTIAL PAIR

References: [1] Electronic circuit Analysis and design by Donald A. Neaman, Tata McGraw-Hill. [2] Microelectronic Circuits by Adel S. Sedra and Kenneth C. Smith, Oxford University Press.

3.1 Differential amplifier

[1]:

 The basic BJT differential-pair configuration is shown in figure below. It is the input stage to an OPAMP and is probably the most widely used amplifier building block in analog integrated circuits.  The output signal is proportional to only the different between the two input signals, and hence it is called a differential amplifier.

 Assumption 1: The transistors Q1 and Q2 are identical and are in the forward-active region.  Assumption 2: The collector resistance (RC) are equal.  Assumption 3: vB1 and vB2 are ideal sources, meaning the output resistances of these sources are negligibly small.  If the input signal voltages (vB1 and vB2) are both zero, Q1 and Q2 are still biased in the active region by the current source IQ. Therefore the common emitter voltage (vE) would be of the order of –0.7V.  Although the differential amplifier contains two-transistors, it is considered as single stage amplifier. Analysis can show that it has characteristics similar to those of common-emitter amplifier.

3.1.1 Response to common mode voltage

[1], [2]:

 Let us consider that the two base terminals are connected together and a common-mode voltage vcm is applied, as shown below. The transistors are biased on by the constant-current source.  The voltage at the common emitter is vE = vcm – VBE (On).  Since the transistors are identical, current IQ splits evenly between the two transistors and we can write iE1 = iE2 = IQ/2.  Since, iC1  αiE1 , iC2  αiE2 we can write vC1  vC2  VCC  αIQRC 2 . Therefore the difference in voltage between the two collectors will be zero.  Now let us change the common mode voltage vcm. Now, as long as both the transistors are in active region and the current source has sufficient voltage across it to operate properly, the current IQ will divide equally between Q1 and Q2 and the voltages at the collectors will not change. Therefore the differential pair does not respond to (or rejects) changes in common mode input voltage.

3.1.2 Response to difference mode voltage input

[1], [2]:

 Assume vB1 increase by few mV and vB2 decreases by same amount or vB1 = +vd/2 and vB2 = -vd/2. Therefore the voltages at the base of Q1 and Q2 are no longer equal.  Since vB1 increases and vB2 decreases, then vB1 > vB2, which means iC1 increases by ΔI above its quiescent value and iC2 decreases by ΔI below its quiescent value. ΔI is proportional to the difference input voltage.  The output voltage at the collectors will be





vC2  VCC  α  IQ 2  I RC .





vC1  VCC  α  IQ 2  I RC and

 Therefore the output voltage taken between the two collectors will be 2α  I  R C , which is proportional to the input voltage difference.

Numerical 1. For a differential amplifier VCC = 10 V, VEE = -10V, IQ = 1 mA, RC = 10 kΩ, β = ∞ (neglect base current), α = 1, VA = ∞, and VBE (ON) = 0.7 V. Determine iC1 and vCE1 for common-mode volages vB1= vB2 = vCM = 0, -5 V and 5 V. iE1 = iE2 = IQ/2 = 0.5 mA





vC1  vC2  VCC  iC1RC  10  0.5 103 104  5 V

Assuming the transistors are on for vCM = 0 V, vE = -0.7 V and vCE1 = vC1-vE = 5 – (-0.7) = 5.7 V. for vCM = -5 V, vE = -5.7 V and vCE1 = vC1-vE = 5 – (-5.7) = 10.7 V. for vCM = 5 V, vE = 4.3 V and vCE1 = vC1-vE = 5 – (4.3) = 0.7 V. 2. Input voltages v1  2  0.005sin  ωt  V and v2  0.5  0.005sin  ωt  V are applied to a differential amplifier. Find the differential and common mode components of the input signal. vd  v1  v2  2  0.005sin  ωt   0.5  0.005sin  ωt   1.5  0.01sin  ωt  V vc 

v1  v 2 2  0.005 sin  ωt   0.5  0.005 sin  ωt    1.25 V 2 2

3. For a differential amplifier VCC = 10 V, VEE = -10V, IQ = 1 mA, RC = 10 kΩ, β = 200, VA = ∞, and VBE (ON) = 0.7 V. Find the voltages vE, vC1, and vC2 for v1= v2 = 0 V. Assuming the transistors are on vE = -0.7 V. β 200 α   0.995 1  β 201 v C1  v C2  VCC 

