# Simplest way to boost voltage from AAAA battery

#### Dritech

Joined Sep 21, 2011
873
Hi all,

Which is the simplest and cheapest way to boost voltage from a AAAA battery with the least possible noise?
I need to boost the voltage from 1.5V to approx 5.5V. I looked as several Switching Voltage Regulators, including the XC6371A651PR-G , which only costs $1.35. From the datasheet, it looks simple and only requires few external components, but the datasheet is a bit vague (ex: it is mentioned that voltage can be set anywhere from 2.0V to 7.0V in increments of 0.01V, but I cannot see from where to set this). Also, are eternal controls required (ex: PWM from a micro-controller) ? I need this to be independent, i.e. AAAA battery, connected to boost converter to get an output of approx 5.5V without any external command lines. Any suggestions would be highly appreciated. jjw #### crutschow Joined Mar 14, 2008 27,008 it is mentioned that voltage can be set anywhere from 2.0V to 7.0V in increments of 0.01V, but I cannot see from where to set this You buy it with the desired output voltage. It's not adjustable. are eternal controls required (ex: PWM from a micro-controller) ? No The device has a maximum output of 100mA. Is that sufficient for your purposes? #### BobTPH Joined Jun 5, 2013 3,111 The AAAA won’t last long if he does need over 100mA. It would pull about 400mA to produce 100 mA at 5.5V. Bob Thread Starter #### Dritech Joined Sep 21, 2011 873 Hi all, Thank you for the prompt feedback. @crutschow , oops, I missed that info. So to get 5.5V, I would need a XC6371A551PR-G. As for output current, only few mA will be consumed (few ICs). I would say not even close to 10mA. Will any schottky diode work for the circuit below? Apparently the MA2Q735 is an obsolete part. #### Papabravo Joined Feb 24, 2006 15,779 You need to remember that no conversion process is 100% efficient. The power out will always be less than the power in. In some cases it will be much less. For example: If your output is 5.5 VDC @ 10 mA, that is 55 milliwatts. Let us ASSUME that the conversion process is 80% efficient (a good place to start in the absence of reliable data) That means you will need to draw 55 milliwatts /0.80 = 69 milliwatts from tha battery Now the battery will supply 69 milliwatts /1.5 VDV = 46 mA for a while, BUT As the battery discharges to say 1.3 VDC it will now require 69 milliwatts/1.3 VDC = 53 mA You can see the problem. As the battery discharges it requires more current to maintain the output voltage The more current that is required from the battery, the faster it will discharge. The thing you should fear the most is what happens as the battery voltages approaches 0. The mathematical answer is that infinite current will be required. In practice the device you have chosen may or may not behave in a sane fashion. You might want an undervoltage warning of some kind. Based on the current required from the battery, you are in a position to estimate the life of the battery before it needs to be replaced or recharged. This figure is expressed in milliampere-hours. Just like the definition of horsepower you need to recognize that some combinations of current and time are impractical. For example if your battery has the capacity of 600 mAh, you could draw 1 mA for 600 hours, but you probably should not try to draw 2.4 Amperes for a quarter of an hour. One more thing about horsepower. I know horses that can lift 550 lbs 1 foot in 1 second. I don't know any horses that can lift 1 lb 550 feet in 1 second. Last edited: #### crutschow Joined Mar 14, 2008 27,008 Will any schottky diode work for the circuit below? Apparently the MA2Q735 is an obsolete part. Yes, any Schottky of similar current and voltage rating should work. #### Papabravo Joined Feb 24, 2006 15,779 How are you fixed for the inductor? Not all 100 μH inductors are created equal. Thread Starter #### Dritech Joined Sep 21, 2011 873 How are you fixed for the inductor? Not all 100 μH inductors are created equal. Hi Papabravo, so are you suggesting that I use a circuit to prevent the battery from dropping to a certain voltage? that will surely complicate the circuit (which i intend to keep it as simple as possible). As inductor, I was planning to use the LB3218T101K, UWT1C221MCL1GS for the 220uF cap, TAJB476M010TNJ for the 47uF and SS16HE as the schottky diode (the cheapest components I found on