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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
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The Differential Amplifier
Asst. Prof. MONTREE SIRIPRUCHYANUN, D. Eng.Dept. of Teacher Training in Electrical Engineering,
Faculty of Technical EducationKing Mongkut’s Institute of Technology North Bangkok
http://www.te.kmitnb.ac.th/msnmts@kmitnb.ac.th
224510 Advanced communication circuit design 2
BJT Differential Pair
• Differential pair circuits are one of the most widely used circuit building blocks. The input stage of every op amp is a differential amplifier
• Basic Characteristics– Two matched transistors with emitters
shorted together and connected to a current source
– Devices must always be in active mode– Amplifies the difference between the
two input voltages, but there is also a common mode amplification in the non-ideal case
• Let’s first qualitatively understand how this circuit works.
– NOTE: This qualitative analysis also applies for MOSFET differential pair circuits
RCRC
VCC
αI/2αI/2
I
VCC-αIRC/2VCC-αIRC/2
I/2 I/2
vCM vCM - 0.7
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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 3
Case 1
• Assume the inputs are shorted together to a common voltage, vCM, called the common mode voltage
– equal currents flow through Q1 and Q2
– emitter voltages equal and at vCM-0.7 in order for the devices to be in active mode
– collector currents are equal and so collector voltages are also equal for equal load resistors
– difference between collector voltages = 0
• What happens when we vary vCM?– As long as devices are in active mode, equal
currents flow through Q1 and Q2
– Note: current through Q1 and Q2 always add up to I, current through the current source
– So, collector voltages do not change and difference is still zero….
– Differential pair circuits thus reject common mode signals
RCRC
VCC
αI/2αI/2
I
VCC-αIRC/2VCC-αIRC/2
I/2 I/2
vCM vCM - 0.7
Q1 Q2
224510 Advanced communication circuit design 4
Case 2 & 3
• Q2 base grounded and Q1 base at +1 V– All current flows through Q1
– No current flows through Q2
– Emitter voltage at 0.3V and Q2’s EBJ not FB– vC1 = VCC-αIRC
– vC2 = VCC
• Q2 base grounded and Q1 base at -1 V– All current flows through Q2
– No current flows through Q1
– Emitter voltage at -0.7V and Q1’s EBJ not FB– vC2 = VCC-αIRC
– vC1 = VCC
RCRC
VCC
0αI
I
VCC-αIRC
I 0
0.3V
Q1 Q2+1V
RCRC
VCC
0 αI
I
VCC
0 I
-0.7V
Q1 Q2-1V
VCC-αIRC
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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 5
Case 4
• Apply a small signal vi– Causes a small positive ΔI to flow in Q1– Requires small negative ΔI in Q2
• since IE1+IE2 = I– Can be used as a linear amplifier for small
signals (ΔI is a function of vi)• Differential pair responds to differences in the input
voltage– Can entirely steer current from one side of the
diff pair to the other with a relatively small voltage
• Let’s now take a quantitative look at the large-signal operation of the differential pair
224510 Advanced communication circuit design 6
• First look at the emitter currents when the emitters are tied together
• Some manipulations can lead to the following equations
• and there is the constraint:
• Given the exponential relationship, small differences in vB1,2 can cause all of the current to flow through one side
BJT Diff Pair – Large-Signal Operation
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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 7
• Notice vB1-vB2 ~= 4VT enough to switch all of current from one side to the other• For small-signal analysis, we are interested in the region we can approximate to
be linear– small-signal condition: vB1-vB2 < VT/2
224510 Advanced communication circuit design 8
BJT Diff Pair – Small-Signal Operation
• Look at the small-signal operation: small differential signal vd is applied
– expand the exponential and keep the first two terms
multiply top and bottom by
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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 9
Differential Voltage Gain
• For small differential input signals, vd << 2VT, the collector currents are…
• We can now find the differential gain to be…
224510 Advanced communication circuit design 10
BJT Diff Pair – Differential Half Circuit
• We can break apart the differential pair circuit into two half circuits – which then looks like two common emitter circuits driven by +vd/2 and –vd/2
Virtual Ground
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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 11
Small-Signal Model of Diff Half Circuit
• We can then analyze the small-signal operation with the half circuit, but must remember
– parameters rπ,gm, and ro are biased at I / 2 – input signal to the differential half circuit is vd/2
– voltage gain of the differential amplifier (output taken differentially) is equal to the voltage gain of the half circuit
rπ vπgmvπ
ro RC
vc1
vd/2
224510 Advanced communication circuit design 12
Common-Mode Gain
• When we drive the differential pair with a common-mode signal, vCM, the incremental resistance of the bias current effects circuit operation and results in some gain (assumed to be 0 when R was infinite)
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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 13
Common Mode Rejection Ratio
• If the output is taken differentially, the output is zero since both sides move together. However, if taken single-endedly, the common-mode gain is finite
