Introduction: the common and the differential mode components of two voltages. differential mode component: v d = v 1 - v 2 common mode component:

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1 EECTONCS-1 CHPTE 3 DFFEENT MPFES ntroduction: the common and the differential mode components of two voltaes v d v c differential mode component: v d = v 1 - v common mode component: v1 v v c = v 1 v vd with these: v1 vc vd v vc (emark: the same definitions are valid for DC as well.) With these the linear system of equations describin the linear operation of differential amplifiers: v od = vdd v ind + vdc v inc v oc = vcd v ind + vcc v inc v in v o1 v o Taret: as hih vdd as possible. Quality factors for this: - discrimination factor vdd D vcc - CM = Common Mode ejection atio vdd CM vdc Sinle ended differential amplifier (symmetrical input, asymmetrical output): vin1 Vin v o = vd v ind + vc v inc Here, the CM definition: CM vd vc Elec1-Part3-Diffamp.doc 7

2 EECTONCS-1 Basic connection of the differential amplifier C B V S+ vod E C B v in a) Pure differential mode input (v inc = 0 and = -v in ) The equal manitude but opposite phase current chanes cancel each-other on E, therefore the potential of the common E point does not chane - virtual round. Because of the symmetry the electrical middle point of the load resistance neither does chane - this also a virtually rounded point. From the aspect of one of the transistors the situation on C is the followin: ccordin to this the differentialdifferential voltae ain by means of the ain formula of the CE stae ( 0): V S- v ind / C v od / t / vdd 1( C x Due to the symmetry, the differential input does not result a common output voltae: vcc = 0 ) b) Pure common input: (v ind = 0, = v in = v inc ) Now the equal current chanes flow toether throuh E, therefore from the aspect of one transistor the situation is such as if E had double value. Due to the symmetry the collectors remain equipotential, no current flows throuh as if it were an open circuit. Therefore the equivalent circuit for one transistor now: The voltae ain referrin to the common output due to the common input can be calculated as the ain of the CE stae with a 'deenerated' transistor: v inc C v oc 1C vcc 1 1 E Because of the symmetry there is no differential mode output voltae: vdc = 0. E The quality factors: Cx D 1 1 E C CM = Elec1-Part3-Diffamp.doc 8

3 EECTONCS-1 (Practice) Numeric example #1 C 7,5k v o1 =10k 10k V S+ = 15V v o C 7,5k Problem: vin1 = 1,005 V and vin = 0,995 V How much are the voltaes v o1 and v o? Solution: (observe DC input couplin) a) Op. point: C0 = 1 m, V C0 = 7.5 V (collector potentials) To this belons 1 = 38 ms. b) nput voltae components: v ind = 0,01 V v inc = 1 V E 7,5k v in V S- = -15,6V c) Gains ( = 0) : vdd = - 1 (C x ) = -114 With this: v od = -1,14 V C vcc 0.5 E With this: v oc = - 0,5 V d) Output voltaes: - eardin only the chane of the collector potentials: v o1 = - 0.5*1,14-0,5 = - 1,07 V v o = + 0.5*1,14-0,5 = 0,07 V - Voltaes that can be measured between the output points and round: V o1 = 7,5 0.5*1,14-0,5 = 6,43 V (V C v 0d + v 0C ) V o = 7, *1,14-0,5 = 7,57 V (V C0 0.5v 0d + v 0C ) End of class 6. Common mode rejection in case of asymmetry C1 v odc C Conditions: 0, and 0. Notations: C = C1 C 1 = 1(1) - 1() C = 0.5( C1 + C ) 1 = 0.5( 1(1)+ 1() ) i C1 = 1(1) v B v B i C = 1() v B v B The differential mode output voltae enerated by the pure common mode input voltaes: v inc E v inc v odc = i C C - i C1 C1 = = v B ( 1 0.5* 1 )( C - 0.5* C ) - - ( * 1 )( C + 0.5* C ) = = -v B ( C 1 + C 1 ) Expressin v B by v inc : v B = v inc - 1 v B E Elec1-Part3-Diffamp.doc 9

4 EECTONCS-1 from where: vinc vb 1 1E The common to differential ain: 1 C 1C v 0dc 1 C vdc vinc 1 1E Since - 1 C = vdd, the common mode rejection: vdd 1 1E CM vdc 1 C 1 C Elec1-Part3-Diffamp.doc 30

5 EECTONCS-1 Versions of circuit arranement a) For the sake of lare b) Usin E-deeneration to stabilise CM (and D) the realisation the ain (by reducin it) of a lare E E E 0 * Here: 1 1 c) Trans-conductance amplifier d) Differential amplifier with symmetric output 1 1 E C1 C C C C C E vin 0 0 E v in The connection can be retraced to the oriinal one (see dashed lines). vin C 1 0 ; So: v o = vd v ind + vc v inc v E One of the collectors is used as a current out- VS- in C 0 vd = vdd = ( C x ) E put. f both are used, then the output sinal is the difference of the two collector vin currents ( ): this is decided by the (not E shown) connection receivin the currents. 1 C x vc vcc 1 1E The notion of invertin and non-invertin inputs. Elec1-Part3-Diffamp.doc 31

6 EECTONCS-1 e) Two-stae differential amplifier with direct couplin C1 C1 E E V S+ N1 0V E1-0.6V E1 E E N 0V B E1 E1 C v 0 V S- Supposin that B 0, find the necessary value of C to et direct output couplin. E1 E C V 1 E1 V E S E1 C1 E S 0.6 Elec1-Part3-Diffamp.doc 3