I need 9V DC & -9V DC - Page 1
Forum section: electronics / projects

author: otpowell
author_rank: Regular Contributor
timestamp: March 28, 2015, 03:48:55 pm
content: I could use two nine volt batteries and then use a voltage divider to get +9 and -9 right. I have a single DC power supply that provides 16.8 volts of power. What's the easiest way to amplify the 16.8 V to 18 V? Would an op amp do fine with amplifying DC Voltage then perhaps a Trimpot to limit the output? I have an LM358 available.. Or would two 9 volt batteries wired in series be the easiest way?

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: March 28, 2015, 04:06:56 pm
content: Take two batteries in series.The ground is in the center.On one side you have +9V, on the other side -9V.You will need an additional DC source if you want to use your opamp.

author: otpowell
author_rank: Regular Contributor
timestamp: March 28, 2015, 04:13:09 pm
content: Quote from: JacquesBBB on March 28, 2015, 04:06:56 pmOk, thanks Jacques, this makes sense that was easy

author: Ian.M
author_rank: Super Contributor
timestamp: March 28, 2015, 04:26:13 pm
content: A good quality alkaline PP3 battery has a capacity of about 600mAH if lightly loaded and its terminal voltage will drop below 8V within the first 10% to 20% of its total discharge life. Therefore if your circuit's current consumption averages more than a few mAH per day, and you don't need wire-free portability, batteries are a really bad choice.I am suspicious of your 16.8V supply - is it a well smoothed and regulated supply with an output that does not drop significantly under load, or is it a cheap wall-wart or similar and that is simply the open-circuit terminal volt you have measured?Lets assume your circuit's current consumption from either rail is in the 1ma to 500mA region - if not, the following suggestions will not be appropriate.If its a regulated supply, you may be able to use a potential divider, unit gain OPAMP buffer and optionally an emitter follower power stage (if you need more current than the OPAMP can handle) to split the supply to get an 8.4V rail that can both sink and source current. If either your application circuit or the PSU doesn't have their 0V rails grounded, you could then feed your circuit +/-8.4V which would probably be enough for most +/-9V applications. If its a cheap unregulated PSU, it will droop too much under load to do this.To get an actual +/- 9.0V supply from what you have would be *MUCH* harder, and I wouldn't even try, as it would be cheaper to get a nominal 12V AC output supply and half-wave recify it to get positive and negative rails then regulate down to +/-9V with linear LM7809 and LM7909 regulators.If your load current is over a few tens of mA, heatsinking for the rail splitter power stage or for the linear regulators will be required.Someone with more experience would almost certainly prefer switched mode regulator circuits , which could convert your 16.8V into +/-9V but your chances of getting something like that designed, built and working properly are very poor.

author: LukeW
author_rank: Frequent Contributor
timestamp: March 28, 2015, 04:32:08 pm
content: Could you tell us a bit more about the application and context?As a general rule, this is really valuable in giving the best possible helpful answer.

author: ajb
author_rank: Super Contributor
timestamp: March 28, 2015, 05:55:27 pm
content: Quote from: Ian.M on March 28, 2015, 04:26:13 pmReally all you need is a pair of identical 9V output buck converters. One gets connected the conventional way to produce +9V, the other has its Vout and Gnd nodes swapped, which gets you -9V. If you really want to avoid designing them yourself, you can use a couple of pre-made modules likethis, but at 9V you don't have that many options to choose from. And of course they'll be more expensive than rolling your own, but that's the cost of having someone else do the engineering and assembly.

author: Zero999
author_rank: Super Contributor
timestamp: March 28, 2015, 06:02:51 pm
content: No you can't amplify 16.8V to 18V. You need to use a boost converter.This isn't needed. You could use a rail splitter to get +/-7.9V which is enough to run most op-amp circuits off.Or use an IC:http://www.ti.com/lit/ds/symlink/tle2426.pdf

author: otpowell
author_rank: Regular Contributor
timestamp: March 30, 2015, 10:50:28 am
content: Quote from: Ian.M on March 28, 2015, 04:26:13 pmQuote from: LukeW on March 28, 2015, 04:32:08 pmThanks all for such great info. My DC supply is a linear 15 V, 2 amp I think supply, bought it for 45 bucks off of amazon. It goes up to 16.8 V though I noticed.The application & context, I'm doing a lab for my analog class, and I prefer to do my labs at home as I have all my own equipment. I am building and testing a Class B Push-Pull Amplifier using two 2N3904 transistors.

