IN3170/4170, Spring 2020 Philipp Hfliger hafliger@ifi.uio.no - PowerPoint PPT Presentation

IN3170/4170, Spring 2020 Philipp Häfliger hafliger@ifi.uio.no Excerpt of Sedra/Smith Chapter 8: Differential and Multistage CMOS Amplifier Basics

Content The MOS Differential Pair (book 8.1) Common Mode Rejection and Random DC offset (book 8.3-8.4) Current Mirror Load (book 8.5) Multi Stage Amplifiers (book 8.6)

Content The MOS Differential Pair (book 8.1) Common Mode Rejection and Random DC offset (book 8.3-8.4) Current Mirror Load (book 8.5) Multi Stage Amplifiers (book 8.6)

The differential pair with resistive loads The resistors turn i d linearly into voltage

Easier (in my opinion): look simply at i d

Analysis for Common Mode Input With ideal current source: common mode voltage V CM has no effect, but beware the range of operation!!! V CMmax = V t + V DD − I 2 R D ( 8 . 7 ) V CMmin = − V SS + V CS + V t + V OV ( 8 . 8 (Note that the book always writes − V SS at the actual terminal, i.e. always expresses V SS as a positive number...)

With current mirror When the output is connected to a voltage source, the output current becomes the difference of the two i d

Large Signal, Weak Inversion Simpler analysis in weak inversion: � � − Vt − nVS V 1 V 2 nVT + e I b = I 1 + I 2 = I S e e nVT nVT � � − Vt − nVS V 1 V 2 nVT − e I out = I 1 − I 2 = I S e nVT e nVT V 1 V 2 nVT − e I out = e nVT V 1 V 2 I b nVT + e e nVT V 1 V 2 nVT − e e = I b tanh V 1 − V 2 nVT I out = I b V 1 V 2 2 nV T nVT + e e nVT

Small Signal, Weak Inversion Since the slope of tanh x for x = 0 is 1, the slope of I b tanh ∆ V 2 nV T with respect to ∆ V (the transconductance g of this transamp) is: I b g = 2 nV T

Large Signal, Strong Inversion (1/2) I b = I 1 + I 2 = k n ( V 2 OV 1 + V 2 OV 2 ) I out = I 1 − I 2 = k n ( V 2 OV 1 − V 2 OV 2 ) = V 2 OV 1 − V 2 I out OV 2 V 2 OV 1 + V 2 I b OV 2 V 2 OV 1 − V 2 OV 2 I out = I b V 2 OV 1 + V 2 OV 2

Large Signal, Strong Inversion (2/2) Rewrite with ˆ V OV = V OV 1 + V OV 2 and ∆ V OV = V OV 1 − V OV 2 2 2 ∆ V OV ˆ V OV I out = I b � � OV + ( 2 ˆ 1 ∆ V 2 V OV ) 2 2 Note that this is not yet a closed solution as i the large signal world V S , and thus ˆ V OV depends on ∆ V OV . Extrema where one transistor conducts the entire I B is where ∆ V OV = 2 ˆ V OV (since then one branch has V OV 2 = 0) and I B = k n ∆ V 2 OV . It follows that: � √ I B ∆ V OV = = 2 V OV k n Where like in the book V OV is the overdrive voltage for ∆ V OV = 0

Normalized I/V Curves and Ranges

I/V Curves for different V OV respectively W L (This is only valid for strong inversion)

Small Signal Analysis on the Half Circuit (1/2) Assuming a ’balanced’ input, i.e. v g 1 = − v g 2 = v id 2 . This results in a virtual small signal Gnd at the source of the transistors.

Small Signal Analysis on the Half Circuit (2/2) Thus one can look at the branches individually: It’s the good old common source amp.

Current Source Load Differential Amplifier

Cascode Differential Amplifier

Content The MOS Differential Pair (book 8.1) Common Mode Rejection and Random DC offset (book 8.3-8.4) Current Mirror Load (book 8.5) Multi Stage Amplifiers (book 8.6)

Common Mode Rejection

Common Mode Rejection i v icm = + 2 iR SS g m v icm ≈ v icm i = 1 2 R SS g m 0 + 2 R SS And R D converts i into the two outout voltages v o 1 and v o 2 . Note that the difference of currents is still 0, i.e. not affected by the comon mode input. However, since a change in i , respectively a change in I b , affects the transconductance, v icm will influence the output difference if the differential input is not zero, and mismatch will lead to common mode gain, i.e. a DC offset with zero input difference that varies with v icm .

Content The MOS Differential Pair (book 8.1) Common Mode Rejection and Random DC offset (book 8.3-8.4) Current Mirror Load (book 8.5) Multi Stage Amplifiers (book 8.6)

Systematic DC offset with current mirror load

Output equivalent circuit A d = v o = G m R o = g m 1 , 2 ( r o 2 || r o 4 ) v id

A more careful deduction of G m v id i o = g m 2 2 − g m 4 v gs 4 ( 8 . 132 ) � 1 v id � v gs 3 = − g m 1 || r o 3 || r o 1 2 g m 3 − g m 1 v id ≈ ( 8 . 134 ) g m 3 2 i o ≈ g m v id ⇒ G m = g m

A more careful deduction of R o i = v x R o 2 R in 1 = r o 1 + R L 1 ≈ g m 1 r o 1 g m 1 R o 2 = R in 1 + r o 2 + g m 2 r o 2 R in 1 ≈ 2 r o 2 ( 8 . 135 )

Common Mode Gain (1/2)

Common Mode Gain (2/2)

For General Current Gain A cm = v o = − ( 1 − A m ) G mcm ( R om || R o 2 ) v icm

For Simple Current Mirror 1 A m i i = v gs 3 g m 4 v gs 3 = i i R im R im = || r o 3 g m 3 1 A m = 1 1 + g m 3 r o 3

Content The MOS Differential Pair (book 8.1) Common Mode Rejection and Random DC offset (book 8.3-8.4) Current Mirror Load (book 8.5) Multi Stage Amplifiers (book 8.6)

Internally All Differential Example Has some advantages, foremost a better CMRR. And with single ended stages you have to care about ’hitting’ the right input DC level of the next stage.

Two Stage CMOS op-amp Example A = A 1 A 2 = g m 1 ( r o 2 || r o 4 ) g m 6 ( r o 6 || r o 7 ) ( W / L ) 6 = 2 ( W / L ) 7 ( W / L ) 4 ( W / L ) 5