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Differential Amplifiers (Chapter 8 in Horenstein)


Differential Amplifiers (Chapter 8 in Horenstein) Differential amplifiers are pervasive in analog electronics Low frequency amplifiers High frequency amplifiers – PowerPoint PPT presentation

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Title: Differential Amplifiers (Chapter 8 in Horenstein)

Differential Amplifiers (Chapter 8 in Horenstein)
  • Differential amplifiers are pervasive in analog
  • Low frequency amplifiers
  • High frequency amplifiers
  • Operational amplifiers the first stage is a
    differential amplifier
  • Analog modulators
  • Logic gates
  • Advantages
  • Large input resistance
  • High gain
  • Differential input
  • Good bias stability
  • Excellent device parameter tracking in IC
  • Examples
  • Bipolar 741 op-amp (mature, well-practiced,
  • CMOS or BiCMOS op-amp designs (more recent,

R. W. Knepper SC412, slide 8-1
Amplifier With Bias Stabilizing Neg Feedback
  • Single transistor common-emitter or common-source
    amplifiers often use a bias stabilizing resistor
    in the common node leg (to ground) as shown below
  • Such a resistor provides negative feedback to
    stabilize dc bias
  • But, the negative feedback also reduces gain
  • We can shunt the common node bias resistor with a
    capacitor to reduce the negative impact on gain
  • Has no effect on gain reduction at low
    frequencies, however
  • Large bypass capacitors are difficult to
    implement in IC design due to large area
  • Conclusion try to avoid using feedback resistor
    R2 in biasing network

R. W. Knepper SC412, slide 8-2
Differential Amplifier Topology
  • In contrast to the single device common-emitter
    (common-source) amplifier with negative feedback
    bias resistor of the previous slide, the
    differential ckt shown at left provides a better
    bypass scheme.
  • Device 2 provides bypass for active device 1
  • Bias provided by dc current source
  • Device 2 can also be used for input, allowing a
    differential input
  • Load devices might be resistors or they might be
    current sources (current mirrors)
  • The basic differential amplifier topology can be
    used for bipolar diff amp design or for CMOS diff
    amp design, or for other active devices, such as

R. W. Knepper SC412, slide 8-3
Differential Amplifier with Two Simultaneous
  • The differential amplifier topology shown at the
    left contains two inputs, two active devices, and
    two loads, along with a dc current source
  • We will define the
  • differential mode of the input vi,dm v1 v2
  • common mode of the input as vi,cm ½ (v1v2)
  • Using these definitions, the inputs v1 and v2 can
    be written as linear combinations of the
    differential and common modes
  • v1 vi,cm ½ vi,dm
  • v2 vi,cm ½ vi,dm
  • These definitions can also be applied to the
    output voltages
  • Differential mode vo,dm vo1 vo2
  • Common mode vo,cm ½ (vo1 vo2)
  • Alternately, these can be written as
  • vo1 vo,cm ½ vo,dm
  • vo2 vo,cm ½ vo,dm

R. W. Knepper SC412, slide 8-4
Bipolar Transistor Differential Amplifier
  • Q1 Q2 are matched (identical) NPN transistors
  • Rc is the load resistor
  • Placed on both sides for symmetry, but could be
    used to obtain differential outputs
  • Io is the bias current
  • Usually built out of NPN transistor and current
    mirror network
  • rn is the equivalent Norton output resistance of
    the current source transistor
  • Input signal is switching around ground
  • Vref 0 for this particular design
  • Both sides are DC-biased at ground on the base of
    Q1 and Q2
  • vBE is the forward base-emitter voltage across
    the junctions of the active devices
  • Since Q1 and Q2 are assumed matched, Io splits
    evenly to both sides
  • IC1 IC2 Io/2

R. W. Knepper SC412, slide 8-5
NPN Bipolar Transistor Physical Structure
  • Dual-Polysilicon Bipolar Transistor Features
  • Two polysilicon layers
  • P for extrinsic base, N for emitter
  • Self-aligned with emitter window opening
  • Trench Isolation
  • Oxide lined, polysilicon filled
  • Shallow Trench Isolation (STI)
  • Isolate base/emitter active region from collector
  • N-type Pedestal Implant for high fT device
  • Self-aligned with emitter opening
  • Limits base push-out (Kirk Effect)
  • Highly doped extrinsic base ? lower Rb
  • Emitter (arsenic) diffused from N poly
  • SiGe Heterojunction BJT
  • Typically 10-15 mole fraction of Ge graded into
    intrinsic base region (as shown), bandgap is
    narrowed in base, adding drift component to
    electron velocity

Yaun Taur, Tak Ning Modern VLSI Devicees
R. W. Knepper SC412, slide 8-6
Bipolar Transistor Operation (1D Device)
  • BJT operation
  • An external voltage (0.75-0.85 V) is applied to
    forward-bias the emitter-base junction
  • Electrons are injected from the emitter into the
    base comprising the majority of the emitter
  • Holes are injected from the base into the
    emitter, as well, but their numbers are much
    smaller, since ND,e gtgt NA,b
  • Since XB ltlt Ln in the base, most of the injected
    electrons get to the collector without
    recombining with holes. Any holes that do
    recombine with electrons in the base are supplied
    as base current.
  • Electrons reaching the collector are collected
    across the base-collector depletion region.
  • Since most of the injected electrons reach the
    collector and only a few holes are injected into
    the emitter, or recombine with electrons in the
    base, IB ltlt IC, implying that the device has a
    large current gain.

