Small-Signal Analysis Of BJT Differential Pairs

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair1/21Small-Signal Analysis ofBJT Differential PairsNow lets consider the case where each input of thedifferential pair consists of an identical DC bias term VB, andalso an AC small-signal component (i.e., v1(t ) and v2(t ) )VCCRCRCvO 1 (t )vO 2 (t )VB v1(t ) Q1iE 2 Q2iE 1 v2(t ) VBvBE 2vBE 1 I VEEAs a result, the open-circuit output voltages will likewise havea DC and small-signal component.

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair2/21Recall that we can alternatively express these two small-signalcomponents in terms of their average (common-mode):vcm (t ) v1(t ) v2(t )2and their differential mode:vd (t ) v1(t ) v2(t )Such that:v1(t ) vcm(t ) vd (t )v2(t ) vcm(t ) 2I.E.:vd (t )2VCCRCRCvO 1 (t )vO 2 (t )VB vcm (t ) vd (t )2 Q1iE 2iE 1Q2vBE 2vBE 1 IVEE vcm (t ) vd (t )2 VB

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair3/21Now, let’s determine the small-signal voltage gain of thisamplifier!Q: What do you mean by gain? Is it:Avo vo 1v1orAvo vo 2v2orAvo vo 1v2or Avo vo 2?v1A: Actually, none of those definitions!This is a differential amplifier, so we typically define gain interms of its common-mode ( Acm ) and differential ( Ad ) gains:Acm So that:vo 1 vo 2 vcm vcmandAd vo 1v o2vdvdvo 1(t ) Acm vcm (t ) Ad vd (t )vo 2(t ) Acm vcm (t ) Ad vd (t )Q: So how do we determine the differential and common-mode gains?A: The first step—of course—is to accomplish a DC analysis;turn off the small-signal sources!

5/11/2011Differential Mode Small Signal Analysis of BJT Diff PairVCCRCVBVO 1Q1VO 2RCI 2 Q2I 2VBIVEEThis DC analysis is quite simple!1. Since the DC base voltage VB is the same for eachtransistor, we know the two emitter currents will each be:IE 1 I E 2 I2We know one current, we know em’ all!IC 1 I C 2 αIB 1 IB 2 II212 ( β 1)4/21

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair5/21Likewise, for the BJTs to be in active mode, we know that:VCB 0VC VB . From KVL, the collector voltage is:VC VCC RC IC VCC αI2RCTherefore, in order for the BJTs to be in the active mode:VCC αI2RC VBV VBI 2 CCα RC 2. Now, we determine the small-signal parameters of eachtransistor:gm 1 gm 2 rπ 1 rπ 2 IC α I 2 VTVTIB1I VT 2 ( β 1 ) VTro 1 ro 2 2 VAαI

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair6/213. Turning off the DC sources:RCRCvo 1 (t )vo 2 (t )vcm (t ) vd (t )Q12ie 1ie 2Q2vcm (t ) vd (t )2And now inserting the hybrid-pi BJT model:vo 2(t )vo 1(t )vcm (t ) vd (t )2 - vbe1RCrπgmv be 1 RCro-Now, tidying this schematic up a bit:rπrogmv be 2vbe2- -vcm (t ) - vd (t )2

5/11/2011Differential Mode Small Signal Analysis of BJT Diff PairRC7/21RCvo 2(t )vo 1(t )rorogmv be 2gmv be 1--vbe1rπvbe2rπ vcm (t ) -vcm (t )vd (t ) - 2vd (t )2 -Q: Yikes! How do we analyze this mess?A: In a word, superposition!Q: I see, we turn off three sources and analyze the circuitwith the one remaining source on. We then move to the nextsource, until we have four separate analysis—then we add theresults together, right?A: It’s actually much easier than that!

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair8/21We first turn off the two differential-mode sources, andanalyze the circuit with only the two remaining (equal valued)common-mode sources.RCRCvo 2(t )vo 1(t )rorogmv be 2gmv be 1-vbe1rπvcm (t ) --vbe2rπ vcm (t ) -From this analysis, we can determine things like the commonmode gain and input resistance!

