DS 1305 Real Time Clock, Pull-up Resistor Difficulties

Thread Starter

rleroy

Joined Mar 14, 2016
3
Hi, I’m hoping for some advice on pull-up resistor values as they apply to the DS1305 real time clock with two programmable alarms.

Background on device: I am trying to create a solar powered audio system in the woods that will turn on at a specific time in the morning and turn off at a specific time at night. I am using a bi-stable two-coil relay (V23079) to control the power to the audio output. The first alarm set on the RTC should switch the relay on, allowing power to the audio device; the second alarm on the RTC should switch the relay off, cutting power to the audio device.

Problem: The alarms are firing in code and seemingly on the RTC; however the power is not being held from the INT1(Interrupt 1 - pin 7) and INT0 (Interrupt 0 - pin 6) on the RTC. I was not able to find the following information in the datasheet: How long does the alarm interrupt last; or does it stay on and you have to deactivate it? In addition, what should the value of the R1 pull-up resistor in the diagram from the datasheet?
 

ericgibbs

Joined Jan 29, 2010
18,848
hi r,
Look at this image for the duration of the INT pin signal, usually this signal is used to trigger the Interrupt pin on a micro-proc, so it can be quite short duration,

A bistable latch could be used to hold the INT signal state for as long as required.
How are the DS1305 INT pins connected to your hardware ??

E
 

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ErnieM

Joined Apr 24, 2011
8,377
The data sheet says the alarm interrupt will be asserted during the time a match ouuures between the current time and the alarm time.

Since you must set the alarm down to one second this implies the interrupt signal will assert for only 1 second when the alarm trips.

The value of R1 mostly depends on what the interrupt pin is driving. For a CMOS logic input values from 10 to 100 K are common.
 

Thread Starter

rleroy

Joined Mar 14, 2016
3
Eric and Ernie, Thanks for the responses and advice.
The DS1305 INT pins are surface mounted on a custom PCB board and coupled with 1K resistors. Attached is a PDF of the schematic.

When I measure the current between R4 & R1 or R2, I get 5V. But when I measure between R1/R2 & INT1/INT0 on the DS1305, I get 0V.

Pin 12 of V23079 (relay) is connected to ground. Is that pulling Pin 1 on relay & INT1 on DS1305 down to ground? Similarly with Pins 7&6 on relay & INT0 on DS1305.

-R
 
Last edited:

ericgibbs

Joined Jan 29, 2010
18,848
hi r,
The INT outputs are open Drain outputs, that means no internal voltage is connected to the Pin, so you will measure 0V.
A pull up resistor on the INT0 and INT1, connected to the positive supply to the DS1305, say,4k7 resistors.

Looking at your circuit, you need a Bistable IC and relay drive transistors between the INT pins and the Relay. A HEF4013 IC would be an option.

One INT pin would Set the Bistable and the other INT would Reset the Bistable.

Do you follow that OK.?

E
 
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ErnieM

Joined Apr 24, 2011
8,377
I trust you have some sort of micro controller/computer attached to conn_10x06 as the ds needs to be configured. Have you given any thought on how you will set (and read) the current time?

I would not depend on the ds part to control the relay, that task is best handled by the micro. I would use the int outputs to send back to the micro to wake it up from a low power sleep mode to engage the relay, then go back to sleep.
 
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