I just want to know why an RC parallel circuit is placed in a DDR reset cicuit
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- 1\$\begingroup\$ Properly to hold in reset state during a boot. \$\endgroup\$Tyassin– Tyassin2025-12-09 10:23:41 +00:00Commented Dec 9 at 10:23
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3 That is a pulldown resistor followed by a low-pass RC filter. The low-pass has a 0 ohm fitted but would be replaced with a chosen value if the low-pass filter is later determined as necessary. Even with 0 ohms the 100n capacitor provides decoupling which is kind of acting as a low pass filter.
This is made much more obvious when the schematic is drawn well:
Pulldown for the usual reasons (starting up in a known low state - in this case reset). Lowpass filter for usual reasons (filtering out high frequency content that might cause a spurious logic low or high).
- 2\$\begingroup\$ yeah that original schematic is terrible! To a degree where it makes me wonder how well the original schematic was designed! \$\endgroup\$Marcus Müller– Marcus Müller2025-12-09 10:34:36 +00:00Commented Dec 9 at 10:34
- \$\begingroup\$ @MarcusMüller This is par for the course in most motherboard schematics I've seen. They tend to pack things pretty tight, so quickly sacrifice things like consistency of ground direction. \$\endgroup\$Polynomial– Polynomial2025-12-11 16:48:06 +00:00Commented yesterday
- \$\begingroup\$ @Polynomial yeah, and while I totally get that, things get dense and the only guy that'll complain is the layouter, who's used to your shenanigans, what confused me about this one really is that they totally lost the "meaning" of "there's a 0 Ω bridge, and something that pulls the line to ground, and a net label that could just as well have fitted next to the 0 Ω resistor without that pitchfork, had you shifted the resistor to the left in an attempt to not hide the fact it's a 0 Ω resistor." \$\endgroup\$Marcus Müller– Marcus Müller2025-12-11 20:27:09 +00:00Commented yesterday