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HC5523IP датащи(PDF) 9 Page - Intersil Corporation |
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HC5523IP датащи(HTML) 9 Page - Intersil Corporation |
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9 / 18 page ![]() 64 The value of RSG should be calculated to allow maximum loop length operation. This requires that the saturation guard reference voltage be set as high as possible without clipping the incoming or outgoing VF signal. A voltage margin of -4V on tip and -4V on ring, for a total of -8V margin, is recommended as a general guideline. The value of RSG is calculated using Equation 6: where: VBAT = Battery voltage. VMAR = Voltage Margin. Recommended value of -8V to allow a maximum overload level of 3.1V peak. For on-hook transmission RL = ∞, Equation 6 reduces to: SLIC in the Standby Mode Overall system power is saved by configuring the SLIC in the standby state when not in use. In the standby state the tip and ring amplifiers are disabled and internal resistors are connected between tip to ground and ring to VBAT. This connection enables a loop current to flow when the phone goes off-hook. The loop current detector then detects this current and the SLIC is configured in the active mode for voice transmission. The loop current in standby state is calculated as follows: where: IL = Loop current in the standby state. RL = Loop resistance. VBAT = Battery voltage. (AC) Transmission Path SLIC in the Active Mode Figure 16 shows a simplified AC transmission model. Circuit analysis yields the following design equations: where: VTR = Is the AC metallic voltage between tip and ring, including the voltage drop across the fuse resistors RF. VTX = Is the AC metallic voltage. Either at the ground referenced 4-wire side or the SLIC tip and ring terminals. IM = Is the AC metallic current. RF = Is a fuse resistor. ZT = Is used to set the SLIC’s 2-wire impedance. VRX = Is the analog ground referenced receive signal. ZRX = Is used to set the 4-wire to 2-wire gain. EG = Is the AC open circuit voltage. ZL = Is the line impedance. (AC) 2-Wire Impedance The AC 2-wire impedance (ZTR) is the impedance looking into the SLIC, including the fuse resistors, and is calculated as follows: Let VRX = 0. Then from Equation 10 R SG 510 5 • V BAT V MAR – () 1 R DC1 R DC2 + () 600R L ---------------------------------------------- + 16.66V – × ---------------------------------------------------------------------------------------------------------------------------------------------- 17300 – = (EQ. 6) R SG 510 5 • V BAT V MAR 16.66V – – ------------------------------------------------------------------ 17300 – = (EQ. 7) I L V BAT 3V – R L 1800 Ω + -------------------------------- ≈ (EQ. 8) V TR V TX I M 2R F • + = (EQ. 9) V TX Z T ----------- V RX Z RX ----------- + I M 1000 ------------- = (EQ. 10) V TR E G I M Z L • – = (EQ. 11) VTX RSN TIP RING IM ZTR VTR EG VTX IM 1000 VTX ZRX 1 HC5523 RF RF A = 4 + - + - + - + - ZT + - VRX + - A = 250 A = 250 IM ZL FIGURE 16. SIMPLIFIED AC TRANSMISSION CIRCUIT V TX Z T I M 1000 ------------- • = (EQ. 12) HC5523 |
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