EECE488: Analog CMOS Integrated Circuit Design

3. Single-Stage Amplifiers

Shahriar Mirabbasi Department of Electrical and Computer Engineering University of British Columbia [email protected]

Technical contributions of Pedram Lajevardi in revising the course notes are greatly acknowledged.

SM 1 Set 3: Single-Stage Amplifiers Overview

1. Why Amplifiers? 2. Amplifier Characteristics 3. Amplifier Trade-offs 4. Single-stage Amplifiers 5. Common Source Amplifiers 1. Resistive Load 2. Diode-connected Load 3. Current Source Load 4. Triode Load 5. Source Degeneration

SM 2 Set 3: Single-Stage Amplifiers Overview

6. Common-Drain (Source-Follower) Amplifiers 1. Resistive Load 2. Current Source Load 3. Voltage Division in Source Followers

7. Common-Gate Amplifiers

6. Cascode Amplifiers

SM 3 Set 3: Single-Stage Amplifiers Reading Assignments

• Reading: Chapter 3 of Razavi’s book

• In this set of slides we will study low-frequency small-signal behavior of single-stage CMOS amplifiers. Although, we assume long-channel MOS models (not a good assumption for deep submicron technologies) the techniques discussed here help us to develop basic circuit intuition and to better understand and predict the behavior of circuits.

Most of the figures in these lecture notes are © Design of Analog CMOS Integrated Circuits, McGraw-Hill, 2001.

SM 4 Set 3: Single-Stage Amplifiers Why Amplifiers?

• Amplifiers are essential building blocks of both analog and digital systems.

• Amplifiers are needed for variety of reasons including:

– To amplify a weak analog signal for further processing

– To reduce the effects of noise of the next stage

– To provide a proper logical levels (in digital circuits)

• Amplifiers also play a crucial role in feedback systems

• We first look at the low-frequency performance of amplifiers. Therefore, all capacitors in the small-signal model are ignored!

SM 5 Set 3: Single-Stage Amplifiers Amplifier Characteristics - 1

• Ideally we would like that the output of an amplifier be a linear function of the input, i.e., the input times a constant gain:

y

y = α1x x

• In real world the input-output characteristics is typically a nonlinear function:

SM 6 Set 3: Single-Stage Amplifiers Amplifier Characteristics - 2

• It is more convenient to use a linear approximation of a nonlinear function. • Use the tangent line to the curve at the given (operating) point. y

x • The larger the signal changes about the operating point, the worse the approximation of the curve by its tangent line.

• This is why small-signal analysis is so popular!

SM 7 Set 3: Single-Stage Amplifiers Amplifier Characteristics - 3

• A well-behaved nonlinear function in the vicinity of a given point can be approximated by its corresponding Taylor series: n f ''(x0 ) 2 f (x0 ) n y ≈ f (x0 ) + f '(x0 ) ⋅ (x − x0 ) + ⋅ (x − x0 ) +L+ ⋅ (x − x0 ) α 2! n! α f n (x ) • Let α = 0 to get: α n n!

2 n y ≈ 0 + 1 (x − x0 ) + 2 (x − x0 ) +L+ α n (x − x0 )

• If x-x0=∆x is small, we can ignore the higher-order terms (hence the name small-signal analysis) to get:

y ≈ α 0 + α1 (x − x0 )

• α0 is referred to as the operating (bias) point and α1 is the small-signal gain.

∆y = y − f (x0 ) = y −α 0 ≈ α1∆x

SM 8 Set 3: Single-Stage Amplifiers Amplifier Trade-offs

• In practice, when designing an amplifier, we need to optimize for some performance parameters. Typically, these parameters trade performance with each other, therefore, we need to choose an acceptable compromise.

SM 9 Set 3: Single-Stage Amplifiers Single-Stage Amplifiers

• We will examine the following types of amplifiers: 1. Common Source 2. Common Drain (Source Follower ) 3. Common Gate 4. Cascode and Folded Cascode

• Each of these amplifiers have some advantages and some disadvantages. Often, designers have to utilize a cascade combination of these amplifiers to meet the design requirements.

SM 10 Set 3: Single-Stage Amplifiers Common Source Basics - 1

• In common-source amplifiers, the input is (somehow!) connected to the gate and the output is (somehow!) taken from the drain.

• We can divide common source amplifiers into two groups: 1. Without source degeneration (no body effect for the main transistor):

2. With source degeneration (have to take body effect into account for the main transistor):

SM 11 Set 3: Single-Stage Amplifiers Common Source Basics - 2

In a simple common source amplifier:

• gate voltage variations times gm gives the drain current variations, • drain current variations times the load gives the output voltage variations. • Therefore, one can expect the small-signal gain to be:

Av = gm ⋅ RD

SM 12 Set 3: Single-Stage Amplifiers Common Source Basics - 3

• Different types of loads can be used in an amplifier: 1. Resistive Load 2. Diode-connected Load 3. Current Source Load 4. Triode Load

• The following parameters of amplifiers are very important: 1. Small-signal gain 2. Voltage swing

SM 13 Set 3: Single-Stage Amplifiers Resistive Load - 1

• Let’s use a resistor as the load.