Mouser, but I believe they should all be good for this application) #### Audioguru again Joined Oct 21, 2019 2,992 Some 9V batteries have 6 little AAAA cells in them. Energizer's datasheet for an alkaline AAAA cell show its voltage dropping to 0.8V in 4.6 hours when it has a current of 100mA. An AAA cell voltage drops to 0.8V in 9 hours at 100mA. An AA cell voltage drops to 0.8V in 25 hours at 100mA. #### Papabravo Joined Feb 24, 2006 15,779 Hi Papabravo, so are you suggesting that I use a circuit to prevent the battery from dropping to a certain voltage? that will surely complicate the circuit (which i intend to keep it as simple as possible). As inductor, I was planning to use the LB3218T101K, UWT1C221MCL1GS for the 220uF cap, TAJB476M010TNJ for the 47uF and SS16HE as the schottky diode (the cheapest components I found on Mouser, but I believe they should all be good for this application) You inductor is rated for 140 mA maximum current. I would cut that number in half and set a design goal of not operating at a lower voltage than 0.986 Volts. Keep an eye on the battery voltage with a meter and decide how much time it takes to get down that far and change (recharge) the battery. No need to add circuitry. At a battery voltage of 0.493 VDC you will be at 140 mA using the assumptions and numbers from my previous post. You must strive to avoid this region of operation, unless you are doing an experiment to find out exactly what happens. #### ronsimpson Joined Oct 7, 2019 1,182 From post #4 From post #8 There is not comparison. One is rated at 1.3A and the other at 0.14A. It is not clear about the IC but I think it can/will pull 400mA peak at start up. Even with no load the start up current will be high. I really think you need a inductor that is designed for at least 400mA. #### Bernard Joined Aug 7, 2008 5,783 What is your load? OK a few ICs but for how long, a few hours a day or continuous ? Would it be possible to use 4 AAAA s? Last edited: #### Papabravo Joined Feb 24, 2006 15,779 From post #4 View attachment 204331 From post #8 View attachment 204332 There is not comparison. One is rated at 1.3A and the other at 0.14A. It is not clear about the IC but I think it can/will pull 400mA peak at start up. Even with no load the start up current will be high. I really think you need a inductor that is designed for at least 400mA. I knew current capacity would be a consideration, but I did forget about the startup phase. Pay attention to this problem. Not all 100 μH inductors are EQUIVALENT. The devil is in the details, #### sparky 1 Joined Nov 3, 2018 496 The AAAA is about 1/8 inch or 3.5mm shorter in length than a AAA. The shorter length serves no practical purpose. In fact the 9V battery should have gone to 12V long ago. Step down beats Step up in efficiency. search ebay for DC-DC 1V-5V To 5V Boost Converter. maybe the usb, be creative put in metal enclosure for lower noise. Replace the old semiconductor company with a business model that is always American is concerned about leveraging our time and our national interests. Last edited: #### MisterBill2 Joined Jan 23, 2018 8,463 The AAAA won’t last long if he does need over 100mA. It would pull about 400mA to produce 100 mA at 5.5V. Bob Certainly correct! AND you will get a bit less power out than goes in. So those veryu small batteries will not last long at all. What do you need the 5.5 volts for?? Thread Starter #### Dritech Joined Sep 21, 2011 873 Hi all, So as per my previous posts on this forum, I am trying to do a simple EMG circuit. The circuit will basically consist of 4 channel op-amp (which fill form a HPF, LPF, amplified) and an instrumentation amplifier. I was planning to use one AAAA battery with a simple boost circuit for the following reasons: 1) I need to use a battery as it is the safest way to go when having surface electrodes connected to your body (fully isolated from any mains voltage) 2) I need the battery to be strapped to the arm together with the EMG circuit. Using four AAA batteries to get the 5V (I need 5V not 3.3V) was my original plan, but having four AAA or AAAA batteries strapped to your arm is not the best option as it is bulky and heavy. Using a single AAAA battery with a simple boost circuit would reduce the weight and discomfort by a lot. Following your feedback, now I have the following concerns: 1) What would possibly go wrong if the battery gets to a low voltage? 