240ln07 - DIFFERENTIAL AMPLIFIERS LN #7 BASIC ANALYSIS VCC...

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1 DIFFERENTIAL AMPLIFIERS LN #7 V CC R c v B1 v o1 R c v B2 v o2 I e1 I e2 I c1 I c2 BASIC ANALYSIS I V o2 = V cc –I c2 R c V o1 = V cc c1 R c I E1 + I E2 = I For matched transistors I E1 = I E2 = 2 I V CC R c v CM v o1 R c v o2 I e1 I e2 I c1 I c2 I V cM –0.7 COMMON MODE INPUT 0 2 2 2 2 1 2 1 2 1 1 = = = = = = = = o o o c cc o o c e c V V V R I V V V I I I i i α
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2 V CC R c 1V v o1 R c v o2 I e1 I e2 I c1 I c2 I ON OFF + - 0.7 0.3 SINGLE SIDE INPUT FINITE IR V V V V V IR V V I I I I c o o CC o c CC o C E c >> = = = = = = = α 2 1 2 1 2 1 0 V CC R c v o1 R c v o2 I E1 I E2 I c1 I c2 I EE -V EE R E + - v i1 + - v i2 EMITTER COUPLED DIFFERENTIAL PAIRS (LARGE SIGNAL) V CC I E1 I E2 + - v be1 + - v be2 V E + v i1 v i2 + Apply KVL v i1 –V be1 + V be2 –v i2 =0 = = = s c T be s c T be V V s c I I V V and I I V V e I I T be 2 2 1 1 / 1 ln ln 1 L L L L L L Now I E1 + I E2 = I EE = (I c1 + I c2 ) / α and
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3 () T id T i i V V V V V c c e e I I / / 2 1 2 1 = = where V id = V i1 - V i2 Æ T id T id T id V V V V V V c c c e e e I I I / / / 2 1 1 1 1 1 + = + = + thus Æ T id T id V V EE F c V V EE F c e I I and e I I / 2 / 1 1 1 + = + = α c V V EE F CC o R e I V V T id / 1 1 + = and c V V EE F CC o R e I V V T id / 2 1 + = + + = = T id T id V V V V c EE F o o od e e R I V V V / / 2 1 1 1 1 1 T id T id T id T id T id T id T id T id T id T id V V V V V V V V c EE F od V V V V V V V V V V V V c EE F od e e e e R I V e e e e e e R I V 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 1 1 1 1 + = + + = and the final result is  − = T id c EE F od V V R I V 2 tanh
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4 I c1 I c2 1 0.5 I c2 , I c1 I EE Currents V id V od -V T -2V T V T 2V T α F I EE R c - α F I EE R c Voltages V id 0 SMALL SIGNAL OPERATION OF THE DIFF AMP (Simple approach derived from large signal equations) V CC R c v D v o1 R c v o2 I e1 I e2 I c1 I c2 I + - + - V be1 + - V be2 i e V BE1 = V BE + v d /2 V BE2 = V BE + v d/2 T d V v F c e I I / 1 1 + = α and T d V v F c e I I / 2 1 + = T d T d T d T d F V v V v V v F c V v if V v V v V v I e e e I I T d T d T d 2 2 1 2 1 2 1 2 / 2 / 2 / 1 << > + + + = + = Q1 Q2 then
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5 1 22 2 d F cF T v I I I V α =+ and 2 2 d F T v I I I V =− Now the collector current that changes with v d which is the “signal” component of collector current is 12 d F cc T v I ii V == from which we can consider a transconductance () T F d c m V I v i g 2 2 / = = CONSIDER ANOTHER POINT OF VIEW The total voltage appears across a total resistance of 2r e where 2 / I V I V r T E T e = = This means that the small signal current i e in above circuit is e d e r v i 2 = which implies that the collector current of Q1 will increase by i c and the collector current of Q2 will decrease by i c where 2 2 d m e d F F c v g r v ie i = = =
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6 MOS DIFFERENTIAL AMPLIFIERS V DD R D v o1 R D v o2 I D1 I D2 I D1 I D2 I T -V SS R + - v i1 + - v i2 + - DC TRANSFER CURVES v i1 –v gs1 + v gs2 i2 = 0 () L W k I V V L W k I V V D th gs D th gs / ' 2 / ' 2 2 2 1 1 + = + = L W k I I v
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240ln07 - DIFFERENTIAL AMPLIFIERS LN #7 BASIC ANALYSIS VCC...

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