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I have been building a portable device which has as its main components a dsPIC33FJ128GP804 and a graphical liquid crystal display, which is working fine.

The PIC voltage range is specified at 3.0v to 3.6v, and an absolute max of 4.0v. The LCD has a very similar voltage range.

While operating, the device uses about 60mA of current and when in sleep mode, both the PIC and LCD can drop their current consumption down to a few micro amps. I was hoping to not need a power switch and just put the device in sleep mode when no buttons have been pressed for a few minutes.

I have been experimenting with different ways of powering the device, initially hoping that I could power it from 2 AA batteries. However it doesn't take long for the battery voltage to drop to a point where the PIC becomes unstable.

I added a surface mount 3.3v regulator (AMS 1117) and tried running the device from 3 AA batteries, but now in sleep mode the current consumption has risen to 3mA which is not ideal.

I experimented removing the regulator and using a few 1N4001 diodes to drop the 4.5v to under 4.0v, but I am not too convinced with the reliability of the results.

I was wondering if anyone could suggest a better way of powering this device to get a good battery life and a still be able to use sleep mode?

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  • \$\begingroup\$ The Regulator Powering would be better option rather than connecting diode and dropping the voltage ! \$\endgroup\$ – Photon001 Feb 25 '16 at 13:27
  • \$\begingroup\$ Look for a better LDO. Most of that additional sleep-mode current is just being wasted by your AMS1117. \$\endgroup\$ – brhans Feb 25 '16 at 13:48
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One way to get better battery life would be to go with an intelligent switching regulator which can transition from buck mode (step-down) to boost-mode (step-up) as the batteries discharge. This allows you to more deeply discharge the batteries vs. a linear regulator solution. Linear Tech and Texas Instruments offer parts that provide this functionality, among others.

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