Get +/- 12V and 5V from 9V battery - Page 1
Forum section: electronics / beginners

author: obiwanjacobi
author_rank: Super Contributor
timestamp: April 28, 2015, 01:56:18 pm
content: I am looking for a way to power a cheap ebay function-gen that takes +/-12V and 5V from a 9V battery. I have a case and want to assemble everything into one case running on the 9V battery.Q1: Is a 5V buck converter more efficient than a (linear?) regulator?Q2: Can I use 2 12V boost converters to make the +/- 12V?I want the battery to drain as much as possible before the device stops working.I am also looking for a (cheap) premade board (kit is ok) that I can use (don't want to have to design a circuit etc).Q3: are there any converters you can recommend?Thanx!

author: mariush
author_rank: Super Contributor
timestamp: April 28, 2015, 03:08:31 pm
content: 9v batteries are pretty lousy choice for powering devices. You'd be better off with using a set of AAA or AA batteries. Ideally, 4 AA rechargeable batteries would be a great choice, as the majority of cheap regulator chips will have enough minimum voltage to work (usually most need at least 3.5v-4v to work) and they'll give you a ton of life between recharges.If you go with better switching regulator chips you could power it from even just 2 AA batteries.q1. It depends on what you choose and what components you choose.A linear regulator dissipates the difference between input and output voltage as heat, so it's basically wasted power. A fresh 9v battery will have about 10v, so you'll output 5v with an efficiency of about 50%. In contrast, even a cheap MC34063 switching regulator chip in combination with properly sized inductor and other parts would be capable of about 65-70% efficiency.Good quality (and a bit more expensive) switching regulator chips can achieve up to 90-92% efficiency.q2. It's probably not a good idea.There switching regulator ICs (again, like MC34063 for example) which can be configured to work in buck, boost or inverting mode (change positive voltage to negative) and there are ready made boards on ebay using better chips like LM2576 or LM2596 (just search for these on eBay) but the majority of the boards are buck or boost. Here's an example for a cheap ready made buck board:http://www.ebay.com/itm/LM2596S-DC-DC-Step-Down-module-Power-Supply-Adjustable-as-LM2576HV-/111492307595?pt=LH_DefaultDomain_0&hash=item19f5759a8bThese boards being "universal"/"adjustable" don't have the inductor and other parts fine tuned to get the best efficiency, but even so you'd still achieve around 75-80%. If you tweak for 6-11v input and 5v output, you could do better.But again, 9v batteries are really a bad choice.

author: jobog
author_rank: Regular Contributor
timestamp: April 29, 2015, 03:27:41 am
content: I have that same function generator and here's a circuit I came up with to power it. It's powered off the mains instead of a battery but this could easily be adapted to work with batteries but a 9 volt is really not a good choice. I have a clock module from ebay that I tried to power with a 9volt battery running through a simple 7805 regulator and it depletes a 9volt in less than a day. Now running it off of a wall wart through the 7805 regulator.

author: hans
author_rank: Super Contributor
timestamp: April 29, 2015, 08:05:34 am
content: I would not even bother in most cases making a dual supply from single. Just get one of these:https://www.conrad.nl/nl/dcdc-converter-print-tracopower-tma-1212d-12-vdc-12-vdc-12-vdc-40-ma-1-w-aantal-uitgangen-2-x-154238.html?sc.queryFromSuggest=trueYou may need to find your suitable one though, because this one powers up from 12V. Most standard modules take 3.3V, 5V, 12V or higher and convert it to an isolated +/- output.As with any DC/DC, there is quite some noise.. so you may want to whack a bit of filtering on the end to reduce noise.To make the 5V supply for least effort I would grab on of those ready made modules. Buck converters are generally more efficient than linear regulators.Generally the efficiency of linear regulators is the ratio of input & output voltages. So 9>5V is about 56% efficient.For DC/DC it's somewhere in the datasheet.. or must be measured if you really want to know it precisely.

author: MMDuino
author_rank: Regular Contributor
timestamp: April 29, 2015, 08:23:52 am
content: How is the size for that plastic Box? is it wide enough? to use 2 * 9v Batteries connected in serial...then all your questions will be answered automatically...

author: obiwanjacobi
author_rank: Super Contributor
timestamp: April 29, 2015, 08:47:16 am
content: Thanks everyone.I don't see why a 9V is such a bad choice - several people say that but do not mention why. There is lots of stuff that powers of a 9V battery (DMM)...?As I thought a linear regulator is not very efficient and (so) I don't see any use for them. I found a cheap (ebay) buck converter for the 5V (from 9V - to as low as 6V) that will probably work well.The case is a only a little bigger (in length) than the PCB. I think I can fit two 9V's in there nut then there is probably no room left for any converters...I really like the suggestion made by @hans - have not taken the time to find the correct one yet. I also wonder how low the input voltage can get before it cuts out...?M.

author: ruffy91
author_rank: Regular Contributor
timestamp: April 29, 2015, 11:21:52 am
content: 9V batteries are such a bad choice because they have a lower energy density (Wh/kg, Wh/cm^3) than AA or AAA cells and they are more expensive than AA/AAA cells per Wh.

