2
\$\begingroup\$

for a project I'm designing I am currently using a simple parallel Resistor Termination on the DDR2-Traces. But I'm wondering, what is the advantage of using a voltage regulator with serial-termination?

  1. Lower BOM/placement-cost?
  2. Lower power requirements?
  3. Something I'm totally missing?

Additional information:

My current design with 4 chips employs 54 terminations in total. This is done with 100 Ohm resistors. So we've got 1.8V/200Ohms = 9mA. 9mA*54 = 486mA plus reference voltage generation etc. it's more than 500mA unless my calculation is totally off. That results in nearly 1W of power dissipation just for the termination.

Are there any good guesstimates what the effect of an active regulation would be in a real world scenario?

| improve this question | |
\$\endgroup\$
3
\$\begingroup\$

The #1 reason is for lower power consumption. With passive termination, on really wide data lines, you can have AMPS worth of power going though the termination resistors-- and all of this is just wasted power.

Power consumption with active termination is harder to calculate, because it depends on the data pattern. But it almost always it consumes less power than passive termination. In the best case scenario (with half of the data bus high, the other half low) the regulator doesn't provide any current at all. There is still power being consumed, but it just doesn't come from the regulator. So there is no "wasted power" in the best case.

Active termination can also be smaller. There are less resistors around your space-constrained chips. The regulator takes up space, but you have more flexibility on where to put it on the PCB. I don't remember the numbers, but passive termination might require higher wattage, a.k.a. bigger, resistors due to the power consumption.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ As DDR2 has ODT for data-lines only the patterns on the control-lines really matter. I'll add some more information too my question. \$\endgroup\$ – Nico Erfurth Aug 26 '11 at 10:32
  • \$\begingroup\$ @Masta79 Ahh, see, you didn't specify DDR2 SDRAM! :) DDR2 signaling is used for other things as well. But yes, you're right for DDR2 SDRAM. \$\endgroup\$ – user3624 Aug 26 '11 at 14:12
  • \$\begingroup\$ Ok I've clarified that a bit. You're right, DDR2-signaling is used for other purposes too, thats the problem when one has his head to deep into some topic. :) \$\endgroup\$ – Nico Erfurth Aug 26 '11 at 14:56
0
\$\begingroup\$
  1. You still need the resistors with the regulator to VTT, so parallel resistors is cheaper.
  2. I'd have to work out the math.
  3. Regulator is quieter, so more noise margin.

Don't worry about the cost of the regulator; go with the regulator for more margin. Make sure it can source and sink current.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ With parallel resistors I need two per dataline, with serial i would only need one. While it's just 4 DDR2-ICs, but together with the placement-cost it might matter. \$\endgroup\$ – Nico Erfurth Aug 25 '11 at 23:46
  • 1
    \$\begingroup\$ @Brian Hmm... Why not use on-die-termination which should be available in DDR2? Or active termination is superior? \$\endgroup\$ – BarsMonster Aug 25 '11 at 23:48
  • 1
    \$\begingroup\$ @BarsMonster, ODT is only used on data lines. Which makes sense for the wide memory-busses usually used in PC-Applications these days. My databus is only 16-bit wide, so there are more control- than data-lines. In total there are 54 parallel terminations in my design for the DDR2-bus. \$\endgroup\$ – Nico Erfurth Aug 26 '11 at 10:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.