Usb 2.0 component selection

[Rajinder]

,
I am using USB 2.0 full speed for a design. Power to the electronics is 5V from the USB, into a FTDI device ( which has an internal 3V3) regulator to power my electronics.

I have a couple of questions, FTDi recommend a LC filter, they have also added an extra 10nF at the input and 4.7uF to the output of the LC filter for VUSB.

My first question is the value of the ferrite is 600R @ 100MHz. This is the DC resistance at 100MHz, will this act as a voltage divider with my cable? Do I have to ensure that the attenuation is not below the acceptable levels for VBUS in terms of voltage? USB 2.0 is 500mA current, the ferrite is from the BLM18SND series, I presume the current handling capability needs to be greater than 500mA? Also would the ferrite prevent inrush current as it has a small DC resistance.

The other question I have is differential pair tracking. I have read that if the wavelength is 1/10th of the signal, then impedance of the tracking is not an issue. Can someone explain this 1/10th rule of thumb at full speed USB 2.0 with calculations as I do not understand it fully.

Lastly I have chosen a TVS array based on protecting 2 I/O lines D+/D- and VBus, the capacitance is less than 2pF, the TVS array also is well above the rating for +/- 4kV / +/-8kV air and contact discharge. Does that seem correct?

My FTDi device FT231x is designed for full speed Comms, it has also extra flexibility of configurable GPIO pins IE. CBuS lines. Is there anything else I need to consider?

Any help would be appreciated.

 

[Harald Kapp]

This is the DC resistance at 100MHz

No, it isn't. simply because 100 MHz isn't DC. This is the impedance (complex resistance) at 100 MHz.

will this act as a voltage divider with my cable?

Yes, but not in the way you imagine. The DC resistance will be << 1 Ω (datasheet) and thus the voltage divider effect will be very small.

I presume the current handling capability needs to be greater than 500mA?

Why? Strictly taken it isn't. USB 2.0 delivers 500 mA max.
Of course, it is always a good idea not to operate components at their limit, so a rated DC current > 500 mA is good for this application.

Can someone explain this 1/10th rule of thumb

See e.g. this discussion.

Lastly I have chosen a TVS array based on protecting ... Does that seem correct?

If the array is specified for use on USB 2.0, why doubt it?

 

[Rajinder]

No, it isn't. simply because 100 MHz isn't DC. This is the impedance (complex resistance) at 100 MHz.


Yes, but not in the way you imagine. The DC resistance will be << 1 Ω (datasheet) and thus the voltage divider effect will be very small.


Why? Strictly taken it isn't. USB 2.0 delivers 500 mA max.
Of course, it is always a good idea not to operate components at their limit, so a rated DC current > 500 mA is good for this application.


See e.g. this discussion.


If the array is specified for use on USB 2.0, why doubt it?

Thanks for the reply.
I can't see to open the link you sent for the discussion on the 1/10 wavelength.
Also if my cable had a impedance of 50 ohms, then would the voltage divider not be
600/(600+50) x 5V = 4.6V?

If the voltage divider affect is very low would that not prevent the circuit from working i.e..attenuating the voltage by too much?

 

[Harald Kapp]

I can't see to open the link

Try copying this url: https://electronics.stackexchange.c...atter-only-when-traces-are-longer-than-half-a

Also if my cable had a impedance of 50 ohms, then would the voltage divider not be
600/(600+50) x 5V = 4.6V?

You haven't understood the difference between the impedance at 100 MHz and the DC resistance.

 

[Rajinder]

Try copying this url: https://electronics.stackexchange.c...atter-only-when-traces-are-longer-than-half-a

You haven't understood the difference between the impedance at 100 MHz and the DC resistance.

Thanks for sending the URL.
Regarding the impedance, is this going to be a resistive loss at 100MHz. I am sorry but not sure. Can you help?

 

[Harald Kapp]

is this going to be a resistive loss at 100MHz.

Not quite. An impedance limits AC currents but does not necessarily involve resistive losses (but it can and it does, especially in a ferrite).
You don't have to worry about the impedance at 100 MHz. This value is useful for the suppression of high frequency signals.
USB full speed (which is btw. USB 1.0, not 2.0) operates at 12 MBit/s which is equivalent to 6 MHz - way lower than the 100 MHz you're looking at.
See the datasheet of the ferrite for the impedance at around 6 MHz. It will be considerably lower.