R. W. Knepper SC412, slide 8-7
  • Shown at left are the effects of different NPN
    bias conditions on the energy bands and the
    electron concentrations
  • (a) No bias (thermal equilibrium)
  • Fermi levels are flat
  • Electron concentration is ND in emitter and
    collector and ni2/NA in the base
  • (b) both junctions reverse biased
  • Increased E-B B-C barriers
  • Increase in depletion regions
  • Electron density in base 0
  • (c) both junctions forward biased
  • Reduced barrier heights
  • Electrons injected into base from both emitter
    and collector
  • (d) forward-biased emitter, reverse-biased
  • Small E-B/large C-B barriers
  • Electrons injected from emitter
  • Electron density 0 at C-B junction and
    appears linear in base region (small WB)

R. W. Knepper, SC412, slide 8-8
BJT Regions of Operation Ebers-Moll DC Model
  • Jim Ebers and John Moll developed a dc model for
    the bipolar transistor which describes the four
    regions of operation on the Vbe vs Vbc voltage
    plot shown at the left
  • Forward active region
  • Emitter-base forward biased, collector-base
    reverse biased
  • Normal useful region for BJT
  • Current Gain ? 100 typically
  • Reverse active region
  • Collector-base forward biased, emitter-base
    reverse biased
  • Transistor is being operated in the inverse mode
  • Inverse ? is usually small 1 or lt 1
  • Saturation region
  • Both E-B and C-B forward biased
  • Base region is flooded with electrons
  • Cut-off region
  • Both junction reverse biased
  • No current flow
  • Both junctions are forward-biased the same
  • No current flows even though the base is
    loaded with
  • charge (electrons).
  • Saturation condition both junctions forward
  • biased. Net electron flow from emitter
    to collector.

R. W. Knepper SC412, slide 8-9
Ebers-Moll BJT DC Model Current Equations
  • The Ebers-Moll model may be used under all
    junction bias conditions (i.e., forward-active,
    inverse, saturation, and cut-off) to estimate the
    terminal currents.

R. W. Knepper SC412, slide 8-10
Bipolar Transistor Collector Characteristics
  • Shown below is a set of BJT (bipolar junction
    transistor) collector characteristics
  • IC versus VCE with IB as the parameter
  • The curves have several regions of operation
  • At low VCE both the emitter-base junction and the
    collector-base junction are forward-biased,
    resulting in what is called saturation in the
    bipolar transistor
  • The base volume is flooded with mobile carriers
    injected from both E-B and C-B junctions
  • At higher (normal) VCE only the emitter-base
    junction is forward-biased, while the
    collector-base junction is reverse-biased,
    resulting in the normal active (forward mode)
  • The carrier concentration is pinned at zero (i.e.
    very small) at the collector junction, resulting
    in a linear (triangular) distribution of charge
    in the base
  • Non-zero slope in normal active region is caused
    by base width narrowing due to increase in VCB
    reverse bias and corresponding increase in C-B
    depletion region (Early Effect named after Jim
  • At even higher VCE the transistor enters the
    onset of avalanche breakdown at the CB junction

The non-zero slope in the forward mode region is
modeled, as shown below, with a linear term
VCE/VA, where VA is the Early Voltage.
R. W. Knepper SC412, slide 8-11
NPN DC Characteristics
  • Top left figure shows a set of collector
    characteristics (common emitter) for base current
    stepped from 0 to 30 uA for a SiGe HBT with
    emitter area of 0.5 x 2.5 um
  • Very flat curves indicate Early voltage greater
    than 70 volts.
  • Gummel plots showing log Ic and log Ib versus VBE
    indicate excellent SiGe NPN behavior and
    extremely low recombination current at low VBE
  • Beta remains constant at about 200 to VBE 0.9
    volt or higher

R. W. Knepper SC412, slide 8-12
Harame, et al., IEEE Trans ED, Vol. 48, No. 11,
Nov. 2001
Definitions of fT and fmax
  • Cuttoff frequency fT can be defined as a series
    of time constants including base storage time ?b,
    emitter storage time ?e, collector storage time
    ?c, and several RC time constants due to emitter
    and collector depletion capacitances and
    collector-to-substrate capacitance
  • Normally the dominant terms in order of
    significance are the base storage time ?b,
    emitter storage time ?e, and the depletion charge
    terms (kT/qIc)(Cje Cjc)
  • For IBM SiGe NPN technology the last several
    terms are usually negligible since Re, Rc, and
    Rns are small

R. W. Knepper SC412, slide 8-13
IBM SiGe Design Kit Training Technology, IBM
Microelectronics, Burlington, VT, July 2002
SiGe NPN Bipolar ? and fT versus Current
  • Plotted at left are the current gain ? and fT
    versus collector current for two different
    emitter width NPN transistors
  • Both ? and fT drop off at high current density
    due to base push-out (called the Kirk Effect)
  • When the number of injected electrons exceeds the
    N type doping of the collector region, the
    base-collector space charge region pushes all the
    way to the heavily-doped N subcollector.
  • The use of a self-aligned collector pedestal N
    implant raises the doping in the intrinsic
    portion of the collector N epi and prevents base
    push-out until very high current (lt1 mA/um2)
  • Use of a self-aligned pedestal implant limits the
    increase in Ccb due to the higher collector
    doping (which is only in the intrinsic portion of
    the device)
  • The two curves in the plots are shifted by the
    area of the emitter.
  • Using minimum width for the emitter improves base
    resistance and therefore improves device

R. W. Knepper SC412, slide 8-14
Harame, et al., IEEE Trans ED, Vol. 48, No. 11,
Nov. 2001
Bipolar Transistor Large-Signal Small-Signal
  • Shown at the left is a simplified dc large-signal
    BJT model for normal forward-mode only
  • The base-emitter junction is modeled as a diode
    with base current IB as an exponential function
    of base-emitter bias VBE
  • The collector current IC is given simply as ?F
    times IB
  • The emitter current IE is given by IC IB
  • A small-signal ac bipolar model is shown at the
  • The base-emitter junction is modeled as a
    parallel RC combination Cbe (stored charge in the
    base B-E junction capacitance) with r? (
    kT/qIB ?o/gm)
  • The collector current is determined by the
    transconductance term gmVbe in parallel with the
    output resistance ro
  • The base resistance is modeled as rb
  • Collector-to-base and collector-to-substrate
    capacitances are shown
  • A simplified expression for fT (shown) can be
    derived by setting ac current gain to unity

