04-20-2021, 10:17 PM
(This post was last modified: 04-21-2021, 01:50 AM by dsimic.
Edit Reason: Small wording cleanup
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(04-20-2021, 07:13 PM)Pine Wrote: That is clever. This is the line you are talking about. An SMD Schottky diode would work. With a microscope, an Exacto knife, and a bit of patience, it will work. One issue I had with soldering is that the whole top layer is a heat conductor, so I had to tin my soldering iron and apply flux to the board and only transfer solder from the tip to the SMD, not heating up the pad first. About one second of touching the soldering iron should do it.
Exactly, I was talking about the vertical PCB trace in your picture above (thanks for the picture, by the way). As a note, there's something even better than an Exacto knife when it comes to removing solder mask: a fiberglass brush pencil. With a stencil made of Kapton tape, it should be possible to use a fiberglass brush pencil to make solder pads that are almost of the same quality and precision as the factory-made ones, with virtually no risk of damaging the traces.
(04-20-2021, 07:13 PM)Pine Wrote: Another idea I was toying with is to uncrimp the +RTC terminal in the holder with a fine jeweler's screwdriver or Exacto knife, pop out the tiny center piece of metal (the battery terminals are rigid loops), and interpose a diode with a new + contact for the holder.
I had another look at the battery holder, the battery terminals and the battery terminal pins, and I'm not really sure how would that work?
The "+" pin is crimped underneath the collar of the "+" battery terminal, so you'd need to squeeze in the radial fins at the other side of the "+" terminal in order to pull the battery terminal out of the battery holder enough to free the "+" pin. Also, the freed "+" pin would stay physically inside the battery holder. Where would you put the additional barrier diode, what would serve the purpose of the new "+" battery terminal, and how would you connect the diode, the pin and the new "+" battery terminal?
(04-20-2021, 07:13 PM)Pine Wrote: Exactly. What I'm actually going to do personally is, I'm going to use the Molex connector where I already supply 5V and add a 3.3V line on Molex pin 2 with barrier diodes to batteries. R351 (placeholder) runs near the +RTC, so another barrier diode will fit nicely there with a jump wire as you described.
Just to make sure, you're referring to the 24-pin ATX connector, right? Also, if you place a barrier Schottky diode on the external 3.3 V line, i.e. between the 24-pin ATX connector and the external 3.3 V power supply, you'd need another Schottky barrier diode on a Clusterboard only if you intend to also use two AA batteries inside the battery holder, to provide the RTC backup power.
(04-20-2021, 07:13 PM)Pine Wrote: Min 1.8V supplying 0.45mA to all 7 SOPines while powered off according to the A64 specs will do it.
Exactly. As a result, the two AA batteries can be used "down to the wire" in a properly modified Clusterboard, ensuring their long usable life.
(04-20-2021, 07:13 PM)Pine Wrote: I popped in 7 SOPines and ran heavy networking and the switch didn't suffer. When powered, the 7 SOPines draw 3mA from RTC. Lots of headroom left.
Excellent, thank you for the experimental verification. In theory, there should even be just enough headroom to power another RTL8370N switch ASIC from the same U1 regulator (2 * ~0.9 A < 2 A), as we may pretty much not count the other components that are fed by the U1.
(04-20-2021, 07:13 PM)Pine Wrote: That will work. You could even solder a thin wire on L1. I only chose D1 because it is after the LC filter (the U1 voltage out waveform is noisy so a big choke is needed), there is a bank of capacitors available after that L1 choke (power cycle but keep RTC powered for a few seconds), and D1 is closest to the +RTC terminal.
Now I see my mistake, thank you very much for pointing it out! Tapping into the output of the U1 before the L1 was a bad idea; actually, I'd never take that route, but the large PCB trace around the pin #3 of the U1 was, visually, so tempting that it became misleading.
After having another close look at the Clusterboard PCB, there is a pretty large PCB trace on the "good" side of the L1, on the top side of the PCB. Even better, there's also a trace in form of a small "island" on the bottom side of the PCB that is also connected to the "good" side of the L1 and serves to accommodate three members of the capacitor bank (C7, PC6 and PC7); I suppose they had to do it that way because the capacitor bank is made of a rather large number of capacitors.
The above-described "3.3 V island" is highlighted in the attached picture. Using that "island" would be even better than using the traces on the top side of the PCB, because the jumper wire would be visible only once on the top side of the PCB.
Moreover, using the "3.3 V island" could allow the entire jumper wire to be placed on the bottom side of the PCB, by placing the additional barrier diode on the bottom side of the PCB, on the short part (before the first via, for the BAT-RTC pin of the SOPine #0) of the horizontal continuation of the long vertical trace in your picture above. However, I'd rather not take that route, as the jumper wire would then go across a zillion of components; the jumper wire will carry DC so it should do no harm, but it's better to stay on the safe side.
(04-20-2021, 07:13 PM)Pine Wrote: I'd personally remove the holder. A super-easy way is to uncrimp the terminals as described before, lift the holder straight up and out, and then clip the terminals or put the holder back if you add another diode. You don't even need to unsolder the battery holder. Pretty neat.
Quite frankly, I don't see that crimping back the original battery terminals would be an easy thing to do. You could use needle-nose pliers to do that, but holding everything together while bending back the radial fins, applying pressure so the crimping makes good electrical contact with the "+" pin, and ensuring that the battery holder doesn't move at the same time, damaging the "+" pin... It might be doable, but I doubt that the final result would be as good as new.
By the way, why would you actually need to remove the battery holder and put it back later? What is not accessible with the holder still in place? Also, I don't think that an additional barrier diode, for example, can be tucked underneath the battery holder.