αIQR C 2

 10 

0.995  10 3  104  5.025 V 2

4. For a differential amplifier VCC = 10 V, VEE = -10V, IQ = 2 mA, β = ∞ (neglect base current), VA = ∞, and VBE (ON) = 0.7 V. Design the circuit such that the commonmode input voltage is in the range 4  vcm  4 V , while Q1 and Q2 remain biased in the forward-active region. iE1 = iE2 = IQ/2 = 1 mA Since β = ∞, we can write iC1 = iE1 = 1 mA and iC2 = iE2 = 1 mA Assuming the transistors are on, for vcm = 4V, v1 = vC1,Min=+4 V. V  v1 10  4 Therefore RC  CC   6 103 Ω  6 kΩ IC 103 3.1.3 Large-Signal operation

[2]:

 Assumption: Early effect can be neglected.  For the differential pair we can write iE1  ISe vB1  vE 

vT

α and iE2  ISe

 From the above two-equations we can write, iE1 iE2  e  Above equations can be manipulated as,

iE2 1 1   vB1  vB2  v T  iE1  iE2 1  e 1  evd

vT

 vB1  vB2 

vT

vB2  vE  v T

α.

.

iE1 1 1   vB2  vB1  v T  iE1  iE2 1  e 1  e vd

vT

and

where vd  vB1  vB2 .

 Since iE1  iE2  IQ , we can write iE1 

IQ

and iE2 

IQ

. 1 1  e vd v T  The above equations of emitter currents revel that the differential amplifier responds only to vd. For common mode excitation vd = 0 and the current IQ divides equally between the two transistors irrespective of the value of the common mode voltage vCM. This is the essence of differential-amplifier operation.  The plot of iC IQ vs. vd v T is shown in the following figure. e  vd v T

 The above figure reveals that a difference voltage vd  4v T  100 mV  is sufficient to switch the current almost one side of the BJT pair. This property can be used to design a fast current switch.

3.1.4 Small-Signal operation

[1][2]:

 The transfer characteristics of a differential amplifier reveals that if vd  v T 2 then the differential amplifier operates in the linear region of its transfer characteristics. When the differential amplifier operates in this region, it is called small-signal operation.  The linear region of the transfer characteristics of a differential pair can be extended by adding to emitter resistance, as shown below. The expansion of the linear range is obtained at the expense of reduced Gm (slope of the curve for vd = 0) and hence reduced gain. Re here performs in exactly the same way as the emitter resistance of CE amplifier with emitter degeneration.

 The transfer characteristics of the above circuit is shown below for different IQRe.

 Let us consider that a differential amplifier, operating in small-signal condition. The quotient collector current in the transistors is αIQ/2 and the magnitude of small-signal collector current in each transistor is αgmvd/2 (See numerical 5).

 The collector currents can be expressed as iC1 

αIQ 1

e  vd v T

and iC2 

αIQ 1  e vd

vT

.

 Multiplying the numerator and denominator of IC1 by evd 2vT we get, αIQevd 2vT iC1  v 2v . e d T  e vd 2vT  Assuming, vd 2v T , we may expand the exponentials in a series and retain only the first two terms.  v  αIQ 1  d  αIQ αIQ vd αIQ g m vd 2v T  g v   Therefore, iC1       IC  m d , 2 2v T 2 2 2 2  vd   vd  1    1   2v T   2v T   where IC  αIQ 2

αIQ

αIQ vd αIQ g m vd g v    IC  m d . 2 2v T 2 2 2 2  Above equations implies that in the absence of vd, the bias current IQ divides equally between the two transistors. But in presence of vd, the collector current of one transistor increases by and amount g mvd 2 whereas the collector current  Similarly, iC2 



of one transistor decreases by and amount g mvd 2 .

 Due to symmetry of the circuit the differential voltage divides equally between the base-emitter junctions of the two transistors and the total base-emitter voltages v v of the junctions are, vBE1  vBE  d and vBE2  vBE  d . 2 2  The collector voltages are, v v vC1   VCC  ICR C   g mR C d and vC2   VCC  ICR C   g mR C d . 2 2  When the output is defined as the difference between the two-collector voltages, we have a two-sided output and is given by vo  vC2  vC1  gmRCvd .  The ratio of the output signal voltage to the differential-mode input signal is called the difference-mode gain, which is Ad  vo vd  gmRC .  If the output voltage is taken at one collector terminal with respect to ground, the resulting voltage is called one-sided output and is given by Ad  vo vd  gmRC 2 . Numerical 5. Determine the maximum differential-mode input signal that can be applied and still maintain linearity (1%) in the differential amplifier. IQ iC1  iE1  1  e  vd v T The slope at vd = 0 is found to be 2  1   v v IQ di 1  IQ 2 1 d T g f  C1  IQ  1 1  e vd v T      gm e dvd v 0 vT  4v 2 v 2 T T  vd  0 d where gf is the forward trans-conductance and g m is the individual transistor transconductance. The linear approximation for iC1 versus vd can be written as, iC1(linear)  0.5IQ  g f vd . The differential-mode input voltage vd(max) that results in a 1% difference between i (linear)  iC1(actual)  0.01 . the ideal linear curve and actual linear curve is, C1 iC1(linear) IQ IQ   0.5IQ  4v v d  max     vd  max  T  1 e Therefore,  IQ   0.5IQ  4v vd  max   T   IQ   1 vd  max    Or, 0.99 0.5   v  max  v T 4v T   1 e d

