Application Report SLYT143 – February 2001 Fully Differential Amplifiers Applications: Line Termination, Driving High-Speed ADCs, and Differential Transmission Lines Jim Karki, Systems Specialist, High-Performance Linear ABSTRACT In high-speed systems, proper line termination requires considering the termination resistors and adjusting the gain-setting resistors to maintain symmetrical feedback. Integrated, fully differential amplifiers are well suited for driving differential ADC inputs. They provide an easy means for anti-alias filtering and for setting the common-mode voltage. 1 2 3 4 5 6 7 8 9 Contents Introduction ................................................................................................................... 2 Terminating the Input Source .............................................................................................. 2 Active Anti-Alias Filtering ................................................................................................... 6 VOCM ............................................................................................................................ 9 Power-Supply Bypass ....................................................................................................... 9 Layout Considerations ...................................................................................................... 9 Using Positive Feedback to Provide Active Termination ............................................................... 9 Conclusion .................................................................................................................. 11 Related Web Sites ......................................................................................................... 11 List of Figures 1 Terminating a Differential Input Signal .................................................................................... 2 2 Differential Termination Impedance ....................................................................................... 2 3 Differential Thevenin Equivalent ........................................................................................... 3 4 Differential Solution for Gain = 1 ........................................................................................... 4 5 Terminating a Single-Ended Input Signal ................................................................................. 4 6 Single-Ended Termination AC Impedance ............................................................................... 4 7 Single-Ended Thevenin Equivalent ........................................................................................ 5 8 Single-Ended Solution for Gain = 1 9 First-Order Active Low-Pass Filter ......................................................................................... 6 10 First-Order Active Low-Pass Filter With Passive Second Pole ........................................................ 7 11 Third-Order Low-Pass Filter Driving an ADC 12 1-MHz, Second-Order Butterworth Low-Pass Filter With Real Pole at 15.9 MHz................................... 9 13 Driving VOCM from ADC Reference Voltage ............................................................................... 9 14 Using Positive Feedback to Provide Active Termination .............................................................. 10 15 Output Waveforms With Active and Standard Termination ........................................................... 11 ....................................................................................... ............................................................................ 