I want to power GSM SIM900A with Lead acid rechargeable battery 12V, 1.2Ah. The voltage regulator I am using is L7805 which gives a regulated output voltage of 4.9 V.

My GSM SIM900A modem is

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My battery is

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How should I design the circuit so that I get 3.2V - 4.8V with 2A for proper GSM module operation without destroying any components ?

  • 1
    \$\begingroup\$ You start by tossing the 7805 back in the parts bin and purchasing a switching regulator to take the 14V from the battery down to the required 3.2 Volts. Using a 7805 to regulate 14V down to 5V at 2A would waste over 18Watts of power, requiring a large heavy heat sink - if the 7805 could provide 2A, which I don't think it can. Your 1.2Ah battery won't power your modem for very long if you are making heavy use of it. Standby could probably run for a good long while (days) but actually making use of the modem would suck the battery dry pretty quick. \$\endgroup\$
    – JRE
    May 27, 2017 at 7:11
  • \$\begingroup\$ Thank you very much for your response. I am going ahead with LM2576. \$\endgroup\$ Jun 1, 2017 at 12:29

1 Answer 1


Before you get too far into designing your circuit, you should calculate your power budget. Your modem draws 2 amps at 5 volts. That is 10 watts of power (2 amps * 5 volts). As JRE recommended, a switching regulator should be used. If you allow for 90% efficiency, you are now up to 11.1 watts (10 watts / .9).

Your battery can supply 1.2 amp hours at 12 volts. That is 14.4 watt hours (1.2 Ah * 12 volts). If you now divide that by your 11.1 watts of power consumption, you can estimate ~1.3 hours of operation. If the average power consumption of the modem is less, this operating time can be extended.

If this is sufficient for your application, start looking at 12 volt in, 5 volt out switching regulator designs. To avoid damaging your battery, the regulator should drop out when the battery reaches ~11 volts.

  • \$\begingroup\$ Okay.....I will try your method. But Right now, I have 2 voltage regulators already with me - LM317, L2576. \$\endgroup\$ May 30, 2017 at 6:59
  • \$\begingroup\$ The LM2576 is a bucking, switching regulator. That is the one to use if you have it in the 5 volt version. The LM317 is a linear regulator that will waste too much power. \$\endgroup\$
    – Glenn W9IQ
    May 30, 2017 at 7:04
  • \$\begingroup\$ Thank you very much for your response. I am going ahead with LM2576. \$\endgroup\$ Jun 1, 2017 at 12:30
  • \$\begingroup\$ A major error in this analysis in terms of battery life is that 2amps is a pulse peak for regulator design, average draw will be much lower. \$\endgroup\$ Sep 23, 2018 at 14:35

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