vT

 0.01

Assuming, vT  26 mV and using trial and error, we found that vd  max   18 mV . 6. Determine the value of differential-mode input signal such that IC2 = 0.99IQ. IQ iC2  iE2  1  e vd v T

 IQ   1 or, vd  v T ln   iC2  Substituting IC2 = 0.99IQ and vT  26 mV we get vd = -119.47 mV 3.1.5 Small-Signal equivalent circuit analysis

[1]:

 The small-signal equivalent circuit of the differential-par configuration is shown below, where RB represents the output resistance of the voltage sources.  Assumption 1: The transistors are identical and biased at the same quotient current, therefore, rπ1  rπ2  rπ and g m1  g m2  g m .  Assumption 2: The constant current-source can be represented by a finite output impedance Ro.

 Writing KCL equation at node Ve, using phasor notation, we have, V Vπ1 V  g mVπ1  g mVπ2  π2  e rπ rπ Ro

1  β   1  β  Ve or, Vπ1  where g mrπ  β .   Vπ2    rπ   rπ  R o

 Vb2  Ve  V  Ve   From the circuit, Vπ1   b1  rπ and Vπ2    R B  rπ  R B  rπ 

  rπ . 

 1  β   Vb1  Ve   1  β   Vb2  Ve  Ve  Therefore,    rπ     rπ  Ro  rπ   R B  rπ   rπ   R B  rπ 

 1 β  Ve Or,    Vb1  Vb2  2Ve   Ro  R B  rπ  Vb1  Vb2 Or, Ve  .  R B  rπ  2    1  β R o   If we consider a one-sided output at the collector of Q2, then,

Vo  VC2  g m Vπ2R C  

βR C  Vb2  Ve  R B  rπ

.

 From the above two equations we get,   R r     1  B π  Vb2  Vb1  βR C     1  β R o   Vo  g m Vπ2R C     R B  rπ   R r     2  B π  1  β R       o    For an ideal current source Ro=∞ and we get, Vo  

βR C  Vb2  Vb1  2  R B  rπ 

.

 For common mode signal Vcm = Vb1 = Vb2 and we get Vo = 0. V βRC  Since Vd = Vb1 – Vb2, we get Ad  o  . Vd 2  R B  rπ   From the definition of common-mode and difference-mode voltages we can write V V Vb1  Vcm  d and Vb2  Vcm  d . 2 2  Using the above two equations we can write  R r  V   V     1  B π   Vcm  d    Vcm  d   2   2  βR C    1  β R o   Vo  g m Vπ2R C     R B  rπ   R B  rπ     2   1  β R       o   βR C βR C Vd  Vcm Or, Vo  2  RB  rπ  R B  rπ  2 1  β Ro  The general form of output voltage is Vo  Ad Vd  AcmVcm .  Comparing the above two equations we get βR C βRC Ad  and Acm   . 2  R B  rπ  RB  rπ  2 1  β Ro  The common mode rejection ratio of the differential amplifier will be βR C Ad 2  R B  rπ  R  r  2 1  β R o 1  2 1  β  R o  CMRR    B π  1   βR A cm 2  R B  rπ  2   R B  rπ   C R B  rπ  2 1  β R o Or, CMRR 

1  β IQRo 2VT  1  1  β IQRo  1  1  β IQRo  1 1    1    1   2  VT  RB  rπ  IQ  2  VT  R B  rπ  g m  2  VTβ 

1   1  β IQR o    CMRRdB  20 log10  1   VTβ   2     Above equation reveal that CMRR increases as Ro increases.

Numerical 7. Assume that for a differential pair VCC = 10 V, VEE = -10 V, IQ = 0.8 mA, RC = 12 kΩ, β = 100, VA = ∞, Ro = 25 kΩ, RB = 0 Ω. Assuming one-sided output at vC2 determine the differential- and common-mode gains of the amplifier. Also calculate the CMRR. What should be the value of Ro so that CMRR increases to 90 dB. 8. In the following differential amplifier neglect the base currents and assume V EB(on) = 0.7 V. Determine vE and vEC1 for common-mode input voltages v1 = v2 = vcm of: (a) 0 V, (b) +2.5 V and (c) -2.5 V.

