5/6/2011 section 7_3 The BJT Differential Pair 1/1 7.3 The BJT Differential Pair Reading Assignment: pp. 704-720 VCC In addition to commonemitter, commonRC RC collector (i.e., the vO 1 (t ) emitter follower), and vO 2 (t ) common-base amplifiers, a fourth important and iE 2 Q2 v v1 Q1 iE 1 + 2 + “classic” BJT amplifier vBE 2 vBE 1 stage is the differential − − I pair. VEE HO: Large Signal Operation of the BJT Differential Pair The differential pair is a differential amplifier—we express its performance in terms of differential and common-mode gains. HO: Small-Signal Analysis of the BJT Differential Pair Jim Stiles The Univ. of Kansas Dept. of EECS 5/6/2011 Large Signal Operation of BJT diff pair 1/7 Large-Signal Operation of BJT Differential Pair Consider the BJT differential pair: VCC Note: vE 1 = vE 2 vE RC RC vO 1 (t ) and: vO 2 (t ) v1 + Q1 iE 2 Q2 iE 1 + v2 iE 1 + iE 2 = I vBE 2 vBE 1 − − I VEE Also, we know that: iE 1 where Is vBE VT = e α vBE 1 = v1 − vE 1 iE 2 and Is vBE = e α 2 VT vBE 2 = v2 − vE 5/6/2011 Large Signal Operation of BJT diff pair 2/7 Therefore, the emitter currents can be written in terms of the base voltages as: iE 1 ⎛ I s −vE VT e =⎜ α ⎝ ⎞ v1VT ⎟e ⎠ iE 2 ⎛ Is −vE VT e =⎜ α ⎝ ⎞ v2VT ⎟e ⎠ (A) Of course, we know that I = iE 1 + iE 2 , thus: v2 ⎛ Is −vE VT ⎞ ⎛ v1VT ⎞ I =⎜ e e + e VT ⎟ ⎟ ⎜ ⎠ ⎝α ⎠⎝ And rearranging we find: Is −vE VT = e α e v1 VT I +e (B) v2 VT Now, we can insert equation (B) into equation (A) and find: iE 1 = I e e v1 VT v1 VT +e v2 VT = I 1+e − (v1 −v2 ) VT 5/6/2011 Large Signal Operation of BJT diff pair as well as: iE 2 = I e e v1 VT v2 VT +e v2 VT I = 1+e + (v1 −v2 ) VT Now, let’s define a differential voltage: vD = v 1 − v 2 We can now (finally) write the emitter currents in terms of this differential voltage vD only! iE 1 = iE 2 = I 1+e −vD I 1+e VT +vD VT In other words, the emitter current in a differential pair is independent of the common-mode voltage vcm = (v1 + v2 ) 2 ! 3/7 5/6/2011 Large Signal Operation of BJT diff pair 4/7 Now, let’s examine this result more closely. We will consider 3 cases, vD = 0, vD VT , and vD −VT . vD = 0 When v1 = v2 (i.e., vD = 0 ), we find that the emitter currents are equal, with each taking half of the available current I: e vD VT =e −vD VT =1 , ∴ iE 1 = iE 2 = I 1+1 = I 2 VCC RC v1 = v 2 RC Q2 Q1 I 2 I 2 VEE I v 2 = v1 . 5/6/2011 Large Signal Operation of BJT diff pair 5/7 vD VT When vD VT (i.e., vD 25mV ), we find that the emitter current iE 1 takes all the available current, leaving emitter current iE 2 with none: e −vD VT ≈0 , ∴ iE 1 = ≈∞ , ∴ iE 2 = and e vD VT I 1+0 I 1+∞ =I =0 VCC RC v1 > v2 +VT RC Q2 Q1 I 0 VEE I v2 < v1 −VT 5/6/2011 Large Signal Operation of BJT diff pair 6/7 vD −VT When vD −VT (i.e., vD −25mV ), we find that the emitter current iE 2 takes all the available current, leaving emitter current iE 1 with none: e −vD VT ≈∞ , iE 1 = ∴ and e vD VT ≈0, ∴ iE 2 = I 1+∞ I 1+0 =0 =I VCC RC v1 < v2 −VT RC Q2 Q1 0 I VEE I v2 > v1 +VT 5/6/2011 Large Signal Operation of BJT diff pair 7/7 Recall that for BJTs in the active mode, the collector current is related to the emitter current by iC = α iE . Thus, we can plot the collector currents iC 1 and iC 2 as a function of differential voltage vD : iC 1 1.0 I iC 2 0.8 I 0.6 0.4 0.2 10 5 vD 5 Note for vD < ≈ 2VT : d iC 1 d vD Q: Is this important? What does this mean? A: 10 VT 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 term VB, and also an AC small-signal component (i.e., v1(t ) and v2(t ) ) VCC RC RC vO 1 (t ) vO 2 (t ) VB + v1(t ) + Q1 iE 2 Q2 iE 1 + v2(t ) +VB vBE 2 vBE 1 − I − VEE As a result, the open-circuit output voltages will likewise have a DC and small-signal component. 