generating -9v rail from 9v battery

Discussion in 'General Electronics Chat' started by DaveH, Jan 1, 2009.

  1. DaveH

    Thread Starter Active Member

    Jan 1, 2009
    53
    0
    Hi AAC forum.

    I have a question about the practical details of using a basic op amp off a battery. For this application I don't think I could use a single rail supply op amp like the LM358N, because the output has to swing + and - about 0v as the offset.

    I'm building a pretty simple 741 circuit, that needs a balanced supply of +9v 0v -9v and want to run it off a 9v battery, not a bench top supply.

    If I had two 9v batteries I could make a balanced supply with that, but it would add a lot of bulk. To do it with one battery I guess the main option is a switching regulator used as an inverter to get the -ve rail. Can someone please recommend something cheap and easy to get that can do that? Surely this is quite a common task when using op amps.

    Or for a basic circuit may be that's over the top and there is a simpler way to do it with a few discrete components?

    The reason I thought about a switching regulator is at first I thought I might need to generate supplies of +15v 0 -15v from one 9v battery. I thought I could use one regulator to boost +9v to +15v and then another regulator to generate -15v from the boosted +15v. I was looking through data sheets to see if there is a device than can do boost and invert functions together in one IC.

    Many thanks for any tips.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    If you get an LMC7660, it will generate the negative equivalent of the input voltage. If your circuit needs much current, don't expect long battery life.
     
  3. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    741 opamps are a 40 year old design. Consider using TL071, TL072, LF351, or LF353 opamps instead; they're about the same price but have 40x the bandwidth for starters, and much better characteristics overall.

    9v batteries have very limited current capability. While you might use a voltage inverter like an ICL7660, a single battery will last much less than half as long as two batteries would. This might not be an issue if you don't mind changing batteries every half-hour.

    Another voltage inverter using a 555 timer IC is here:
    http://www.ikalogic.com/shm_voltage_inv.php
    Similar:
    http://www.electronic-circuits-diagrams.com/psimages/1.gif
    It will be less efficient than the ICL7660 though.
     
    Last edited: Jan 1, 2009
  5. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Correct me if I'm wrong, but the design could be improved by using Schottkey diodes and a 7555 (CMOS 555). The diodes would drop less voltage, and the CMOS chip wouldn't have the 1.4 drop on the postive output voltage swing.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Agreed on the Shottky diodes. Axial leaded Shottky diodes seem to be getting somewhat harder to come by; Radio Shack sure doesn't carry them. :rolleyes: 1N5817's would be a good candidate. BAT54 Shottkys would too, if one can handle the tiny SMT package - but the OP would have to make a PCB to use the SMT package.

    But for a one-off deal, 1N4148's or 1N914's should work, as well as 1N4000 series diodes - with a penalty that's the sum of the Vf of the diodes.

    I briefly considered suggesting a CMOS 555, but the source/sink current capability is abysmal when compared to the bjt 555.

    Another solution could use an inductor driven by an NPN transistor, but he'd have to make his own toroidal inductor; do-able but not so simple for a n00b.
     
  7. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    How about this for a driver for the 555?

    [​IMG]

    You need to use red LEDs for 5V, higher voltages can use other colors. The LEDs are probably better than zeners, but with the right selection of diodes you can eliminate the shoot through problem.
     
    Last edited: Jan 2, 2009
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    The problem is size, Bill.
    Our OP initially objected to using two 9v batteries due to the added bulk.
    However, even just a simple 555 inverter circuit is going to drain his battery pretty quickly, as well as adding bulk to his project.

    If he doesn't care about regulation of the negative voltage, he might use a modified Joule Thief circuit, but then he's into winding center-tapped inductors on ferrite beads or toroids. Seems like every solution brings a new problem to deal with.
     
  9. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Ain't that the truth. I've tried to do something similar for an RF project over 20 years ago, a microwave LNA needed a ± power supply, and it was on a pole. I really didn't want to run a 2nd wire, so I tried something similar. Problem was, the noise from the oscillator defeated the idea.
     
  10. italo

    New Member

    Nov 20, 2005
    205
    1
    9V BATTERIES ARE NOT 9V but more like 8.4v and if you expect to get 9v out of anything you put in there will never make it to 9v. IC all IC have a tendency to limit the +/- swing. So now to get 9v you will ned at least 11-13v.
     
  11. DaveH

    Thread Starter Active Member

    Jan 1, 2009
    53
    0
    This is interesting feedback, thanks.

    I'm now thinking of running off 4xAA cells giving a nominal 6v to start with, because of the increased total energy capacity, which everyone is indicating is an issue.

    It's not critical to me that the batteries give 9v, the total energy of the cells is more important and I guess I can always use a regulator to boost it up to whatever voltage I need.
     
  12. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Actually, it depends upon the load. You're right, most typical '9v transistor" batteries will measure 8.3-8.6v under a light load. There are also "industrial" 9v batteries available which actually put out over 10v with the same loading; they have 7 internal cells rather than the standard 6 cells. Radio Shack even sells an industrial version. However, the case is actually slightly larger than a standard 9v "transistor" battery, so they don't fit in all cases.

    That's if he tries to use a linear regulator, which all have a "dropout" voltage (even the "low dropout" types).
     
  13. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    It certainly is an issue. Look for the 2500mAh rechargeables and use the manufacturer's recharger; more expensive to begin with but a great bargain over the long haul.

    A standard "linear regulator" can only limit voltage and/or current; it won't boost the voltage.

    A switching buck/boost regulator can increase the available output voltage, at the expense of increased current draw on the source. A typical "boost" configuration may be around 80%-90% efficient depending upon how well it's designed for the intended application.

    So, if you want to get 9v @ 10mA from four 1.5v batteries, you'll need 10mA/6v*9v/80%, or about 18.75mA current at 6v. If your rechargeables are 1.2v, you'll need 10mA/4.8v*9v/80% = 23.44mA current. The lower your remaining battery charge, the more current the boost regulator will draw until the batteries are dead.

    Batteries will last much longer when discharged at a slow rate. This is due to the internal resistance of the batteries themselves. The higher the current draw, the more power is lost via the internal resistance.
     
  14. engcharlo

    New Member

    Jan 11, 2009
    1
    0
    How cn use buck boost regulator effectively to a low rpm wind mill?
     
  15. papercutt

    New Member

    Oct 2, 2009
    1
    0
    Art of Electronics Paul Horowitz Page 225 fig 4.53 gives you a general schematic...i ahve built thecircuit it works...my phone line was very soft so i built this amp for boosting...
     
  16. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The opamp circuit in The Art of Electronics has only a positive supply voltage and biases the input at half the supply voltage then uses an input coupling capacitor and output coupling capacitor so that their inputs can swing above and below 0V.
    another coupling capacitor blocks DC to ground for the negative feedback.
     
  17. rogs

    Active Member

    Aug 28, 2009
    279
    37
    As the device is to be battery powered, why not use a simple resistive divider to generate a 'half line' votage, and then use that point as the external common (or 0v) reference to the outside world. Output voltage would then swing plus and minus to that reference?

    Use a low powered opamp -- TLO61 or something similar -- make the reference divider chain high resistance, to keep the quiescent current down.

    Got to be simpler, and less current hungry, than creating a real negative rail I would have thought?

    Unless of course that won't give you enough output voltage swing.
     
Loading...