author: Ian.M
author_rank: Super Contributor
timestamp: March 30, 2015, 12:56:43 pm
content: As you are trying to follow the labs from a formal course, although +/-8.4V would make the circuit work well enough, you don't want to be having to explain all the way through your report why your results are not quite as expected because the supply was 7% low. Rail splitters are out, even though they would be satisfactory if this was only for your own understanding and use!You've got an adjustable regulated supply that I presume could be set to 9.0V.The problem then becomes providing a regulated -9.0V rail. The simplest would be to use another PSU with a floating output. Even a cheap regulated wall-wart would do as long as its is short-circuit proof. If you have an unregulated supply that outputs at least 11V under load, use it with a LM7909 regulator wired using the application circuit in its datasheet.You would loose the convenience of an adjustable current limit, which can save your circuit if you have made a mistake and have it set to less than the lowest Ic_max for the transistors you are using. If you can afford it, you may want to consider a second supply of the same type (provided its negative output isn't grounded).I'd also get a suitable DPDT switch and wire it in the lead from the supplies to your circuit so both the positive and negative rails can be switched on and off at the same time.

author: otpowell
author_rank: Regular Contributor
timestamp: March 30, 2015, 01:27:48 pm
content: Quote from: Ian.M on March 30, 2015, 12:56:43 pmThank you for such great info. I may just grab another power supply, you can never have to many power supplies. I have an old Dell PC PSU that I was going to turn into a bench supply but I think I may have blew something in the supply. I'm gonna try to trouble shoot that and use the -12 V, wire it up to a potentiometer to get -9V. Or maybe I'll just use two 9 V batteries. Do you see any potential issues with any of my methods described? I'd like to fix my PC power supply if I can. I'll open it up and see if I can trace where the problem is.

author: Ian.M
author_rank: Super Contributor
timestamp: March 30, 2015, 03:07:14 pm
content: PC PSUs are fairly unfriendly as bench PSUs unless you need 'brute force' high current 12V, 5V or 3.3V supplies. All high current rails must be loaded or they wont regulate properly, they can easily deliver enough current to melt a solderless breadboard or burn tracks off PCBs and their negative rails are weak and usually poorly regulated. A modern ATX class PSU may not even have a -12V rail. Also their 0V rail is grounded, (and cannot be safely floated without a *LOT* of knowlage and extensive modifications).They are also a P.I.T.A to repair, even for an experienced technician, as simple faults often cause extensive damage to the surrounding circuit due to the power levels involved, and our general policy was to only attempt repair if the supply had a proprietary form factor and was not readily available as otherwise replacement with a good quality new generic PSU of the appropriate type was far more cost-effective.You cant sensibly use a potential divider to supply a lower voltage power rail in any real circuit needing more than a few mA. The output voltage will change significantly due to the load on it and to be even somewhat usable, the standing current through it needs to be an order of magnitude greater than the maximum load current. That means it would need expensive power resistors which will run hot. A potentiometer just doesn't have the required power rating unless its a large open frame wirewound one.It is possible to use a potentiometer to set the base voltage of an emitter follower stage, but by the time you've added a power transistor and heatsinking for it, you might as well have used an adjustable regulator IC and got better regulation + short circuit protection without increasing the part count or significantly increasing the cost. The classic adjustable positive regulator is the LM317 and it's negative regulator compliment is the LM337.A LM337 in the typical application circuit from its datasheet and a 12V to 18V wall-wart with a floating output (i.e 2 pin mains in) would solve your problems. Caution: its heatsink tab is connected to the input and will be at -12V relative to the 0V rail in your final circuitI should have said DPST switch - a DPDT one would of course work but it would be wired as DPST.