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I read a schematic where it has a Z80 CPU tied to a 78LS138. The CPU's WR line is tied to pin 1 (A) of the 138. Also, the schematic shows a 42uH inductor tied to pin 1 of the 138 as well.

What is the purpose of this? As of my understanding, inductors are primarily used to filter AC current no?

And if the Z80 <-> 74LS138 is DC only (I am assuming) then wouldn't the inductor really just act as a small piece of wire and if anything, a super small resistor?

Could it be because the entire circuit (this is for a Colecovision, BTW) is expected to have AC noise? I see several examples of using inductors on various pins in the circuit.

Thanks.

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    \$\begingroup\$ atarihq.com/danb/files/colecovision.pdf page 2, for those wondering. \$\endgroup\$
    – pjc50
    Commented Mar 23, 2015 at 15:01
  • \$\begingroup\$ It isn't DC, it usually is square waves \$\endgroup\$
    – PlasmaHH
    Commented Mar 23, 2015 at 15:16

4 Answers 4

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My guess is that the inductor (L17) is actually on pin 16 (Vcc), not pin 1. Having an inductor on the ~WR line would really mess up the ~WR signal everwhere on the board.

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  • \$\begingroup\$ I agree, the inductor's position at the top of the chip makes it clear that that's supposed to be the power connection for the chip. I'm not clear why they would use an inductor, though. That's like the reverse of a decoupling capacitor. The LC filter on the 12V supply on page 1 also confuses me. \$\endgroup\$
    – Adam Haun
    Commented Mar 23, 2015 at 15:48
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Assuming this isn't simply an error, given that it's a reverse engineered schematic, it looks like an inductor tied from ~WR to +5v. That would have the effect of de-asserting WRITE except for a brief time after a falling edge output from ~WR. My guess is that the CPU is capable of doing multi-cycle writes but this part of the circuit is intended to act differently on the first cycle of such a write.

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  • \$\begingroup\$ Oh yeah, sorry, forgot to mention this was a reverse engineered schematic. :-) But still, I thought an inductor would basically have no effect on a DC/TTL operation? \$\endgroup\$
    – cbmeeks
    Commented Mar 23, 2015 at 15:14
  • \$\begingroup\$ TTL isn't DC, it consists of changing values (edges). More like a square wave that ends up rounded off or with ringing artefacts at transitions. \$\endgroup\$
    – pjc50
    Commented Mar 23, 2015 at 15:30
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And if the Z80 <-> 74LS138 is DC only (I am assuming) then wouldn't the inductor really just act as a small piece of wire and if anything, a super small resistor?

To answer this part of your question, digital logic definitely not DC. The CPU in the ColecoVision runs at 3.58 MHz. The external signals probably toggle more slowly, but for digital bandwidth, the rise and fall times matter more than the absolute frequency. I found a datasheet that says that max rise time for the Z80 clock is 30 ns. The formula for approximating the bandwidth is:

$$BW = \frac{0.35}{t_r}$$

For the Z80, this suggests that the external signals will have a bandwidth on the order of 11.7 MHz at minimum.

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I think that's an error in the reverse-engineered schematic.

  1. Why would anyone deliberately put 2 pins with the same pin number on an IC? That's asking for confusion.
  2. Even if that strange '2 pins with 1 number' situation was intended, then that inductor is not just pulling the WR signal to that one IC. Its connected to the entire 'WR' net on the board, so everything will be affected. That poor Z80 would have to be able to sink a significant amount of current if it ever tried to drive WR low for more than a few microseconds.
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  • \$\begingroup\$ Well, the Colecovision did have one gigantic power supply. And it's wildly known that it pulls a lot of current. Not that it's relevant, but I thought I would mention it. :-) \$\endgroup\$
    – cbmeeks
    Commented Mar 23, 2015 at 15:33

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