6 7 All trademarks are the property of their respective owners. SLYT143 – February 2001 Submit Documentation Feedback Fully Differential Amplifiers Applications: Line Termination, Driving HighSpeed ADCs, and Differential Transmission Lines Copyright © 2001, Texas Instruments Incorporated 1 Introduction 1 www.ti.com Introduction The August 2000 issue of Analog Applications Journal introduced the fully differential amplifiers from Texas Instruments and illustrated their basic operation. The November 2000 issue delved into the topic more deeply, by analyzing gain and noise.This issue investigates some typical applications, such as transmission lines and driving ADC inputs. To simplify calculations and formulas, assume that the amplifier is used at frequencies where the openloop gain is very large (AF >> 1) and will not include its effects in the analysis. The circuit analysis assumes that symmetrical feedback is being used (β1 = β2). Before going into the application circuits, this document will detour briefly into how termination affects the feedback factor, and how to account for it. 2 Terminating the Input Source Double termination is typically used in high-speed systems to reduce transmission-line reflections. With double termination, the transmission line is terminated with the same impedance as the source. Common values are 50 W, 75 W, 100 W, and 600 W. When the source is differential, the termination is placed across the line. When the source is single-ended, the termination is placed from the line to ground. Figure 1 shows an example of terminating a differential signal source. The situation depicted is balanced so that half of VS and half of RS is attributed to each input, with VIC being the center point. RS is the source impedance and Rt is the termination resistor. The circuit is balanced, but there are two issues to resolve: (1) proper termination and (2) gain setting. RS /2 R2 R1 VN –VS 2 Rt VIC +VS 2 – + VOUT+ + – VOUT– VP R /2 S R3 VOCM R4 Figure 1. Terminating a Differential Input Signal As long as AF >> 1 and the amplifier is in linear operation, the action of the amplifier keeps VN ≈ VP. Thus, to first-order approximation, a virtual short is seen between the two nodes as shown in Figure 2. The termination impedance is the parallel combination: Rt || (R1+R3). The value of Rt for proper termination is calculated as shown in Figure 2. R1 Zt Rt R3 Virtual Short Rt = 1 1 1 − RS (R1+ R3) Figure 2. Differential Termination Impedance Once Rt is found, the required gain is found by generating a Thevenin equivalent circuit. The circuit is broken between Rt and the amplifier input resistors R1 and R3. VIC is not a concern at this point, so leave it out. Then: VTH = VS × Rt / (Rt + RS) (1) and RTH = RS || Rt (half is attributed to each side). The resulting Thevenin equivalent is shown in Figure 3. 2 Fully Differential Amplifiers Applications: Line Termination, Driving HighSpeed ADCs, and Differential Transmission Lines Copyright © 2001, Texas Instruments Incorporated SLYT143 – February 2001 Submit Documentation Feedback Terminating the Input Source www.ti.com The proper gain is calculated by: VOUT = VTH RF R ||Rt RG + S 2 where • where VOUT = (VOUT+) – (VOUT–). (2) R2 RS||R t 2 VTH R1 RS||R t 2 R3 – + VOUT+ + – VOUT– R4 VOCM Figure 3. Differential Thevenin Equivalent Substituting for VTH, this becomes: VOUT = VS Rt RF × RS||Rt RS + Rt RG + 2 where • • RF is the feedback resistor (R2 or R4) RG is the input resistor (R1 or R3) (3) Remember: for symmetry, keep the gain equal on the two sides with R2 = R4 and R1 = R3. As an example, suppose you are terminating a 50-Ω differential source that is balanced, and you want an overall gain of 1 from the source to the differential output of the amplifier. Start the design by first choosing the values for R1 and R3, then calculate Rt and the feedback resistors. With the voltage divider formed by the termination, assume that a gain of about 2 will be required in the amplifier. Also, feedback resistor values of approximately 500 Ω are reasonable for a high-speed amplifier. Using these starting assumptions, choose R1 and R3 equal to 249 Ω. Next calculate Rt from the formula: Rt = 1 1 = 55.6 Ω = 1 1 1 1 − − RS ( R1 + R3) 50 (249 + 249) (4) (the closest standard 1% value is 56.2 Ω). Then set the gain by calculating the value of the feedback resistors: RF = ) ( VOUT R ||R t R + Rt × RG + S × S VS Rt 2 ( ) 0 + 56.2 50||56.2 50 = 495.5Ω = 1 × 249 + × 56. 