HC55185 Ringing SLIC & the AK2306/2306LV Dual PCM CODEC

Transcription

1 TM HC55185 inging SLC & the AK206/206LV Dual PCM CODEC Application Note December 2001 AN9991 Author: Don LaFontaine eference Design using the HC55185 and the AK206/206LV Dual PCM CODEC The purpose of this application note is to provide a reference design for the HC55185 and AK206/206LV Dual PCM CODEC. The network requirements of many countries require the analog subscriber line circuit (SLC to terminate the subscriber line with an impedance for voiceband frequencies which is complex, rather than resistive (e.g. 600Ω. The HC55185 accomplishes this impedance matching with a single network ( Figure 1 connected between the VTX pin and the N pin. The AK206/206LV Dual PCM CODEC includes Selectable Alaw/µlaw function, nternal Gain Adjustment from 6dB to 18dB by 1dB steps control and a selectable 16Hz/20Hz ing Tone Generator. Discussed in this application note are the following: 2wire impedance matching eceive gain (4wire to 2wire and transmit gain (2wire to 4wire calculations eference design for both 600Ω and nF Complex mpedance mpedance Matching mpedance matching of the HC55185 to the subscriber load is important for optimization of 2 wire return loss, which in turn cuts down on echoes in the end to end voice communication path. mpedance matching of the HC55185 is accomplished by making the SLC s impedance (, Figure 1 equal to the desired terminating impedance, minus the value of the protection resistors ( P. The formula to calculate the proper for matching the 2wire impedance is shown in Equation 1. = 1. ( 2 P (EQ. 1 Equation 1 can be used to match the impedance of the SLC and the protection resistors (Z T to any known line impedance (. Figure 1 shows the calculations of to match a resistive and 2 complex loads. EXAMPLE 1: Calculate to make Z T = 600Ω in series with 2.16µF. P =. 1 = jω2.16x10 6 ( 2 ( 49 (EQ. 2 = 66.9kΩ in series with 16.2nF. Note: Some impedance models, with a series capacitor, will cause the op amp feedback to behave as an open circuit DC. A resistor with a value of about 10 times the reactance of the capacitor (2.16µF/1. = 16.2nF at the low frequency of interest (200Hz for example can be placed in parallel with the capacitor in order to solve the problem (491kΩ for a 16.2nF capacitor. EXAMPLE 2: Calculate to make Z T = //115nF P =. 820 Z T = jω820( 115X10 9 ( 2 ( 49 (EQ. = 16.26kΩ in series with the parallel combination of 109.kΩ and 862pF. V2W E G P P ZT V T NTESL HC55185 NG V X ESSTVE = Z T = 600Ω = 1.(600 2*49 C TX 66.9kΩ STD VALUE 66.5kΩ COMPLEX = Z T = 600Ω 2.16µF = 1.(600 2* µF/ kΩ 491kΩ 16.2nF COMPLEX = Z T = 2 820//115nF = 1.(220 2*49 1.(820 // 115nF/1. Z T 16.26kΩ = 2 P N 862pF 109.kΩ FGUE 1. MPEDANCE MATCHNG NTESL or ntersil (and design is a trademark of ntersil Americas nc. Copyright ntersil Americas nc All ights eserved