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 2/21 Recall that we can alternatively express these two small-signal components in terms of their average (common-mode): vcm (t ) v1(t ) + v2(t ) 2 and their differential mode: vd (t ) v1(t ) − v2(t ) Such that: v1(t ) = vcm(t ) + vd (t ) v2(t ) = vcm(t ) − 2 I.E.: vd (t ) 2 VCC RC RC vO 1 (t ) vO 2 (t ) VB +vcm (t ) + vd (t ) 2 + Q1 iE 2 iE 1 Q2 vBE 2 vBE 1 − I VEE − + vcm (t ) − vd (t ) 2 +VB 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 3/21 Now, let’s determine the small-signal voltage gain of this amplifier! Q: What do you mean by gain? Is it: Avo vo 1 v1 or Avo vo 2 v2 or Avo vo 1 v2 or Avo vo 2 ??? v1 A: Actually, none of those definitions! This is a differential amplifier, so we typically define gain in terms of its common-mode ( Acm ) and differential ( Ad ) gains: Acm So that: vo 1 vo 2 = vcm vcm and Ad vo 1 v = − o2 vd vd vo 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! 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair VCC RC VB VO 1 Q1 VO 2 RC I 2 Q2 I 2 VB I VEE This DC analysis is quite simple! 1. Since the DC base voltage VB is the same for each transistor, we know the two emitter currents will each be: IE 1 = I E 2 = I 2 We know one current, we know em’ all! IC 1 = I C 2 = α IB 1 = IB 2 = I I 2 1 2 ( β + 1) 4/21 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 5/21 Likewise, for the BJTs to be in active mode, we know that: VCB > 0 VC >VB . ⇒ From KVL, the collector voltage is: VC = VCC − RC IC = VCC − α I 2 RC Therefore, in order for the BJTs to be in the active mode: VCC − α I 2 RC >VB V −VB I < 2 CC α RC ⇒ 2. Now, we determine the small-signal parameters of each transistor: gm 1 = gm 2 = rπ 1 = rπ 2 = IC α I = 2 VT VT IB 1 I = VT 2 ( β + 1 ) VT ro 1 = ro 2 = 2 VA αI 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 6/21 3. Turning off the DC sources: RC RC vo 1 (t ) vo 2 (t ) vcm (t ) + vd (t ) Q1 2 ie 1 ie 2 Q2 vcm (t ) − vd (t ) 2 And now inserting the hybrid-pi BJT model: vo 2(t ) vo 1(t ) vcm (t ) +vd (t ) 2 + + - + vbe1 RC rπ gmv be 1 + RC ro - Now, tidying this schematic up a bit: rπ ro gmv be 2 vbe2 - + - vcm (t ) + - −vd (t ) 2 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair RC 7/21 RC vo 2(t ) vo 1(t ) ro ro gmv be 2 gmv be 1 - - vbe1 rπ vbe2 rπ + + vcm (t ) + - vcm (t ) vd (t ) + - − + 2 vd (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 circuit with the one remaining source on. We then move to the next source, until we have four separate analysis—then we add the results together, right? A: It’s actually much easier than that! 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 8/21 We first turn off the two differential-mode sources, and analyze the circuit with only the two remaining (equal valued) common-mode sources. RC RC vo 2(t ) vo 1(t ) ro ro gmv be 2 gmv be 1 - vbe1 rπ vcm (t ) + - - vbe2 rπ + + vcm (t ) + - From this analysis, we can determine things like the commonmode gain and input resistance! 