author: Mechatrommer
author_rank: Super Contributor
timestamp: April 29, 2015, 11:25:37 am
content: Quote from: obiwanjacobi on April 29, 2015, 08:47:16 amDMM s different, it has "hi impedance input", along with other T&M devices. FG has lowish 50ohm impedance output.

author: Zero999
author_rank: Super Contributor
timestamp: April 29, 2015, 11:37:30 am
content: Do you know how much power the function generator uses?The problem with small 9V batteries is they can't provide much power and have a lower energy density, compared to other batteries.

author: Ian.M
author_rank: Super Contributor
timestamp: April 29, 2015, 01:22:37 pm
content: The PP3 9V battery is a really expensive way of buying energy.To put some figures on it, Kodak claim a capacity of 570mAH for their K9V battery discharging to 4.8V and 2900mAh for their AA cell discharging to 0.8V.http://www.kodak.com/eknec/PageQuerier.jhtml?pq-path=1694&&pq-locale=en_US&_requestid=5746Taking the nominal initial voltages of 9V and 1.5V respectively and approximating the discharge as linear gives total energies of 3.93 Joules for the 9V battery and 3.34 Joules for the AA cell respectively.However the K9V battery weighs 45 g and the AA cell weighs 23 g so in the interests of fairness one should compare one K9V with two AA cells, giving 6.67 Joules for 2x AA.Results from other manufacturers and/or if one properly integrates the area under the discharge curve remain in a similar ratio.Here in the UK, the pound shops and some major supermarkets stock Kodak brand alkaline batteries. One 9V PP3 or six AA cells for one pound. One can find some better deals on AA cells but AFAIK that's the cheapest brand name 9V PP3 alkaline battery commonly available.When one does the maths, buying energy as PP3 alkalines typically works out five or more times more expensive per Joule than as AA cells

author: mariush
author_rank: Super Contributor
timestamp: April 29, 2015, 01:23:06 pm
content: obi, by design a 9v battery can not output a lot of power.. multimeters use them because they only consume something like 2-6 mA constantly and it's a cheap for them, they just put a small linear regulator if needed on the board and they're done.Look at a datasheet for a regular 9v battery, here's an example:http://data.energizer.com/PDFs/522.pdfLook at page 2 on the "Constant current performance" graph, the one on top right. There, you can see that if your device uses 10mA no matter what voltage the battery has, the battery will slowly go down from a bit over 9v down to 6v in about 60 hours. However, if your device uses 20 mA the life goes down to 30 hours... with 100 mA, the battery will go down to 6v in about 3 hours if you're lucky.In real world however, your device won't draw a constant current, as you create several voltages from battery, as the battery voltage goes down, the current consumption will increase.You need to figure out how much power your function generator needs.Let me give you a fictional example. Let's say for example your function generator needs 5mA at 12v (60mW) and 5mA at -12v (60mW) and 20mA at 5v (100mW).When the battery is new and outputs 9v, your 9v->5v may have a 90% efficiency, your 9v->-12v may be 80% efficient and your 9v->12v may be 85% efficient so your regulators will draw 100mW/0.9 + 60mW/0.8 + 60mW/0.85 = 111 + 75 + 70 = 256 mW. At 9v, that means current drawn is 28.5 mA.But, when the battery voltage goes down to 7v, your 7v->5v may only be 85% efficient, your 7v->-12v may only be 60% efficient, your 7->12v may only be 70% efficient so now your regulators will draw :100mW / 0.85 + 60mW / 0.6 + 60mW / 0.7 = ~117.5 + 100 + ~85.5 = 303mW which at 7v equals to 43 mA.So you see, as the voltage goes down, the current goes up, the battery discharges much faster.In comparison, a rechargeable AA battery has usable voltage between 1.35v and 1.15v and about 2000-2500mAh so with two in series, you have 2.2v-2.7v. Of course, the current drawn from these batteries in my fictional example would be higher, but the energy in AA batteries is much denser so they'll last longer. With 4 batteries in series, it gets even better.ps. And since the min-max voltage range is much narrower, you can really tweak your switching regulators and inductors and so on to get the peak efficiency... with 6v-10v you get from a 9v it's harder. Even harder with 2 x 9v batteries in series, as you'd have 12v-20v voltages to work with.

author: obiwanjacobi
author_rank: Super Contributor
timestamp: April 30, 2015, 07:45:05 am
content: Ok I get it.Question remains on how low the input voltage can get before the converters cut out...

author: Zero999
author_rank: Super Contributor
timestamp: April 30, 2015, 10:33:23 am
content: Quote from: obiwanjacobi on April 30, 2015, 07:45:05 amAnd the answer to that will be on the datasheets.

author: Ian.M
author_rank: Super Contributor
timestamp: April 30, 2015, 02:19:58 pm
content: Some boost converter chips are designed to run from a single AA battery and discharge it to its 0.8V end of life voltage, however layout and component choices tend to be pretty critical if you need higher output currents, and you will find a lot more possible chips if you use two cells in series giving you a minimum input voltage of 1.6V. Once you have found a likely candidate, there is no substitute for studying the controller chip datasheet in detail as there are a lot of pitfalls - e.g. The full rated output current may not be possible at the minimum input voltage.