2 2 (5) (the closest standard 1% value is 499 Ω). The solution is shown in Figure 4 with standard 1% resistor values. SLYT143 – February 2001 Submit Documentation Feedback Fully Differential Amplifiers Applications: Line Termination, Driving HighSpeed ADCs, and Differential Transmission Lines Copyright © 2001, Texas Instruments Incorporated 3 Terminating the Input Source www.ti.com 25 499 249 –Vs 2 – + VOUT+ + – VOUT– 56.2 VIC +Vs 2 VOCM 25 499 249 Figure 4. Differential Solution for Gain = 1 Figure 5 shows an example of terminating a single-ended signal source. RS is the source impedance, and Rt is the termination resistor. The circuit is not balanced, so there are three issues to resolve: • Proper termination • Gain setting • Balance . R2 R1 VN VIN RS VS – + VOUT+ + – VOUT– R3 Rt VP VOCM R4 Figure 5. Terminating a Single-Ended Input Signal To determine the termination impedance seen from the line looking into the amplifier’s input at VIN, remove VS and RS and short all other sources. As long as AF >> 1 and the amplifier is in linear operation, the action of the amplifier keeps VN ≈ VP. VN will see the voltage at VOUT+ divided by the resistor ratio R1 / (R1 + R2). Assuming that the amplifier is balanced, VOUT+ = K × VIN / 2 where • K is the closed-loop gain of the amplifier (VOCM = 0) (6) The termination impedance is the parallel combination: Rt in parallel with: VIN = IR3 R3 . K 1− 2 × (1 + K) (7) The value of Rt for proper termination is then calculated as shown in Figure 6. R3 VIN Zt R VP = VIN × K 2 × ( 1+K ) t V − VP IR3 = IN R3 1 Rt = 1 − RS 1− K 2 × (1 + K) R3 Figure 6. Single-Ended Termination AC Impedance 4 Fully Differential Amplifiers Applications: Line Termination, Driving HighSpeed ADCs, and Differential Transmission Lines Copyright © 2001, Texas Instruments Incorporated SLYT143 – February 2001 Submit Documentation Feedback Terminating the Input Source www.ti.com Once Rt is found, the required gain is found by generating a Thevenin equivalent circuit. The circuit is broken between Rt and the amplifier’s input resistor R3. VTH = VS × Rt Rt + RS (8) and RTH = RS || Rt. The resulting Thevenin equivalent is shown in Figure 7. The gain is set by: VOUT/VTH = RF/RG where • • • RF = R2 = R4 RG = R1 = R3 + RS || Rt VOUT = (VOUT+) – (VOUT–) (9) Substituting for VTH, this becomes: VOUT RF Rt = × VS RG RS + Rt (10) Remember, for symmetry: R2 = R4 and R1 = R3 + (RS || Rt). R1 RS||Rt R2 – + VOUT+ + – VOUT– R3 VTH R4 VOCM Figure 7. Single-Ended Thevenin Equivalent As an example, suppose you are terminating a 50-Ω single-ended source and want an overall gain of 1 from the source to the differential output of the amplifier. Start the design by first choosing the value for R3, then calculate Rt and the feedback resistors. This will be an iterative process, starting with some initial assumptions that are then refined. Start with the assumptions that Rt = 50 Ω and that a gain of 2 will be required in the amplifier. Also, feedback resistor values of approximately 500 Ω are reasonable for a high-speed amplifier. Using these starting assumptions, choose R1 = 249 Ω and R3 = R1 – RS || Rt = 249 Ω – 25 Ω = 224 Ω. Next calculate Rt from the formula: 1 Rt = 1 − RS 1− K 2 (1+ K ) R3 1 = 1 − 50 1− 2 2(1 + 2) 224 = 58.7 (11) Then calculate the value of the feedback resistors: R2 = VOUT R + Rt 50 + 58.7 × R1 × S = 1× 249× = 460.9Ω , VS Rt 58. 7 and R4 = VOUT R + Rt × ( R3 + RS||Rt) × S VS Rt = 1 × (224 + 50||58.7)× SLYT143 – February 2001 Submit Documentation Feedback 50 + 58 .7 =464.7 Ω 58. 7 Fully Differential Amplifiers Applications: Line Termination, Driving HighSpeed ADCs, and Differential Transmission Lines Copyright © 2001, Texas Instruments Incorporated (12) 5 Active Anti-Alias Filtering www.ti.com The process is iterative because the gain is not 2, but rather 460.9/249 = 1.85; and Rt is calculated to be 58.7 Ω, not 50 Ω. Iterating through the calculations two more times results in: R3 = 221.9 Ω (the closest standard 1% value is 221 Ω), Rt = 59.0 (which is a standard 1% value), and R2 = R4 = 460.9 (the closest standard 1% value is 464 Ω). The solution is shown in Figure 8 with standard 1% resistor values. 