2 SLC in the Active Mode Figure 2 shows a simplified AC transmission model of the HC55185 and the connection of the AK206 to the SLC. Figure shows a simplified AC transmission model of the HC55185 and the connection of the Low Voltage AK206LV to the SLC. The Low Voltage AK206LV CODEC requires a different connection to the HC55185 to achieve the voltage gain required at tip and ring without clipping the output signal of the CODEC. The following analysis is performed with the AK206 CODEC connection. Circuit analysis of the Low Voltage circuit is left for the reader. Circuit analysis of the HC55185 yields the following design equations: The Sense Amplifier is configured as a 4 input differential amplifier with a gain of /4. The voltage at the output of the sense amplifier ( is calculated using superposition. 1 is the voltage resulting from V1, 2 is the voltage resulting from V2 and so on (reference Figure 2. 1 = ( V 4 1 (EQ. 4 2 = ( V 4 2 (EQ. 5 = ( V 4 (EQ. 6 4 = ( V 4 4 (EQ. 7 = [( V 1 ( V 4 V ] 4 = [ V V] 4 (EQ. 8 Where V is equal to M ENSE (ENSE = = 2( M 20 4 = M 0 (EQ. 9 The voltage at VTX is equal to: V SA S = = 8K 8K M 0 (EQ. 10 V T is defined in Figure 2, note polarity assigned to V T V T = 2V ( X (EQ. 11 Setting V X equal to zero, substituting Equation 10 into Equation 11 and defining = V T / M will enable the user to determine the require feedback to match the line impedance at. 1 = 1. S (EQ. 12 is the source impedance of the device and is defined as = 2 p. ZL is the line impedance. is defined as: = 1.( 2 P (EQ. 1 Node Equation V X = at HC55185 V X input (EQ. 14 Substitute Equation 10 into Equation 14 V X M 0 = 8K (EQ. 15 Loop Equation V T = 0 Substitute Equation 15 into Equation 16 Substitute Equation 12 for and w / for M into Equation 17. Substitute Equation 18 into Equation 19 and combine terms where: V X = The input voltage at the V X pin. = An internal node voltage that is a function of the loop current and the output of the Sense Amplifier. = nternal current in the SLC that is the difference between the input receive current and the feedback current. M = The AC metallic current. P = A protection resistor (typical 49.9Ω. = An external resistor/network for matching the line impedance. V T = The tip to ring voltage at the output pins of the SLC. = The tip to ring voltage including the voltage across the protection resistors. = The line impedance. at HC55185 feed amplifiers and load = The source impedance of the device. HC55185 eceive Gain (V X to (EQ. 16 M 60 V T = 2V X 8K (EQ. 17 V T = 2V X (EQ. 18 Loop Equation at Tip/ing interface M 2 P V T = 0 (EQ P = 2V (EQ. 20 Z X L 4wire to 2wire gain across the HC55185 is equal to the divided by the input voltage V X, reference Figure 2. The receive gain is calculated using Equation 20. Equation 21 expresses the receive gain (V X to in terms of network impedances. From Equation 1, the value of was set to match the line impedance ( to the HC55185 plus the protection resistors (Z 0 P. This results in a 4wire to 2wire gain of 1, as shown in Equation 21. G 42 = = 2 V X Z L 2 2 = = 1 (EQ. 21 P 2

3 M E G NG M P V T M P Z 0 ENSE M ENSE V V4 M V 1 AMPLFE AMPLFE NTESL HC55185 (1 OF 2 ECEVE BLOCK 1:1 TA BACK AMPLFE V N V X CX V X K CTX VO 7 VF 6 GS A F GST AK206 ECEVE PATH AMP AMPLFE PCM /F AMPT AMPLFE TANSMT PATH D FS BCLK K 66.5K N 4 = M 0 SENSE AMPLFE FGUE 2. HC55185 SMPLFED AC TANSMSSON CCUT AND AK206 M M P V T E G M NG P Z 0 ENSE M ENSE V V4 M V 1 AMPLFE AMPLFE NTESL HC55185 (1 OF ECEVE BLOCK 1:1 TA BACK AMPLFE 8K = M 0 SENSE AMPLFE V N V X CX V X 8 6.5K CTX 66.5K N N 45.K CN 7 6 A 42.2K F VO VF GS GST 0.1K AK206LV AK206LV ECEVE PATH AMP AMPLFE PCM /F AMPT AMPLFE TANSMT PATH D FS BCLK ECEVE FOM D TO T/ S.dB TANSMT FOM T/ TO S 9.dB LOW VOLTAGE CONNECTON FGUE. HC55185 SMPLFED AC TANSMSSON CCUT AND AK206LV

