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High Efficiency Battery Boost Regulator using the MCP1640Author
Giorgos Lazaridis
February 8, 2012

Here is what... I love camping - which means that in the summer i find time to go camp somewhere with my friends, usually in the middle of nowhere, get a fire going, grill sausages and play the guitar. But when the night falls, we need light. Full moon cannot be considered as a sufficient light source. Some years ago i made an LED camping light which works still today (and frankly i'm surprised how well it works).

What i want now, is something to spice up this hack. So here is what - I used the MCP1640 boost converter to drain the last electron from the batteries. This chip can work with a ridiculous low voltage and provide enough power to drive a couple LEDs. Which means the 2 AA batteries will operate even longer and the LEDs will be much brighter.

Moreover, rechargeable batteries can be used. Each rechargeable battery provides 1.2 volts. So 2 batteries provide at best 2.4 volts, which is not enough to drive effectively an LED. The MCP1640 chip boosts this voltage to the required levels of an LED, and maintains this voltage (roughly) even when each battery is as low as 0.8 volts, less than half its capacity!

Here is a photo from the circuit:


This is the bottom side of the PCB. Do you see the MCP1640? No? Of course not, it is tiny! Look here:


The PCB is extremely small, and could become smaller if i had used ceramic capacitors. The size can be trimmed down to 23x23mm (that is 2.5cm each side, or about 0.1 inch!). The chip and the inductor that I've used are ultra small! 3x3mm is the overall size of the inductor and the chip!

And here is the top side with the terminals:


I've used a jumper to bypass the enable signal (middle position of the 3-way terminal) in case that you want to have the regulator always running, otherwise remove the jumper (blue plastic HDD-like jumper) and the regulator will only work if the middle terminal of the power supply is pulled high.

Here is a photo with the boost supply in action:


As you can see, only 2 batteries (that are half dead) can provide 135mA of current to 2 super bright LEDs (each one can draw 100mA).

And finally, a video

Now you may wanna go to the worklog tab and find the schematics and the PCB designs for this circuit, or go to the references tab to find out my resources.

   Continue reading. Click here to view the worklog.



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  • At 12 January 2015, 21:21:09 user 123 wrote:   [reply @ 123]
    • VFB is always 1.12V -> should be corrected to 1.21 V
      Data Sheet MCP 1640

  • At 9 July 2013, 16:29:40 user Angel G. wrote:   [reply @ Angel G.]
    • I experimented today with a MCp1640 and also got around 140ma over range of Vin=2.7 .. 4.5V.
      Divider is set so the output to be 5.5V, but it can't get over 4V.
      I think there's a bug in the chip, because: When I disconnected the load (a resistor) & powered up w/o load. It managed to hold 5.5V with over 200ma to the resistor. So...

  • At 15 February 2013, 22:11:29 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @Ben Choy It is designed for 3.3 and 5 volts, but i'm quite sure it can do well at 1.2 as well. Or you can use 2 diodes to drop with little power loss.

  • At 14 February 2013, 4:35:19 user Ben Choy wrote:   [reply @ Ben Choy]
    • Just wonder, can this be adopted as a laser LED PSU, since I have a need to drive an laser LED on/off with logic voltage. The laser should take about 200mA to 300mA @ 1.2V.

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