Amplifiers: Op Amps Texas Instruments Incorporated Analysis of fully differential amplifiers By Jim Karki Systems Specialist, High-Speed Amplifiers Introduction The August issue of Analog Applications Journal introduced the fully differential amplifiers from Texas Instruments and illustrated their basic operation (see Reference 1). This article explores the topic more deeply by analyzing gain and noise. The fully differential amplifier has multiple feedback paths, and circuit analysis requires close attention to detail. Care must be taken to include the VOCM pin for a complete analysis. diagram from which specific circuit configurations can be easily solved. The voltage definitions are required to arrive at practical solutions. AF is used to represent the open-loop differential gain of the amplifier such that (VOUT+) – (VOUT –) = AF(VP – VN). This assumes that the gains of the two sides of the differential amplifier are well matched and that variations are insignificant. With negative feedback, this is typically the case when AF >> 1. Input voltage definitions: Circuit analysis Circuit analysis of fully differential amplifiers follows the same rules as normal single-ended amplifiers, but subtleties are present that may not be fully appreciated until a full analysis is done. The analysis circuit shown in Figure 1 is used to calculate a generalized circuit formula and block VID = ( VIN + ) − ( VIN − ) (1) ( VIN + ) + ( VIN − ) 2 (2) VIC = Output voltage definitions: Figure 1. Analysis circuit R2 R1 – R3 + VOUT+ – VOUT – AF VP VIN + + VOCM R4 Figure 2. Block diagram β2 VIN – 1–β 2 – + VIN + 1–β1 – Σ (3) ( VOUT + ) + ( VOUT − ) 2 (4) VOC = VN VIN – VOD = ( VOUT + ) − ( VOUT − ) VP– VN AF + β1 ( VOUT + ) − ( VOUT − ) = A F ( VP − VN ) (5) VOC = VOCM (6) There are two amplifiers: the main differential amplifier (from VIN to VOUT) and the VOCM error amplifier. The operation of the VOCM error amplifier is the simpler of the two and will be considered first. It may help to review the simplified schematic shown in Reference 1. VOUT+ and VOUT – are filtered and summed by an internal RC network. The VOCM amplifier samples this voltage and compares it to the voltage applied to the VOCM pin. An internal feedback loop is used to drive “error” voltage of the VOCM error amplifier (the voltage between the input pins) to zero, so that VOC = VOCM. This is the basis of the voltage definition given in Equation 6. There is no simple way to analyze the main differential amplifier except to sit down and write some node equations, then do the algebra VOUT+ to massage them into practical form. We will first derive a solution based solely on nodal VOUT– analysis. Then we will make use of the voltage definitions given in Equations 1–6 to derive solutions for the output voltages, looking at them single-ended; i.e., VOUT+ and VOUT –. These are then used to calculate VOD. 48 Analog and Mixed-Signal Products November 2000 Analog Applications Journal Amplifiers: Op Amps Texas Instruments Incorporated Solving the node equations at VN and VP yields ⎛ R3 ⎞ ⎛ R4 ⎞ ⎛ R1 ⎞ ⎛ R2 ⎞ VN = ( VIN − )⎜ ⎟. ⎟ + ( VOUT − )⎜ ⎟ and VP = ( VIN + )⎜ ⎟ + ( VOUT + )⎜ ⎝ R3 + R4 ⎠ ⎝ R3 + R4 ⎠ ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠ By setting ⎛ R1 ⎞ ⎛ R3 ⎞ β1 = ⎜ ⎟ , VN and VP can be rewritten as ⎟ and β 2 = ⎜ ⎝ R1 + R2 ⎠ ⎝ R3 + R4 ⎠ VN = ( VIN − )(1 − β 2 ) + ( VOUT + )( β 2 ), and (7) VP = ( VIN + )(1 − β 1 ) + ( VOUT − )( β 1 ). (8) With Equations 7 and 8, a block diagram of the main differential amplifier can be constructed, like that shown in Figure 2. Block diagrams are useful tools for understanding circuit operation and investigating other variations. By using the block diagram, or combining Equations 7 and 8 with Equation 5, we can find the input-to-output relationship: (VOUT + )(1 + A F β 2 ) − ( VOUT − )(1 + A F β 1 ) = A F [( VIN + )(1 − β 1 ) − ( VIN − )(1 − β 2 )] . (9) Although accurate, Equation 9 is somewhat cumbersome when the feedback paths are not symmetrical. By