Aly El-Osery Electrical Engineering Department, New Mexico Tech Socorro, New Mexico, USA September 8, 2009 1 / 17
1 Op-Amps - Handbook 2 Differential Amplifiers (DA) CMRR - Measurement Source Resistance Asymmetry 3 Operational Amplifiers Broadband Amplifier 2 / 17
Handbook of Operational Amplifier http://focus.ti.com/lit/an/sboa092a/sboa092a.pdf 3 / 17
Overview Used for a wide variety of applications. Responds to the use difference in the input signals. Discriminates against changes in the dc power supply voltage. 4 / 17
DA Basics Given two inputs v i,1 and v i,2 define difference mode and common-mode input and output signals as v id = (v i,1 v i,2 )/2, v ic = (v i,1 + v i,2 )/2 (1) v od = (v o,1 v o,2 )/2, v oc = (v o,1 + v o,2 )/2 (2) vi,1 vi,2 Differential Amplifier vo,1 vo,2 Middlebrook Equations Figure: Differential amplifier block v od = A dd v id + A dc v ic, v oc = A cd v id + A cc v ic (3) 5 / 17
Output Equations For single output v o = v o,1 v o = A dd + A cd + A dc + A cc 2 = A D v id + A C v ic v i,1 + A cc + A dc A dd A cd v i,2 2 (4) Ideal DA A cc = A cd = A dc = 0 (5) A C = 0, A D = A dd (6) 6 / 17
Common-Mode Rejection Ratio (CMRR) How close is the real differential amplifier close to an ideal one? CMRR ( v ic to give a certain v o ) ( v id to give the same v o ) (7) CMRR = A dd A cc = A D (8) A C 7 / 17
Measurement of A D and A C First case, connect both inputs to v s1 generating a common mode input signal which results in an output v o. Second Case, connect the + input to v s2 and negative input to ground. v id = v s2 /2, v ic = v s2 /2 Adjust v s2 such that the output is equal to v o generated in the common mode input. A C = v o /v ic = v o /v s1 also consequently, and v o = A D v s2 /2 + A C v s2 /2 A D = v o (2/v s2 1/v s1 ) (9) CMRR = A D /A C = (2v s1 /v s2 1) (10) 8 / 17
DC Equivalent Input Circuit Most frequently, common-mode input resistance, R ic, measured between one input to ground using common mode input, and a difference-mode input resistance, R id, measured using difference mode input from either inputs to ground, are provided by the manufacturer. 9 / 17
DA Input Circuit R s V s,1 R 1 R ic V s,2 R s + δr R ic Figure: Differential amplifier input circuit at dc showing source resistance unbalance. 10 / 17
CMRR-Computation CMRR sys = A D/A C + δr/[2(r ic + R s )] [A D /A C ]δr/[2(r ic + R s )] + 1 CMRR A CMRR A δr/[2(r ic + R s )] + 1 (11) CMRR A is specified by the manufacturer 11 / 17
CMRR - Special Cases If the thevenin source resistance are matched CMRR sys = CMRR A (12) When then δr/r s = 2(R ic /R s + 1)/CMRR A CMRR sys 12 / 17
Source Resistance Asymmetry CMRR sys CMRR A (R ic /R s+1) CMRR A 0 δr/r s Figure: Differential amplifier CMRR magnitude vs fractional unbalance in source resistance. 13 / 17
Open-loop Transfer Function Typical Op-Amp A D = v o v i v i = k ov (τ 1 s + 1)(τ 2 s + 1) (13) f 1 = 1/(2πτ 1 ) and f 2 = 1/(2πτ 2 ). To ensure stability A D (jf 2 ) 1. f 1 occurs at a relatively low value. Small-signal gain bandwidth product GBWP k ov /(2πτ 1 ) (14) The unity gain of 0dB frequency of the open loop, f T, is approximately equal to the GBWP. Slew rate η. 14 / 17
Notes Two pieces of information define are required to define the op-amp s open-loop characteristics. DC open-loop gain k ov, and 0dB frequency, f T. Closed-loop gain is limited by the open-loop gain. The higher the closed-loop gain is the smaller the bandwidth. To overcome this problem, it is better to cascade identical amplification stages rather than having one op-amp with a high gain. 15 / 17
Non-inverting Amplifier (A) (B) Vs R1 Vi V i + Vo Vs Vi V i + Vo RF Figure: Non-inverting op-amp circuits v o /v s = k ov /(τs + 1) 1 + βk ov /(τs + 1) (15) 16 / 17
Inverting Amplifier and Summer R2 RF Vs2 Vs1 R1 + Vo Figure: Summing op-amp 17 / 17