464 249 50 Vs – + VOUT+ + – VOUT– 221 VOCM 59.0 464 Figure 8. Single-Ended Solution for Gain = 1 Using a spreadsheet makes the iterative process very simple. Also, component values can be easily adjusted to find a better fit to the standard available values. 3 Active Anti-Alias Filtering A major application for fully differential amplifiers is lowpass anti-alias filters for ADCs with differential inputs. Creating an active first-order low-pass filter is easily accomplished by adding capacitors in the feedback, as shown in Figure 9. With balanced feedback, the transfer function is: 1 VOUT RF = × VIN R G 1+ j2πf (R FCF ) where • • VOUT = (VOUT+) – (VOUT–) VIN = (VIN+) – (VIN–) (13) The pole created in the transfer function is a real pole on the negative real axis in the s-plane. CF RF VCC 0.1 RG VIN– RG VIN+ + 10 – + VOUT+ + – VOUT– VOCM 0.01 VEE 0.1 0.1 + 10 RF CF Figure 9. First-Order Active Low-Pass Filter 6 Fully Differential Amplifiers Applications: Line Termination, Driving HighSpeed ADCs, and Differential Transmission Lines Copyright © 2001, Texas Instruments Incorporated SLYT143 – February 2001 Submit Documentation Feedback Active Anti-Alias Filtering www.ti.com To create a two-pole low-pass filter, a passive real pole can be created by placing RO and CO in the output, as shown in Figure 10. With balanced feedback, the transfer function is: 1 1 VOUT RF = × × 1+j2πf × 2× R C RG 1+ j2πf (R C ) VIN O O F F where • • VOUT = (VOUT+) – (VOUT–) VIN = (VIN+) – (VIN–) (14) CF RF VCC + 10 0.1 RG 2Co VOUT+ Ro – + + – VIN– Co RG VIN+ Ro VOUT– VOCM 0.01 VEE 0.1 + 0.1 10 2Co RF CF Figure 10. First-Order Active Low-Pass Filter With Passive Second Pole The second pole created in the transfer function is also a real pole on the negative real axis in the s-plane. The capacitor CO can be placed differentially across the outputs as shown in solid lines; or two capacitors (of twice the value) can be placed between each output and ground as shown in dashed lines. Typically, RO will be a low value; and, at frequencies above the pole frequency, the series combination with CO will load the amplifier. The extra loading will cause extra distortion in the amplifier’s output. To avoid this, you might stagger the poles so that the ROCO pole is placed at a higher frequency than the RFCF pole. The classic filter types like Butterworth, Bessel, Chebyshev, and so forth (second-order and greater), cannot be realized by real poles—they require complex poles. The multiple feedback (MFB) topology creates a complex pole pair, and is easily adapted to fully differential amplifiers, as shown in Figure 11. A third-order filter is formed by adding R4s and C3 at the output. R2 C1 2C2 VCC 0.1 + 10 2C3 R1 R3 R4 – VIN– VOUT+ + VIN + C3 R1 R3 R4 – + VIN+ VOUT– VIN – VCM VOCM 0.01 VEE 2C2 0.1 + 10 0.1 2C3 C1 R2 Figure 11. Third-Order Low-Pass Filter Driving an ADC SLYT143 – February 2001 Submit Documentation Feedback Fully Differential Amplifiers Applications: Line Termination, Driving HighSpeed ADCs, and Differential Transmission Lines Copyright © 2001, Texas Instruments Incorporated 7 Active Anti-Alias Filtering www.ti.com Capacitors C2 and C3 can be placed differentially across the inputs and outputs, as shown in solid lines. Alternatively, for better common-mode noise rejection, two capacitors of twice the value can be placed between each input or output and ground, as shown in dashed lines. The transfer function for this filter circuit is: [ VOUT = VIN K 2 f 1 jf (FSF× f ) +Q FSF× f − ] × C C +1 1 + j2πf × 2 × R4C3 where • • VOUT = (VOUT+) – (VOUT–) VIN = (VIN+) – (VIN–) (15) 1 R2 K= , FSF× fC = , R1 2π 2 × R2R3C1C2 and Q= 2 × R2R3C1C2 . R 3C1+ R2C1+ KR3 C1 (16) K sets the pass-band gain, fC is the cut-off frequency of the filter, FSF is a frequency scaling factor, and Q is the quality factor. 