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair9/21We then turn off the two common-mode sources, and analyzethe circuit with only the two (equal but opposite valued)differential-mode sources.RCRCvo 2(t )vo 1(t )rorogmv be 2gmv be 1-vbe1rπ vd (t )2- -vbe2rπ vd (t )2 -From this analysis, we can determine things like thedifferential mode gain and input resistance!Q: This still looks very difficult! How do we analyze these“differential” and “common-mode” circuits?A: The key is circuit symmetry.We notice that the common-mode circuit has a perfect planeof reflection (i.e., bilateral) symmetry:

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pairic 1RCRC10/21ic 2vo 2(t )vo 1(t )rogmv be 1ve 1(t ) vce 1vce 2 ie 1ie 2-rπvbe1iin 1 vcm (t ) -rogmv be 2 ve 2(t )-vbe2rπ -iin 2vcm (t )The left and right side of the circuit above are mirror imagesof each other (including the sources with equal value vcm ).The two sides of the circuit a perfectly and preciselyequivalent, and so the currents and voltages on each side ofthe circuit must likewise be perfectly and precisely equal!For example:vbe 1vo 1vce 1ve 1 vbe 2 vo 2 vce 2 ve 2andiin 1 iin 2gmvbe 1 gmvbe 2ic 1 ic 2ie 1 ie 2

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair11/21Q: Wait! You say that—because of “circuit symmetry”—that:ie 1 ie 2 .But, just look at the circuit; from KCL it is evident that:ie 1 ie 2How can both statements be correct?A: Both statements are correct!In fact, the statements (taken together) tell us what thesmall-signal emitter currents must be (for this common-modecircuit).There is only one possible solution that satisfies the twoequations—the common-mode, small-signal emitter currentsmust be equal to zero!ie 1 ie 2 ie 2 0Hopefully this result is a bit obvious to you.If a circuit possess a plane of perfect reflection (i.e.,bilateral) symmetry, then no current will flow across thesymmetric plane. If it did, then the symmetry would bedestroyed!Thus, a plane of reflection symmetry in a circuit is known asvirtual open—no current can flow across it!

5/11/2011Differential Mode Small Signal Analysis of BJT Diff PairicRCRC12/21icvo (t )vo (t )rogmv beve (t )rπiinvcm (t ) vcevce 00rogmv beve (t )--vbevberπ iin -The Virtual Open -vcm (t )Thus, we can take pair of scissors and cut this circuit into twoidentical half-circuits, without affecting any of the currentsor voltages—the two circuits on either side of the virtual openare completely independent!RCvo (t )rogmv be-rπvbeiin vcm (t ) -The Common-ModeHalf Circuit

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair13/21Now, since ro rπ and ro RC , we can simplify the circuit byapproximating it as an open circuit:RCvo (t )gmv be-rπvbeiin vcm (t ) -Now, let’s analyze this half-circuit!From Ohm’s Law:vbe rπ iinAnd from KCL:iin gmvbeThus combining:vbe ( gm rπ )vbe βvbe

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair14/21Q: Yikes! How can vbe βvbe ? The value β is not equal to 1 !!A: You are right ( β 1 )!Instead, we must conclude from the equation:vbe βvbethat the small-signal voltage vbe must be equal to zero !vbe 0Q: No way! If vbe 0 , then gmvbe 0 . No current isflowing, and so the output voltage vo must likewise be equal tozero!A: That’s precisely correct! The output voltage isapproximately zero:vo (t ) 0Q: Why did you say “approximately” zero ?A: Remember, we neglected the output resistance ro in ourcircuit analysis. If we had explicitly included it, we would findthat the output voltage would be very small, but not exactlyzero.

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair15/21Q: So what does this all mean?A: It means that the common-mode gain of a BJTdifferential pair is very small (almost zero!).Acm vo 0vcmLikewise, we find that:iin 0Such that the common-mode input resistance is really big:Rincm !!!The common-mode component of inputs v1(t ) and v2(t ) havevirtually no effect on a BJT differential pair!Q: So what about the differential mode?A: Let’s complete our superposition and find out!