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 9/21 We then turn off the two common-mode sources, and analyze the circuit with only the two (equal but opposite valued) differential-mode sources. RC RC vo 2(t ) vo 1(t ) ro ro gmv be 2 gmv be 1 - vbe1 rπ + vd (t ) 2 - + - vbe2 rπ + − vd (t ) 2 + + - From this analysis, we can determine things like the differential 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 plane of reflection (i.e., bilateral) symmetry: 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair ic 1 RC RC 10/21 ic 2 vo 2(t ) vo 1(t ) ro gmv be 1 ve 1(t ) + + vce 1 vce 2 − ie 1 ie 2 - rπ vbe1 iin 1 + vcm (t ) + - ro gmv be 2 − ve 2(t ) - vbe2 rπ + + - iin 2 vcm (t ) The left and right side of the circuit above are mirror images of each other (including the sources with equal value vcm ). The two sides of the circuit a perfectly and precisely equivalent, and so the currents and voltages on each side of the circuit must likewise be perfectly and precisely equal! For example: vbe 1 vo 1 vce 1 ve 1 = vbe 2 = vo 2 = vce 2 = ve 2 and iin 1 = iin 2 gmvbe 1 = gmvbe 2 ic 1 = ic 2 ie 1 = ie 2 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 11/21 Q: 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 2 How can both statements be correct? A: Both statements are correct! In fact, the statements (taken together) tell us what the small-signal emitter currents must be (for this common-mode circuit). There is only one possible solution that satisfies the two equations—the common-mode, small-signal emitter currents must be equal to zero! ie 1 = ie 2 = −ie 2 = 0 Hopefully 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 the symmetric plane. If it did, then the symmetry would be destroyed! Thus, a plane of reflection symmetry in a circuit is known as virtual open—no current can flow across it! 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair ic RC RC 12/21 ic vo (t ) vo (t ) ro gmv be ve (t ) rπ iin vcm (t ) + + vce vce − − 0 0 ro gmv be ve (t ) - - vbe vbe rπ + + iin + - The Virtual Open + - vcm (t ) Thus, we can take pair of scissors and cut this circuit into two identical half-circuits, without affecting any of the currents or voltages—the two circuits on either side of the virtual open are completely independent! RC vo (t ) ro gmv be - rπ vbe iin + vcm (t ) + - The Common-Mode Half Circuit 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 13/21 Now, since ro rπ and ro RC , we can simplify the circuit by approximating it as an open circuit: RC vo (t ) gmv be - rπ vbe iin + vcm (t ) + - Now, let’s analyze this half-circuit! From Ohm’s Law: vbe = rπ iin And from KCL: iin = − gmvbe Thus combining: vbe = −( gm rπ )vbe = −βvbe 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 14/21 Q: 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 = −βvbe that the small-signal voltage vbe must be equal to zero ! vbe = 0 Q: No way! If vbe = 0 , then gmvbe = 0 . No current is flowing, and so the output voltage vo must likewise be equal to zero! A: That’s precisely correct! The output voltage is approximately zero: vo (t ) ≅ 0 Q: Why did you say “approximately” zero ?? A: Remember, we neglected the output resistance ro in our circuit analysis. If we had explicitly included it, we would find that the output voltage would be very small, but not exactly zero. 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 15/21 Q: So what does this all mean? A: It means that the common-mode gain of a BJT differential pair is very small (almost zero!). Acm = vo ≅0 vcm Likewise, we find that: iin ≅ 0 Such that the common-mode input resistance is really big: Rincm ≅ ∞ !!! The common-mode component of inputs v1(t ) and v2(t ) have virtually no effect on a BJT differential pair! Q: So what about the differential mode? A: Let’s complete our superposition and find out! 