• The region of operation of M1 depends on its size and the values of Vin and R. • We are interested in the small-signal gain and the headroom (which determines the maximum voltage swing). • We will calculate the gain using two different methods 1. Small-signal model 2. Large-signal analysis

SM 14 Set 3: Single-Stage Amplifiers Resistive Load - 2

Gain – Method 1: Small-Signal Model

• This is assuming that the transistor is in saturation, and channel length modulation is ignored.

• The current through RD:

iD = g m ⋅ vIN • Output Voltage:

vOUT = −iD ⋅ RD = −g m ⋅ vIN ⋅ RD • Small-signal Gain:

vOUT Av = = −g m ⋅ RD vIN

SM 15 Set 3: Single-Stage Amplifiers Resistive Load - 3

Gain – Method 2: Large-Signal Analysis

• If VIN

VOUT = VDD − RD ⋅iD = VDD

∂VOUT Av = = 0 ∂VIN

• As VIN becomes slightly larger than VTH, M1 turns on and goes into saturation (VDS≈ VDD > VGS-VTH ≈0). 1 W V = V − R ⋅i = V − R ⋅ ⋅ ⋅C ⋅ ⋅ (V −V )2 OUT dd D D dd D 2 n ox L IN TH µ ∂VOUT W Av = = −RD ⋅ µn ⋅Cox ⋅ ⋅ (VIN −VTH ) = −RD ⋅ g m ∂VIN L

• As VIN increases, VDS decreases, and M1 goes into triode when VIN-VTH = VOUT. We can find the value of VIN that makes M1 switch its region of operation. 1 W V = V − R ⋅i = V − R ⋅ ⋅ µ ⋅C ⋅ ⋅ (V −V )2 = (V −V ) OUT dd D D dd D 2 n ox L IN TH IN TH

SM 16 Set 3: Single-Stage Amplifiers Resistive Load - 4

Gain – Method 2: Large-Signal Analysis (Continued)

• As VIN increases, VDS decreases, and M1 goes into triode. ⎡ 2 ⎤ W VOUT VOUT = VDD − RD ⋅iD = VDD − RD ⋅ n ⋅Cox ⋅ ⋅ ⎢(VIN −VTH )⋅VOUT − ⎥ L ⎣⎢ 2 ⎦⎥ µ ∂VOUT W ⎡ ∂VOUT ∂VOUT ⎤ = −RD ⋅ µn ⋅Cox ⋅ ⋅ ⎢(VIN −VTH )⋅ +VOUT −VOUT ⋅ ⎥ ∂VIN L ⎣⎢ ∂VIN ∂VIN ⎦⎥

• We can find Av from above. It will depend on both VIN and VOUT.

• If VIN increases further, M1 goes into deep triode if VOUT<< 2(VIN-VTH).

W V = V − R ⋅i = V − R ⋅ ⋅C ⋅ ⋅(V −V )⋅V OUT DD D D DD D n ox L IN TH OUT

VDD µ VDD RON VOUT = = = VDD ⋅ W 1 RON + RD 1+ RD ⋅ µn ⋅Cox ⋅ ⋅(VIN −VTH ) 1+ RD ⋅ L RON

SM 17 Set 3: Single-Stage Amplifiers Resistive Load - 5

Example: Sketch the drain current and gm of M1 as a function of VIN.

•gm depends on VIN, so if VIN changes by a large amount the small- signal approximation will not be valid anymore. • In order to have a linear amplifier, we don’t want gain to depend on

parameters like gm which depend on the input signal.

SM 18 Set 3: Single-Stage Amplifiers Resistive Load - 6 µ

• Gain of common-source amplifier:

W VRD W VRD − 2⋅VRD Av = −gm ⋅ RD = − nCox (VIN −VTH ) ⋅ = − 2µnCox ⋅ = L ID L ID Veff • To increase the gain:

1. Increase gm by increasing W or VIN (DC portion or bias). Either way, ID increases (more power) and VRD increases, which limits the voltage swing.

2. Increase RD and keep ID constant (gm and power remain constant). But, VRD increases which limits the voltage swing.

3. Increase RD and reduce ID so VRD remains constant.

¾ If ID is reduced by decreasing W, the gain will not change.

¾ If ID is reduced by decreasing VIN (bias), the gain will increase. Since RD is increased, the bandwidth becomes smaller (why?). • Notice the trade-offs between gain, bandwidth, and voltage swings.

SM 19 Set 3: Single-Stage Amplifiers Resistive Load - 7

• Now let’s consider the simple common-source circuit with channel length modulation taken into account.

• Channel length modulation becomes more important as RD increases (in the next slide we will see why!).

• Again, we will calculate the gain in two different methods 1. Small-signal Model 2. Large Signal Analysis

SM 20 Set 3: Single-Stage Amplifiers Resistive Load - 8

Gain – Method 1: Small-Signal Model

• This is assuming that the transistor is in saturation.

• The current through RD:

iD = gm ⋅ vIN • Output Voltage:

vOUT = −iD ⋅ (RD ro )= −g m ⋅ vIN ⋅ (RD ro )

• Small-signal Gain:

vOUT Av = = −g m ⋅ ()RD ro vIN

SM 21 Set 3: Single-Stage Amplifiers Resistive Load - 9

Gain – Method 2: Large-Signal Analysis

• As VIN becomes slightly larger than VTH, M1 turns on and goes into saturation (VDS≈ VDD > VGS-VTH ≈0).

