LOW VOLTAGE OP AMPS We will cover: –Low voltage input stages –Low voltage bias circuits –Low voltage op amps –Examples Methodology: –Modify standard circuit blocks for reduced power supply voltage –Explore new circuits suitable for low voltage design

ITRS Projection – near term

ITRS Projection – longer term

Low-Voltage, Strong-Inversion Operation Reduced power supply means decreased dynamic range Nonlinearity will increase because the transistor is working close to V DS(sat) Large values of λ because the transistor is working close to V DS(sat) Increased drain-bulk and source-bulk capacitances because they are less reverse biased. Large values of currents and W/L ratios to get high transconductance Small values of currents and large values of W/L will give smallV DS(sat) Severely reduced input common mode range Switches will require charge pumps

Input common mode range drop V DD – V DS3sat + V T1 > v icm > V DS5sat + V T1 + V on > v icm > , unsymmetric!

p-n complementary input pairs n-channel: v icm > V DSN5sat + V TN1 + V onN1 p-channel: v icm <V DD - V DSP5sat - V TP1 - V onP1

Non-constant input g m N

constant input g m solution Let Vb1 depends on Vicm so that Mb1 is turned on when MN1,2 are turned off, and Ip becomes 4 times. Similarly when MP1,2 are off, In becomes 4 times. When both pair on, In and Ip are bothe 1 times

Set VB1 = Vonn and VB2 = Vonp

Rail-to-rail constant g m input As Vin+ and Vin- reduce, MN1,2 begins to turn off, MNC1,2 also begins to turn off. I7 reduces, so does I8. I9 = I12-I8 increases, so does I10, which is 3(I12-I8)=3(Ip-In), which becomes 3Ip when n-pair turns off. When both on, I5=I1=I12=IBP=Ip; I11=I7=I6=IBN=In

V i+ V i- Complementary input stage with rail-to-rail Vicmr and constant g m V i+ V i- 3:   :3 V i- V i+ IpIp InIn 3(I p -I n ) IpIp InIn InIn V bp V bn IpIp 3(I n -I p ) InIn IpIp  (I p -I n )  (I n -I p ) i n+ i n- i p+ i p- N1N2P1P2PC1,2 NC1,2 BN BP 1234

V o1+ V o1- Folded cascode stage: summing current and convert to voltage i n+ i n- i p+ i p- -i n+ -i p- i o1- =i p- - i n+ i n+ =+g mn1 *v id /2 i n- = -g mn1 *v id /2 i p+ =+g mp1 *v id /2 i p- = -g mp1 *v id /2 i o1- =i p- - i n+ = -(g mp1 +g mn1 )*v id /2 i o1+ =i p+ - i n- = (g mp1 +g mn1 )*v id /2 V o1- = i o1- /g o1 V o1+ = i o1+ /g o1 V o1d = (V o1+ - V o1+ ) = v id *(g mp1 +g mn1 )/g o1 V xx V yy V zz V bb

Folded cascode stage: feedback to reduce go1 V xx V yy V zz M1 M2a M2b V bb M3 M4 i g mn1 v d /2 g dsn1 g ds4 g ds3 v cp v cn g ds1 g ds2a g ds2b g m2a v cn +g mb2a v cn g m2b v cn +g m2b v cp +g mb2b v cn v o1+ g m3 v cp Show that it is possible to make gain (vo1/vd) infinity by proper sizing.

Folded cascode stage: feedback to reduce go1, alternative V xx V yy V zz M1a M2 M1b V bb M3 M4 i g mn1 v d /2 g dsn1 g ds4 g ds3 v cp v cn g ds1 g ds2a g ds2b g m2a v cn +g mb2a v cn g m2b v cp v o1+ g m3 v cp

In either case, you can set vd=0, write KCL’s for the vn, vn and vo1 nodes, eliminate vn and vp, obtain expression in vo1 alone, set coefficient to zero, this gives conditions for go=0. From that, you can solve for gm for the feedback transistor and see how that can be realized. For example, in the first choice, if you make gds of M1 and M3 4 times as large as the other transistors, it becomes relatively simpler to meet the conditions. Small signal analysis for 2 nd choice is easier, but quiescent voltage a concern. You can also feedback to PMOS transistors. Feeding back to top or bottom transistors faces big challenges when supply voltage increases. If the whole M2 (or M3 in the P version) is controlled by feedback, then the VgsQ of M2 is independent of supply.

