Lect. 11: Differential

Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

Differential

If vG1 = vG2, vo = vD1-vD2 = 0

If vG1 > vG2, vo < 0

If vG1 < vG2, vo > 0

Non-zero output only for input difference Î

Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

Consider Common-Mode and Differential-Mode separately. vv vV=+id, vV =− id GCMGCM1122 vv+ Vvvv==−GG12, CM2 id G12 G

Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

Common-Mode: vG1= vG2 = vCM

υ DD12−υ = 0 (No CM output)

vCM, max?

vvVDS≥− GS t I ()()VRvvvV− −−≥− DD2 D CM GS GS t I VRVv−+≥ DD2 D t CM I ∴υ =+VV − R CM,max t DD2 D

vCM, min? Î Input common-mode range

Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

Differential-mode small-signal analysis (assume vid < VCM)

vv No change vvv=−=− gRgRid − id = − gRv od o12 o mD22 mD mDid since left and right are anti-symmetric vod AgRdmD==− vid

With ro , AdmDo=−g (Rr|| ) Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

Differential-mode small-signal half-circuit Î Consider only half circuit

vo1 =−gRmDo(||) r vid /2 v v = od o1 2

vod ∴ =−gRmDo(||) r vid

Differential pair acts as CS amplifier for input difference!

Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

Why a differential amplifier?

- Input stage for operational amplifier (Op-Amp)

- Not sensitive to noises that are common to both input signals.

Common-Mode Rejection Ratio

CMRR = |Ad/Acm|

Ideally, infinite

Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

What if the current source is NOT ideal?

Common-mode

open circuit (Left and Right are symmetric)

Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

Common-Source with source resistance Common-mode half circuit

vo1 = ? vomgsD1 = -g vR vicm

vvgsicmmgsSS= − g vR(2 )

vicm vgs = 12+ gmSSR

vgR RD omD1 =−  − 2R vicm12+ g mR SS SS

Î common-mode gain due to Rss for single-ended output

Differential output ( vo = vo1 –vo2), vo= 0

Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

What if the current source is NOT ideal? Differential-Mode

With anti-symmetric input, the mid-point is still small-signal ground!

AdmDo= −g (Rr|| ) vid/2 -vid/2

CMRR: Common-Mode Rejection Ratio (Single-ended output) AgRr(||) CMRR= dmDo R Acm D

2Rss CMRR for differential output?

Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

MOS differential pair is based on the symmetry

Anything that breaks the symmetry affects the circuit performance (CMRR, input offset)

Device mismatches cause non-ideal performance ÎEliminate RD

Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

Active-loaded MOS differential pair (single-ended)

vicm vicm

Common mode

It can be shown with finite Rss ,

vo 1 Acm =− vgRicm2 m ss due to non-ideal current source.

Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

Differential mode small-signal analysis

vo =−grmo(||) r o −vid /2 vgr ∴ omo= vid 4

But with more detailed analysis, vgr A ==omo Assume it is (almost) symmetric d vid 2 Î Source terminals are grounded What is wrong with the half-circuit analysis? Half circuit analysis is possible

Electronic Circuits 2 (07/1) W.-Y. Choi Lect. 11: Differential Amplifiers

The action is not considered!

Current mirror doubles the

Î Twice voltage gain!

vo 1 ∴ Admo==gr vid 2

1 gr A mo ∴CMRR==d  2 g2 r R 1 moss Acm

2gRmss

Electronic Circuits 2 (07/1) W.-Y. Choi