R. W. Knepper SC412, slide 8-15
BJT SPICE Model Parameters
  • Typical SPICE circuit model parameters for a
    vintage 1 um silicon bipolar technology are given
    below (from Johns and Martin, Analog Integrated
    Circuit Design, 1997, p. 65)
  • The fT would be about 13 GHz, based on the
    forward base transit time ?F of 12 ps
  • Reverse current gain-bandwidth product would be
    about 40 MHz based on ?R of 4 ns
  • Rb of 500 ohms and Ccb of 18 fF suggest a
    relatively low fmax of about 7-8 GHz
  • fmax fT / 8?RbCcb½

R. W. Knepper SC412, slide 8-16
Small-Signal Model Analysis for Single Input Diff
  • Consider transistor Q2 with grounded base
  • dc small-signal model shown in top-left figure
  • Use the test voltage approach to calculate Q2s
    input impedance looking into emitter
  • Using KCL equations, we can write
  • itest vtest/r? ?oib2 where ib2 - vtest/r?
  • Rearranging and solving for vtest/itest, we have
  • rth2 vtest/itest r?/(?o 1) r?/?o
  • Generally gm2 is large, causing rth2 to act like
    an ac short
  • Consider transistor Q1 with Q2 replaced by rth2
  • Since rth2 is much smaller than rn (output
    impedance of Io), we will neglect rn
  • Writing KCL, we have
  • vin ib1r?1 ib1(?o 1) rth2 ib12 r?1
  • where we assumed r?1 r?2
  • We can now find vout as a function of vin
  • vout - ic1Rc - ?oibRc - ?ovinRc/2r?1 - ½
  • where we have used gm ?o/r?1
  • Small signal gain Av vout/vin - ½ gmRc

R. W. Knepper, SC412, slide 8-17
Bipolar Diff Amp with Differential Inputs
  • At left is a bipolar differential amplifier
    schematic having two inputs that are differential
    in nature, i.e. equal in magnitude but opposite
    in phase
  • The differential input v1 v2 va(t) (-va(t))
  • The common mode input va (-va)/2 0
  • A small-signal model for the diff amp is shown
    below, where the Tx output collector resistance
    ro is assumed to be gtgt RC (in parallel) and is
  • We can derive the small-signal gain due to the
    differential input by applying KVL to loop A
  • va(t) (-va(t)) 2va(t) ib1r?1 ib2r?2
  • since ib1 -ib2 and r?1 r?2
  • Or, ib1 va(t)/r? and ib2 - va(t)/r?

R. W. Knepper SC412, slide 8-18
Bipolar Diff Amp with Differential Inputs
  • Solving for the output voltages we can obtain
  • vo1 -ic1RC - ?oib1RC - (?o/r?)va(t)RC and
    v02 (?o/r?)va(t)RC
  • We can now find the gain with differential-mode
    input and single-ended output or with
    differential-mode input and differential output
  • Adm-se1 v01/vidm -gmRC/2 and
    Adm-se2 gmRC/2
  • Adm-diff (v01 v02 )/ vidm - gmRC
  • Since corresponding currents on the left and
    right side of the differential small-signal model
    are always equal and opposite, implying that no
    current ever flows throw rn
  • Node E acts as a virtual ground
  • If the output resistances of Q1 and Q2 are low
    enough to require keeping them in the analysis,
    we simply replace RC with the parallel
    combination of RCro for transistor Q1 and Q2

R. W. Knepper SC412, slide 8-19
Small-Signal Model of BJT Diff Amp with CM Inputs
  • The figure below is the small-signal model for
    the diff amp with common-mode inputs
  • v1 v2 vb(t) and vicm ½ (v1 v2) vb(t)
  • The common-mode currents from both inputs flow
    through rn as shown by the two loops
  • in 2(?o 1) ib1 2 (?o 1) ib2
  • and therefore, vb ibr? 2(?o 1)ibrn or ib
    vb/r? 2(?o 1)rn
  • The collector voltages can be found as
  • v01 v02 - ?oRCvb/r? 2(?o 1)rn -
    gmRCvb/ 1 2gmrn
  • The common-mode gain with single-ended output is
    given by
  • Acm-se1 Acm-se2 vo1/vicm vo2/vicm -
    gmRC/1 2gmrn -RC/2rn
  • The common-mode gain with differential output is
    Acm-diff (vo1 vo2)/vicm 0
  • Do Example 8.1, p. 488

R. W. Knepper SC412, slide 8-20
BJT Diff Amp Circuit with Both Diff CM Inputs
  • The example below illustrates the principle of
    superposition in dealing with both differential
    mode and common mode inputs to a diff amp
  • v1 vx cos ?1t vy sin ?2t and v2 vx
    cos ?1t vy sin ?2t
  • Using the definitions of differential mode and
    common mode inputs, respectively,
  • vidm v1 v2 2vy sin ?2t and vicm
    (v1 v2)/2 vx cos ?1t ,
  • we can obtain
  • vo1 Adm-se1 vidm Acm-se1 vicm
  • - ?oRC (vy/ r?) sin ?2t (vx/r? 2
    (?o 1) rn) cos ?1t
  • The expression for v02 is similar except that the
    first term (differential mode) has a minus sign
  • Note that the common mode output is reduced by
    the factor (?o 1) in the denominator