using the voltage definitions given in Equations 1–4 and Equation 6, we can derive more practical formulas. Substituting (VOUT –) = 2VOC – (VOUT+), and VOC = VOCM, we can write ( VOUT + )( 2 + A F β 1 + A F β 2 ) − 2 VOCM (1 + A F β 1 ) = A F [( VIN + )(1 − β 1 ) − ( VIN − )(1 − β 2 )] , or ( VOUT + ) = 1 (β1 + β 2 ) ⎛ 1 ⎞ + β1 ⎟ ( VIN + )(1 − β 1 ) − ( VIN − )(1 − β 2 ) + 2 VOCM ⎜ ⎝ AF ⎠ ⎛ ⎞ 2 ⎜1+ ⎟ A F β1 + A F β 2 ⎠ ⎝ . (10) With the “ideal” assumption AFβ1 >> 1 and AFβ2 >> 1, this reduces to ( VOUT + ) = ( VIN + )(1 − β 1 ) − ( VIN − )(1 − β 2 ) + 2 VOCM β 1 . (β1 + β 2 ) (11) VOUT – is derived in a similar manner: ( VOUT − ) = 1 (β1 + β 2 ) ⎛ 1 ⎞ − ( VIN + )(1 − β 1 ) + ( VIN − )(1 − β 2 ) + 2 VOCM ⎜ + β2 ⎟ A ⎝ F ⎠ ⎛ ⎞ 2 ⎜1+ ⎟ A F β1 + A F β 2 ⎠ ⎝ . (12) Again, assuming AFβ1 >> 1 and AFβ2 >> 1, this reduces to ( VOUT − ) = −( VIN + )(1 − β 1 ) + ( VIN − )(1 − β 2 ) + 2 VOCM ( + β 2 ) . (β1 + β 2 ) (13) To calculate VOD = (VOUT+) – (VOUT –), subtract Equation 12 from Equation 10: VOD = 2[( VIN + )(1 − β 1 ) − ( VIN − )(1 − β 2 )] + 2 VOCM ( β 1 − β 2 ) 1 (β1 + β 2 ) ⎛ ⎞ 2 ⎜1+ ⎟ A F β1 + A F β 2 ⎠ ⎝ (14) Continued on next page 49 Analog Applications Journal November 2000 Analog and Mixed-Signal Products Amplifiers: Op Amps Texas Instruments Incorporated Continued from previous page Again, assuming AFβ1 >> 1 and AFβ2 >> 1, this reduces to VOD = 2[( VIN + )(1 − β 1 ) − ( VIN − )(1 − β 2 )] + 2 VOCM ( β 1 − β 2 ) . (β1 + β 2 ) It can be seen from Equations 11, 13, and 15 that even though the obvious use of a fully differential amplifier is with symmetrical feedback, the gain can be controlled with only one feedback path. Using matched resistors R1 = R3 and R2 = R4 in the analysis circuit of Figure 1 balances the feedback paths so that β1 = β2 = β, and the transfer function is ( VOUT + ) − ( VOUT − ) AF 1− β 1 . × = = β ( VIN + ) − ( VIN − ) (1 + A F β ) ⎛ 1 ⎞ ⎜1+ ⎟ A Fβ ⎠ ⎝ (15) Figure 3. Single-ended to differential amplifier R1 R2 – + VOUT+ AF R3 VIN + – + VOUT– VOCM The common-mode voltages at the input and output do not enter into the equation, VIC is rejected, and VOC is set by the voltage at VOCM. The ideal gain (assuming AFβ >> 1) is set by the ratio R4 1 − β R2 = . β R1 Note that the normal inversion we might expect, given two balanced inverting amplifiers, is accounted for by the output voltage definitions, resulting in a positive gain. Many applications require that a single-ended signal be converted to a differential signal. The circuits in Figures 3–7 show various approaches. Using Equations 11, 13, and 15, we can easily derive circuit solutions. With a slight variation of Figure 1 as shown in Figure 3, single-ended signals can be amplified and converted to differential signals. VIN – is now grounded and the signal is applied to VIN+. Substituting VIN – = 0 in Equations 11, 13, and 15 results in ( VOUT + ) = ( VIN + )(1 − β 1 ) + 2 VOCM β 1 , (β1 + β 2 ) ( VOUT − ) = 2 VOCM β 2 − ( VIN + )(1 − β 1 ) , and (β1 + β 2 ) VOD = Figure 4. β1 = 0 R1 R2 – + VOUT+ AF VIN + – + VOUT– VOCM Figure 5. β2 = 0 2( VIN + )(1 − β 1 ) + 2 VOCM ( β 1 − β 2 ) . (β1 + β 2 ) If the signal is not referenced to ground, the reference voltage will be amplified along with the desired signal, reducing the dynamic range of the amplifier. To strip unwanted dc offsets, use a capacitor to couple the signal to VIN+. Keeping β1 = β2 will prevent VOCM from causing an offset in VOD . The circuits in Figures 4–7 have nonsymmetrical feedback. This causes VOCM to influence VOUT+ and VOUT – differently, making VOCM show up in VOD . This will change the operating points between the internal nodes in the – VOUT+ AF R3 VIN + + + – VOUT– VOCM R4 50 Analog and Mixed-Signal Products November 2000 Analog Applications Journal Amplifiers: Op Amps Texas Instruments Incorporated Figure 7. β1 = 0, and β2 = 1 Figure 6. β2 = 1 – + VOUT+ – AF R3 – + VIN + VOUT+ + AF VOUT– VIN + VOCM + – VOUT– R4 VOCM differential amplifier, and matching of the open-loop gains will degrade. CMRR is not