1 W 2 VOUT = VDD − RD ⋅ ID = VDD − RD ⋅ ⋅µn ⋅Cox ⋅ ⋅(VIN −VTH ) ⋅()1+ λ ⋅VOUT 2 L

∂VOUT W ⎡ 1 ∂VOUT ⎤ = −RD ⋅µn ⋅Cox ⋅ ⋅(VIN −VTH ) ⋅ ⎢()1+ λ ⋅VOUT + ⋅(VIN −VTH ) ⋅λ ⋅ ⎥ ∂VIN L ⎣ 2 ∂VIN ⎦ W − RD ⋅µn ⋅Cox ⋅ ⋅(VIN −VTH ) ⋅()1+ λ ⋅VOUT L − RD ⋅ gm − RD ⋅ gm Av = = = 1 W 2 1+ RD ⋅ ID ⋅λ 1 1+ ⋅ RD ⋅µn ⋅Cox ⋅ ⋅(VIN −VTH ) ⋅λ 1+ RD ⋅ 2 L ro − ro ⋅ RD ⋅ gm = = −gm ⋅()RD ro ro + RD

SM 22 Set 3: Single-Stage Amplifiers Resistive Load - 10

Example:

• Assuming M1 is biased in active region, what is the small-signal gain of the following circuit?

•I1 is a current source and ideally has an infinite impedance.

vOUT Av = = −g m ⋅ ()∞ ro = −g m ⋅ ro vIN • This is the maximum gain of this amplifier (why?), and is known as the intrinsic gain. µ

• How can VIN change if I1 is constant? 1 W I = ⋅ ⋅C ⋅ ⋅ (V −V )2 ⋅ ()1+ λ ⋅V D 2 n ox L IN TH OUT

• Here we have to take channel-length modulation into account. As VIN changes, VOUT also changes to keep I1 constant.

SM 23 Set 3: Single-Stage Amplifiers Diode Connected Load - 1

• Often, it is difficult to fabricate tightly controlled or reasonable size resistors on chip. So, it is desirable to replace the load resistor with a MOS device. • Recall the diode connected devices:

Body Effect RX (when λ≠0) RX (when λ=0)

1 1 NO R = r R = X o X g g m m

1 1 R = r RX = YES X o g + g g m + g mb m mb

SM 24 Set 3: Single-Stage Amplifiers Diode Connected Load - 2

• Now consider the common-source amplifier with two types of diode connected loads: 1. PMOS diode connected load: (No body effect)

2. NMOS diode connected load: (Body effect has to be taken into account)

SM 25 Set 3: Single-Stage Amplifiers Diode Connected Load - 3

PMOS Diode Connected Load: • Note that this is a common source configuration

with M2 being the load. We have: v ⎛ 1 ⎞ A = OUT = −g ⋅ ()R r = −g ⋅⎜ r r ⎟ v m1 X o1 m1 ⎜ o 2 o1 ⎟ vIN ⎝ gm 2 ⎠

• Ignoring the channel length modulation (ro1=ro2=∞), we can write:

⎛ W ⎞ ⎛ W ⎞ 2µn ⋅Cox ⋅⎜ ⎟ ⋅ ID1 µn ⋅⎜ ⎟ ⎛ 1 ⎞ g ⎝ L ⎠ L ⎜ ⎟ m1 1 ⎝ ⎠1 Av = −gm1 ⋅⎜ ∞ ∞⎟ = − = − = − ⎝ gm2 ⎠ gm2 ⎛ W ⎞ ⎛ W ⎞ 2µ ⋅C ⋅⎜ ⎟ ⋅ I µ p ⋅⎜ ⎟ p ox D2 ⎝ L ⎠ ⎝ L ⎠2 2 2⋅ ID1 gm1 VGS1 −VTH1 VSG2 − VTH 2 Av = − = − = − gm2 2⋅ ID2 VGS1 −VTH1 VSG2 − VTH 2

SM 26 Set 3: Single-Stage Amplifiers Diode Connected Load - 4

NMOS Diode Connected Load:

• Again, note that this is a common source configuration

with M2 being the load. We have: v ⎛ 1 ⎞ A = OUT = −g ⋅ ()R r = −g ⋅⎜ r r ⎟ v m1 X o1 m1 ⎜ o 2 o1 ⎟ vIN ⎝ g m 2 + g mb 2 ⎠

• Ignoring the channel length modulation (ro1=ro2=∞), we can write:

⎛ 1 ⎞ g g ⎜ ⎟ m1 m1 Av = −gm1 ⋅⎜ ∞ ∞⎟ = − = − ⎝ gm2 + gmb2 ⎠ gm2 + gmb2 gm2 ⋅(1+ ) ⎛ W ⎞ η ⎜ ⎟ 1 ⎝ L ⎠1 1 VGS2 −VTH Av = − = − ⋅ η 1+ ⎛ W ⎞ 1+η V −V ⎜ ⎟ GS1 TH ⎝ L ⎠2

SM 27 Set 3: Single-Stage Amplifiers Diode Connected Load - 5

• For a diode connected load we observe that (to the first order approximation): 1. The amplifier gain is not a function of the bias current. So, the change in the input and output levels does not affect the gain, and the amplifier becomes more linear.

