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DIFFERENTIAL AMPLIFIER, SINGLE-ENDED OUTPUT Analysis of a symmetrical differential single-ended output amplifier THE CIRCUIT V CC RC vi1 Q1 Q2 IBIAS -V EE DC ANALYSIS RC vout vi2 SINGLE-ENDED OUTPUT AMPLIFIER Differential amplifiers are used in integrated circuit design. In IC design, large capacitors and large (Ω) resistors are impractical. R V CC RC C IB B E 0.5 IBIAS -V EE IC IE Since the circuit is symmetrical, we are looking at only ½ the circuit, through which ½ the bias current flo
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  Tom Penick tomzap@eden.com www.teicontrols.com/notes 12/06/98 DIFFERENTIAL AMPLIFIER, SINGLE-ENDED OUTPUT Analysis of a symmetrical differential single-ended output amplifier THE CIRCUIT  out i2 1 2  R-V   EE   I   BIAS v i1 Q R C  Qvv R C  V  CC  SINGLE-ENDED OUTPUT AMPLIFIER Differential amplifiers are used in integrated circuitdesign. In IC design, large capacitors and large ( Ω )resistors are impractical. DC ANALYSIS   I   B BIAS 0.5  I -V   EE  EB I   E  C  I  C   R C  V  CC  Since the circuit is symmetrical, we are looking at only ½ thecircuit, through which ½ the bias current flows. The biasresistance  R is ignored. We find the value of  r  e for our AC circuitmodel using these formulas:  BIAS E C  I  I  I  15.01 +ββ=+ββ= C m I g 40 = m gr  β= π 1 +β= π r r  e AC ANALYSIS  For the AC analysis, we divide the input signals v i1 and v i2 into differential and common-modecomponents. The inputs are written as 2 1 id cmii vvv += 2 2 id cmii vvv −= We solve for the differential output v d and the common mode output v cm separately using differentmodels and then combine the results.  Tom Penick tomzap@eden.com www.teicontrols.com/notes 12/06/98 DIFFERENTIAL MODE ANALYSIS  - solving for the differential mode voltage gain  A d DM MODEL out  vv id  2 e r  i e  Ri cb β i C  Since the circuit is symmetrical and the differential mode inputs are180° out of phase, a ground potential results at the emitters. The twohalves of the circuit are isolated. Since there is an output only on theright-hand half, this is the only part of the circuit we include in thisanalysis. ed ied ie r vr vi 22 −=÷−=       −+ββ= ed ic r vi 21 C ed iC cout d  Rr v Riv       +ββ=−= 21 d iout d d  vv A =  A d  is the differential mode gain [ V   /  V  ] COMMON MODE ANALYSIS  - solving for the common mode voltage gain  A cm AC ANALYSIS 1 2 2R 2RQ R C  Q R C  Because of the symmetry of thecircuit, we can write  R as twoparallel 2  R resistors and observethat no current flows betweenthem. Therefore this connectioncan be eliminated and weseparate the circuit into halves.Once again, we need only look atthe right-hand half since thatcontains the output.Even though we have looked onlyat the right-hand side in bothphases of the analysis, thepresence of the left-hand sidedoes significantly affect theoutcome. CM MODEL i e v i cm e r 2R out  v Ri cb β i C   Rr vi ecmie 2 +=      ++ββ=  Rr vi ecmic 21 C eid C cout cm Rr v Riv       +ββ=−= 21 cmiout cmcm vv A = COMMON MODE REJECTION RATIO (CMRR)  cmd   A A = CMRR cmd   A A 10dB log20CMRR =
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