Lecture 14:

More MOS Circuits and the Differential

Gu-Yeon Wei Division of Engineering and Applied Sciences Harvard University [email protected]

Wei 1 Overview

• Reading – S&S: Chapter 5.10, 6.1~2, 6.6

• Background – Having seen some of the basic amplifier circuits we can build with MOSFETs, conclude material in Chapter 5 of S&S by looking at the high- frequency model for MOSFETs. We will use this high-frequency model when analyzing the high-frequency operation of MOSFETs in amplifier circuits. The performance of amplifier circuits can be improved by using a differential pair topology. Differential pair have two inputs – positive and negative terminals. The differential pair amplifier is what we assume for the ideal amplifier when we learned about op amp circuits. We will now investigate how to build these amplifiers. The reading in Sections 6.1~2 provides important and necessary background information for differential amplifiers. Then, we will skip to 6.6 which investigates building with MOSFETs.

Wei ES154 - Lecture 14 2 MOSFET Internal Capacitances

• From our study of the physical operation of MOSFETs, we can see that there are internal capacitances – Gate capacitance • from gate oxide (parallel plate and fringing capacitors)

• Cgs, Cgd, Cgb – Junction capacitances • from source-body and drain-body depletion layer capacitances (reverse biased PN junctions)

• Csb, Cdb Cgs, Cgd, Cgb

Cdb

Wei ES154 - Lecture 14 3 MOS Gate Cap

• The three gate capacitances (Cgs, Cgd, Cgb) depend on the ’s mode (region) of operation

– In triode (linear) region (vDS = small), channel has uniform depth 1 C =C = WL C gs gd 2 ox – In saturation region, channel is tapered and pinched off near the drain. We can approximate the capacitances as follows: 2 C = WL C gs 3 ox

C gd = 0

– In cut off, no channel but model capacitance between bulk and gate

C gb = WL C ox

C gd = C gs = 0

– There is also an overlap capacitance that should be added to Cgs and Cgd

C ov = WL ov C ox

Wei ES154 - Lecture 14 4 MOS Junction Cap

• The depletion-layer capacitances of the two reverse-biased pn junctions are governed by the following equation:

C C C = sb 0 C = db 0 sb V db V 1 + SB 1 + DB V 0 V 0

– where V0 is the built-in potential of the pn junctions (approx. 0.6~0.8V, a function of the NA and ND doping concentrations)

Wei ES154 - Lecture 14 5 High-Frequency Model

• We must augment our (low-frequency) small-signal model of the MOS transistor with these capacitors in order to accurately model its operation at high frequencies

complete simplified

– when source is connected to the body, model is simplified (remove Csb)

– in saturation, Cgb ~= 0

– further simplify model by removing Cdb for hand calculations

Wei ES154 - Lecture 14 6 MOSFET fT

• We can again find the fT as a figure of merit for the transistor’s high-frequency operation (unity current gain frequency) – Solve for the short circuit output current w.r.t an input current

Io = gmVgs − sCgdVgs ≅ gmVgs

assumes Cgd is small and drops out in above eq. I I g g V = i o = m ω = m gs () T s Cgs + Cgd Ii s()Cgs + Cgd Cgs + Cgd

– note that fT is a function of gm

Wei ES154 - Lecture 14 7 BJT Differential Pair

• Differential pair circuits are one of the most widely used circuit building blocks. The input stage of every op amp is a differential amplifier • Basic Characteristics – Two matched with emitters shorted together and connected to a current source – Devices must always be in active mode – Amplifies the difference between the two input voltages, but there is also a common mode amplification in the non-ideal case

• Let’s first qualitatively understand how this circuit works. – NOTE: This qualitative analysis also applies for MOSFET differential pair circuits

Wei ES154 - Lecture 14 8 Case 1

• Assume the inputs are shorted together to a common voltage, vCM, called the common mode voltage – equal currents flow through Q1 and Q2 – emitter voltages equal and at vCM-0.7 in order for the devices to be in active mode – collector currents are equal and so collector voltages are also equal for equal load resistors – difference between collector voltages = 0

• What happens when we vary vCM? – As long as devices in active mode, equal currents flow through Q1 and Q2 – Note: current through Q1 and Q2 always add up to I, current through the current source – So, collector voltages do not change and difference is still zero…. – Differential pair circuits thus reject common mode signals

Wei ES154 - Lecture 14 9 Case 2 & 3

• Q2 base grounded and Q1 base at +1 V

– All current flows through Q1

– No current flows through Q2

– Emitter voltage at 0.3V and Q2’s EBJ not FB

– vC1 = VCC-αIRC

– vC2 = VCC

• Q2 base grounded and Q1 base at -1 V

– All current flows through Q2

– No current flows through Q1

– Emitter voltage at -0.7V and Q1’s EBJ not FB

– vC2 = VCC-αIRC

– vC1 = VCC

Wei ES154 - Lecture 14 10 Case 4

• Apply a small signal vi – Causes a small positive ∆I to flow in Q1 – Requires small negative ∆I in Q2

•since IE1+IE2 = I – Can be used as a linear amplifier for small signals (∆I is a function of vi) • Differential pair responds to differences in the input voltage – Can entirely steer current from one side of the diff pair to the other with a relatively small voltage

• Let’s now take a quantitative look at the large-signal operation of the differential pair