4 eceive Gain Across the System The receive gain across the system is defined as the gain from D to the phone (. With the receive gain through the HC55185 set to 1, the receive gain across the system is entirely controlled by programming the AK206. The AK206 can program the receive gain across the system from 6dB to 18dB in 1 db increments (reference Figure 4. f more precise gain increments are required, the AMP amplifier can be used to adjust the overall eceive gain (6/7. Transmit Gain Across HC55185 (E G to The 2wire to 4wire gain is equal to /E G with V X = 0, reference Figure 2. Loop Equation E G M 2 P M V T = 0 From Equation 18 with V X = 0 Substituting Equation 2 into Equation 22 and simplifying. Substituting Equation 12 into Equation 10 and defining M = / results in Equation 25 for VTX. (EQ. 22 V T = (EQ. 2 2 P E G = (EQ P = (EQ Combining Equations 24 and 25 results in Equation 26. G 24 = 2 P (EQ. 26 = = E G 2Z ( L 2 P 2Z ( L 2 P A more useful form of the equation is rewritten in terms of /. A voltage divider equation is written to convert from E G to as shown in Equation P (EQ. 27 = EG 2 P Substituting = 2 P and rearranging Equation 27 in terms of E G results in Equation 28. E G = 2 (EQ. 28 Substituting Equation 28 into Equation 26 results in an equation for 2wire to 4wire gain that s a function of the synthesized input impedance of the SLC and the protection resistors. G 24 = = = (EQ. 29 ( 2 P is set to 600Ω, is programmed with to be Ω (66.5kΩ/1., and P is equal to 49.9Ω. This results in a 2wire to 4wire gain of or 7.6dB. Transmit Gain Across the System The transmit gain across the system is defined as the gain from the phone or 2wire side ( to the PCM highway (. Setting the gain of the AK206 will have to account for the attenuated signal through the HC The system gain is entirely controlled by programming the AK206. The AK206 can program the transmit gain across the system from 6dB to 18dB in 1 db increments (reference Figure 4. f more precise gain increments are required, the AMPT amplifier can be used to adjust the overall Transmit gain (f/8. V2W L EGZ = 2 P P P ZT _ V T NTESL HC55185 (1 OF 2 V X V NG TX N 66.5K 4.7µF CX K CTX VO 7 VF 6 GS A F GST AK206 ECEVE PATH 6dB to 18dB AMP AMPLFE TANSMT PATH 6dB to 18dB AMPT AMPLFE D FGUE 4. ECEVE G(42, TANSMT (24 4

5 Transhybrid Balance G(44 Transhybrid balance is a measure of how well the input signal is canceled (that being received by the SLC from the transmit signal (that being transmitted from the SLC to the CODEC. Without this function, voice communication would be difficult because of the echo. The signals at V GS and (Figure 4 are opposite in phase. Transhybrid balance is achieved by summing two signals that are equal in magnitude and opposite in phase into the AMPT amplifier inside the AK206. Transhybrid balance is achieved by summing the V GS signal with the output signal from the HC55185 when proper gain adjustments are made to match V GS and magnitudes. For discussion purpose, the AMPT amplifier is redrawn with the external resistors in Figure 5. V GS A 8 AMPT Transhybrid balance is achieved by adjusting the magnitude from both and V GS so their equal to each other. The gain across the system is set by the gain through the SLC (0.416 and the AMPT amplifier through F/8. F is randomly selected to be 120kΩ. To achieve a gain F FGUE 5. TANSHYBD BALANCE CCUT GAOT across the system, with the transmit gain of the AK206 set to, we set 8 equal to 49.9kΩ. as shown in Equation 0. G VTX G F k = = G = 0.416( = k (EQ. 0 The gain through the AMPT amplifier from V GS must equal the gain from to achieve transhybrid balance. A is therefore equal to F, as shown in Equation 1. F G VGS = V GS A = V 120k = 1 GS 120k (EQ. 1 eference Design of the HC55185 and the AK206 With a 600Ω Load The design criteria is as follows: 4wire to 2wire gain (D to equal 2wire to 4wire gain ( to equal p = 49.9Ω Figure 6 gives the reference design using the ntersil HC55185 and the AK206 Dual PCM CODEC. Also shown in Figure 6 are the voltage levels at specific points in the circuit. mpedance Matching The 2wire impedance is matched to the line impedance Z 0 using Equation 1, repeated here in Equation 2. = 1. ( 2 P (EQ. 2 For a line impedance of 600Ω, equals: = 1. ( = 66.9kΩ (EQ. The closest standard value for would be 66.5kΩ. V2W L EGZ m0 (600Ω V MS P P ZT = 2 P V T SYSTEM EQUEMENTS: MPEDANCE: 600Ω TANSMT (A/D: 5. ECEVE (D/A: NTESL HC55185 (1 OF 8 NG V X 66.5K G42 m0(600ω V MS CX CTX N m0 (600Ω V MS 4.7µF 7.619dBm0 (600Ω V MS m0 (600Ω V MS m0 (600Ω V MS G K 7 6 A F VO VF GS GST AK206 ECEVE PATH AMP AMPLFE AMPT AMPLFE TANSMT PATH FGUE 6. EFEENCE DESGN OF THE HC55185 AND THE AK206/206LV WTH A 600Ω LOAD MPEDANCE m0 (600Ω V MS D PCM BUS 5