R. W. Knepper SC412, slide 8-21
Common-Mode Rejection Ratio
  • In a differential amplifier we typically want to
    amplify the differential input while, at the same
    time, rejecting the common-mode input signal
  • A figure of merit Common Mode Rejection Ratio is
    defined as
  • CMRR Adm/Acm
  • where Adm is the differential mode gain and Acm
    is the common mode gain
  • For a bipolar diff amp with differential output,
    the CMRR is found to be
  • CMRR Adm-diff/Acm-diff - gmRC / 0
  • In the case of the bipolar diff amp with
    single-ended output, CMRR is given by
  • CMRR Adm-se/Acm-se ½gmRC / ?oRC/r?
    2(?o 1)rn
  • r? 2(?o 1)rn/2r? ?orn/r?
    gmrn ICrn/?VT
  • Iorn/2?VT
  • since ?o gmr? and VT is defined as kT/q
  • CMRR is often expressed in decibels, in which
    case the definition becomes
  • CMRR 20 log (Adm/Acm)

R. W. Knepper SC412, slide 8-22
BJT Diff Amp Input and Output Resistance
  • Input Resistance
  • For differential-mode inputs, the input
    resistance can be found as
  • rin-dm (v1 v2)/ib1 (va (-va)) / (va/r?)
    2var?/va 2r?
  • For common-mode inputs, the input resistance is
    quite different
  • rin-cm ½(v1 v2)/ib1 vb / vb /(r? 2(?o
    1)rn) r? 2(?o 1)rn
  • Output Resistance
  • For differential outputs, we can use the test
    voltage method (below) for deriving the output
    resistance where all inputs are set to zero
  • Since ib1 and ib2 are both zero, we have itest
    vtest/(RC RC) vtest/2RC or rout-diff 2RC
  • For single-ended outputs, rout-se RC ro

R. W. Knepper SC412, slide 8-23
Bipolar Diff Amp Biasing Considerations
  • A bipolar differential amplifier with ideal
    current source and resistor loads is shown
  • It is assumed that components are matched
    sufficiently such that bias current Io is split
    evenly between the left and right-hand legs
  • Node E will take a voltage value such that
  • IC1 IC2 Io/2 when v1 v2 0
  • By using the Ebers-Moll dc model for the NPN
    transistors, we can determine the voltage at node
  • IE IEO exp (qVBE/?kT) 1
  • IEO exp (qVBE/?kT)
  • Io/2
  • or, VBE (?kT/q) ln (IE/IEO)
  • Typically, VBE 0.75-0.85 V in modern NPN
  • It is important to design RC such that vout never
    drops so low so as to force Q1 or Q2 into

R. W. Knepper SC412, slide 8-24
BJT Diff Amp with Simple Resistor Current Source
  • The simplest approach to building a current
    source is with a resistor
  • Given that node E is one VBE drop below GND, we
    can choose RE to provide the desired bias current
  • RE (0 VBE VEE) / Io
  • Preventing saturation in Q1 and Q2 provides an
    upper bound for RC
  • RC lt (VCC 0)/(Io/2) 2 VCC / Io
  • Look at Example 8.3 in text.
  • Do problem 8.31 in class.

R. W. Knepper SC412, slide 8-25
Example 8.3 Diff Amp with Complete Bias Design
  • Design Conditions
  • Differential-mode, single-ended gain gt 50
  • Common-mode, single-ended gain lt 0.2
  • Completed design is shown above
  • In class Exercise 8.4, 8.5, 8.6

R. W. Knepper SC412, slide 8-26
BJT Diff Amp with BJT Current Source
  • The expression for common-mode gain on slide 8-20
    (-RC/2rn) shows that in order to reduce Acm, we
    want to make the effective impedance of the
    current source very high
  • Using a resistor to generate the current source
    limits our design options in making rn (RE in
    this case) high
  • An alternate method of generating Io is to use an
    NPN transistor current source similar to that
    shown at the left
  • Q3 is an NPN biased in the forward active region
    so that rn (given by the inverse slope of the
    collector characteristics) is very high
  • RA and RB form a voltage divider establishing VB
    VEE x RA/(RA RB) where VEE is lt0
  • The voltage across RE can be used to find Io
  • Io (VB Vf VEE)/RE is the bias current
    provided to the diff amp

R. W. Knepper SC412, slide 8-27
Small Signal Model of BJT Current Source
  • Find the small-signal resistance looking into the
    collector of Q3 on slide 8-27 diff amp
  • If RE were 0, then the solution becomes simply
    ro, since the incremental base current ib3 would,
    in fact, be 0
  • With a finite feedback resistor RE, we need to
    use KVL and KCL to derive an expression for rn
    (See Example 8.4 in text)
  • Apply a test current itest and find vtest
  • Obtain v?3 by applying KVL to the 3 left-most
    resistors to obtain ib3 and multiply by r?3
  • v?3 -itest RE r?3 /RE r?3 RP
  • If we multiply this result by gm3 and substract
    from itest, we obtain io3 which can be used to
    find vo3 by multiplying by r03
  • vo3 itest1 gm3RE r?3 /RE r?3 RPro3
  • ve can be found as (itest ib3) x RE
  • ve itest (r?3 RP) RE/(RE r?3 RP)
  • Adding vo3 ve vtest, we obtain rn
  • rn RE (r?3 RP) r03 1 ?oRE/(RE
  • Do Exercise 8.8 and 8.9 in class.

