Need help with an odd design construct

[dwight]

Why not just measure the voltage across the resistor. That will tell you the amount of current flowing.

Dwight


________________________________
From: cctalk <cctalk-bounces at classiccmp.org> on behalf of Brent Hilpert via cctalk <cctalk at classiccmp.org>
Sent: Tuesday, November 14, 2017 6:29:47 PM
To: General Discussion: On-Topic and Off-Topic Posts
Subject: Re: Need help with an odd design construct


On 2017-Nov-14, at 4:51 PM, Jim Brain via cctalk wrote:

> On 11/14/2017 6:32 PM, Brent Hilpert via cctalk wrote:
>>
>> Once the cart is pulled the 4008 chip should end up in standby mode - no enables asserted.
> The 3K3 ties both !CROM and !CRAM high, and they are both open collector outputs on the port.
>
> But, I forgot to put one important detail in the original note.  I so apologize.  The original cart has a HM62256LP-12, but it is evident from the traces being cut on the cart that this was not even the original SRAM.  I do believe the original SRAM was 32kB in size, though.
>
> SInce the original SRAM was not even original, I replaced with the '4008 for various reasons.
>>
>> In standby a CMOS chip like this will appear as a near infinite impedance, so there isn't much voltage dividing going on with a 6.8K R.
>> The full battery voltage (minus epsilon) will be across the chip.
> Then, why do the 6K8 there?
>>
>> The datasheet specs standby current Isb1 at typically 4 uA (50 max).
>> Ohm's law will get you an idea of the effective resistance of the chip if you really want to calculate what epsilon is here.
> I'll try and calculate it for the HM62256LP, which seems to be .5mA under no load:
>
> http://www.farnell.com/datasheets/97905.pdf
>
> (2.7/.0005) = 5K4?    That seems incomplete, like I need to also take into account that .5mA flows through the 6800 resistor, but that means (6800*.0005) = 3.4V, which can't be right.
>
> If I put that into a Voltage Divider (2.7V Vih, R1 = 5400, R2 = 6800, output is 1.5V, implying the GND line is at 1.5V relative to the battery...
>>
>>
>> Why the circuit might be as you have it with the 6.8K I'm not sure, perhaps for some current limiting or glitch suppression as the cart is pulled out/in.
>> Doesn't really matter much as the only other thing on that gnd side is the switches.
> And they aren't there on the original.
>>
>> Those 4 10K resistors are kind of weird, each 'on' switch is chewing up orders of magnitude more battery current than the chip,
> Good point.  I copied the design from the "ROM cart" schematic, where the battery is not in place, and the switches were there to choose the bank of 32K that would be visible in the address space. I should reconsider the schematic, and...
>> might make more sense if they went to Vcc where they would be diode blocked when the cart is on battery.
>>
>>
> Excellent suggestion.  I will do so.


With the 62256 drawing consequential standby current, you can't ensure an accurate calculated value for
the voltage division from specs as the chip current probably isn't a linear relation to Vcc.

But going from what we have:
        Isb @ 5V Vcc = 0.5ma  then chip R = 5V / 0.0005A = 10 Kohms

        2.7V Vbatt * 10000 / (10000+6800) = 1.6V across the chip supply pins.

But from the datasheet it also looks like Isb goes down considerably (40uA) if the voltage on nCS approaches Vcc
and other inputs pins go to 0V, as it might/would/should with the cart pulled.
That would be a higher effective R for the chip and so a higher V supply for the chip.

There is another potential problem with those switches though,
it's not clear to me whether you resolved whether X1P pins 1 & 27 connect together when plugged into the computer.
If not, then when plugged in and operational, the switch 10K Rs form a varying voltage divider with the 6.8K and the
input levels on bank-select address pins would be all over the place.

I would suggest you redraw the schematic with GND as the chip/computer GND
rather than establishing a second GND, and have a bus line across the bottom for the battery negative.

If you move the switches R supply it may be a good idea to put another diode in series to give a similar V drop as the main supply diode
so the V on the bank address pins doesn't go above the chip Vcc.

I'm not clear on what all is going on there between original design and modifications and additions.