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I have an alarm device. It convert 12v to 4,3V to the phone. It's outputs are 4.7V, GND, and two output to the phone buttons

If the inputs loop of the alarm device is opened that means the wire have been cut, so it gives a signal to the buttons, so the mobile is start dialing a given phone number. Now because the something went wrong with the previously phone. It was needed to replace the phone so I am trying to interface a motorola t191 to the alarm curcuit. If I give 4.3V direct to the phone it doesnt accept and show an "invalid battery" message. The battery has 4 pin. I measured the battery curcuit to figure it out how to attach my supply.

The pins of battery are:

  • pin1 B- Battery negative
  • pin2 BATID Battery ID
  • pin3 TBAT TemperatureBat(thermistor)
  • pin4 B+ Battery positive

I found the the schematics of the battery. http://monitor.net.ru/forum/files/motorola_t191_-_dorabotka_v_cepi_raz_ema_akkumulyatora_546.png I took apart the battery I have. http://imgur.com/W73Y7K5 R1 is the battery id resistor R2 is the thermistor It is different then the curcuit above.

Is there a way to power this phone with my 4.3V external power, or a simple way to solve this? Any help is appreciated.

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  • \$\begingroup\$ So, have you tried to power the phone using the schematic you've found? \$\endgroup\$ – Dmitry Grigoryev Mar 30 '16 at 13:38
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    \$\begingroup\$ Just I didn't understand the russian text on the schematics. I tried just without the capacitor and with potentiometers. Now it sometimes swithces off randomly. but when it was switched on for a few seconds the phone accepted the power. i will give a try with resistors \$\endgroup\$ – Kombye Mar 30 '16 at 15:41
  • \$\begingroup\$ Random switching off may indicate that the phone wants more current than your alarm device can provide. If the current peak is short, a 1000uF capacitor between terminals 1 and 4 may help. \$\endgroup\$ – Dmitry Grigoryev Mar 30 '16 at 15:49
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    \$\begingroup\$ Thank you for your fast answer as I see the alarm devices has a capacitor. Now I set the potentiometer values well, and didn't switch off and accept the power supply too (only vibrate the screen I think because I didn't put C1 capacitor in the curcuit and used a lot of jumper cable to test. ) Thank you for your help and fast answer. \$\endgroup\$ – Kombye Mar 30 '16 at 16:02
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Well it is obvious that when you have 4.7V V between pins 1 and 4 while the phone accepts 3.7V then you'll have problems.

The solution would be to step down the voltage to 3.7V using various different methods. A voltage regulator would do the job.

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