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair RC 16/21 RC vo 2(t ) vo 1(t ) ro ro gmv be 2 gmv be 1 - vbe1 rπ + vd (t ) 2 - vbe2 + + - rπ + + - − vd (t ) 2 Q: Hey, it looks like we have the same symmetric circuit as before—won’t we get the same answers? ic 1 RC RC ic 2 vo 2(t ) vo 1(t ) ro gmv be 1 ve 1(t ) + vce 1 vce 2 − - vbe1 rπ iin 1 + + vd (t ) 2 + - + ie 1 ie 2 ro gmv be 2 − ve 2(t ) - vbe2 rπ + + - iin 2 v (t ) − d 2 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 17/21 A: Not so fast! Look at the two-small signal sources—they are “equal but opposite”. 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 the symmetric plane being equal to the other, we find that each current and voltage must be “equal but opposite”! For example: vbe 1 vo 1 vce 1 ve 1 = −vbe 2 = −vo 2 = −vce 2 and = −ve 2 iin 1 = −iin 2 gmvbe 1 = − gmvbe 2 ic 1 = −ic 2 ie 1 = −ie 2 This type of circuit symmetry is referred to as odd symmetry; the common-mode circuit, in contrast, possessed even 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 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 18/21 How can both statements be correct? A: Both statements are correct! In fact, the statements (taken together) tell us what the small-signal emitter voltages must be (for this differentialmode circuit). There is only one possible solution that satisfies the two equations—the differential-mode, small-signal emitter voltages must be equal to zero! ve 1 = −ve 2 = ve 2 = 0 More generally, the electric potential at every location along a plane of odd reflection symmetry is zero volts. Thus, the plane of odd circuit symmetry is known as virtual ground! ic RC RC ic −vo (t ) vo (t ) − + ro gmv be ve = 0 rπ iin + vd (t ) 2 + - vce vce 1 − ie ie ro − gmv be + - + vbe vbe + - The Virtual Ground ve = 0 rπ + - −iin − vd (t ) 2 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 19/21 Again, the circuit has two isolated and independent halves. We can take our scissors and cut it into two separate “halfcircuits”: RC vo (t ) ro gmv be - vbe rπ iin vd (t ) 2 + + - The Differential-Mode Half Circuit Note the only difference (aside from the small-signal source) between the differential half-circuit and its common-mode counterpart is that the emitter is connected to ground Æ it’s a common-emitter amplifier! Let’s redraw this half-circuit and see if you recognize it: iin vo (t ) + vd (t ) 2 + - rπ vbe - gmv be ro RC 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 20/21 Q: Hey, we’ve seen this circuit (about a million times) before! We know that: g R v ( ) vo (t ) = − gm ( ro RC ) d t ≅ − m C vd (t ) 2 And also: iin (t ) = 2 1 vd (t ) rπ 2 Right? A: Exactly! From this we can conclude that the differential-mode smallsignal gain is: Ad vo (t ) = − 21 gm RC vd (t ) And the differential mode-input resistance is: Rind vd (t ) = 2rπ iin (t ) In addition, it is evident (from past analysis) that the output resistance is: d Rout = ro RC ≅ RC 5/11/2011 Differential Mode Small Signal Analysis of BJT Diff Pair 21/21 Now, putting the two pieces of our superposition together, we can conclude that, given small-signal inputs: v1(t ) = vcm(t ) + vd (t ) 2 v2(t ) = vcm(t ) − vd (t ) 2 The 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 difference amplifier—the kind of gain stage that is required in every operational-amplifier circuit!