2. The amplifier gain is not a function of the input signal (amplifier becomes more linear).

3. The amplifier gain is a weak function (square root) of the transistor sizes. So, we have to change the dimensions by a considerable amount so as to increase the gain.

SM 28 Set 3: Single-Stage Amplifiers Diode Connected Load - 6

4. The gain of the amplifier is reduced when body effect should be considered.

5. We want M1 to be in saturation, and M2 to be on (M2 cannot be in triode (why?)):

6. The voltage swing is constrained by both the required overdrive voltages and the threshold voltage of the diode connected device.

M1:VOUT > VGS1 −VTH1 = Veff 1 , M 2 :VOUT < VDD − VTH 2

7. A high amplifier gain leads to a high overdrive voltage for the diode connected device which limits the voltage swing.

SM 29 Set 3: Single-Stage Amplifiers Diode Connected Load - 6

Example: • Find the gain of the following circuit if M1 is biased in saturation and Is=0.75I1. v ⎛ 1 ⎞ ⎛ 1 ⎞ A = OUT = −g ⋅ ()R r r = −g ⋅⎜ r ∞ r ⎟ = −g ⋅⎜ r r ⎟ v m1 X Is o1 m1 ⎜ o 2 o1 ⎟ m1 ⎜ o 2 o1 ⎟ vIN ⎝ gm 2 ⎠ ⎝ gm 2 ⎠

• Ignoring the channel length modulation (ro1=ro2=∞) we get:

⎛ W ⎞ 2⋅ I 2 n ⋅Cox ⋅⎜ ⎟ ⋅ ID1 D1 ⎛ 1 ⎞ g ⎝ L ⎠ V −V ⎜ ⎟ m1 1 GS1 TH1 Av = −gm1 ⋅⎜ ∞ ∞⎟ = − = − = − ⎝ gm2 ⎠ gm2 ⎛ W ⎞ 2⋅ ID2 µ 2 ⋅C ⋅⎜ ⎟ ⋅ I p ox D2 VSG2 − VTH 2 ⎝ L ⎠2

µ ⎛ W ⎞ n ⋅⎜ ⎟ µ ⎝ L ⎠1 VSG2 − VTH 2 Av = −2⋅ = −4⋅ ⎛ W ⎞ VGS1 −VTH1 µ p ⋅⎜ ⎟ ⎝ L ⎠2

SM 30 Set 3: Single-Stage Amplifiers Diode Connected Load - 7

Example (Continued):

• We observe for this example that:

1. For fixed transistor sizes, using the current source increases the gain by a factor of 2.

2. For fixed overdrive voltages, using the current source increases the gain by a factor of 4.

3. For a given gain, using the current source allows us to make the diode connected load 4 times smaller.

4. For a given gain, using the current source allows us to make the overdrive voltage of the diode connected load 4 times smaller. This increases the headroom for voltage swing.

SM 31 Set 3: Single-Stage Amplifiers Current Source Load - 1

• Note that current source M2 is the load. •

• Recall that the output impedance of M2 seen from Vout:

vX RX = = ro 2 iX

vOUT Av = = −g m1 ⋅ ()RX ro1 = −g m1 ⋅ ()ro 2 ro1 vIN 1 1 L2 • For large gain at given power, we want large r and r = ∝ = o o 1 W λ ⋅ ID ⋅ W L L Increase L and W keeping the aspect ratio constant (so ro increases

and ID remains constant). However, this approach increases the capacitance of the output node.

• We want M2 to be in saturation so

VSD2 = VDD −VOUT > VSG2 − VTH = Veff 2 →VOUT < VDD −Veff 2

SM 32 Set 3: Single-Stage Amplifiers Current Source Load - 2

• We also want M1 to be in saturation:

VDS1 = VOUT > VGS1 −VTH = Veff 1 → VOUT > Veff 1

• Thus, we want Veff1 and Veff2 to be small, so that there is more headroom for output voltage swing. For a constant ID, we can increase W1 and W2 to reduce Veff1 and Veff2.

• The intrinsic gain of this amplifier is: Av = −gm ⋅ ro • In general, we have: W L2 g ∝ , r ∝ → A ∝ L mµ L o W v • But since current in this case is roughly constant:

W W 1 gm = 2 n ⋅Cox ⋅ ⋅ I D ∝ , ro = ∝ L → Av ∝ LW L L λ ⋅ I D

SM 33 Set 3: Single-Stage Amplifiers Triode Load

• We recognize that this is a common source configuration with M2 being the load. Recall that if

M2 is in deep triode, i.e., VSD<<2(VSG-|VTH|), it behaves like a resistor.

If VSD << µ2(VSG − VTH ): 1 1 R = = , A = −g ⋅ ()R r ON 2 W W v m1 ON 2 o1 ⋅C ⋅ ⋅ ()V − V µ ⋅C ⋅ ⋅ ()V −V − V p ox L SG TH p ox L dd b TH

•Vb should be low enough to make sure that M2 is in deep triode region and usually requires additional complexity to be precisely generated.

•RON2 depends on µp, Cox, and VTH which in turn depend on the technology being used. • In general, this amplifier with triode load is difficult to design and use!

