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Differential Amplifiers (Chapter 8 in Horenstein)

- Differential amplifiers are pervasive in analog

electronics - 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

implementation - Examples
- Bipolar 741 op-amp (mature, well-practiced,

cheap) - CMOS or BiCMOS op-amp designs (more recent,

popular)

R. W. Knepper SC412, slide 8-1

Amplifier With Bias Stabilizing Neg Feedback

Resistor

- 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

accordingly - 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

JFETs

R. W. Knepper SC412, slide 8-3

Differential Amplifier with Two Simultaneous

Inputs

- 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

reach-thru - 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

current - 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

collector - 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

amount. - 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)

region - 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

Early) - 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

performance.

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

Models

- 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

left - 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

Amp

- 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

1/gm2 - 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

gmRcvin - 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))

2va(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

neglected - 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

2ib1r? - 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

(continued)

- 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

infinity - 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

RC

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

E - 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

transistors - It is important to design RC such that vout never

drops so low so as to force Q1 or Q2 into

saturation.

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

Io - 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
- VRE VB Vf VEE
- 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

Transistor

- 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

vtest/itest - rn RE (r?3 RP) r03 1 ?oRE/(RE

r?3RP) - 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

circuit - In the bipolar design arena, the method is as

follows - 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

transistors - 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

transistor. - 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

Q8) - 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

Q1 - 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)

ln(Iref/Io) - 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

enough!!

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)

RA - Thus, the output impedance seen looking into the

collector of the Widlar Current Source is given

by - rn R2 (r?2 RP) r02 1 ?o2R2/(R2

r?2RP) - 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

transistors - fed by a current mirror Q7 and Q5
- Q3 and Q4 are NMOS enhancement mode saturated

loads - Q6 and Qref are used for biasing the NMOS current

mirror - 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

models

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

as - 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

assumption. - 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

velocity - 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

write - Vdsat (Vgs Vt)(LEsat) / (Vgs Vt)

(LEsat) - 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

below - 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

MOSFET SPICE Model

- 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

drain - 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

potential.

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

exists - 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

have - 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

saturation - 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

voltage - 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

source-to-substrate). - 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

?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

transistors - 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