M1 V G3 V G2 M1 M3 M2     M4 M3 M2     M4     Regulated Cascode for gain improvement V xx V zz k VDVD VSVS VSVS VDVD VGVG VGVG If you regulate, you have to regulate all four.

Second stage V xx V yy V zz M1 M2 M6 V bb M3 M4 M7

Second stage push pull: Monticelli style V xx V yy V zz M1 M2 M6 V bb M3 M4 M7 Requires: VDD-VSS > Vgs6+Vgs7+Vdssat_floating_CS VbpVbn D. M. Monticelli, “A quad CMOS single-supply Op Amp with rail-to-rail output swing,” IEEE J.Solid-State Circuits, no. 6, pp. 1026–34, Dec

Don’t do V xx V yy V zz M1 M2 M6 V bb M3 M4 M7 Unpredictable current in second stage

Vzz Vbn Vxx Vbp So, Vg6Q = Vzz This sets Id6.Q So, Vg7Q = Vxx This sets Id7Q. These circuits can come from the same biasing circuit for the main amplifier. So, no extra current, power consumption, noise, and offset introduced.

g ds4 g ds3 g ds1 g ds2a g m2a v cn +g mb2a v cn v o1+ g m3 v cp Floating CS do not change ro1 or DC gain g dsn g dsp g mn v o1+ -g mp v’ o1+ V’ o1+ The impedance looking down from Vo1+ is Rn To find impedance looking up from Vo1+, inject a test current i up. V’ o1+ = i*Rp i_g dsn/p = (i-g mn v o1+ +g mp v’ o1+ ) V o1+ =V’ o1+ + i_g dsn/p /(g dsn + g dsp ) V o1+ =i*Rp+(i-g mn v o1+ +g mp i*Rp) /(g dsn +g dsp ) V o1+ (1+g mn /(g dsn +g dsp )) =i*{Rp[1+g mp /(g dsn +g dsp )] +1/(g dsn +g dsp )} V o1+ /I =Rp g mp /g mn RnRn RpRp So, size them so that g mp ≈ g mn Note: gmn and gmp include possible body effects.

To the two gate terminals of M6 and M7, the two floating CS appears as a voltage source providing a voltage offset between the gates. The impedance seen by the two gate terminals can be calculated by: Rs = (Vgp – Vgn)/(current through the floating CS) = (Vgp – Vgn)/(gmp*Vgp – gmn*Vgn + (Vgp – Vgn)*(gdsn+gdsp)) ≈ 1/(gmp + gdsn+gdsp) ≈ 1/gmp The above assumed that the NMOS and PMOS are sized to have the same gm. Also, the calculation is only valid when both NMOS and PMOS are fully on and both in saturation. When Vo is experiencing large swings, these conditions are not met. And the voltage difference between the two gate terminals no longer remain constant.

Differential signal path compensation V xx V yy V zz M1 M2 M6 V bb M3 M4 M7

In order for M6 and M7 to have well defined quiescent current, we have to bias the circuit so that at Q, Vd2 = Vzz. This is naturally provided by the usual bias generator: Vzz Vbb Problem: Vd2 is a high impedance node, small current mismatch in M1 and M4 leads to significant voltage change at vd2, which in turn changes the biasing current in the output stage. Solution: use feedback to stabilize common mode of Vd2.

Vzz Vd2LVd2R Feedback to M4 or part of it Since Vd2L and Vd2R are normally nearly constant. We do not need to worry about the input range accommodation for this circuit. Size the circuit so that, when v id =0, Id6 remain near desired level over all process variations in M1 and M4.