Wei ES154 - Lecture 14 11 BJT Diff Pair – Large-Signal Operation

• First look at the emitter currents when the emitters are tied together

v −v vB1 −VE vB 2 −VE B1 B 2 i V I S VT I S VT E1 T iE1 = e iE 2 = e = e α α iE 2

• Some manipulations can lead to the following equations i 1 i 1 E1 = E 2 = vB 2 −vB1 vB1 −vB 2 iE1 + iE 2 iE1 + iE 2 1+ e VT 1+ e VT

• and there is the constraint: iE1 + iE 2 = I I I i = i = E1 vB 2 −vB1 E 2 vB1 −vB 2 1+ e VT 1+ e VT

• Given the exponential relationship, small differences in vB1,2 can cause all of the current to flow through one side

Wei ES154 - Lecture 14 12 • Notice vB1-vB2 ~= 4VT enough to switch all of current from one side to the other • For small-signal analysis, we are interested in the region we can approximate to be linear

– small-signal condition: vB1-vB2 < VT/2

Wei ES154 - Lecture 14 13 BJT Diff Pair – Small-Signal Operation

• Look at the small-signal operation: small

differential signal vd is applied αI v − v = v i = B1 B2 d C1 −vd 1+ e VT

vd multiply v 2VT d αIe top and 2VT e iC1 = v −v bottom by d d e 2VT + e 2VT

– expand the exponential and keep the first two terms

αI()1+ vd 2VT αI αI vd αI v I αI 2 iC1 ≅ = + d C ic = gm = = ()1+ vd 2VT + ()1− vd 2VT 2 2VT 2 2VT 2 VT VT v αI αI d i = g v 2 iC 2 ≅ − c m d 2 2VT 2

Wei ES154 - Lecture 14 14 Differential Voltage Gain

• For small differential input signals, vd << 2VT, the collector currents are… v v i = I + g d i = I − g d C1 C m 2 C 2 C m 2 v v v = ()V − I R − g R d v = ()V − I R + g R d C1 CC C C m C 2 C 2 CC C C m C 2

• We can now find the differential gain to be…

vc1 − vc2 Ad = = −gm RC vd

Wei ES154 - Lecture 14 15 BJT Diff Pair – Differential Half Circuit

• We can break apart the differential pair circuit into two half circuits – which then

looks like two circuits driven by +vd/2 and –vd/2

Wei ES154 - Lecture 14 16 • We can then analyze the small-signal operation with the half circuit, but must remember

– parameters rπ,gm, and ro are biased at I/2

– input signal to the differential half circuit is vd/2

vc1

gmvπ

vd/2 rπ vπ rο RC

– voltage gain of the differential amplifier (output taken differentially) is equal to the voltage gain of the half circuit

vc1 Ad = = −gm ()RC ro vd 2

Wei ES154 - Lecture 14 17 Common-Mode Gain

• When we drive the differential pair with a common-mode signal, vCM, the incremental resistance of the bias current effects circuit operation and results in some gain (assumed to be 0 when R was infinite)

αRC αRC vC1 = −vCM ≅ −vCM 2R + re 2R αR v ≅ −v C C 2 CM 2R

Wei ES154 - Lecture 14 18 • If the output is taken differentially, the output is zero since both sides move together. However, if taken single-endedly, the common-mode gain is finite αR A = − C cm 2R • If we look at the differential gain single-endedly, we get…

Ad = gm RC

• Then, the common rejection ratio (CMRR) will be

Ad 1 CMRR = ≅ gm R Acm 2

– which is often expressed in dB

Ad CMRR = 20log10 Acm

Wei ES154 - Lecture 14 19 CM and Differential Gain Equation

• Input signals to a differential pair usually consists of two components: common

mode (vCM) and differential(vd) v + v v = 1 2 v = v − v CM 2 d 1 2

• Thus, the differential output signal will be in general…

v + v v = A ()v − v + A 1 2 o d 1 2 cm 2

Wei ES154 - Lecture 14 20 MOS Diff Pair

• The same basic analysis can be applied to a MOS differential pair 1 W i = µ C ()v −Vt 2 D1,2 2 n ox L GS1,2 1 W i = µ C ()v −Vt D1,2 2 n ox L GS1,2 – and the differential input voltage is

vGS1 − vGS 2 = vid iD1 + iD2 = I 1 W i − i = µ C v D1 D2 2 n ox L id – With some algebra (detailed in S&S p.529)

I  I  v gm = 2I D ()VGS −Vt = 2(I 2) (VGS −Vt )= I (VGS −Vt )   id iD1,2 ≅ ±   2 VGS −Vt  2 v i = g id d m 2

Wei ES154 - Lecture 14 21 • We get full switching of the current when… v = 2(v −Vt) id max GS

Wei ES154 - Lecture 14 22 Next Time

• We will finish off S&S 6.6 by looking at the sources of offset voltages in MOS differential pair circuits. Then, we will revisit current sources yet again and introduce schemes that improve their performance (higher output resistance). Then, we will finish off MOS differential pair circuits with an active-loaded differential amplifier. • Then, we will shift gears a little and look at a circuit technique called cascoding that helps to improve the performance of amplifier circuits. The book covers this technique with BJTs. We will look at them in the context of MOSFETs.

Wei ES154 - Lecture 14 23