R. W. Knepper SC412, slide 8-28
Bipolar Current Mirror Circuit
  • A method used pervasively in analog IC design to
    generate a current source is the current mirror
  • In the bipolar design arena, the method is as
  • A reference current is forced through an NPN
    transistor connected as a base-emitter diode
    (base shorted to collector), thus setting up a
    VBE in the reference transistor
  • This VBE voltage is then applied to one or more
    other identical NPN transistors which sets up
    the same current Iref in each one of the bias
  • As long as the bias transistor(s) is (are)
    identical to the reference transistor, and as
    long as the bias transistor(s) is maintained in
    its normal active region (where collector current
    is independent of the collector-emitter voltage),
    then the current in the bias transistor(s) will
    be identical to the current in the reference
  • Variations on the basic current mirror circuit
    can be used to generate 2X or 3X or maybe 10X the
    original reference current by using several bias
    NPN transistors in parallel
  • Or alternately, by using an emitter that has 2X
    or 3X or 10X emitter stripes and is otherwise
    identical to the reference transistor
  • Advantages
  • One reference current generator can be used to
    provide bias to several stages
  • Very high incremental output impedance can be
    obtained from the current mirror
  • The technique can be used in both bipolar and in
    CMOS/BiCMOS technologies

R. W. Knepper SC412, slide 8-29
Bipolar Current Mirror Bias Circuit Design
  • Design procedure
  • Given RA and the IC vs VBE characteristics of the
    NPN reference device, we can determine IA, or
  • Given the desired IA and the IC vs VBE
    characteristics of the NPN reference device, we
    can choose RA
  • We can find IA by dividing the voltage drop
    across RA by the resistance value
  • IA (VCC VBE1 VEE) / RA
  • Assuming that the two base currents are small, we
    can say IA Iref
  • Because of the current mirror action, the VBE1
    set up in Q1 to sustain current Iref will be
    equal to VBE2, the base-emitter voltage in Q2
  • Therefore, Io Iref IA
  • Note corrections for IB1 and IB2 can easily be
    made is needed
  • Note 2 Q2 must be maintained in its forward
    active region

R. W. Knepper SC412, slide 8-30
BJT Diff Amp with Current Mirror Bias (Ex. 8.5)
  • Design Objectives
  • Diff amp with 1.5 mA in each leg
  • 5V drop across load resistors
  • VCC 10V, VEE -10V
  • Design Procedure
  • Set Io IA 3 mA
  • RA (0 VBE VEE)/3mA 3.1K
  • where we used VBE 0.7 volt
  • RC1 RC2 can be found as follows
  • RC1 RC2 5V/1.5 mA 3.3K
  • Check VCE of Q2, Q3, and Q4 to see if they are in
    normal active region
  • VC VCC 1.5 mA x 3.3K 5V
  • VE 0 VBE -0.7V
  • VCE 5 (-0.7) 5.7V for Q2 and Q3
  • For Q2 VCE -0.7V (-10) -9.3V
  • Calculate power in each device
  • PQ3 PQ4 1.5mA x 5.7V 8.6 mW
  • PQ2 3 mA x 9.3V 28 mW
  • PQ1 3 mA x 0.7V 2.1 mW

R. W. Knepper SC412, slide 8-31
BJT Current Mirror Feeding 2-stage Diff Amp
  • The example below shows a 2-stage bipolar diff
    amp fed from two current sources with a single
    current mirror
  • Reference current 0.93mA is determined by placing
    (0 VBE VEE) across a 10K bias resistor
  • The reference current is used for the first
    differential stage with 0.47 mA on each leg
  • The second differential stage is to have double
    the bias current of the first stage
  • This is accomplished by using two bias NPN
    transistors in parallel giving 1.86 mA bias
    current with 0.93 mA flowing on each leg (Q7 and
  • Check the VCE of each device to check for normal
    active region and calculate power in circuit.
  • The total circuit power is found by computing the
    sum of the three current source currents
    multiplied by the source-sink voltage
    differential for each.
  • Q1 0.93mA x 10V 9.3mW
  • Q2 0.93mA x 20V 18.6mW
  • Q3/Q4 1.86mA x 20V 37.2 mW
  • Total circuit power 65.1 mW

R. W. Knepper SC412, slide 8-32
Bipolar Widlar Current Source
  • A special use of the current mirror is the Widlar
    Current Source (shown at left)
  • A resistor in the emitter of Q2 is used to reduce
    the current Io in Q2 to a value less than that in
  • Io can be set to a very small value by increasing
    the R2 value
  • Design procedure
  • As in the standard current mirror, we can find
    Iref as follows
  • Iref (VCC VEE VBE1)/RA
  • But, in contrast to the standard current mirror,
    VBE2 will not be equal to VBE1
  • VBE1 VBE2 IE2R2
  • Using the Ebers-Moll model for emitter current
  • IE IEO (expVBE/?VT 1) IEO expVBE/?VT
  • We can invert this expression and insert it into
    the above equation for VBE1 to obtain
  • IE2 (?VT/R2) ln(IE1/IE2) Io (?VT/R2)
  • Since this is not a closed form solution, an
    iterative approach can be used to solve for Io by
    starting with a best guess.
  • Example iteration procedure
  • Assume that Iref 1 mA and R2 500 ohms. Guess
    Io inside ln term. Find LHS Io.
  • Initial guess 0.5 mA, then Io 0.036mA
  • Try a guess of 0.2 mA, then Io 0.083mA
  • Try a guess of 0.1mA, then Io 0.119mA
  • Try a guess of 0.11mA, then Io 0.114mA Close

R. W. Knepper SC412, slide 8-33
Small-Signal Model for Widlar Current Source Q1
  • The incremental output impedance (looking into Q2
    collector) of Widlar Current Source is similar to
    the expression derived for the BJT current source
    (slide 8-28) except that RP must be replaced by
    the incremental resistance at the base of Q1
  • From the model below, the incremental resistance
    at the base of Q1 is given by
  • r?1 1/gm1 ro1 RA r?1/(?o1 1)
  • Thus, the output impedance seen looking into the
    collector of the Widlar Current Source is given
  • rn R2 (r?2 RP) r02 1 ?o2R2/(R2
  • where the above expression is to be used in place
    of RP
  • However, with a number of approximations and
    using the relation IoR2/?VT ln (Iref/Io), the
    expression may usually be simplified to
  • rn r02 1 ln (Iref/Io)
  • Look over Example 8.9 in text.