• However, compared to diode-connected load, triode load consumes less headroom: M1 :VOUT > VGS1 −VTH = Veff 1 , M 2 :VOUT ≈ VDD

SM 34 Set 3: Single-Stage Amplifiers Source Degeneration - 1

• The following circuit shows a common source configuration with a degeneration resistor in the source.

• We will show that this configuration makes the common source amplifier more linear.

• We will use two methods to derive the gain of this circuit: 1. Small-signal Model 2. Using the following Lemma

Lemma:

In linear systems, the voltage gain is equal to –GmRout.

SM 35 Set 3: Single-Stage Amplifiers Source Degeneration - 2

Gain – Method 1: Small Signal Model

vOUT − iOUT ⋅ RS −vOUT iOUT = gm ⋅v1 + gmb ⋅vBS + , iOUT = rO RD

vOUT vOUT v1 = vIN − iOUT ⋅ RS = vIN + ⋅ RS , vBS = −iOUT ⋅ RS = ⋅ RS RD RD v v + OUT ⋅ R − v ⎛ v ⎞ ⎛ v ⎞ OUT R S OUT ⎜ OUT ⎟ ⎜ OUT ⎟ D = gm ⋅⎜vIN + ⋅ RS ⎟ + gmb ⋅⎜ ⋅ RS ⎟ + RD ⎝ RD ⎠ ⎝ RD ⎠ rO ⎛ R R ⎞ v ⎜1 g R g R D S ⎟ g v R OUT ⋅⎜ + m ⋅ S + mb ⋅ S + + ⎟ = − m ⋅ IN ⋅ D ⎝ rO rO ⎠

vOUT − gm ⋅ rO ⋅ RD Av = = vIN rO()⋅ 1+ ()gm + gmb ⋅ RS + RD + RS

SM 36 Set 3: Single-Stage Amplifiers Source Degeneration - 3

Gain – Method 2: Lemma

• The Lemma states that in linear systems, the voltage gain is

equal to –GmRout. So we need to find Gm and Rout.

1. Gm: Recall that the equivalent transconductance of the above Circuit is:

iOUT gm ⋅ rO gm ⋅ rO Gm = = = vIN rO + rO ⋅()gm ⋅ RS + gmb ⋅ RS + RS rO[1+ ()gm + gmb ⋅ RS ] + RS

SM 37 Set 3: Single-Stage Amplifiers Source Degeneration - 4

Gain – Method 2: Lemma (Continued)

1. ROUT: We use the following small signal model to derive the small signal output impedance of this amplifier:

v1 = −iX ⋅ RS , vBS = −iX ⋅ RS

vX = iX ⋅ RS + ()iX − gm ⋅v1 − gmb ⋅vBS ⋅ rO ()()() = iX ⋅ RS + iX − gm ⋅ − iX ⋅ RS − gmb ⋅ − iX ⋅ RS ⋅ rO

vX RX = = RS + ()()1+ gm ⋅ RS + gmb ⋅ RS ⋅ rO = RS + 1+ (gm + gmb ) ⋅ RS ⋅ rO iX

()RS + ()1+ (gm + gmb ) ⋅ RS ⋅ rO ⋅ RD ROUT = RX RD ()= () RS + 1+ (gm + gmb ) ⋅ RS ⋅ rO + RD

• Since typically rO>>RS:

RX = RS + (1+ (gm + gmb ) ⋅ RS )⋅ rO = (1+ (gm + gmb ) ⋅ RS )⋅ rO = (gm + gmb )⋅ RS ⋅ rO

SM 38 Set 3: Single-Stage Amplifiers Source Degeneration - 5

Gain – Method 2: Lemma (Continued)

gm ⋅ rO Gm = rO (1+ ⋅()gm + gmb ⋅ RS ) + RS

()RS + ()1+ (gm + gmb ) ⋅ RS ⋅ rO ⋅ RD ROUT = () RS + 1+ (gm + gmb ) ⋅ RS ⋅ rO + RD

gm ⋅ rO (rO + (1+ gm ⋅ rO + gmb ⋅ rO )⋅ RS )⋅ RD Av = −Gm ⋅ ROUT = − ()⋅ rO + rO ⋅()gm ⋅ RS + gmb ⋅ RS + RS rO + 1+ gm ⋅ rO + gmb ⋅ rO ⋅ RS + RD − g ⋅ r ⋅ R = m O D rO + ()1+ (gm + gmb ) ⋅ RS ⋅ rO + RD

SM 39 Set 3: Single-Stage Amplifiers Source Degeneration - 6

• If we ignore body effect and channel-length modulation:

Method 1 – Small-signal Model:

vOUT = −gm ⋅ vGS ⋅ RD , vGS = vIN − g m ⋅ vGS ⋅ RS

1 vOUT − g m ⋅ RD vGS = vIN ⋅ → Av = = 1+ g m ⋅ RS vIN 1+ g m ⋅ RS Method 2 – Taking limits: g ⋅ r g g G = lim m O = m = m m ro→∞ gmb→0 rO + rO ⋅ ()g m ⋅ RS + gmb ⋅ RS + RS 1+ ()g m + g mb ⋅ RS 1+ gm ⋅ RS

