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I've built a RAM expansion cartridge for my VIC-20 computer.

It uses the CY62256L 32KB SRAM chip. It has been working fine except there's a quirk.

If I power off the computer, but leave the external disk drive powered, there is a small amount of current (apparently through the drive's reset line into the main Vcc) supplied to the expansion board still.

It is in the micro-watts but it is enough to keep the RAM chip in low-power standby.

Is there a way I can prevent this small amount of current from flowing?

Here is my circuit (please excuse its messiness): enter image description here

EDIT: Here's a clarification of what is going on. Here is a page from the schematic of the computer, which is powered off when the current leak happens.

The external drive is powered externally. The red circle is where it is sending +5V into the serial port while the computer is turned off, i.e. the reset line. The blue circle is where the reset line connects to my board.

However, the green circle is where the reset line is usually pulled up by the computer, and here I think is where it is leaking onto the Vcc. Which enters my expansion board at the purple circle.

It's Vcc that has current which is powering the RAM. The board's reset line is high too, but it's only connected to a switch.

Does that help explain?

enter image description here

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  • \$\begingroup\$ You say "apparently." Can you confirm this by opening up reset to verify that the current goes to zero? If that is the case, it is pretty easy to add a simple circuit which disconnects reset when VCC to the expansion card is low. I can draw it if you are interested, but it is worth it to confirm the source of the micro-current first. In general, once the RAM VCC is zero, all IO connections to the RAM need to be low or you will have something like this happen. There is a fix for that, too, but it is more complicated. \$\endgroup\$ – mkeith Jul 16 '15 at 18:35
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    \$\begingroup\$ What is the schematic of? Where is the external hard disk in the schematic? Why doesn't the RAM have a connection to VCC? How does the external hard disk connect to the computer? \$\endgroup\$ – AngryEE Jul 16 '15 at 18:35
  • \$\begingroup\$ @AngryEE, This is the schematic of the RAM expansion cartridge. The IC chips are connected to Vcc and GND it's just not shown in the circuit. If you really want to dig though the computer and drive schematics they are found here: zimmers.net/anonftp/pub/cbm/schematics/computers/vic20 Mine is the NTSC version "324001" zimmers.net/anonftp/pub/cbm/schematics/drives/new/1541/… \$\endgroup\$ – Sam Washburn Jul 16 '15 at 18:53
  • \$\begingroup\$ @mkeith, I have confirmed it is the drive's reset line. I am interested in your ideas. Thanks! \$\endgroup\$ – Sam Washburn Jul 16 '15 at 18:56
  • \$\begingroup\$ Something is not making sense. In your schematic, reset goes from CN1 to a switch. Is CN1 the connector? How is the drive connected to reset? If reset only goes from the connector to a switch, I don't see how it can be powering up the RAM. Can you clarify who is responsible for driving reset, and who treats it as an input? \$\endgroup\$ – mkeith Jul 16 '15 at 23:19
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Seeing as you are certain it's the reset, and the reset is active low, you can re-create the reset. Before you do that you should verify that the reset when disconnected on the external drive pulls nicely high.

It should, because it is "sourcing power" to your expansion board, so it very likely has an internal pull-up causing all your troubles. If there is a pull-up inside the drive, you can just use two MOSFETs:

schematic

simulate this circuit – Schematic created using CircuitLab

They can be simple low power MOSFETs.


You can also disconnect the reset from its origin with board VCC with a little trick:

schematic

simulate this circuit

If the Board Vcc is high and the Board Reset is low, the MOSFET has a positive gate-source voltage and it will conduct and also allow the drive Reset to be pulled low by the reset signal.

If the Drive Reset is high, but Board Reset and Board VCC are low, the MOSFET has 0V gate-source voltage, and thus be turned off. The internal body diode in that case is in the blocking direction and in modern MOSFETs should not leak enough anymore.

If the Board Reset is high (regardless of the Board VCC) the body diode of the MOSFET will be put in the forward direction, and conduct, pulling the Drive Reset high, but of course with the voltage drop across the body diode subtracted. The very likely pull-up in the drive will undoubtedly take it the rest of the way.


If none of that "clicks" for you (pun intended), this certainly will:

schematic

simulate this circuit

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Here is my idea. I am not sure it is better than Asmyldof's various ideas. There are some caveats with this, which I will add in the description.

schematic

simulate this circuit – Schematic created using CircuitLab

RESET is the reset from the drive. SRAM_RESET is the reset on your expansion cartridge. SRAM_VCC is VCC from your expansion cartridge. When SRAM_VCC is high, Q1 and M1 are on and the two resets are connected together. When SRAM_VCC goes low, Q1 will turn off, and R1 will then pull up the gate of M1, thus turning off M1, and preventing current flow into the SRAM_RESET net.

The drawback to this circuit is that it loads the reset line with 470k when Q1 is on. This could actually activate reset, which would be bad. So, if 470k is too much load, you could change R1 to a much higher value (even 5 Meg). It should be OK to load reset with 5M.

Or, if the VCC from the drive is available (let's call it VCC_DRIVE), you could pull up the gate of M1 to VCC_DRIVE instead of pulling it up to RESET. Then the pullup value could be 470k or whatever.

You could change Q1 to an N-channel mosfet, also, if you want. If you do that, I suggest you change R2 to 1k.

Good luck, and we would love to hear back how you make out!

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