2) Will the booster circuit create a lot of noise and jeopardize the performance of the IA? (Please note that following the booster circuit which will supply 5.5V, I was planning to use a LDO regulator to regulate the voltage to 5V) 3) How high can the start-up current of the IA and 4-ch op-amp possibly be? Is this listed in the datasheet? (I cannot find any reference) #### sparky 1 Joined Nov 3, 2018 496 you may first find the routine of adjustng like an " AD620 microvolt amplifying module ". I recall that the board required much experience to design. It was very sensitive and input needed to be well isolated. yes when a stable constant current and voltage regulated source gets near the low drop out voltage the circuit becomes unstable, however the unit itself does not draw much current that may not happen often. #### ci139 Joined Jul 11, 2016 1,696 the simplicity is very much dependent on your output -POWER , -VOLTAGE(Stability) , Peak-/AVG. -CURRENT . . . dependent about : if your output consumes less than 1mA the full AAA gets a slight voltage drop on it's 500mΩ internal resistance --versus-- if your output is say ±15V and consumes 10mA + your efficiency is 70% then the input receives 420mW demand the best transfer of which occurs at R.ext=R.int (which we can't use because it already would cut the efficiency down to 50% without the conversion losses) . . . so say $$\frac1{1+\frac{R_{INT}}{R_{EXT}}}≈84\% → R_{EXT}=\frac{R_{INT}}{\frac1{84\%}-1}≈2.6Ω$$ that sets the "RMS" terminal voltage of the AAA to apx. $$\frac{V_{EMF}}{1+\frac{R_{INT}}{R_{EXT}}}=\frac{1.47V}{1+\frac{500mΩ}{2.6Ω}}≈1.2V$$ . . . and less as the battery discharges and its internal resistance increases Last edited: #### Alec_t Joined Sep 17, 2013 12,004 I suspect that the use of a booster will create intolerable noise levels, considering that for EMG use you will be trying to measure microvolts. Screening the booster would add weight, which you are hoping to avoid. Given that patients routinely have to put up with BP cuffs and the like attached to various bits of their anatomy, I'm sure they'd hardly notice 4 AAA alkaline cells. If those proved too heavy, then a small-capacity 2S lithium battery would be lighter. #### MrAl Joined Jun 17, 2014 8,150 Hi all, Which is the simplest and cheapest way to boost voltage from a AAAA battery with the least possible noise? I need to boost the voltage from 1.5V to approx 5.5V. I looked as several Switching Voltage Regulators, including the XC6371A651PR-G , which only costs$1.35. From the datasheet, it looks simple and only requires few external components, but the datasheet is a bit vague (ex: it is mentioned that voltage can be set anywhere from 2.0V to 7.0V in increments of 0.01V, but I cannot see from where to set this). Also, are eternal controls required (ex: PWM from a micro-controller) ? I need this to be independent, i.e. AAAA battery, connected to boost converter to get an output of approx 5.5V without any external command lines.
Any suggestions would be highly appreciated.
Hi,

First, to get a range of versions that have output voltage from 2v to 7v in increments of 0.01v they would have to have provided 500 different versions. While this is possible it is not likely especially with the given tolerance so i would guess that it is more like 2 to 7v with 0.1v increments. The way you set that would usually be with a part number suffix or somewhere else within the part number you order.

Second, with some versions you can probably set the output with a voltage divider or additional series resistor but only to raise the output voltage from what it was meant to be. It is better to get the required version however.

Third, with boost converters the inductor ESR is more important than with other converters so be sure to get a low enough ESR type. If the ESR is too high it will not be able to put out the required voltage with normal load current even with the right part number.

Fourth, a quad A battery (AAAA) will run down rather quickly. Move to a triple A (AAA) if possible. Check quiescent current and normal load current and typical milliampere hour ratings of AAAA vs AAA to estimate run time.
A standard alkaline AA has about 2000mAh capacity when new,
a standard alkaline AAA has about 1000mAh capacity,
a standard alkaline AAAA has about 500mAh capacity.

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