• If we look at the differential gain on one side (single-ended), we get…
• Then, the common rejection ratio (CMRR) will be
– which is often expressed in dB
224510 Advanced communication circuit design 14
CM and Differential Gain Equation
• Input signals to a differential pair usually consists of two components: common mode (vCM) and differential(vd)
• Thus, the differential output signal will be in general…
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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 15
MOS Diff Pair
• The same basic analysis can be applied to a MOS differential pair
– and the differential input voltage is
– With some algebra…
224510 Advanced communication circuit design 16
• We get full switching of the current when…
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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 17
Another Way to Analyze MOS Differential Pairs
• Let’s investigate another technique for analyzing the MOS differential pair
• For the differential pair circuit on the left (driven by two independent signals), compute the output using superposition
– Start with Vin1, set Vin2=0 and first solve for X w.r.t. Vin1
– Reduces to a degenerated common-source amp
– neglecting channel-length modulation and body-effect, RS = 1/gm2
– so…
X Y Vout2Vout1
Vin1 Vin2
I
RDRD
X Y Vout2Vout1
Vin1
I
RDRD
X Y Vout2Vout1
Vin1
RDRD
RS
XVout1
Vin1
RD
RS
M1M2
224510 Advanced communication circuit design 18
Cont’d
• Now, solve for Y w.r.t. Vin1
• Replace circuit within box with a Thevenin equivalent– M1 is a source follower with VT=Vin1
– RT=1/gm1
• The circuit reduces to a common-gate amplifier where…
• So, overall (assuming gm1 = gm2)
by symmetry
X Y Vout2Vout1
Vin1
RDRD
Y Vout2
RD
VT
RT
M1 M2
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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 19
Differential Pair with MOS loads
• Can use load resistors or MOS devices as loads– Diode-connected nMOS loads = 1/gm load resistance
• Load resistance looking into the source– Diode-connected pMOS loads = 1/gm load resistance
• Load resistance looking into diode connected drain– pMOS current source loads = ro load resistance
• Has higher gain than diode-connected loads– pMOS current mirror
• Differential input and single-ended output
224510 Advanced communication circuit design 20
Differential Pair with MOS Loads
• Consider the above two MOS loads used in place of resistors• Left:
– a diode connected pMOS has an effective resistance of 1/gmP
• Right:– pMOS devices in saturation have effective resistance of roP
Vout
Vin
I
Vout
Vin
I
Vb
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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 21
Active-Loaded CMOS Differential Amplifier
• A commonly used amplifier topology in CMOS technologies• Output is taken single-endedly for a differential input
– with a vid/2 at the gate of M1, i1 flows
– i1 is also mirrored through the M3-M4 current mirror– a –vid/2 at the gate of M2 causes i2 to also flow through M2
• Given that ID= I / 2 (nominally)
• The voltage at the output then is given by…
vo
I
M1 M2vid
M3 M4
i1
i1
i2
224510 Advanced communication circuit design 22
Differential Amp with Linearized Gain
• Use source generation to make the gain linear with respect to the differential input and independent of gm
– Can build in two ways…
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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 23
• Assuming a virtual ground at node X, we can draw the following small-signal half circuit.
– Assume ro is very large (simplifies the math)
vid vπ
gmvπro RD
RS
vο
is vS
224510 Advanced communication circuit design 24
Offsets in MOS Differential Pair
• There are 3 main sources of offset that affect the performance of MOS differential pair circuits– Mismatch in load resistors– Mismatch in W/L of differential pair devices– Mismatch in Vth of differential pair devices
• Let’s investigate each individually
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Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 25
Resistor Mismatch
• For the differential pair circuit shown, consider the case where
– Load resistors are mismatched by ΔRD
– All other device parameters are perfectly matched• With both inputs grounded, I1 = I2= I/2, but VO is not zero
due to differences in the voltages across the load resistors
– It is common to find the input-referred offset which is calculated as
– since Ad = gmRD
VO
RD1 RD2
I
I1 I2
224510 Advanced communication circuit design 26
W/L Mismatch
• Now consider what happens when device sizes W/L are mismatched for the two differential pair MOS devices M1 and M2
• This mismatch causes mismatch in the currents that flow through M1 and M2
– This mismatch results in VO
– So in the input referred offset is…
14
Asst. Prof. Dr. MONTREE SIRIPRUCHYANUN
224510 Advanced communication circuit design 27
Vt Mismatch
• Lastly, consider mismatches in the threshold voltage
• Again, currents I1 and I2 will differ according to the following saturation current equation
– For small ΔVt << 2(VGS-Vth)
– Again, using VOS=VO/Ad (Ad = gmRD and VO =2ΔIRD) we get…
224510 Advanced communication circuit design 28
Mismatch Summary
• The 3 sources of mismatch can be combined into one equation:
– arising from Vt, RD, and W/L mismatches• Notice that offsets due to ΔRD and ΔW/L are functions of the overdrive
voltage VGS – Vt
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