()r + ()1+ g ⋅ r + g ⋅ r ⋅ R ⋅ R (1+ ()g + g ⋅ R )⋅ R R = lim O m O mb O S D = m mb S D = R OUT ro→∞ ()() D gmb→0 rO + 1+ gm ⋅ rO + g mb ⋅ rO ⋅ RS + RD 1+ gm + gmb ⋅ RS

− gm ⋅ RD Av = −Gm ⋅ ROUT = 1+ gm ⋅ RS

SM 40 Set 3: Single-Stage Amplifiers Source Degeneration - 7

Obtaining Gm and Rout directly assuming λ=γ=0:

1. Gm:

iD = gm ⋅ vGS , vGS = vIN − gm ⋅ vGS ⋅ RS

1 iD g m vGS = vIN ⋅ → Gm = = 1+ g m ⋅ RS vIN 1+ g m ⋅ RS

2. ROUT:

vGS = −gm ⋅ vGS ⋅ RS → vGS = 0

vX vX iX = + gm ⋅ vGS = RD RD

vX ROUT = = RD iX

− gm ⋅ RD Av = −Gm ⋅ ROUT = 1+ gm ⋅ RS

SM 41 Set 3: Single-Stage Amplifiers Source Degeneration - 8

• If we ignore body effect and channel-length modulation:

g m − g m ⋅ RD Gm = , ROUT = RD → Av = 1+ g m ⋅ RS 1+ gm ⋅ RS

• We Notice that as RS increases Gm becomes less dependent on gm:

gm 1 lim Gm = lim = Rs→∞ Rs→∞ 1+ gm ⋅ RS RS

iOUT 1 • That is for large RS: Gm = ≈ → vIN ≈ RS ⋅iOUT vIN RS

• Therefore, the amplifier becomes more linear when RS is large enough. Intuitively, an increase in vIN tend to increase ID, however, the voltage drop across RS also increases. This makes the amplifier less sensitive to input changes, and makes ID smoother! • The linearization is achieved at the cost of losing gain and voltage headroom.

SM 42 Set 3: Single-Stage Amplifiers Source Degeneration - 9

• We can manipulate the gain equation so the numerator is the resistance seen at the drain node, and the denominator is the resistance in the source path. − g ⋅ R − R A = m D = D v 1 1+ g m ⋅ RS + RS g m

• The following are ID and gm of a transistor without RS.

•ID and gm of a transistor considering RS are:

• When ID is small such that RS gm<<1, Gm ≈ gm.

• When ID is large such that RS gm>>1, Gm ≈1/ RS.

SM 43 Set 3: Single-Stage Amplifiers Alternative Method to Find the Output-Resistance of a Degenerated Common-Source Amplifier

SM 44 Set 3: Single-Stage Amplifiers Why Buffers?

• Common Source amplifiers needed a large load impedance to provide a large gain.

• If the load is small but we need a large gain (can you think of an example?) a buffer is used.

• Source-follower (common-drain) amplifiers can be used as buffers.

RIN = ∞ , ROUT = 0 , AV = 1 Ideal Buffer:

1. RIN=∞: the input current is zero; it doesn’t load the previous stage.

2. ROUT=0: No voltage drop at the output; behaves like a voltage source.

SM 45 Set 3: Single-Stage Amplifiers Resistive Load - 1

• We will examine the Source follower amplifier with two different loads: 1. Resistive Load 2. Current Source Load

– Resistive Load: • As shown below the output (source voltage) will follow the input (gate voltage). We will analyze the following circuit using large-signal and small-signal analysis.

SM 46 Set 3: Single-Stage Amplifiers Resistive Load - 2 µ Large Signal Analysis: µ • The relationship between V and V is: IN OUT λ 1 W 2 V = R ⋅ I = C ()()V −V ⋅ 1+ ⋅V ⋅ R OUT S D 2 n ox L GS TH DS S λ 1 W 2 V =µ C ()()V −V −V ⋅ 1+ ⋅V − λ ⋅V ⋅ R OUT 2 n ox L IN OUT TH DD OUT S

• Differentiate with respect to VIN: ∂V µW ⎛ ∂V ∂V ⎞ OUT = C ()V −V −V ⋅⎜1− OUT − TH ⎟ ⋅ ()1+ ⋅V ⋅ R n ox IN OUT TH ⎜ ⎟ DS S ∂VIN L ⎝ ∂VIN ∂VIN ⎠

2 λ 1γ W ∂VOUT + nCox ()()VGS −VTH ⋅ RS ⋅ − λ ⋅ 2 L ∂VIN

• Need to find the derivative of VTH with respect to VIN: γ

VTH = VTH 0 + ⋅ ( 2 ⋅ Φ F +VSB − 2 ⋅ Φ F ) , VSB = VOUT ∂V ∂V ∂V ∂V ∂V TH = TH ⋅ OUT = ⋅ OUT =η ⋅ OUT

∂VIN ∂VOUT ∂VIN 2 2 ⋅ Φ F +VSB ∂VIN ∂VIN

SM 47 Set 3: Single-Stage Amplifiers Resistive Load - 3

Large Signal Analysis (Continued): • The small signal gain ηcan be found:

