Parallel termination resistor on the Rx end

OK, so you suggest a series 1K termination between Rx and Tx.
Suppose such a resistor is sufficient for SI issues.
But how does it help setting a known and constant voltage level at the Rx side when the Tx is high 'Z' ?

shaiko said:

OK, so you suggest a series 1K termination between Rx and Tx.
Suppose such a resistor is sufficient for SI issues.
But how does it help setting a known and constant voltage level at the Rx side when the Tx is high 'Z' ?

Click to expand...

- The 1K would be between the net and ground (or Vcc if you want). It would not be between the Rx and Tx of the net
- A series resistor between Rx and Tx would be on the order of 20-30 ohms for a 50 ohm transmission line.

The 1K to ground would supply the known voltage, the 20-30 ohm would supply the termination.

Kevin Jennings

Kevin,

Why not an RC to ground at the Rx end ?
It should eliminate DC current (as soon as the small capacitor gets fully charged).
What do you think ?

shaiko said:

Kevin,

Why not an RC to ground at the Rx end ?
It should eliminate DC current (as soon as the small capacitor gets fully charged).
What do you think ?

Click to expand...

- Depending on what frequency we're talking about (kHz? MHz?), when the oscillator is running, it's not really 'DC' very long anyway. When the oscillator is stopped the 1K pulldown will pull the net down, no current. At the end of the day, adding the capacitor adds cost but is of debatable value.
- If you're concerned about DC current, then take a look at the leakage current at 'RX' and use as large of a resistor that you can that won't develop a voltage greater than VIH for that part. That will minimize the current used.
- Another approach is a 'bus keeper' where you actively drive the net through another pin on the receiving device through a relatively large (1K or larger) series resistor. The intent is that the resistor is large enough that the 'real' driver (in this case the oscillator) can easily overcome the bus keeper driver, but when the other part is not there or otherwise not driving the net, then the bus keeper will keep the net in its last state.

Kevin Jennings

OK, so you suggest a series 1K termination between Rx and Tx.

Click to expand...

I really wonder if it's necessary to draw a schematic for this simple thing. I clearly suggested "a combination of source side series termination and a pull-down resistor (e.g. 1 k)". That's the same what Kevin said in post #22. The 20-30 ohm value accounts for the expectable driver output impedance so that the sum fits the transmission line impedance.

It's clearly the best option for a point-to-point connection. The RC series termination would come into play if the line has taps and you want a square waveform not only at the RX end.

The pull-down resistor or other means to prevent the signal from floating (e.g. the said bus hold circuit) are required in any case where no permanent driver or static parallel termination is present.

I understand.
You suggest to use a series termination at the driver where :
R_series_termination = Z_transmittion_line - R_driver

Together with a parrallel 1K pull-down at the Rx side.

This means that the Rx end will have an impedance of around 1K.
That's twenty times the value of the transmittion line's impedance...Wouldn't this missmatch cause a lot of signal distortion ?

In order to have a good transmission of signal quality, the source , load and transmission lines ought to be matched over the frequency range of interest.

50Ω cable is common for coax and is easily done on strip line or microstrip layout with track geometry and dielectric constant known within 10%. This gives the best SNR and signal integrity as well as maximum power transfer from transmission Line Theory.

Since LVCMOS clock drivers can handle ~ 30mA, an "active termination" is used and not a standard gate with 3mA drive. ( i believe "dynamic termination is used by some folks" ) but I always referred top it as "active termination" which means the Load resistor equivalent circuit is connect to the mid-range of the voltage swing, which is Vdd/2. So guy from FLA (who's name I just forgot) was right on with his solution to use 100Ω pullup and 100Ω to gnd to get an equivalent 50Ω to Vdd/2 active termination.. The same was used for SCSI and many other data lines. This is how it's done.

By active termination you mean thevenin termination?

No but it is, I mean a powered active biased load.

Thevenin termination is the same thing as what you describe as active termination with two resistors. For a clock signal with constant duty cycle, a RC termination can achieve the same effect with less power dissipation. But I don't see the necessity for a point-to-point board level connection at low and medium speed.