Current sensors Quiescent biasing IEEE JOURNAL OF SOLID-STATE CIRCUITS, VOL. 33, NO. 10, OCTOBER 1998

Vo+ and Vo- also need common mode stabilization M6 M7 V ocmd M7t M7t’s are in triode Choose R to a couple times bigger than Ro Choose C to be near or a couple times larger than Cgs of CMFB circuit. R R CC

M6 M7 V ocmd M7t M7t’s are in triode R R CC Insert these

Why RC in common mode detector V1 V2 R R Cgs V KCL at V: (V1 – V)/R + (V2 – V)/R = V * sCgs (V1 + V2)/R = V(sCgs + 2/R) V = (V1 + V2)/2 * 1/(1 + sRCgs/2) When |s| =|jw| << 2/RCgs, V ≈ (V1 + V2)/2 Otherwise V is not close to common mode To have CM detector work up to GB, R << 2/(2  GB*Cgs)

Why RC in common mode detector V1 V2 R R Cgs V KCL at V: (V1 – V)(1/R +sC) + (V2 – V) (1/R +sC) = V * sCgs (V1 + V2) (1/R +sC) = V(sCgs + 2/R +2sC) V = (V1 + V2)/2 * (1 + sRC)/(1 + sRC + sRCgs/2) As long as Cgs/2 < C, at all freq: V ≈ (V1 + V2)/2 C C Hence, the RC network acts as a better CM detector

Why CMBF to M7t instead of first stage: Vo1+ Vo1- Vo+ For CM behavior, assume DM=0. Vo1+=Vo1-, and Vo+=Vo-=Vocm. Without CMFB effect, at Q, Vo+ will be equal to Vg, which may be far below desired Vocm level. With CMFB connected, the feedback effect will drive Vo1 so as to move Vo+ up to the desired Vocm level. Since Vo1+ and Vo1- have a competing action on Vo+, it may take quite bit Vo1 movement to achieve the desired Vo+ movement, causing the biasing current in the second stage to be much larger than what is intended. Vg

Compensation for diff signal path closed-loop stability V xx V yy V zz M1 M2 M6 V bb M3 M4 M7 C C

At relatively low frequency: Because of gain from Vo1 to Vo, small signal Vo1 is much smaller than small signal Vo. Small signal current in compensation network is approximately Vo/(1/gmz+1/sCc). This current is injected to the Vo1 node. Alternatively, a similar current can be injected: Impedance looking into a cascode node is about 1/gm Connecting Cc to a cascode node generates a current of the form Vo/(1/gm +1/sCc) Because the base transistor is a current source, this small signal current goes to the Vo1 node Even at high frequency, the current form is still valid.

Alternative compensation V xx V yy V zz M1 M2 M6 V bb M3 M4 M7 C C

In the Cc+Mz connection, Bias voltage of Mz can be matched to track bias voltage of M6  robustness to process and temperature variations Size of Mz can be parameter scanned so as to place zero to cancel the secondary pole of the amplifier In the Cc to cascode connection, Bias voltage can still be derived using current mirrors from a single current source,  still have process and temperature tracking But size of cascode transistor is determined based on folded cascode stage design Cannot arbitrarily choose its size without considerations for output impedance at Vo1, gain of op amp, and so on.

