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ELECTRONICS-1 ZOLTAI CHAPTER 3 DIFFERENTIAL AMPLIFIERS Introduction: the common and the differential mode components of two voltages differential mode component: vd v1 common mode component: vc vd = v1 - v2 v v2 vc = 1 2 v2 vd 2 vd v 2 vc 2 (Remark: the same definitions are valid for DC as well.) with these: v1 v c With these the linear system of equations describing the linear operation of differential amplifiers: vod = Avdd vind + Avdc vinc voc = Avcd vind + Avcc vinc vin1 vin2 vo1 vo2 Target: as high Avdd as possible. Quality factors for this: - discrimination factor A D vdd A vcc - CMRR = Common Mode Rejection Ratio A CMRR vdd Avdc Single ended differential amplifier (symmetrical input, asymmetrical output): vin1 Vin2 vo = Avd vind + Avc vinc Here, the CMRR definition: CMRR Elec1-Part3-Diffamp.doc Avd Avc 27 ELECTRONICS-1 ZOLTAI Basic connection of the differential amplifier VS+ RC RC RL vin1 vin2 vod RB RB RE a) Pure differential mode input (vinc = 0 and vin1 = -vin2) The equal magnitude but opposite phase current changes cancel each-other on RE, therefore the potential of the common E point does not change - virtual ground. Because of the symmetry the electrical middle point of the load resistance neither does change - this also a virtually grounded point. From the aspect of one of the transistors the situation on AC is the following: According to this the differentialdifferential voltage gain by means of the gain formula of the CE stage (g22 0): vod/2 RC VS- Avdd g 21 ( RC x Rt/2 vind/2 RL ) 2 Due to the symmetry, the differential input does not result a common output voltage: Avcc = 0 b) Pure common input: (vind = 0, vin1 = vin2 = vinc) Now the equal current changes flow together through RE, therefore from the aspect of one transistor the situation is such as if RE had double value. Due to the symmetry the collectors remain equipotential, no current flows through R L as if it were an open circuit. Therefore the equivalent circuit for one transistor now: The voltage gain referring to the common output due to the common input can be calculated as the gain of the CE stage with a 'degenerated' transistor: RC voc vinc 2RE g 21 RC 1 g 21 2 RE Because of the symmetry there is no differential mode output voltage: Avdc = 0. Avcc The quality factors: R R Cx L 2 D 1 g 21R E RC CMRR = ∞ Elec1-Part3-Diffamp.doc 28 ELECTRONICS-1 ZOLTAI (Practice) Numeric example #1 Problem: vin1 = 1,005 V and vin2 = 0,995 V How much are the voltages vo1 and vo2 ? Solution: (observe DC input coupling) a) Op. point: IC0 = 1 m A, VC0 = 7.5 V (collector potentials) To this belongs g21 = 38 mS. VS+ = 15V RC 7,5k RC 7,5k RL =10k 10k vo1 b) Input voltage components: vind = 0,01 V vinc = 1 V vo2 c) Gains (g22 = 0) : vin2 vin1 RE 7,5k RL ) = -114 2 vod = -1,14 V Avdd = - g21(RC x With this: VS- = -15,6V RC 0.5 2 RE voc = - 0,5 V Avcc With this: d) Output voltages: - Regarding only the change of the collector potentials: vo1 = - 0.5*1,14 - 0,5 = - 1,07 V vo2 = + 0.5*1,14 - 0,5 = 0,07 V - Voltages that can be measured between the output points and ground: Vo1 = 7,5 – 0.5*1,14 - 0,5 = 6,43 V (VC0 + 0.5v0d + v0C) Vo2 = 7,5 + 0.5*1,14 - 0,5 = 7,57 V (VC0 – 0.5v0d + v0C) End of class 6. Common mode rejection in case of asymmetry RC1 RC2 vodc Conditions: R g 0, R L and g 22 0 . Notations: RC = RC1 – RC2 g21 = g21(1) - g21(2) RC = 0.5(RC1+RC2) g21 = 0.5(g21(1)+g21(2)) iC1=g21(1)vB vB iC2=g21(2)vB vinc RE vB vinc The differential mode output voltage generated by the pure common mode input voltages: vodc = iC2 RC2 - iC1 RC1 = = vB (g21 – 0.5*g21)(RC - 0.5*RC) - (g21 + 0.5*g21)(RC + 0.5*RC) = = -vB (RC g21 + RC g21) Expressing vB by vinc: vB = vinc - 2g21 vB RE Elec1-Part3-Diffamp.doc 29 ELECTRONICS-1 ZOLTAI from where: vinc 1 2 g 21 RE The common to differential gain: g RC g 21 RC 21 g 21 RC v0 dc Avdc vinc 1 2 g 21 R E Since -g21 RC = Avdd , the common mode rejection: A 1 2 g 21 R E CMRR vdd g 21 RC Avdc g 21 RC vB Elec1-Part3-Diffamp.doc 30 ELECTRONICS-1 ZOLTAI Versions of circuit arrangement a) For the sake of large CMRR (and D) the realisation of a large RE b) Using E-degeneration to stabilise the gain (by reducing it) RE RE 2I0 g 21 1 g 21 RE d) Differential amplifier with symmetric output * Here: g 21 c) Trans-conductance amplifier IC1 IC2 RC RC ∞ RL RL RE ∞ CC ∞ RL vin I0 vin1 I0 VS- vin ; RE v I 0 in RE vin2 RE The connection can be retraced to the original one (see dashed lines). I C1 I 0 So: IC2 Avd = - 0.5Avdd = 0.5g21(RC x RL) One of the collectors is used as a current output. If both are used, then the output signal is the difference of the two collector v currents ( 2 in ): this is decided by the (not RE shown) connection receiving the currents. Elec1-Part3-Diffamp.doc vo = Avd vind + Avc vinc g 21 RC xRL 1 2 g 21 RE The notion of inverting and non-inverting inputs. Avc Avcc 31 ELECTRONICS-1 ZOLTAI e) Two-stage differential amplifier with direct coupling RC1 RC1 0V IE1 -0.6V IE1 IN1 IN2 2IE2 IE2 VS+ RE2 RL IE2 RC2 0V 2I E1 IB v0 RE1 VS- Supposing that IB ≈ 0, find the necessary value of RC2 to get direct output coupling. I E1 1 0.6 VS 2 R E1 I E2 1 I E1 R C1 0.6 2 R E2 R C2 VS I E2 Elec1-Part3-Diffamp.doc 32