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It's a general question about series resistors in this type of application but I will use an example to make it easier to understand.

I have a LCD controller which uses an SDRAM, we can read a lot on the internet and it is common knowledge that you can put series resistors on numeric signals to increase the rising and falling time. For example, if my controller emits a clock in the direction of the SDRAM, the resistor has to be close to the controller. Since this resistor is meant to limitate the current which will flow through the parasitic capacitor at the input,I don't understand why the placement is important?

The question is : Why does this resistor has to be as close as possible to the source of the signal? What is the difference if it is close to the destination?

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    \$\begingroup\$ Because the capacitance of the line is part of that parasitic capacitance that the resistor is supposed to separate from the output. \$\endgroup\$ – Dmitry Grigoryev Sep 8 '16 at 15:19
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The purpose of a resistor placed here is not to limit (or limitate) the current, nor to increase the rise and fall time of the signal (though that's what it would do if it were feeding a lumped capacitor).

The purpose is to match the source end of the transmission line formed by the track. This means that when the unterminated far end of the line reflects the signal, those reflections will be absorbed in the source resistor, resulting in a clean signal received at the far end. Without that resistor, there could be multiple transitions at the far end, distorting the signal.

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  • \$\begingroup\$ I also put some logical buffer for digital clocks that need to travel to multiple places i.e. for SPI buses, etc. \$\endgroup\$ – lucas92 Sep 8 '16 at 14:26
  • \$\begingroup\$ Even with your explanation if the resistor is at the end the result would be the same no? \$\endgroup\$ – damien Sep 8 '16 at 14:53
  • \$\begingroup\$ it would be even better if it were at the end no? because the reflected signal would be absorbed right after it is reflected \$\endgroup\$ – damien Sep 8 '16 at 15:07
  • \$\begingroup\$ No, a series resistor has to be at the source end. Look up some trnsmission line explanations. You can use a shunt resistor at the receive end, without a series resitor at the source. \$\endgroup\$ – Neil_UK Sep 8 '16 at 15:27
  • \$\begingroup\$ @damien AFAIK, the signal just reflects off the series resistor instead of reflecting off the end of the line. Moving the resistor closer to the source means the reflection "dies" sooner. \$\endgroup\$ – immibis Sep 8 '16 at 23:37
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Not considering what Neil_UK tells us (which is likely to be more relevant than this) - it would still make a difference for a simple reason:

The resistor will divide your line in two parasitic capacitors. If you place it close to the source, you will have a "large" parasitic capacitance behind the resistor and such the time constant will increase for the receiving side.

If you place it close to the receiver, you will have a large parasitic capacitance directly connected to the source, which will charge nearly instantly. The capacitance on the receiver is now close to the input capacitance of the pin you are toggling, so the time constant is smaller and you don't get as much smoothing.

Depending on the layout this effect might be negligible, but you can get some picofarads quite quickly and they might matter.

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