R. W. Knepper SC412, slide 8-34
NMOS Differential Amplifier Circuit
  • Shown below is a differential amplifier circuit
    built with NMOS technology
  • Q1 and Q2 comprise the diff amp active gain
  • fed by a current mirror Q7 and Q5
  • Q3 and Q4 are NMOS enhancement mode saturated
  • Q6 and Qref are used for biasing the NMOS current
  • Current Io is presumed to split equally on the
    left and right legs of the diff amp
  • The voltage rails are now called VDD and VSS
  • Before going into the biasing and small signal
    models, we will take a look at MOSFET devices and

R. W. Knepper SC412, slide 8-35
MOSFET Transistor DC Current Modes
  • DC Current Modes
  • Cut-off Vgs ? VT
  • Ids 0 (interface is depleted)
  • Linear Region Vgs gt VT, Vgs VT gt Vds
  • Ids ?N Vds (Vgs VT Vds/2)
  • interface is inverted and not pinched off at
    drain (Fig. a)
  • Pinch-off Point Vgs gt VT, Vds Vdsat
  • channel pinches off at the drain junction
  • simple theory Vdsat Vgs VT (Fig. b)
  • Saturated Region Vgs gt VT, Vgs VT lt Vds
  • Ids ½?N (Vgs VT)2
  • interface is inverted and pinched off at drain
  • further increase in Vds occurs across pinch-off
    depletion region (Fig. c)
  • ?N ?NCox (W/L)N where ?N is the mobility of
    electrons in the channel, Cox is the gate
    capacitance per unit area, W is the device width
    and L is the device effective channel length

R. W. Knepper SC412, page 8-36
MOSFET DC Characteristics linear vs saturation
  • If the linear Ids expression from the previous
    chart is plotted with increasing Vds, one
    observes a maximum at Vds Vgs Vt after the
    current reduces in a parabolic fashion.
  • In fact, the charge in the channel Qn(y) goes to
    zero at Vds Vgs Vt
  • The voltage Vds Vgs Vt Vdssat is the
    pinch-off voltage where the channel pinches off
    at the drain junction.
  • Further increase in Vds simply increases the
    voltage between the drain and the channel
    pinch-off point, and does not increase the
    voltage V(y) along the channel.
  • Therefore, IDS remains constant for further
    increases in Vds and we say the device is in the
    saturation region (or active region) with
  • IDS ½ ?n Cox (W/L) (Vgs Vt)2
  • The transconductance in saturation
  • can be found by differentiating the
  • expression for IDS with Vgs, giving
  • gm ?nCox (W/L) (Vgs Vt)
  • 2 ?nCox (W/L) IDS½

R. W. Knepper SC412, slide 8-37
IDS versus VDS for A Real Device
  • Channel length modulation
  • As VDS increases, in fact, there is some non-zero
    slope on the IDS vs VDS characteristic
  • The increase in IDS with VDS is caused by a
    shortening of the effective channel length Leff
    with increasing VDS due to an increase in the
    depletion region thickness from the channel tip
    to the drain junction
  • Substituting for L, from the expression for the
    thickness of an abrupt junction (with a square
    root dependence on reverse biased voltage), one
    can obtain a modified expression for the current
    IDS in saturation (active region) as shown below.
  • From this new expression one can derive an
    expression for the output drain-source resistance
    of the NMOS transistor in the saturation region
  • rds 1/(dIDS/dVDS) 1/(?IDS)
  • where ? is defined below and kds 2?SI?o/qNA½

R. W. Knepper SC412, slide 8-38
MOSFET Capacitance Model
  • The MOSFET capacitances (gate-to-source,
    gate-to-drain, gate-to-substrate,
    source-to-substrate, and drain-to-substrate) are
    illustrated in the drawing at left and summarized
    in the table below
  • Cov is an overlap capacitance due primarily to
    lateral diffusion of the source and drain
    junctions, but also includes fringing capacitance
  • Cov 2/3 Cox W xj
  • where xj is the junction depth
  • The gate-to-channel capacitance is evenly divided
    between source and drain in the linear (triode)
    region, but is effectively connected only to the
    source at pinchoff
  • Integration of the channel charge shows that only
    2/3 of Cgc becomes part of Cgs in the saturation
    (active) region
  • A similar reasoning is used to partition Ccx (CCB
    in picture) between Csx and Cdx
  • Cgx (gate-to-substrate) is zero when Vgs gt Vt,
    but increases to CoxW(L - 2?L) in the
    accumulation region.

R. W. Knepper SC412, slide 8-39
MOSFET High Frequency Figures of Merit
  • Unity gain bandwidth product fT (frequency where
    current gain falls to 1)
  • Assume that a small signal sinusoidal source vGS
    Vpsin(?t) is applied to the gate
  • Input current is given by iG CG (dvGS/dt) CG
    ? Vpcos(?t) (Cgs Cgd) ? Vpcos(?t)
  • Output current is given by iDS gm vGS from the
    definition of gm
  • If we write the magnitude of the ac current gain,
    we have
  • iDS/iG gmvGS / (Cgs Cgd)?Vpcos(?t)
    gm/?(Cgs Cgd)
  • Where we have replaced Vpcos(?t) with vGS since
    we are using only the magnitudes
  • Setting the magnitude of the current gain to
    unity, we obtain
  • fT gm/2?(Cgs Cgd)
  • Because of the manner in which it is derived, fT
    neglects series gate resistance rg and
    capacitance on the output, such as Cgd.
  • Unity power gain bandwidth product fmax
    (frequency where power gain falls to 1)
  • A useful expression for the unity power gain
    point fmax is given by
  • fmax fT / 8?rgCgd½
  • These figure of merits are useful for technology
    comparisons and are also often used in high
    frequency amplifier design