5/11/2011Differential Mode Small Signal Analysis of BJT Diff PairRC16/21RCvo 2(t )vo 1(t )rorogmv be 2gmv be 1-vbe1rπ vd (t )2-vbe2 -rπ - vd (t )2Q: Hey, it looks like we have the same symmetric circuit asbefore—won’t we get the same answers?ic 1RCRCic 2vo 2(t )vo 1(t )rogmv be 1ve 1(t ) vce 1vce 2 -vbe1rπiin 1 vd (t )2 - ie 1ie 2rogmv be 2 ve 2(t )-vbe2rπ -iin 2v (t ) d2

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair17/21A: Not so fast!Look at the two-small signal sources—they are “equal butopposite”. The fact that the two sources have opposite “sign”changes the symmetry of the circuit.Instead of each current and voltage on either side of thesymmetric plane being equal to the other, we find that eachcurrent and voltage must be “equal but opposite”!For example:vbe 1vo 1vce 1ve 1 vbe 2 vo 2 vce 2and ve 2iin 1 iin 2gmvbe 1 gmvbe 2ic 1 ic 2ie 1 ie 2This type of circuit symmetry is referred to as oddsymmetry; the common-mode circuit, in contrast, possessedeven symmetry.Q: Wait! You say that—because of “circuit symmetry”—that:ve 1 ve 2 .But, just look at the circuit; from KVL it is evident that:v e 1 ve 2

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair18/21How can both statements be correct?A: Both statements are correct!In fact, the statements (taken together) tell us what thesmall-signal emitter voltages must be (for this differentialmode circuit).There is only one possible solution that satisfies the twoequations—the differential-mode, small-signal emittervoltages must be equal to zero!ve 1 ve 2 ve 2 0More generally, the electric potential at every location along aplane of odd reflection symmetry is zero volts. Thus, theplane of odd circuit symmetry is known as virtual ground!icRCRCic vo (t )vo (t ) rogmv beve 0rπiin vd (t )2 -vcevce 1 ieiero gmv be - vbevbe -The Virtual Groundve 0rπ - iin vd (t )2

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair19/21Again, the circuit has two isolated and independent halves.We can take our scissors and cut it into two separate “halfcircuits”:RCvo (t )rogmv be-vberπiinvd (t )2 -The Differential-ModeHalf CircuitNote the only difference (aside from the small-signal source)between the differential half-circuit and its common-modecounterpart is that the emitter is connected to ground Æ it’sa common-emitter amplifier!Let’s redraw this half-circuit and see if you recognize it:iinvo (t ) vd (t )2 -rπvbe-gmv beroRC

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair20/21Q: Hey, we’ve seen this circuit (about a million times) before!We know that:g Rv ( )vo (t ) gm ( ro RC ) d t m C vd (t )2And also:iin (t ) 21 vd (t )rπ 2Right?A: Exactly!From this we can conclude that the differential-mode smallsignal gain is:Ad vo (t ) 21 gm RCvd (t )And the differential mode-input resistance is:Rind vd (t ) 2rπiin (t )In addition, it is evident (from past analysis) that the outputresistance is:dRout ro RC RC

5/11/2011Differential Mode Small Signal Analysis of BJT Diff Pair21/21Now, putting the two pieces of our superposition together, wecan conclude that, given small-signal inputs:v1(t ) vcm(t ) vd (t )2v2(t ) vcm(t ) vd (t )2The small-signal outputs are:vo 1(t ) Acm vcm (t ) Ad vd (t ) Ad vd (t )vo 2(t ) Acm vcm (t ) Ad vd (t ) Ad vd (t )Moreover, if we define a differential output voltage:vod (t ) vo 1(t ) vo 2(t )Then we find it is related to the differential input as:vod (t ) 2Ad vd (t )Thus, the differential pair makes a very good differenceamplifier—the kind of gain stage that is required in everyoperational-amplifier circuit!

5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 1/21 Small-Signal Analysis of BJT Differential Pairs Now lets consider the case where each input of the differential pair consists of an identical DC bias