6 eference Design of the HC55185 and the AK206 With a Complex Load The design criteria for a Complex load solution are as follows: Desired line circuit impedance is //115nF eceive gain / D is.5db Transmit gain / is m0 is defined as 1mW into the complex impedance at 1020Hz p = 49.9Ω Figure 7 gives the reference design using the ntersil HC55185 and the AK206 Dual PCM CODEC. Also shown in Figure 7 are the voltage levels at specific points in the circuit. Note: The transmit gain of the system is (1.79dB (897Ω =.5dB (600Ω as explained in the following section. Adjustment to Get.5dBm0 at the Load eferenced to 600Ω The voltage equivalent to m0 into 897Ω (m0 (897Ω is calculated using Equation 4 (897Ω is the impedance of complex load at 1020Hz. m ( 897Ω = 10log = V 897( MS (EQ. 4 The gain referenced back to m0 (600Ω is equal to: V MS = 20log = 1.747dB (EQ V MS The adjustment to get.5dbm0 at the load referenced to 600Ω is: Adjustment =.5dBm dBm0 = 1.75dB (EQ. 6 The voltage at the load (referenced to 600Ω is given in Equation dBm ( 600Ω = 10log = 0.606V (EQ ( MS Setting the eceive Path Gain equal to 1dB and adjusting 6/7 with standard resistor values results in a voltage of Vrms or 1.7m0 (600Ω. V2W L EGZ = 2 P 1.79dBm0 (600Ω V MS P P ZT V T 1.79dBm0 (600Ω NG NTESL HC55185 V X N 1.79dBm0(600Ω V MS 90.9kΩ 4.7µF G42 CX CTX 1dBm0 (600Ω V MS 8 1K 9.41dBm0 (600Ω V MS K A 1.2K F VO VF GS GST 25.5K.56dBm0 (600Ω V MS AK206 ECEVE PATH 1dB AMP AMPLFE AMPT AMPLFE TANSMT PATH.56dBm0 (600Ω V MS m0 (600Ω V MS D PCM BUS V MS FGUE 7. EFEENCE DESGN OF THE HC55185 AND THE AK206 WTH A COMPLEX LOAD MPEDANCE All ntersil U.S. products are manufactured, assembled and tested utilizing SO9000 quality systems. ntersil Corporation s quality certifications can be viewed at ntersil products are sold by description only. ntersil Corporation reserves the right to make changes in circuit design, software and/or specifications at any time without notice. Accordingly, the reader is cautioned to verify that data sheets are current before placing orders. nformation furnished by ntersil is believed to be accurate and reliable. However, no responsibility is assumed by ntersil or its subsidiaries for its use; nor for any infringements of patents or other rights of third parties which may result from its use. No license is granted by implication or otherwise under any patent or patent rights of ntersil or its subsidiaries. For information regarding ntersil Corporation and its products, see 6