R. W. Knepper SC412, slide 8-40
Long-Channel versus Short-Channel Considerations
  • Consider the current gain bandwidth product fT as
    a function of device parameters
  • fT gm / 2?Cgs µnCox(W/L)(Vgs Vt) /
    (2/3)WLCox 1.5µn(Vgs Vt) / L2
  • where we have assumed the gate capacitance is
    predominantly the Cgs portion
  • Note that fT increases with small L (inverse with
    L2) and with increasing Vgs
  • This is a result based only on a long-channel
  • As the channel shortens, the electric field
    increases beyond the point where mobility is
    constant any longer (typical of todays advanced
    CMOS technology)
  • Scattering of electrons by optical phonons causes
    the drift velocity to saturate at about 1E7
    cm/sec, occurring at an electric field Esat
    1E4 V/cm.
  • Beyond this point further increases in E field
    result in diminishing increases in carrier
  • This effect represents itself in the IDS current
    equation by a reduction in Vdssat below Vgs Vt
    thus reducing IDSsat to less than ½
    µnCox(W/L)(Vgs Vt)2
  • If we redefine Vdsat to be determined by the
    minimum of (Vgs Vt) and LEsat (i.e. sort of
    having (Vgs Vt) and LEsat in parallel), we can
  • Vdsat (Vgs Vt)(LEsat) / (Vgs Vt)
  • We can then rewrite the current equation as
  • IDS ½ µnCox(W/L)(Vgs Vt)(Vdsat) WCox(Vgs
    Vt) vsat 1 LEsat/(Vgs Vt)1
  • where vsat is the saturation velocity given by
    ½µnEsat and µn is the low field mobility
  • At short L, the equation becomes IDS ½
    µnCoxW(Vgs Vt)Esat which is independent of L

R. W. Knepper SC412, slide 8-41
Small Signal Model for a MOSFET
  • The main contribution to the output current is
    the source gmvgs and is given by the expression
  • The current source gsvsx is due to the
    possibility that the source and substrate (bulk)
    voltages may not be the same.
  • gs ?IDS/?VSX ? gm / 2(Vsx 2?F)½
  • where ? ?2q?NA/Cox and 2?F is the band
    bending at strong inversion (from Vt equation)
  • In essence gs is a back-gate transconductance
    which contributes current due to bulk-charge
    voltage change
  • rds accounts for the finite output impedance and
    is given by
  • rds 1 / ? IDS
  • where ? is the output impedance constant (defined
    on slide 8-38)

R. W. Knepper SC412, slide 8-42
  • Level 3 SPICE model parameters are shown in the
    table at the left.
  • The following parameters are given in text for a
    0.5 um technology
  • (PMOS in parentheses if different than NMOS)
  • PHI0.7, TOX9.5E-9, XJ0.2U
  • TPG1 (-1), VTO0.7 (-0.95)
  • DELTA0.88(0.25), LD5E-8 (7E-8)
  • KP1.56E-4 (4.8E-5), UO420 (130)
  • THETA0.23 (0.20), RSH2.0 (2.5)
  • GAMMA0.62 (0.52)
  • NSUB1.4E17 (1.0E17)
  • NFS7.2E11 (6.5E11)
  • VMAX1.8E5 (3E5)
  • ETA0.02125 (0.025)
  • KAPPA0.1 (8)
  • CGDOCGSO3.0E-10 (3.5E-10)
  • CGBO4.5E-10, CJ5.5E-4 (9.5E-4)
  • MJ0.6 (0.5), CJSW3E-10 (2E-10)
  • MJSW0.35 (0.25), PB1.1 (1.0)

R. W. Knepper SC412, slide 8-43
MOSFET Transistor at Threshold (N-FET)
Kang Leblebici, CMOS Digital Integrated
Circuits, 1999
  • MOSFET with Vgs gt VT causes formation of a
    channel (inversion layer) connecting source to
  • With Vds gt 0, a positive current Ids flows from
    drain to source (N-FET)
  • Depletion layer exists from source, drain,
    channel N region to P substrate
  • At Vgs VT, bands are bent by 2?F at
    oxide-silicon interface
  • Threshold definition (by summation of charges in
    gate, oxide, channel, and bulk
  • VTN - Qfc/Cox ?2q?NA(2?F Vsx)/Cox ?MS
    2?F for N-FETs
  • VTP - Qfc/Cox - ?2q?NA(2?F Vsw)/Cox ?MS
    2?F for P-FETs

R. W. Knepper SC412, page 8-44
CMOS NFET and PFET Transistors
  • VTN - Qfc/Cox ?2q?NA(2?F Vsx)/Cox ?MS
    2?F for N-FETs
  • VTP - Qfc/Cox - ?2q?NA(2?F Vsw)/Cox ?MS
    2?F for P-FETs
  • Threshold Voltage is a square root function of
    source-to-substrate per chart at left. Applies
    to both N and P devices using Vsx2?F
  • Implications for circuit applications where the
    source voltage rises significantly above ground

R. W. Knepper SC412, page 8-45
DC Bias Considerations for NMOS Diff Amp
  • It is desired to design the NMOS diff amp (below)
    with device symmetries in such a way that VDS3
    VDS4 VDSref
  • Since Io/2 Iref/2 flows through Q3 and Q4, and
    since Qref, Q3, and Q4 are all biased in their
    saturation (active) regions where IDS ½ ?nCox
    (W/L)(VGS VT)2 K(VGS VT)2 , we can obtain
    VGS3 VGS4 VGSref if W3 W4 ½ Wref
  • This condition will be met independent of other
    device parameter values as long as their ratios
    remain fixed, i.e. good tracking between devices
  • Assuming that the W of Q6 and Q7 are identical to
    that of Qref, then we can see that the above
    current equation will require that VGS6 VGS7
    VGSref 1/3 (VDD VSS), where we have neglected
    any dependence of VT on Vsx.