a real issue with single-ended inputs, but the analysis points out that CMRR is severely compromised when nonsymmetrical feedback is used. In the discussion of noise analysis that follows, it is shown that nonsymmetrical feedback also increases noise introduced at the VOCM pin. For these reasons, even though the circuits shown in Figures 4–7 have been tested to prove they work in accordance with the equations given, they are presented mainly for instructional purposes. They are not recommended without extensive lab testing to prove their worthiness in your application. In the circuit shown in Figure 4, VIN – = 0 and β1 = 0. The output voltages are ( VOUT + ) = ( VIN + )(1 − β 1 ) + 2 VOCM , β1 2( VIN + )(1 − β 1 ) + 2 VOCM . β1 ( VOUT + ) = ( VIN + )(1 − β 1 ) + 2 VOCM β 1 , β1 + 1 ( VOUT − ) = 2 VOCM − ( VIN + )(1 − β 1 ) , and β1 + 1 VOD = With β1 = 0, this circuit is similar to a noninverting amplifier. In the circuit shown in Figure 5, VIN – = 0 and β2 = 0. The output voltages are −( VIN + )(1 − β 1 ) , and β1 With β2 = 0, the gain is twice that of an inverting amplifier (without the minus sign). In the circuit shown in Figure 6, VIN – = 0 and β2 = 1. The output voltages are ( VIN + ) , and β2 2( VIN + ) − 2 VOCM . β2 ( VOUT + ) = VOD = ( VIN + ) , β2 ( VOUT − ) = 2 VOCM − VOD = ( VOUT − ) = 2( VIN + )(1 − β 1 ) + 2 VOCM ( β 1 − 1) . ( β 1 + 1) The gain is 1 with β1 = 0.333; or, with β1 = 0.6, the gain is 1/2. In the circuit shown in Figure 6, VIN – = 0, β1 = 0, and β2 = 1. The output voltages are ( VOUT + ) = ( VIN + ), ( VOUT − ) = 2 VOCM − ( VIN + ), and VOD = 2[( VIN + ) − VOCM ] . This circuit realizes a gain of 2 with no resistor. Continued on next page 51 Analog Applications Journal November 2000 Analog and Mixed-Signal Products Amplifiers: Op Amps Texas Instruments Incorporated Continued from previous page Figure 8. Noise analysis circuit Noise analysis The noise sources are identified in Figure 8, which will be used for analysis with the following definitions. EIN is the input-referred RMS noise voltage of the amplifier: EIN ≈ eIN x √ENB (assuming the 1/f noise is negligible), where eIN is the input white noise spectral density in volts per square root of the frequency in Hertz, and ENB is the effective noise bandwidth. EIN is modeled as a differential voltage at the input. IIN+ and IIN – are the input-referred RMS noise currents that flow into each input. They are taken as equal and called IIN. IIN ≈ iIN x √ENB (assuming the 1/f noise is negligible), where iIN is the input white noise spectral density in amps per square root of the frequency in Hertz, and ENB is the effective noise bandwidth. IIN develops a voltage in proportion to the equivalent input impedance seen from the input nodes. Assume the equivalent input impedance is dominated by the parallel combination of the gain setting resistors: R EQ1 = R1R2 R1 + R2 and R EQ2 = R3R4 . R3 + R4 ECM is the RMS noise at the VOCM pin, taking into account the spectral density and bandwidth as with the inputreferred noise sources. Noise current into the VOCM pin will develop a noise voltage across the impedance seen from the node. It is assumed that proper bypassing of the VOCM pin is done to reduce the effective bandwidth, so this voltage is negligible. If this is not the case, the added noise should be added to ECM in a similar manner, as shown below. ER1 through ER4 are the RMS noise voltages from the resistors, calculated by ERn = √4kTR x ENB, where n is the resistor number, k is Boltzmann’s constant (1.38 x 10–23j/K), T is the absolute temperature in Kelvin (K), R is the resistance in ohms (Ω), and ENB is the effective noise bandwidth. E R1 R1 E R2 R2 IIN – – E IN + VOUT+ AF IIN + + – VOUT– VOCM E R3 R4 R3 E R4 ECM EOD is the differential RMS output noise voltage. EOD = A(EID), where EID is the input noise source, and A is the gain from the source to the output. Half of EOD is attributed to the positive output (+EOD/2), and half is attributed to the negative output (–EOD/2). Therefore, (+EOD/2) and (–EOD/2) are correlated to one another and to the input source, and can be directly added together; i.e., ⎛ + EOD ⎞ ⎛ − EOD ⎞ ⎜ ⎟ −⎜ ⎟ = EOD = A( EID ). ⎝ 2 ⎠ ⎝ 2 ⎠ Independent noise sources typically are not correlated. To combine noncorrelated noise voltages, a sum-of-squares technique is used. The total RMS voltage squared is equal to the square of the individual RMS voltages added together. The output noise voltages from the individual noise sources are calculated one at a time and then combined in this fashion. The block diagram shown in Figure 9 helps in analyzing the amplifier’s noise sources. Considering only EIN, from the block diagram we can write: ( − EOD ) β 1 ( + EOD ) β 2 ⎡ − EOD = A F ⎢ EIN + 2 2 ⎣ Figure 9. Block diagram of the amplifier’s input-referred noise ⎤ ⎥. ⎦ Solving yields β2 I IN x REQ1 – E IN + – Σ + VP – VN + EOD 2 AF + I IN x REQ2 β1 – EOD 2 E CM EOD ⎛ ⎞ ⎜ ⎟ ⎛ 2EIN ⎞ ⎜ 1 ⎟. =⎜ ⎟ ⎟ 2 ⎝ β1 + β 2 ⎠ ⎜ ⎜ 1+ ⎟ A F ( β1 + β 2 ) ⎠ ⎝ Assuming AFβ1 >> 1 and AFβ2 >> 1, EOD = 2EIN . (β1 + β 2 ) 52 Analog and Mixed-Signal Products November 2000 Analog Applications Journal Amplifiers: Op Amps Texas Instruments Incorporated Given β1 = β2 = β (symmetrical feedback), EOUT = EIN , β the same as a standard single-ended voltage feedback op amp. Similarly, the noise contributions from IIN x REQ1 and IIN x REQ2 will be 2I IN × R EQ1 (β1 + β 2 ) and 2I IN × R EQ2 (β1 + β 2 ) , respectively. The VOCM error amplifier will produce a common-mode noise voltage at the output equal to ECM. Due to the feedback paths, β1 and β2, a noise voltage is seen at the input that is equal to ECM(β1 – β2 ). This is amplified, just as an input, and seen at the output as a differential noise voltage equal to 2ECM ( β 1 − β 2 ) . (β1 + β 2 ) Noise gain from the VOCM pin ranges from 0 (given β1 = β2) to a maximum absolute value of 2 (given β1 = 1 and β2 = 0, or β1 = 0 and β2 = 1). Noise from resistors R1 and R3 appears like signals at VIN+ and VIN – in Figure 1. From the circuit analysis presented earlier, the differential output noise contribution is 2( E R1 )(1 − β 2 ) and (β1 + β 2 ) 2( E R3 )(1 − β 1 ) (β1 + β 2 ) for each resistor respectively. Noise from resistors R2 and R4 (ER2 and ER4, respectively) is imposed directly on the output with no amplification. Adding the individual noise sources yields the total output differential noise: EOD = ( 2EIN ) 2 + ( 2I IN × R EQ1 ) 2 + ( 2I IN × R EQ2 ) 2 + [ 2ECM ( β 1 − β 2 )] 2 + [ 2( E R1 )(1 − β 2 )] 2 + [ 2( E R3 )(1 − β 1 )] 2 (β1 + β 2 ) 2 The individual noise sources are added in sum-of-squares fashion. Input-referred terms are amplified by the noise gain of the circuit: Gn = 2 . β1 + β 2 If symmetrical feedback is used where β1 = β2 = β, the noise gain is Gn = R 1 = 1+ F , β RG + E R2 2 + E R4 2 . Reference For more information related to this article, you can download an Acrobat Reader file at www-s.ti.com/sc/techlit/ litnumber and replace “litnumber” with the TI Lit. # for the materials listed below. Document Title TI Lit. # 1. Jim Karki, “Fully differential amplifiers,” Analog Applications Journal (August 2000), pp. 38-41 . . . . . . . . . . . . . . . . . . . . . . .slyt165 Related Web site amplifier.ti.com where RF is the feedback resistor and RG is the input resistor, the same as a standard single-ended voltage feedback amplifier. 53 Analog Applications Journal November 2000 Analog and Mixed-Signal Products IMPORTANT NOTICE Texas Instruments Incorporated and its subsidiaries (TI) reserve the right to make corrections, modifications, enhancements, improvements, and other changes to its products and services at any time and to discontinue any product or service without notice. Customers should obtain the latest relevant information before placing orders and should verify that such information is current and complete. All products are sold subject to TI's terms and conditions of sale supplied at the time of order acknowledgment. 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