I am using this RTC clock module (http://datasheets.maximintegrated.com/en/ds/DS3231.pdf). The DS3231 has alarm function that can be programmed twice a day. When the alarm goes off it makes the interrupt pin an open collector (according to the data sheet). The particular board I am using can be found on eBay , and here is the schematics for the board I am using https://edwardmallon.files.wordpress.com/2014/05/rtc-goodparts.jpg?w=640 as you can see, the interrupt pin "SQW" is not pulled high by default). I have already verified on a 3V circuit (with arduino) that the interrupt pin works as expected. For my use case, I am using the DS3231 from the battery Vcc , and not steady 3.3V vcc from a regulator.

What I am trying to achieve is to utilize the alarm function to drive a 9v circuit. I have a few questions:

1) My first attempt is to use the IRF9510 P channel MOSFET. The result is MOSFET is always on . Pull up 9V resistor is used at the gate. Gate-Ground voltage is 0.19v when interrupt is active low. Gate-Ground voltage is 3.5v when alarm is not triggered (interrupt is not open collector) , which explains why the MOSFET is on, but I don't understand why, as the pin should be high impedance when not active low ? (note my battery is at 8.65V, spec sheet says maximum input range to be 5V and 5V + 3.5V is 8.5V quite close to 8.65V)

2) My second attempt is to use a small 3v cr2032 battery and a P-channel MOSFET (I have also tried the same setup with a PNP transistor and that works too). The active low interrupt can drive a 2V+ signal to another N-channel MOSFET gate. (ofcourse, the common ground are established.) That N-channel MOSFET has its source connected to 9V. Setup 2) works except for the current consumption for the 3V cr2032 battery. I measure 0.11mA when the interrupt is not activated, and 0.4mA when interrupt is low. Given 200mAh typical capacity, this only amounts to approximately 1/2 a month of runtime (assuming average 0.55 mA draw) I should mention, both the MOSFET-MOSFET setup and PNP-MOSFET setup draws about the same amount of currents, (MOSFET-MOSFET is better, at 0.1mA when not activated and 0.25mA when active low)

3) for setup 2) there are only a handful of pull-up resistors and gate resistor pairs that result in a working system. I have tried many combinations and my conclusion is anything from 1k ~ 4.7k works very reliably. The current draw is a function of the resistors used but at higher resistance value, the N-channel mosfet (the second mosfet) will not turn on appropriately, why is that?

any suggestions ? thanks

  • \$\begingroup\$ When you say "drive a 9V circuit" can you be a little more explicit as to what your 9V interface needs to be? Do you need a signal that simply swings from GND to 9V and back? Or do you need this to switch on 9V to a load? Or something else? \$\endgroup\$ Commented Dec 14, 2014 at 7:11
  • \$\begingroup\$ According to Maxim's data sheet, INT is active low and open-drain, not active high. Also according to Maxim's data sheet, the absolute maximum Vcc/Vbat allowed is 6VDC referenced to ground, so how can you expect it to work properly when your Vbat is 8.65V? \$\endgroup\$
    – EM Fields
    Commented Dec 14, 2014 at 7:29
  • \$\begingroup\$ A really basic diagram would help. What you are doing sounds very convoluted \$\endgroup\$
    – KyranF
    Commented Dec 14, 2014 at 8:21
  • \$\begingroup\$ thanks @MichaelKaras , my wording is bad. By driving I meant to switch the 9V circuit on/off (with the MOSFET). \$\endgroup\$
    – nonokunono
    Commented Dec 14, 2014 at 17:13
  • \$\begingroup\$ thanks @EMFields I am not connecting the 9V to the Vcc of the chip. I am using the interrupt line (~INT/SQW) to the gate of a P-MOSFET. That P-MOSFET is then connected to 9V at its Source. \$\endgroup\$
    – nonokunono
    Commented Dec 14, 2014 at 17:15

2 Answers 2


I've solved problems like that in the past with a Sziklai pair. It's like a Darlington, but with two complementary transistors. In your case, use a PNP transistor with its emitter connected to the alarm chip's VCC, and the base, through a resistor, to the open drain alarm output. The collector goes to the base of an NPN transistor (again, through a bias resistor) wired as a low-side switch for your 9 volt circuit.


You could devise a scheme where you put the burden of bias for the switching circuit onto the 9V supply instead of having to introduce the coin cell. The circuit design idea below would support switching your 9V supply to a load. As shown the load could be up to ~400mA. The bias load on the 9V supply when the alarm output of the RTC is off is about 2mA with the shown component values.

enter image description here


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