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4/28/2017 769832736 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 v1t Q1 iE 2 Q2 iE 1 v2t VB vBE 2 vBE 1 I VEE As a result, the open-circuit output voltages will likewise have a DC and small-signal component. 4/28/2017 769832736 2/21 Recall that we can alternatively express these two small-signal components in terms of their average (common-mode): vcm t v1t v2t 2 and their differential mode: vd t v1t v2t Such that: v1(t ) = vcm(t ) + vd (t ) 2 I.E.: v2(t ) = vcm(t ) - 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 4/28/2017 769832736 3/21 Now, let’s determine the small-signal voltage gain of this amplifier! Q: What do you mean by gain? Is it: Avo B vo 1 v1 or Avo B vo 2 v2 or Avo B vo 1 v2 or Avo B 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 B So that: vo 1 v = o2 vcm vcm and Ad B 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 commonmode gains? A: The first step—of course—is to accomplish a DC analysis; turn off the small-signal sources! 4/28/2017 769832736 4/21 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 IE 2 I 2 We know one current, we know em’ all! IC 1 IC 2 IB 1 IB 2 I 2 I 1 2 1 4/28/2017 769832736 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 I R 2 C Therefore, in order for the BJTs to be in the active mode: VCC I RC VB 2 V VB I 2 CC RC 2. Now, we determine the small-signal parameters of each transistor: gm 1 gm 2 rπ 1 rπ 2 IB VT IC VT ro 1 ro 2 I 2 VT 1 I 2 β 1 VT 2 VA a I 4/28/2017 769832736 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 2t vo 1t vcm t vd (t ) 2 + + - + vbe1 RC r + RC ro gmv be 1 - Now, tidying this schematic up a bit: r ro gmv be 2 vbe2 - + - vcm t + - vd (t ) 2 4/28/2017 769832736 RC 7/21 RC vo 2t vo 1t ro ro gmv be 2 gmv be 1 - - vbe1 r vbe2 r + vcm t vd (t ) 2 + + - vcm t + - 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! 4/28/2017 769832736 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 2t vo 1t ro ro gmv be 2 gmv be 1 - vbe1 r vcm t + - - vbe2 r + + vcm t + - From this analysis, we can determine things like the commonmode gain and input resistance! 4/28/2017 769832736 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 2t vo 1t ro ro gmv be 2 gmv be 1 - vbe1 r v t d 2 - + - vbe2 r + 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: 4/28/2017 769832736 ic 1 10/21 RC RC ic 2 vo 2t vo 1t ro gmv be 1 ve 1t + + vce 1 vce 2 - ie 1 ie 2 - r vbe1 iin 1 + vcm t + - ro gmv be 2 - ve 2t - vbe2 r + 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: iin 1 = iin 2 vbe 1 = vbe 2 vo 1 = vo 2 vce 1 = vce 2 ve 1 = ve 2 and gmvbe 1 = gmvbe 2 ic 1 = ic 2 ie 1 = ie 2 4/28/2017 769832736 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! 4/28/2017 769832736 ic 12/21 RC RC ic vo t vo t ro gmv be ve t r iin vcm t + - + + vce vce - - 0 0 ro gmv be ve t - - vbe vbe r + + 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 - r vbe iin + vcm t + - The Common-Mode Half Circuit 4/28/2017 769832736 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 - r 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 4/28/2017 769832736 14/21 Q: Yikes! How can vbe = - βvbe ?? The value β is not equal to - 1 !! A: You are right ( β ¹ - 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. 4/28/2017 769832736 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! 4/28/2017 769832736 RC 16/21 RC vo 2t vo 1t ro ro gmv be 2 gmv be 1 r v t d 2 - - vbe1 vbe2 + + + - r + - 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 2t vo 1t ro gmv be 1 ve 1t r iin 1 vd t 2 + - + + vce 1 vce 2 - ie 1 ie 2 ro gmv be 2 - ve 2t - - vbe1 vbe2 r + + iin 2 v t d 2 + - 4/28/2017 769832736 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: iin 1 = - iin 2 vbe 1 = - vbe 2 vo 1 = - vo 2 vce 1 = - vce 2 and ve 1 = - ve 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: ve 1 = ve 2 4/28/2017 769832736 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 vce + ro vce 1 gmv be ve 0 r iin v t d 2 + - - ie ie ro gmv be + ve 0 - + vbe vbe r + - iin v t d 2 The Virtual Ground + - 4/28/2017 769832736 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 - r vbe 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 it’s a common-emitter amplifier! Let’s redraw this half-circuit and see if you recognize it: iin vo t + vd t 2 + - r vbe - gmv be ro RC 4/28/2017 769832736 20/21 Q: Hey, we’ve seen this circuit (about a million times) before! We know that: vo (t ) = - gm (ro RC g R v ( ) ) d t @ - m C vd (t ) And also: iin (t ) = 2 2 1 vd (t ) rπ 2 Right? A: Exactly! From this we can conclude that the differential-mode smallsignal gain is: Ad B vo (t ) = vd (t ) 1 2 gmRC And the differential mode-input resistance is: Rind B vd (t ) = 2rπ iin (t ) In addition, it is evident (from past analysis) that the output resistance is: d Rout = ro RC @ RC 4/28/2017 769832736 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 1t Acm vcm t Ad vd t Ad vd t vo 2t Acm vcm t Ad vd t Ad vd t Moreover, if we define a differential output voltage: vod (t ) B 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!

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