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THVD8000: THVD8000 inductor calculation

1958835
Prodigy 10 points
Part Number: THVD8000
Other Parts Discussed in Thread: THVD8010

Hello everyone!

I'm glad to be able to use the THVD8000 chip, our team absolutely love it.

But I can't understand how to calculate the inductance correctly for THVD8000 / THVD8010 chips.

I need transmit data via the power line (about 20 meters of non-twisted cable, AWG14) at low speed (9600 b/s is enough for me) between two devices. At the same time, a large current sometimes flows along the line, 12 volts 5 amperes.

Initially I planned to use the minimum frequency of 125 kHz. However, when calculating the inductances, a very large value is obtained. Inductors of this rating with a current of 5a or more take up a lot of space on the board, and I simply cannot place them, they will not fit.

What are the disadvantages when choosing a frequency for example 1 MHz or 5 MHz? I am thinking to choose 5 MHz.

And I cannot understand the datasheet corrrectly: how to calculate the inductance correctly.

L = 750 / (2 * 3.15 * 5) = 24 uH

I choose a frequency, look at the graph.

1) Will I get the total inductance of the two coils or the inductance of each of them?
For example, for a frequency of 5 MHz, should I use two chokes at 48 μH or two at 24 μH (on the each side)?

2) I don't understand what this phrase means:
If no termination is used in theapplication, then the total equivalent impedance at the carrier frequency could be reduced to Z ≥ 60 Ω

3) How can I lower the inductance requirements?

Thanks!

  • 0
    TI__Mastermind 25705 points

    Your calculation looks right to me. The intention of limiting the load impedance to 375Ohm is to make sure that the bus voltage is high enough to exceed the receiver threshold with the cable attenuation. If the cable loss is small, you might be able to decrease 375Ohm to a smaller value. With termination and load impedance, if the total impedance is bigger than 60Ohm, the driver is guaranteed to generate 1.5V bus voltage. Also this 375Ohm is the total impedance of one wire (one side of the bus, A or B). Generally two inductor are attached to one wire (one at each transmission end), whose impedance are parallel to each other.

    With higher carrier frequency, the main drawback is resulting in shorter communication distance. 

    Another point is that the impedance calculation is first order that doesn't take the node resistance and capacitor into account. It'd be better to measure the signal amplitude in the system.

  • 0 in reply to Hao L
    Prodigy 10 points

    for a frequency of 5 MHz, should I use four chokes at 48 μH or four at 24 μH (two chokes on the each side)?

  • 0 in reply to 1958835
    TI__Mastermind 25705 points

    You need 4 24uH individual inductor, two for each node, in which one is for Vcc and one for GND.