2 2 FSF = Re + Im and Q = 2 Re + Im 2 2Re where • • Re is the real part of the complex pole pair Im is the imaginary part (17) Setting R2 = R, R3 = mR, C1 = C, and C2 = nC results in: FSF× fC = Q= 1 and 2πRC 2 × mn 2 × mn 1 + m (1 − K) (18) Start by determining the ratios, m and n, required for the gain and Q of the filter type being designed, then select C and calculate R for the desired fC. R4 and C3 are chosen to set the real pole in a third-order filter. Exercise care when setting this pole. Typically, R4 will be a low value; and, at frequencies above the pole frequency, the series combination with C3 loads the amplifier. The extra loading causes extra distortion in the amplifier output. To avoid this, place the real pole at a higher frequency than the cut-off frequency of the complex pole pair. Figure 12 shows the gain and phase response of a second-order Butterworth low-pass filter, with corner frequency set at 1 MHz and the real pole set by R4 and C3 at 15.9 MHz. The components used are: R1 = 787 Ω, R2 = 787 Ω, R3 = 732 Ω, R4 = 50 Ω, C1 = 100 pF, C2 = 220 pF, C3 = 100 pF, and the THS4141 fully differential amplifier. At higher frequencies, parasitic elements allow the signal to feed through. 8 Fully Differential Amplifiers Applications: Line Termination, Driving HighSpeed ADCs, and Differential Transmission Lines Copyright © 2001, Texas Instruments Incorporated SLYT143 – February 2001 Submit Documentation Feedback VOCM www.ti.com 0 0 Gain (dB) -20 -90 Phase -40 -180 -60 -270 -80 100 k 1M 10 M 100 M Phase (degrees) Gain -360 1G Frequency (Hz) Figure 12. 1-MHz, Second-Order Butterworth Low-Pass Filter With Real Pole at 15.9 MHz 4 VOCM The proper VOCM is provided as an output by many ADCs with differential inputs. Typically, all that needs to be done is to provide bypass capacitors; 0.1 μF and 0.01 μF are useful choices. If VOCM is not provided, it can be created by forming a summing node with the plus and minus reference voltages of the ADC to drive VOCM, as shown in Figure 13. The voltage at the summing node is the midpoint value between +VREF and –VREF. Depending on the loading of the VOCM input, the summing node voltage may need to be buffered. +VREF R C Summing node to VOCM R C –VREF Figure 13. Driving VOCM from ADC Reference Voltage 5 Power-Supply Bypass Each power rail should have 6.8-μF to 10-μF tantalum capacitors located within a few inches of the amplifier, to provide low-frequency power-supply bypassing. A 0.01-μF to 0.1-μF ceramic capacitor should be placed within 0.1 inch of each power pin on the amplifier to provide high-frequency power-supply bypassing. 6 Layout Considerations As with all high-speed amplifiers, minimize parasitic capacitance at the input of the amplifier by removing the ground plane near the pins and near any circuit traces. Also, make trace routing as direct as possible, and use surface-mount components. 7 Using Positive Feedback to Provide Active Termination Driving transmission lines differentially is a typical use for fully differential amplifiers. Using positive feedback with amplifiers can provide active termination, as shown in Figure 14. Because of the positive feedback, the output line impedance appears larger than the value of output resistor RO. The voltage dropped across the resistor depends on its actual value, resulting in increased efficiency. SLYT143 – February 2001 Submit Documentation Feedback Fully Differential Amplifiers Applications: Line Termination, Driving HighSpeed ADCs, and Differential Transmission Lines Copyright © 2001, Texas Instruments Incorporated 9 Using Positive Feedback to Provide Active Termination www.ti.com RP RF VCC VO+ 0.1 RG VIN– IOUT+ ZO ZO+ RO – + RG VOUT+ ZO– RO – + VIN+ + 10 R t VOUT– VOCM IOUT– VEE 0.1 RF +10 VO – RP Figure 14. Using Positive Feedback to Provide Active Termination Use symmetrical feedback with this application. With double termination, the output impedance of the amplifier, ZO, equals the characteristic impedance of the transmission line; and the far