Alternative compensation V xx V yy V zz M1 M2 M6 V bb M3 M4 M7 C C

Vicm Vo+, Vo- Use open loop Vicm sweep to find a “sweet spot” for your Vicm

Vicm Vin Vo+ With Vicm at “sweet spot”, sweep Vin near Vicm with very fine steps (uV) Vin Vo- d(V o+ -V o- ) dV in V o+ -V o- V in

V b2 V CMFB V b1 C V o+ V o- V o+ V o- Folded cascode stage: summing current and convert to voltage V i- 3:1

Summing circuit to add n-signal and p-signal together

Rail-to-rail constant g m input Coban and Allen, 1995

The composite transistor

Bulk-Driven MOSFET

Bulk-Driven, n-channel Differential Amplifier I1=I2=I5/2 As Vic varies, V d5 changes and g mb varies  Varied gain, slew rate, gain bandwidth; nonlinearity; and difficulty in compensation

Bulk-driven current mirrors Increased vin range and vout range

Traditional techniques for wide input and output voltage swings 1 Io = Iin+Ib Iin Ib /4 V T +V on V T +2V on V on >2V on + – V T +V on

Traditional techniques for wide input and output voltage swings 1 Io Iin Ib 1 1 1/4 V T +V on V T +2V on V on >2V on + – V eb Io = Iin

A 1-Volt, Two-Stage Op Amp Uses a bulk-driven differential input pair, wide swing current mirror load, and emitter follower level shifter

Op Amp Performance

Frequency Response

Low voltage VBE and PTAT reference

Threshold Voltage Tuning for low power supply voltages operation

Implementation of the voltage sources

A low voltage Op Amp core

Op Amp Implementation Clock boosterBias voltage generator

Clock booster (doubler) C B1 >> C BL

Experimental Results Power supply750mV Slew Rate3.1V/uS GB3.2MHz DC gain62dB Input offset voltage2.2mV Input common mode range0.1V-0.58V Output swing for linear operation0.31V-0.58V PSRR at DC82dB CMRR at DC56dB Total power consumption38.3uW

Common mode feedback for low voltage

1.5v op amp for 13bit 60 MHz ADC

Output Stage and CMFB

Folded cascode with AB output Lotfi 2002

Simulated performance 0.25 um process 1.5 V power supply 82 dB DC gain 2 V p-p diff output swing 170 MHz 10 pF load 77 o PM with  = 1/5 0.02% 1V step settling time: 8.5 ns Full output swing Op Amp power: 25 mW

Differential difference input AB output Alzaher 2002

Nested Miller Cap Amplifier Not much successes

Low voltage amp

LOW POWER OP AMPS Op Amp Power = (VDD-VSS)*Ibias –Reduce supply voltage: effect is small Many challenges in low voltage design same as before –Reduce bias: factor of hundred reduction Weak inversion operation Nano-amp to small micro-amp currents Needs small current biasing circuits and small current reference generators Needs output stage to drive the load –Design it so that it consume tiny quiescent power –But generate sufficient current for large signals –Tradeoff speed for reduced power

Sub-threshold Operation Most micro-power op amps use transistors in the sub-threshold region. n p ~1.5; n n ~2.5

Two-Stage, Miller Op Amp in Weak Inversion At VDD-VSS=3V, ID5=0.2uA, ID7=0.5uA, got A=92dB, GB=50KHz, P=2.1uW

Push-Pull Output in Weak Inversion First stage gain Total gain S=W/L

Increasing gain gogo Gain=gm/go What is V ON ? L 5 =L 12, W 12 =W 5 /2 S 13 <<S 4

Increasing I out with positive feedback When v i1 >v i2 i 2 >i 1 i 26 =i 2 -i 1 >0 i 27 =0 i 28 =A*i 26 i tail =I 5 +i 28 =i 1 +i 2 i 2 /i 1 =e (vi1-vi2)/nvt =e vin/nvt i 2 =i 1 e vin/nvt I 5 + A*(e vin/nvt -1)i 1 = (e vin/nvt +1)i 1 i 1 =I 5 /{A+1-(A-1)e vin/nvt )}

A=0 is normal case A > 0 can greatly enhance available output current for load driving

A=0 A=1 A=2 A=3 i1 i2 i 1 =i 2 i 2 =i 1 e vin/nvt as v in  I5 i 1 +i 2 =I 5 New i1+i2 i 1 +i 2  much faster than i 2 -i 1 

DC Offset (Self-mixing) A D A D  c 0