∂VOUT ⋅ ()1+ gm ⋅ RS + g m ⋅ ⋅ RS + I D ⋅ RS ⋅ λ = g m ⋅ RS ∂VIN ∂V g ⋅ R g ⋅ R A = OUT = m S = m S V R ∂VIN S ⎛ 1 ⎞ 1+ g m ⋅ RS + gmb ⋅ RS + 1+ ⎜ g + g + ⎟ ⋅ R ⎜ m mb ⎟ S ro ⎝ ro ⎠

• If channel-length modulation is ignored (ro=∞) we get:

∂VOUT g m ⋅ RS AV = = ∂VIN 1+ ()gm + g mb ⋅ RS

SM 48 Set 3: Single-Stage Amplifiers Resistive Load - 4

Small Signal Analysis: • We get the following small signal model:

vOUT = ()g m ⋅ vGS + g mb ⋅ vBS ⋅ RS rO , vGS = vIN − vOUT , vBS = −vOUT

RS ⋅ rO vOUT =()g m ⋅ ()()vIN − vOUT + g mb ⋅ − vOUT ⋅ RS + rO

vOUT ⋅ ()RS + rO + gm ⋅ RS ⋅ rO + g mb ⋅ RS ⋅ rO = g m ⋅ RS ⋅ rO ⋅ vIN

vOUT g m ⋅ RS ⋅ rO g m ⋅ RS ⋅ rO Av = = = vIN RS + rO + g m ⋅ RS ⋅ rO + gmb ⋅ RS ⋅ rO RS ⋅ ()1+ g m ⋅ rO + g mb ⋅ rO + rO g ⋅ R A = m S v ⎛ 1 ⎞ R ⋅⎜ + g + g ⎟ +1 S ⎜ m mb ⎟ ⎝ rO ⎠

SM 49 Set 3: Single-Stage Amplifiers Resistive Load - 5

• Graph of the gain of a source-follower amplifier:

1. M1 never enters the triode region as long as VIN

2. Gain is zero if VIN is less than VTH (because gm is 0).

3. As VIN increases, gm increases and the gain becomes:

gm 1 Av ≈ = g m + g mb 1+η

4. As VOUT increases, η decreases, and therefore, the maximum gain increases.

5. Even if RS=∞, the gain is less than one: g A ≈ m < 1 V 1 g m + g mb + ro 6. Gain depends heavily on the DC level of the input (nonlinear amplifier).

SM 50 Set 3: Single-Stage Amplifiers Current Source Load

• In a source follower with a resistive load, the drain current depends on the DC level of VIN, which makes the amplifier highly nonlinear.

• To avoid this problem, we can use a current source as the load.

• The output resistance is: 1 1 1 1 RM 1 = ro1 , RI 1 = ro 2 → ROUT = RM 1 RI 1 = ro1 ro 2 gm1 gmb1 gm1 gmb1

• If channel length modulation is ignored (ro1=ro2=∞) : 1 1 1 1 1 ROUT = ∞ ∞ = = gm1 gmb1 gm1 gmb1 gm1 + gmb1

• Note that the body effect reduces the output impedance of the source follower amplifiers.

SM 51 Set 3: Single-Stage Amplifiers Voltage Division in Source Followers - 1

• When Calculating output resistance seen at the source of M1, i.e., RM1,, we force vIN to zero and find the output impedance:

1 1 RM 1 = ro1 g m1 g mb1

• However, if we were to find the gain of the amplifier, we would not

suppress vIN.

• Here, we would like to find an equivalent circuit of M1, from which we can find the gain.

• Consider the small-signal model of M1:

SM 52 Set 3: Single-Stage Amplifiers Voltage Division in Source Followers - 2

• For small-signal analysis vBS= vDS, so gmbvBS dependant current source can be replaced by a resistor (1/gmb) between source and drain. • Note that, when looking at the circuit from the source terminal, we can

replace the gmvGS dependant current source with a resistor (of value 1/gm) between source and gate.

• Simplified circuit:

SM 53 Set 3: Single-Stage Amplifiers Voltage Division in Source Followers - 3

Example: • Find the gain of a source follower amplifier with a resistive load. • We draw the small signal model of this amplifier as shown below to get: 1 1 R r R r S O g S O g mb vOUT mb vOUT = ⋅ vIN → Av = = 1 1 vIN 1 1 RS rO + RS rO + g mb g m g mb gm • We can show that this is equal to what we obtained before: 1

1 1 RS ⋅ rO + + g mb RS rO RS + rO + RS ⋅ rO ⋅ g mb RS ⋅ rO ⋅ g m Av = = = 1 1 RS ⋅ rO 1 R + r + R ⋅ r ⋅ g + R ⋅ r ⋅ g + + S O S O mb S O m 1 1 gm RS + rO + RS ⋅ rO ⋅ g mb g m + + g mb RS rO SM 54 Set 3: Single-Stage Amplifiers Voltage Division in Source Followers - 4

Example: • Find the gain of a source follower amplifier with a current source load. • Small-signal model of this amplifier is:

1 1 r r r r O 2 O1 g O 2 O1 g mb1 vOUT mb1 vOUT = ⋅ vIN → Av = = 1 1 vIN 1 1 rO 2 rO1 + rO 2 rO1 + gmb1 g m1 g mb1 gm1

• If we ignore channel length modulation:

1 1 g v g v = mb1 ⋅ v → A = OUT = mb1 OUT 1 1 IN v v 1 1 + IN +

g mb1 gm1 gmb1 gm1

SM 55 Set 3: Single-Stage Amplifiers Voltage Division in Source Followers - 5

Example: • Find the gain of a source follower amplifier with a resistive load and biased with a current source.

