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CCS/AFE031: AFE031

Part Number: AFE031

Tool/software: Code Composer Studio

Hi People,

  I have a doubt on coupling the signal into powerline (phase and neutral).I have a circuit .plz help me to decode the circuit  and also help me to choose the component .

 my FSK signal 2Vrms,0.3A Irms

I need to calculate the resistor wattage and package of this ciruit and capacitor also,

plz help me its very immediate requirement 

I need to couple the signal to power line.

  • Hi Gobinath,

    Q: I need to calculate the resistor wattage and package of this circuit and capacitor also.

    I need more information from you to calculate the wattage of resistor in above circuit. 

    1. What is rail voltage of AFE031 side?

    2. What is the AC line voltage and frequency at N and P node? what types of load on the other side of N and P?

    3. What was the transformer ratios and types of transformer? I need to know the worst case scenario during operating, including transient conditions?

    4. What are the symbols in JP2 and JP3? are they jumpers?

    5. which side of R and C calculations do you need help?

    6.  My FSK signal 2Vrms,0.3A Irms, what is FSK's operating frequency? Are you using CENELEC B/C/D bands or something else? 

    I am asking the above because I have questions about the values of certain components. For instance, 150nF capacitor, if the FSK transmission frequency is 150kHz, the impedance will be 1/sC or 7.1Ohm. In our reference design, we use 10uF or 0.11Ohm. In other words, you will have more FSK signal losses across 150nF at 150kHz. 

    Enclosed is our reference design, see the link below. 

    https://www.ti.com.cn/cn/lit/ug/sbou223/sbou223.pdf?ts=1605023416512&ref_url=https%253A%252F%252Fwww.google.com%252F

    Best,

    Raymond

  • 1. rail voltage   ===>2Vrms ,0.3A

    2. AC line voltage 230V and 50HZ frequency. load means like motor,TV.I am just transmitting the signal and receiving the signal other using same boards.

    3 . transformer ration is 1:1 and audio or signal transformer./That D6 diode used to clamp the voltage to 230v to 6.1V,The transformer get 6.1 at the input side the other side also having the same voltage signal 

    4. yes, they are jumpers to activate passive receive filter  to make them active R18,L7,C16,C20 [At the time of receiving JP1 2 and 3 and JP2 ,2 and 3 must be closed . 

    5. both side of the filter calculation and package selection, component  withstand voltage  because 230V how much withstand voltage resistor and cap going to use .no idea about that so need help to design both side.

    6. yes, CENELEC B/C/D bands I used to transmit the signal. Currently I am using 132 KHZ signal  for TX and RX

     and how you calculate that 150nF GIVES THAT MUCH LOSS .and plz help me to design the coupling circuit and how to calculate them .because i am very beginner to AC signals.

                                        thank you

    Regards,

    Gobinath P 

         

     

  • Hi Gobinath,

    1. rail voltage   ===>2Vrms ,0.3A

    I was asking about the the supply voltage rail for AFE031 PLC controller (12-24Vdc single supply rail for AFE031). The output FSK's amplitude is 2.83Vp-p and 0.3Arms. I would like to know the worst case scenario where output signals and across the transformer. 

    The voltage, current and impedance will couple the transmission signal from primary to secondary. The reverse is also true. The 230Vrms, peak current and impedance at secondary windings can also reflect back to the primary side. So we need to minimize and reduce the signal reflection through the transformer windings. Therefore, I will need all these parameters to calculate the component ratings. 

    2. AC line voltage 230V and 50HZ frequency. load means like motor,TV.I am just transmitting the signal and receiving the signal other using same boards.

    ok, I understood. 

    3 . transformer ration is 1:1 and audio or signal transformer./That D6 diode used to clamp the voltage to 230v to 6.1V,The transformer get 6.1 at the input side the other side also having the same voltage signal 

    The link below is the transformer that we recommended for PLC application.  A regular iron/Si magnetic core  for audio application is not going to work, because it has too much core losses due to hysteresis and eddy currents losses (please check on the frequency response of your transformer). You will need ferrite based magnetic core material to transmit high frequency signals (e.g. 143.75kHz and 131.25kHz) with minimum magnetic core losses.  

    https://www.we-online.de/katalog/datasheet/750510476.pdf

    4. yes, they are jumpers to activate passive receive filter  to make them active R18,L7,C16,C20 [At the time of receiving JP1 2 and 3 and JP2 ,2 and 3 must be closed . 

    Our reference design is tailored for both Tx and Rx signals. No switches or jumpers are needed. PLC communication works in Half-Duplex mode. Although the Tx/Rx can be sent bidirectionally, it can only transmit or receive modulated signal at direction at a time. Full-Duplex mode can transmit and receive at the same time. Examples are landline telephone communication is operating in Full Duplex mode. 

    5. both side of the filter calculation and package selection, component  withstand voltage  because 230V how much withstand voltage resistor and cap going to use .no idea about that so need help to design both side.

    I will help you to calculate the component value once I have a complete design information. From the BOM of the reference design, only certain components have to withstand 230Vrms and current ratings. The AFE031 EVM reference design is tailored for 115/230Vrms AC coupling applications.   

    6. yes, CENELEC B/C/D bands I used to transmit the signal. Currently I am using 132 KHZ signal  for TX and RX

    ok, you are using 131.25kHz and 143.75kHz frequencies to represent "1" and "0" in modulated signal via FSK protocol. 

    7.  how you calculate that 150nF GIVES THAT MUCH LOSS .and plz help me to design the coupling circuit and how to calculate them .because i am very beginner to AC signals.


    impedance for capacitor is Xc = 1/sC= 1/(2*pi*f*C). With 150nF, the impedance at  f=143.75kHz is 7.38Ohm. In our reference design, we use 10uF in C2, its impedance at 143.75kHz is 0.111Ohm. You want to transmit high frequency signal at low IR losses at high frequency, but attenuate low frequency signal at 50Hz. At 50Hz, 10uF will have impedance of 1/(2*pi*50*10uF) = 318.3 ohm. The impedance of 150nF at 50Hz is approx. 21.2kOhm. but you want to place such C values at the AC powerline side, however, you still want to minimize Tx.Rx attenuation. So you need to choose capacitor value accordingly. At AC powerline side, 0.47-1uf range may be adequate. At 143.75kHz, 0.47uF will have impedance of 2.36Ohm. At 50Hz, its impedance is 6.77kOhm, where majority of AC voltage will drop across the capacitor and block most of low frequency signal.   

    Inductor's impedance is X_L= sL=2*pi*f*L. Its impedance is low at low frequency, and high at high frequency. 

    I will try to send you relevant reference design information when I search for a folder in my drive. 

    https://www.ti.com/lit/an/sboa130a/sboa130a.pdf?ts=1605203006859&ref_url=https%253A%252F%252Fwww.google.com%252F

    Best,

    Raymond