9. Design a differential amplifier circuit such that the differential-mode voltage gain at vC2 is +150 and differential-mode voltage gain at vC1 is -100. Assume IQ = 1 mA and bae current in negligible. 10. For a differential amplifier VCC = 15 V, VEE = -15 V, β = 200. The range of commonmode voltage is 5  vcm  5 V. (a) Redesign the circuit to produce maximum onesided differential-mode voltage gain at vC2. (b) If Ro = 100 kΩ for the current source, determine the resulting common-mode gain and CMRRdB for IQ = 0.5 mA and RC = 40 kΩ. 11. Consider the differential-amplifier with modifier Widlar current source. Assume β = 200, Vd = 125 V for Q3 and Q4, and VA = ∞ for Q1 and Q2. Design the circuit such that the common-mode input voltages is in the range 5  vcm  5 V, the CMRRdB = 95 dB, and the maximum differential-mode voltage gain is achieved. Assume IQ = 1 mA and I1 = 1 mA.

3.1.6 Differential and common mode gains

[1]:

 To understand the mechanism behind the difference- and common-mode gains let us consider the differential amplifier with pure differential- and common-mode signals.  The differential amplifier with pure difference mode signals are shown below. The current direction, shown in the figure, assumes vb1 is in positive half-cycle.  The input voltages are of same magnitude and180 degree out of phase. Therefore, as the base voltage of Q1 goes into positive-half cycle, the base voltage of Q2 is in negative half-cycle and vice-versa. Therefore for the circuit vb1+vb2=0 and the common emitters of Q1 and Q2 remain as signal ground.  Since ve is always at ground potential, each half of the differential half-circuits can be assumed as common-emitter configuration, as shown.

 For pure common-mode excitation the equivalent circuit is shown below. Here the current source is represented by an ideal source with output resistance Ro in parallel.

Since the circuit is perfectly symmetrical, it can be split into two identical common mode half circuit, as shown.  Consequently, a common-mode sinusoidal input signal produces a sinusoidal output, which means that differential amplifier has a non-zero common mode voltage gain. However due to presence of the emitter resistance the gain is reduced.  If Ro increases, the magnitude of iq decreases for a given common-mode input signal, producing a smaller output voltage and hence a smaller common-mode gain. 

3.1.7 Differential and common mode input impedance

[1]:

 The input impedance is an important parameter of an amplifier as it determines the loading effect of the circuit on signal source.  For a differential amplifier two types of input impedance may be defined – differential-mode input impedance and common-mode input impedance.  The differential-mode input impedance is the effective resistance between the two input base terminals when a differential-mode signal is applied.  The differential amplifier with pure differential input signal is shown below.

 For this circuit we have,

vd 2  rπ . ib

 Therefore, the differential mode input impedance is Rid  vd ib  2rπ .  For a differential-amplifier with emitter resistance, shown below, with pure applied differential-mode voltage we have vd 2 Or, Rid  vd ib  2 rπ  1  β RE  .  rπ  1  β RE ib

 Above equation implies that differential-mode input impedance increases significantly when emitter resistors are included. The emitter resistors, however, decreases the gain of the amplifier.  For a differential-amplifier with an applied common-mode voltage, shown below, we have, 2R icm  rπ  2 1  β R o  2 1  β R o . This is a first approximation for determining common-mode input resistance

 Normally, Ro is large and Ricm is typically in the MΩ range. Therefore transistor output resistance ro and the base collector resistance rμ should be included in the calculation.  The following figure shows more complete equivalent half-circuit model.  For this model, we have, 2Ricm  rμ 2 1  β Ro  1  β ro  .

Numerical: 12. Find the differential and common mode input resistance of the following differential-amplifier. Assume: β = 100, VBE(on) = 0.7 V, and VA = 100 V.

13. If the differential-mode gain of a differential-amplifier is 60 and the common-mode gain is 0.5, determine the output voltages for input signals of (a) v1  0.505sin  ωt  V , v2  0.495sin  ωt  V and (b) v1  0.5  0.005sin  ωt  V , v2  0.5  0.005sin  ωt  V . 14. For a differential amplifier VCC = 10 V, VEE = - 10 V, IQ = 2 mA, RC = 5 kΩ, Ro = 50 kΩ, β = 150, VBE(on) = 0.7 V and VA = ∞. Determine, (a) the dc input base currents, (b) the differential input signal currents if a differential-mode input voltage vd  10sin  ωt  mV is applied. If a common-mode input voltage vcm  3sin  ωt  V is applied, determine the common-mode signal input currents.