I am using the SIM900A GSM module which requires 3.2V to 4.8V with max current draw of 2A.

In my board, I have a 5V, 3A supply which will be used to power the SIM900A.

Generally, in my search I found 2 ways to power it:

  1. Use a LM317 regulator to produce around 3.9-4V
  2. Just simply use a high current series diode (like the 1N4007, with forward voltage = 0.6v) to produce 5V-0.6V = 4.4V

My question is Is there any advantage of 1 over 2 (or disadvantage of 2 over 1 for that matter) as 2. is much simpler and cheaper.

I want to use 2 if there shouldn't be any major disadvantage.

I have seen both in commercial Chinese SIM900A modules, so I guess there should not be any problem with 2. But then why use the LM317?


  • 5
    \$\begingroup\$ An LM317 won't work to regulate 5V down to 3.9-4V since it has a minimum dropout voltage (difference between input and output) of at least 1.5V. 2A is also pushing your luck since the LM317's I'm familiar are only guaranteed to supply up to 1.5A \$\endgroup\$
    – brhans
    Jul 27, 2016 at 2:05
  • \$\begingroup\$ A 1N4007 can handle 1A max, and at 1A its voltage drop will be significantly higher than 0.6V (I'd expect about 0.9V). \$\endgroup\$
    – marcelm
    Jul 27, 2016 at 8:09
  • \$\begingroup\$ @marcelm: thanks for pointing that out. Well, 0.9V drop would be even better as the recommended voltage for the SIM900A is around 3.9-4V. While I still believe the 1N4007 can handle the job (as I have seen in commercial modules), I will look for an alternative with higher current rating. Any suggestions you have in mind? \$\endgroup\$
    – Dave
    Jul 28, 2016 at 2:50
  • \$\begingroup\$ @dave 1N400x diodes are really only rated for 1A. Consider that at ±1V drop and 3A it'll dissipate ±3W. For more current, you could look at the 1N540x series which is rated at 3A. \$\endgroup\$
    – marcelm
    Jul 28, 2016 at 6:34
  • \$\begingroup\$ @marcelm: Thanks very much for the advice. Yes, I did do a quick search for high forward current diodes to replace the 1N4007 and the 1N540x came out. Just that from datasheet they said max voltage drop at 3A is 1.2V (which is still good), but not sure how much it is in lower load... Need the voltage drop to be at least 0.3V in any case. I think I will connect some tiny load and check out with a multimeter \$\endgroup\$
    – Dave
    Jul 29, 2016 at 7:34

1 Answer 1


The simple series diode is just fine IF you can guarantee that the input voltage will never exceed 5.2 Vdc.

The LM317 regulator has the advantage that it will protect the GSM module if the input voltage is excessive.

For example, if someone connected the power supply to 12 Vdc instead of 5V, the LM317 regulator would simply get very hot, then shut down because of the internal thermal protection. This would save your (expensive) GSM module from being damaged.

Either method is fine.

  • \$\begingroup\$ The 5V is provided by the LM2576-5V buck converter (down from 12V) and I think the quality of the output is very good. I don't think it will ever exceed 5.1V \$\endgroup\$
    – Dave
    Jul 27, 2016 at 2:02
  • \$\begingroup\$ In that case, two diodes in series would make sure it stays in range even if the input goes from 5V to 6V. \$\endgroup\$
    – Passerby
    Jul 27, 2016 at 2:14

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