Steve Barnes wrote:

The keypad "input" rows are currently setup as quasi-bidirectional to take advantage of the weak pullup. So are you saying that when I go into low power down mode, I should not use quasi-bidirectional and instead use external pullups?

Steve,

you fully misunderstood my reply! And again we are in a situation, where big confusion takes place, only because original poster is not willing to reveal relevant details: I do not know anything about your circuitry that is switched-off and I do not know anything of your circuitry that is powered together with micro. So, my recommendation cannot be more concrete than information given by you.

Ok, let's discuss the subject point by point:

I wrote:

"Whenever extremely little power supply current is wished from a CMOS chip, you must take care that applied voltage to inputs must not float!! Allowed potentials are only the rails, means exact Vcc or excat 0V. If applied voltage is floating even only very slightly from these fixpoints drastical increase of supply current is to be expected!"

Assume you have a CMOS chip, which you want to power by a battery and which shall draw as little power supply current as possible. Then, at each available INPUT a voltage must be applied, which is either exact Vcc or exact 0V. Otherwise PMOS and NMOS transistors are partially turned-on and power supply current will increase. THIS IS OF GENERAL VAILIDITY!!

How high this supply current increase actually is, highly depends on situation. It cannot be predicted in detail. But it's a wise design practice to guarantee, that input voltage is as near to Vcc or 0V as possible. I gave you a link to demonstrate how high this increase of supply current CAN be. And that it is wise to limit deviation of input voltage to power supply rails to less than about 200mV.

BUT: HOW YOU REALIZE THIS, DEPENDS ON YOU!! And on your actual application.

Now let's think about, what this can mean for you:

If the assumed CMOS chip is your 'LPC932, then, where can be such inputs, where you should take care????? Well, AT EACH PORT PIN! Why?? Because, due to 'quasi-bidirectionality' of port pins, at each port pin you will find a very special pair of an input and an output. Please have a look into the 'bibles' to learn more about this very special port pin circuitry, which is so typical for all '51 derivatives.

Now there are two possible configurations with your 'LPC932:

1. Either a port pin of 'LPC932 (configured as input) is connected to an output of 'LPC932, or

2. a port pin of 'LPC932 (configured as input) is NOT connected to an output of 'LPC932, BUT to an output of ANOTHER chip, which is NOT powered by the same power supply voltage, that is used for the 'LPC932, but is powered by THE certain power supply voltage, which is switched-off by you to save total power supply current.

Your keypad matrix is an example for the first point. And, that we talk about the same subject,
if your keypad matrix looks like that published in

http://www.semiconductors.philips.com/acrobat/a...0184_1.pdf

then, you must of course use some pull-up configuration for the case that no key is pressed, because inputs otherwise would float. And in THIS case, the configuration of according inputs as 'quasi-bidirectional' is of benefit, because then already internal pull-ups are connected and there's no further need for the add of external parts. This need would exist, if inputs were configured as 'input-only'!

But, what I was referring to with the following lines was point 2!

"So, you should always look for a good path to 0V or Vcc at inputs. In your case, where other circuitry is powered-off, only a path to 0V is possible, of course. But keep in mind, that there's no natural path to 0V at output of an off-powered chip! Mostly there will be protection diodes (or pn-junctions of parasitic thyristor, see another thread) connected from output to Vcc and 0V.
So, floating input current of your 'LP932 micro will bias these pn-junctions and cause a floating potential of about +0.5V!! From the above it's clear that this cannot be accepted. The only remedy is to connect a 100kOhm pull-down resistor (yes, pull-DOWN!) from each input to 0V.

Keep in mind, that port pins must be configured as 'Input-only', NOT 'Quasi-bidirectional'!!! Whenever you configure a port pin as input, this should be kept in mind."

Again: I was referring to point 2!!! Let's have a concrete example:
You feed the output of a 74HC04 inverter stage to an input of 'LPC932. 74HC04 shall be powered by the extra power supply voltage, which is switched-on and -off by you to save total power supply current. But 'LPC932 shall be permanently powered. THEN, in the case that 'LPC932 is powered, but 74HC04 is NOT powered, you can get two problems:

a. If input of 'LPC932 is configured as 'quasi-bidirectional', then all the time current of the internal active pull-up would flow to ground via parasitic pn-junctions at the output of 74HC04 inverter stage. This can be dozens of µA, which must be delivered by power supply voltage of 'LPC932!!

b. If input of 'LPC932 is configured as 'input-only', unavoidable input leakage current will cause a voltage drop of about 0.5V when flowing accross parasitic pn-junctions at the output of 74HC04 inverter stage. And this voltage drop leads to a deviation of input voltage from supply voltage rails (here 0V) which can cause drastical increase of power supply current of 'LPC932. So, also a pull-down resistor of about 100kOhm is needed from each port pin to 0V.

Again: If point 2 is valid, according inputs of 'LPC932 must be configured as 'input-only' AND 100kOhm pull-down resistors should be connected from these inputs to 0V.

Kai