Help with uC external pull-down resistor

Discussion in 'Embedded Systems and Microcontrollers' started by av8or1, Jul 6, 2013.

  1. av8or1

    Thread Starter New Member

    Jul 6, 2013

    I am developing a microcontroller (uC) based hand-held device. I'd like to include a self-power down capability. The idea being that after so-much-time of inactivity, the uC sends a signal through a digital output to physically power down the device. This is functionally and electrically equivalent to the momentary power pushbutton that I will have mounted on the front face of the device. Keep in mind that I am a software guy, so this hardware/circuitry stuff is familiar to me but not transparent, thus my difficulty.

    But I'm running into a bit of difficulty with the implementation of that concept. Specifically, the problem is trying to determine the value of a pull-down resistor. First though, the implementation strategy that I have is to use a "latching circuit" to do the dirty work, or more specifically a chip in the form of a power on/off controller. The Maxim 16054 is what I found to do the job:

    And the uC that I am using is the Amtel 1284P:

    And the configuration that I have planned for these two is the same as shown in the MAX16054's datasheet under "Typical Operating Circuits" on page 1, although I plan to put the MAX16054 downstream of the voltage regulator, thus powering it with +5V. Otherwise the schematic I had in mind is the same as shown.

    So the problem, again, is how to calculate the value for the external pull-down resistor as shown in that schematic. I've done a fair amount of research (including this forum) and have discovered that external pull-downs are typically about 10kΩ to 1MΩ or more, depending. I understand the concept as to why they are of such large values, BTW.

    But I am having difficulty calculating the exact value I should use with the 1284. Based on the datasheet for the uC on page 329, I calculated a value for the pull-down that is much lower than what I was expecting, based on the rules-of-thumb that I found (1/10th for pull-up, 10X for pull-down of the internal uC pull-up). Based on those rules, it seems that I should use a 0.5MΩ pull-down.

    So. I dunno. How far off base am I here? Anyone have any feedback? I'd appreciate it.

  2. MrChips


    Oct 2, 2009
    There are no exact values for pullup and pulldown resistors. You may find that resistances that differ by orders of magnitude still do the job. It also depends on what you are trying to achieve.

    In your case you may want to select a resistance that results in the lowest power consumption while the unit is in the power down mode. In this mode the regulator and mcu are off and hence there is no voltage from the mcu side.

    The criteria is entirely from the MAX16054 requirements and that is to present a voltage that is less than 0.7V at the CLEAR input while conducting 1μA max.

    Since the current drain is less than 1μA you really don't have much to worry about. The pulldown resistor can be anything from 1k to 700kΩ.

    I would narrow it down to between 10k and 100kΩ.

    100kΩ will present 0.1V max at the input which is well below the required 0.7V.

    Go with 100kΩ.
    Last edited: Jul 6, 2013
    av8or1 likes this.
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    Agreed. Just about any value will work here. My prejudice is to use 100K max as the "feels" right but the math behind using 500K is correct. I just don't like going too high in case there's some noise floating by in the ether.

    I did something similar last year for a low voltage timing circuit I published here. The idea was to have a single external push button to start/stop/(plus other) a micro controller, then let the micro time out and turn itself off. This way there is no need for an external IC to latch a power on signal (the "latch" resides inside the micro you've already paid for).

    Basically the power is controlled by a diode-OR gate so either the button or a micro pin turns the LDO on. On power up the micro drives the pin high to keep itself on, and later on can turn itself off.
    av8or1 likes this.
  4. MrChips


    Oct 2, 2009
    MCU manufacturers today have a wide selection of ultra low power MCUs. I was going to suggest as Ernie did that there is no need for an external chip. You can put the push button directly on an I/O pin and do the power ON/OFF in software.
  5. av8or1

    Thread Starter New Member

    Jul 6, 2013

    Thank you much for the feedback, I appreciate it. Finding and joining this forum will be an incredible help!

    About 5 am this morning I realized that it was the input of the MAX16054 that I should be looking at, not the output of the 1284. When I did that I came up with the same numbers as MrChips.

    Nice design Ernie, I liked it. The cost of using the Maxim chip versus the other components isn't that great, so I will probably just stick with using the chip.

    Thanks again!

    Last edited: Jul 7, 2013
  6. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    If all this goes after the regulator, what is the quescent current usage of the regulator?

    And you can get regulators with a shutdown pin, which fixes the need for an external shutdown device AND reduces regulator quiescent current to practically zero.
  7. John P

    AAC Fanatic!

    Oct 14, 2008
    You haven't said what your battery voltage is, but if it's hand-held, I'd try to run the device in the most common way, with a pair of AA or AAA cells. That might allow you to operate off the battery voltage directly with no regulator, in which case you can probably just leave the processor powered all the time, and put it in Sleep mode instead of using any kind of turn-off circuit. If it's like the PIC chips I use, you can have a "wakeup on change" feature which lets a pushbutton trigger a restart. If you need a step-up voltage converter (if you can't live without 5V, perhaps) then the processor can shut it down before going to sleep itself.