If we set the current in Q3 to that in Qref, we
can obtain the following expression VDS3
?(VDD VSS)/3 1 - ?VT where ?
(Kref/2Kpu)½ Thus, setting Kref 2Kpu leads
to VDS3 1/3 (VDD VSS) or Vout1
VDD VDS3 2/3 VDD 1/3 VSS
R. W. Knepper SC412, slide 8-46
Modified MOSFET Current Mirror Reference Ckt
  • At left is a modified current mirror reference
    circuit in which four saturated NMOS transistors
    split the voltage between VDD and VSS
  • Assuming that each transistor is designed with
    the same W/L ratio, the reference device VDS will
    be ¼ of VDD VSS
  • Assuming we design transistor Q3 with ½ the W of
    Qref (as on the previous chart), then we will
  • VDS3 ¼ (VDD VSS) and Vout1 (3VDD VSS)/4
  • For n reference devices in series, we can
    generalize the above to
  • Vout1 (n-1)/nVDD VSS/n
  • Exercises 8.15 and 8.16

R. W. Knepper SC412, slide 8-47
Small-Signal Model for the NMOS Diff Amp Ckt
  • The small signal model below is the starting
    point for deriving the gain expressions for the
    NMOS differential amplifier
  • Each transistor is modeled by the gate
    transconductance current source, the back-gate
    transconductance current source, and the
    incremental ac impedance of the device in
  • Note the opposite direction of the back-gate
    (body effect) term is due to the use of
    bulk-to-source voltage rather than source-to-bulk
  • The current mirror current source is modeled
    simply by its output impedance (in saturation)
  • Each transistor is presumed to be in its
    saturation (constant current) region

R. W. Knepper SC412, slide 8-48
Small-Signal Model for NMOS Diff Amp Load Imp
  • We can simplify the equivalent circuit of the
    previous chart by replacing the load transistors
    by their Thevenin equivalent resistance looking
    into their source nodes.
  • Using the test voltage method (shown at left), we
    can obtain
  • rth 1/gm gmb (1/ro)
  • 1/gm(1 ?) ro
  • where gmb ?gm
  • and ? (2?/qNA)/(Vsx 2?F)½
  • Generally we can assume ? 0.2, I.e. the
    back-gate (body) effect adds about 20 to the
    gate transconductance (if we define the voltage
    as bulk-to-source) or reduces the gate
    transconductance about 20 (using the voltage as
  • With the above approximation for the load device,
    we can simplify the NMOS diff amp incremental
    model to that shown on the following slide

R. W. Knepper SC412, slide 8-49
Simplified Small-Signal Circuit Model of NMOS
Diff Amp
  • Differential mode gain can be found from the
    small signal circuit below
  • Adm-se1 -Adm-se2 - ½ gm1 (1/gm3(1 ?3))
    ro3 ro1
  • Adm-diff - gm1(1/gm3(1 ?3)) ro3 ro1
    and gm 2?nCox (W/L) IDS½
  • where we have assumed matched pairs Q1 Q2 and
    Q3 Q4
  • Resistances r01 and r03 are often large enough to
    be neglected relative to 1/gm
  • Common mode gain can also be found from the small
    signal circuit below
  • Acm-se1 Acm-se2 -gm1rth3/1 2ro5gm1(1

  • rth3/2ro5(1 ?1)
  • Input impedance is assume to be infinite
  • Output impedance is given by
  • rout-se rth3 1/gm3(1 ?3) and rout-diff
    2/gm3 (1 ?3)

R. W. Knepper SC412, slide 8-50
Generic CMOS Differential Amplifier
  • A simple version of a CMOS differential amplifier
    is shown at the left
  • The load devices Q3 and Q4 are built with PMOS
  • Q3 and Q4 operate as a form of current mirror, in
    that the small signal current in Q4 will be
    identical to the current in Q3
  • Q3 has an effective impedance looking into its
    drain of 1/gm ro3 since its current will be a
    function of the voltage on node vd1
  • Q4 has an effective impedance looking into its
    drain of ro4 only, since its current will be
    constant and not a function of vout
  • The gain of the right hand (inverting) leg will
    be higher than the gain of the left side
  • Since all transistors have grounded source
    operation, there is no body effect to worry about
    with this CMOS diff amp circuit

R. W. Knepper SC412, slide 8-51
CMOS Diff Amp Equivalent Circuit Model
R. W. Knepper SC412, slide 8-52
CMOS Diff Amp with Current Mirror Sources
R. W. Knepper SC412, slide 8-53
BiCMOS Differential Amplifier
R. W. Knepper SC412, slide 8-41
Small Signal Model of BiCMOS Diff Amp
R. W. Knepper SC412, slide 8-41
BiCMOS Diff Amp with Cascode BJT EF Devices
R. W. Knepper SC412, slide 8-41
JFET Differential Amplifier Circuit
R. W. Knepper SC412, slide 8-41
Large Signal Analysis of Bipolar Diff Amp
R. W. Knepper SC412, slide 8-41
Large Signal Analysis of MOSFET Diff Amp
R. W. Knepper SC412, slide 8-41
Large Signal Analysis of CMOS Diff Amp
R. W. Knepper SC412, slide 8-41
Bipolar Diff Amp DC Design Analysis
R. W. Knepper SC412, slide 8-41
Bipolar Diff Amp DC Design Example 8.11
R. W. Knepper SC412, slide 8-41
NMOS Diff Amp SPICE Simulation Example
R. W. Knepper SC412, slide 8-41
R. W. Knepper SC412, slide 8-41
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