• Small-signal model of this amplifier is:

1 1 r R r r R r O 2 L O1 g O 2 L O1 g mb1 vOUT mb1 vOUT = ⋅ vIN → Av = = 1 1 vIN 1 1 rO 2 RL rO1 + rO 2 RL rO1 + g mb1 gm1 g mb1 gm1

SM 56 Set 3: Single-Stage Amplifiers Voltage Division in Source Followers - 6

Example: • Find the gain of a source follower amplifier with a resistive load.

• Small-signal model of this amplifier is:

1 1 1 1 1 1 r r r r O 2 O1 g g g O 2 O1 g g g mb1 m 2 mb 2 vOUT mb1 m 2 mb 2 vOUT = ⋅ vIN → Av = = 1 1 1 1 vIN 1 1 1 1 rO 2 rO1 + rO 2 rO1 + gmb1 g m 2 gmb 2 g m1 g mb1 g m 2 g mb 2 gm1

SM 57 Set 3: Single-Stage Amplifiers Advantages and Disadvantages - 1

1. Source followers have typically small output impedance.

2. Source followers have large input impedance.

3. Source followers have poor driving capabilities..

4. Source followers are nonlinear. This nonlinearity is caused by: ¾ Variable bias current which can be resolved if we use a current source to bias the source follower.

¾ Body effect; i.e., dependence of VTH on the source (output) voltage. This can be resolved for PMOS devices, because each PMOS transistor can have a separate n-well. However, because of low mobility, PMOS devices have higher output impedance. (In more advanced technologies, NMOS in a separate p-well, can be implemented that potentially has no body effect)

¾ Dependence of ro on VDS in submicron devices.

SM 58 Set 3: Single-Stage Amplifiers Advantages and Disadvantages - 2

5. Source followers have voltage headroom limitations due to level shift. Consider this circuit (a common source followed by a source follower):

• If we only consider the common source stage, VX>VGS1-VTH1.

• If we only consider the source follower stage, VX>VGS3-VTH3+ VGS2.

• Therefore, adding the source follower will reduce the allowable voltage swing at node X.

• The DC value of VOUT is VGS2 lower than the DC value of VX.

SM 59 Set 3: Single-Stage Amplifiers Common-Gate

Av = (gm + gmb )RD = gm (1 + η)RD

SM 60 Set 3: Single-Stage Amplifiers Common-Gate

(gm + gmb )ro +1 Av = RD ro + (gm + gmb )roRS + RS + RD

SM 61 Set 3: Single-Stage Amplifiers Common-Gate

(gm + gmb )ro +1 Av = RD ro + (gm + gmb )roRS + RS + RD

for RS = 0 : Av ≈ (gm + gmb )(ro || RD )

SM 62 Set 3: Single-Stage Amplifiers Common-Gate Input Impedance

RD + ro RD ro Rin = = + 1+ (gm + gmb )ro 1+ (gm + gmb )ro 1+ (gm + gmb )ro

RD 1 1 Rin = + (ro || || ) 1+ (gm + gmb )ro gm gmb

SM 63 Set 3: Single-Stage Amplifiers Common-Gate Input Impedance

• Input impedance of common-gate stage is relatively low only if

RD is small

• Example: Find the input impedance of the following circuit.

SM 64 Set 3: Single-Stage Amplifiers Example

• Calculate the voltage gain of the following circuit:

Av =1+ (gm + gmb )ro

SM 65 Set 3: Single-Stage Amplifiers Common-Gate Output Impedance

Rout = {[1+ (gm + gmb )RS ]ro + RS }|| RD

SM 66 Set 3: Single-Stage Amplifiers Example

• Compare the gain of the following two circuits (λ=γ=0 and 50Ω transmission lines!)

SM 67 Set 3: Single-Stage Amplifiers Cascode Stage

• Cascade of a common-source stage and a common-gate stage is called a “cascode” stage.

Rout = {[1+ (gm2 + gmb2 )ro1]ro2 + ro1}|| RD

≈ [(gm2 + gmb2 )ro1ro2 ]|| RD

AV ≈ gm1{[ro1ro2 (gm 2 + gmb2 )] || RD ]}

SM 68 Set 3: Single-Stage Amplifiers Cascode Stage

AV ≈ gm1[(ro1ro2 gm2 ) || (ro3ro4gm3 )]

SM 69 Set 3: Single-Stage Amplifiers Output Impedance Comparison

SM 70 Set 3: Single-Stage Amplifiers Shielding Property

SM 71 Set 3: Single-Stage Amplifiers Board Notes

SM 72 Set 3: Single-Stage Amplifiers Triple Cascode

• What is the output resistance of this circuit?

• Problem?

SM 73 Set 3: Single-Stage Amplifiers Folded Cascode

SM 74 Set 3: Single-Stage Amplifiers Output Impedance of a Folded Cascode

Rout = [1+ (gm2 + gmb2 )ro2 ](ro1 || ro3 ) + ro2

SM 75 Set 3: Single-Stage Amplifiers