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I have a keychain-sized China-made camera that has an internal LiPo 3.7v dc battery, and a TI MSP430 microcontroller that operates between 1.5v and 3.6v dc. I want to use the microcontroller on the same battery as the camera, and use the controller with transistors to digitally "press" the on/off and start/stop buttons on the camera. The end goal is to take periodic photos for time lapse usage.

My question is: How can I drop the voltage from a 3.7v battery to about 3v so I can use the same battery for both?

A CS professor I talked to suggested in passing that I can use a resistor. Will that work, and if so where would I need to insert it? How do I determine what resistance the resistor would need to be?

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  • \$\begingroup\$ I might mention that the china-made camera can run on as little as 3.45v, but has been known to malfunction under 3.6v, so I would like to keep it above that point. I would also like to eventually convert the final device to run on 3 AA batteries, but one thing at a time. \$\endgroup\$ – mouseas Aug 21 '12 at 17:50
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The cheapest way is a diode in series with the 3.7 V LiPO. That will drop about 0.6 V (the MSP430 will probably not consume much power). A resistor's voltage drop will be less constant. The MSP430 will operate at voltages as low as 1.8 V, and then the camera will have given up already.

To simulate a button-push you can control a transistor which you mount parallel to the button, from the MSP430.

The diode will give you a constant voltage drop, so if the battery's voltage sags to 3.5 V the MSP430 will get 2.9 V instead of 3.1 V. It doesn't matter much for its operation, like I said it will keep working at a voltage as low as 1.8 V. (At work I tested how low it would go, and it still worked at 1.3 V, but that's no longer guaranteed.)

If you want to be a Good Boy you would use a voltage regulator, in this case an LDO, for Low Drop-Out. Many voltage regulators are three pin devices: 1 pin input, 1 output, and a common ground, that's the negative pole of your voltage. Standard regulators need a few volt more input than output, so for 3 V out you may need 5 V in, more than the battery can supply. An LDO will work with less voltage difference. For instance a 2.5 V LDO may need only 2.7 V input. The regulator will keep the output at 2.5 V as long as the input is higher than the 2.7 V. You'll also need a couple of capacitors, one for the input, one for the output.

Oli's MCP1700 is a good choice. It has a very low ground current, together with the MSP430 you should be able to stay below 10 µA, so you'll barely discharge the battery. 44 cent at Digikey (Farnell is 55 cent).

enter image description here

A 1N4148 diode costs 10 cent.

enter image description here

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  • \$\begingroup\$ Any diode will work? \$\endgroup\$ – mouseas Aug 21 '12 at 18:00
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    \$\begingroup\$ @mouseas - Almost any silicon diode. A 1N4148 is the standard for low currents, a 1N4001 is higher current, but both are suitable. \$\endgroup\$ – stevenvh Aug 21 '12 at 18:01
  • \$\begingroup\$ Ok, cool. Would two in series drop about 1.3v, or is there only one voltage drop? \$\endgroup\$ – mouseas Aug 21 '12 at 18:03
  • \$\begingroup\$ Two will give you a 1.2 V drop, three 1.8 V, etc. until you've used all the 3.7 V available. \$\endgroup\$ – stevenvh Aug 21 '12 at 18:04
  • \$\begingroup\$ Awesome. Just for kicks, what is the normal way to reduce a given voltage to a given lower voltage? (ie the way you're "supposed" to do it?) \$\endgroup\$ – mouseas Aug 21 '12 at 18:05
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Yes a resistor would work, but you could also use a Linear Voltage Regulator. It's actually very simple if you want to do it using a resistor as well:

R = V / I

R = 0.7 (Potential difference across resistor) / I

What this mean's is that the resistor you'll want to use will change based on the current being supplied from the battery. This is why I'd suggest using a linear voltage regulator. To a certain extent, it will deal with a range of input currents and it will automatically regulate voltage for you.

Note that if you choose to use a linear regulator then you must pay attention to the so called 'dropout' voltage. Essentially this value is the minimum input voltage above the specified output voltage in order to produce the desired output. For instance, if you were to use a 3.0V linear regulator with a dropout voltage of 0.5V, you'd have to supply 3.5V in order to reliably get the full 3V.

So the maximum dropout value of linear voltage regulator you could use would be 0.7V as that is what is being provided by your battery(though in reality you'd want it to be a bit less than this to account for battery voltage variation). At my usual electronics site I found a 3V regulator that has a dropout voltage of 0.38V @ 200mA within a minute or so, selling 50 for £8.85, so it's entirely possible to get them cheap as well.

Regarding efficiency, the efficiency drops as the input voltage increases relative to the output voltage.

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    \$\begingroup\$ Maybe you could tell us what regulator that is. \$\endgroup\$ – stevenvh Aug 21 '12 at 18:29
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A resistor won't work very well, as the voltage will then depend on how much current the microcontroller is drawing.
A very simple method would be to use a series diode to drop ~0.6V from the supply (e.g. 3.7V - 0.7V = 3.1V), but the most accurate/stable way would be to use an inexpensive 3V low dropout linear regulator (LDO) like this (also has an SOT-23 version):

 3V LDO 3V LDO SOT

It costs £0.35 qty 1, can handle a current up to 250mA, and has a dropout voltage of just 178mV (this means it will stop regulating properly at 178mV above it's setpoint, so ~3.178V, which is well below the battery nominal voltage)

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It's been a few months since OP asked, but frankly, the msp430 can take more than 3.6v. The datasheet for most of the valueline shows 4.1v as the maximum rated voltage, with 3.6 as the recommended. Giving it 0.1v more than recommended won't kill it. Between the low current draw of the msp430 and the voltage drop that the camera will cause, you will be beyond fine.

tl:dr; there is a difference between recommended voltage, and maximum voltage.

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  • \$\begingroup\$ thank you, i was about to buy a lipoly 1s battery and this was my hunch but everybody seems to be rather "politically correct" these days :) \$\endgroup\$ – necromancer Jul 30 '14 at 10:57

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