end of the line will be terminated with the same value resistor, that is, Rt = ZO. For proper balance, half of ZO is placed in each half of the differential output, so that ZO = 2 x ZO±. To calculate the output impedance, ground the inputs and insert either a voltage or current source between VOUT+ and VOUT–. Due to symmetry, ZO+ = ZO–, VOUT+ = –(VOUT–), and VO+ = –(VO–). Calculating the impedance of one side provides the solution. VOUT + (V + ) − ( VO+) || RP,IOUT+ = OUT , IOUT + RO ( ZO + = ) ( ) and VO+ = ( VOUT−) × −RF RP (19) The output impedance of the amplifier on each side of the line is RO divided by 1 minus the gain from the opposite line: ZO ± = ( RO ||RP R 1− F RP ) (20) The positive feedback also affects the forward gain. Accounting for this effect and the voltage divider between RO and Rt || 2RP, the gain from VIN = (VIN+) – (VIN–) to VOUT = (VOUT+) – (VOUT–) is: A= VOUT RF 1 = × VIN RG 2RO + Rt||2 RP RF − Rt||2RP RP (21) Design is easily accomplished if by first choosing the value of RF and RO. Then, calculate the required value of RP to give the desired ZO. Then calculate RG for the required gain. For example: if you want a gain of 1, and to terminate a 100-Ω line properly with RF = 1 kΩ and RO = 10 Ω, the proper value for ZO and Rt is 100 Ω (ZO± = 50 Ω). Rearranging Equation 20 yields: RP = 10 RF − RO 990 Ω = = 1.24 kΩ RO 10 Ω 1− 1− ZO ± 50 Ω Fully Differential Amplifiers Applications: Line Termination, Driving HighSpeed ADCs, and Differential Transmission Lines Copyright © 2001, Texas Instruments Incorporated (22) SLYT143 – February 2001 Submit Documentation Feedback Conclusion www.ti.com Then, rearranging Equation 21 gives: RG = = RF 1 × A 2RO + Rt||2 RP RF − Rt||2RP RP 1 kΩ = 2.49 kΩ 20 Ω + 100Ω||2.48 kΩ 1 kΩ − 1.25 kΩ 100Ω||2.48 kΩ (23) The circuit is built and tested with the nearest standard values to those previously computed: RF = 1 kΩ, RP = 1.24 kΩ, RG = 2.49 kΩ, Rt = 100 Ω, and RO = 10 Ω. Compare the output voltage waveforms (VOUT = 2VPP) with active termination and standard termination shown in Figure 15: [VO = (VO+) – (VO–), and VOUT = (VOUT+) – (VOUT–)]. 2V VO with standard termination 1.5 V VO with active termination 1V 0.5 V VOUT 0V -0.5 V -1 V -1.5 V -2 V Figure 15. Output Waveforms With Active and Standard Termination For standard termination, RF = 1 kΩ, RP = open, RG = 499 Ω, Rt = 100 Ω, and RO = 50 Ω. With standard termination, 20 mW of power is dissipated in the output resistors, as opposed to 6.25 mW with active termination, which wastes 69% less power. Another attractive feature about active termination, especially in low-voltage applications, is the effective increase in output-voltage swing for a given supply voltage. 8 Conclusion In high-speed systems, proper line termination requires considering the termination resistors and adjusting the gain-setting resistors to maintain symmetrical feedback. Integrated, fully differential amplifiers are well suited for driving differential ADC inputs. They provide an easy means for anti-alias filtering and for setting the common-mode voltage. Integrated, fully differential amplifiers are also well suited for driving differential transmission lines, and active termination provides for increased efficiency. 9 Related Web Sites www-s.ti.com/sc/techlit/slyt018 amplifier.ti.com SLYT143 – February 2001 Submit Documentation Feedback Fully Differential Amplifiers Applications: Line Termination, Driving HighSpeed ADCs, and Differential Transmission Lines Copyright © 2001, Texas Instruments Incorporated 11 IMPORTANT NOTICE Texas Instruments Incorporated and its subsidiaries (TI) reserve the right to make corrections, enhancements, improvements and other changes to its semiconductor products and services per JESD46, latest issue, and to discontinue any product or service per JESD48, latest issue. Buyers should obtain the latest relevant information before placing orders and should verify that such information is current and complete. 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