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The DiSEqC™ 2.x protocol supports two way communications for receiver or controller and accessory or peripheral modes. Figure 1-1 shows a typical DiSEqC 2.x implementation using TPS65235.
In accessory or peripheral mode a 22-kHz tone is received at VOUT. The tone is blocked from reaching VLNB by the LCR filter. This is a band stop filter presenting 15-Ω impedance at 22 kHz. The tone is passed via the 10 nF, 10-kΩ RC filter to the DIN pin. The DC current for the LNB power flows from VLNB to VOUT via the 220-µH inductor which has low DC resistance.
In receiver or controller mode, the LNB power supply transmits a 22-kHz tone on VOUT. When transmitting, a FET is turned on to bypass the LCR filter and avoid unwanted attenuation of the 22-kHz tone. The lowest resistance DC path is through this FET when on. However, when the FET is turned off, the current cannot ramp instantly in the inductor and initially the current flows through the 15-Ω resistor. The voltage dropped across the resistor leads to a sudden drop or negative spike in the VOUT signal, which subsequently decreases as the inductor current ramps. For a 600-mA LNB current, the drop across the resistor could be as high as 9 V but it is clamped at approximately 700 mV by the body diode of the FET as seen in Figure 1-2. This negative spike can be incorrectly interpreted by some LNBs and can lead to incorrect communication and missed commands.
To avoid this negative spike, a couple of options are available. One is to add a Schottky diode across the FET as shown in Figure 2-1. This diode clamps the spike similar to the FET body diode, but with a lower forward voltage. Figure 2-2 shows the output with a B320A-13-F Schottky diode added. The spike is still present but is reduced to approximately 330 mV.
Another solution is to add a DC blocking capacitor in series with the FET as shown in Figure 2-3. In this way, when the FET is turned on, the 22-kHz tone passes through the capacitor whose impedance is approximately 330 mΩ at 22 kHz. However, the DC current is blocked and continues to flow via the inductor. So when the FET is turned off there is no change in the DC current flow and no negative spike is generated (Figure 2-4). A 22-µF, 25-V with 0805 capacitor was used for this test, but similar results were achieved with a 10-µF, 25-V capacitor. In normal working conditions, the voltage across the capacitor is less than 2 V but, as the output may be subjected to surges, a 25-V rated capacitor is recommended.
Figure 3-1 shows the equivalent circuit of the typical implementation, where VLNB is the output voltage of VLNB pin and it is equivalent to a voltage source supplying to the LRC network and the load resistor Ro. Vo is the output voltage, io is the load current, isw is the current of the FET and Vsw is the voltage drop across the FET.
To reflect the attenuation of data transmission, the transfer function of Vo to VLNB is defined as follows in Equation 1
When the FET is on, assuming all component are ideal, then isw = io and Vo = VLNB. Therefore, G(s) = 1, which means there is no attenuation of the 22-kHz tone.
When the FET is off, isw = 0, the tone signal is attenuated due to LCR network. And G(s) can be expressed as Equation 2.
Where Z(s) can be expressed as Equation 3:
Hence the attenuation of 22-kHz tone can be expressed as Equation 4:
Figure 3-2 shows the simulation waveforms of the equivalent circuit, where GDR is the gate drive signal of the FET. It can be seen from the waveforms that there is no attenuation of the tone signal when FET is on, and when the FET is off, the attenuation of 22-kHz tone is –4.437 dB.
In practice, when the 22-kHz tone ends, VLNB is a constant voltage, but the FET will not turn off at once. So there is a constant current flowing through the FET and isw = io. After a fixed delay time tdelay, the FET turns off and isw = 0. The switching action causes a current transient ΔIsw to the FET path. At the moment the FET turns off, there will be a voltage drop across the FET due to the current transient. To better understand, the control circuit of the typical implementation can be represented by the blocks in Figure 3-3 based on Figure 3-1.
The transfer function of Vsw(s) to isw(s) can be generated as Equation 5:
The inverse Laplace transform of Equation 5 allows the voltage to be expressed as Equation 6.
where: x1 and x2 can be expressed as Equation 7 and Equation 8.
The maximum value of the voltage drop can be calculated at the extremum point tEP when the derivative of vsw(t) equals 0, and can be expressed as Equation 9:
According to Equation 9, tEP becomes Equation 10:
Considering that Ro = 22.3 Ω, so ΔIsw = 0.6 A, then Vsw_dropmax can be expressed as Equation 11:
Figure 3-4 shows the simulation results of the voltage drop across the FET, and it can be found that Vsw_dropmax = 4.59 V.
However, in practice, the voltage drop is clamped by the Vf of body diode of the FET. So the voltage drop Vsw_drop = min[Vf, Vsw_dropmax].
As previously described, adding a capacitor in series with the FET blocks the DC current path. The DC current flows continuously in the inductor and there is no transient dip when the FET is turned off. Figure 3-5 shows the equivalent circuit with the added capacitor.
The value of the capacitor will affect the attenuation of 22-kHz tone when FET is on, and G(s) will be changed as Equation 12:
where Z1(s) can be expressed as Equation 13:
Hence the attenuation of the 22-kHz tone can expressed as Equation 14:
Based on Equation 14, Cadd can be selected to meet the maximum attenuation. For example, if the amplitude of 22-kHz tone is 650 mV, and maximum attenuation is 100 mV, then Cadd of at least 1 µF is required. Figure 3-6 shows the experimental results for 1-µF Cadd. Results for a 22-µF Cadd have already been seen in Figure 2-4.