Replacing 9v battery with a wall plug sanity check

Discussion in 'The Projects Forum' started by tsservo, Jun 15, 2008.

  1. tsservo

    Thread Starter New Member

    Jun 15, 2008
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    Hi all. I have to admit up front that I know staggeringly little about wiring, electricity, etc. I have soldered before, but just to connect some insulated wires together... I also know that I need a sanity check before trying something like this out...

    I have a particular piece of audio gear, an XLR converter for some microphones that I use to record test sessions - one of these: http://www.beachtek.com/dxa6vu.html

    It takes 9v batteries for the phantom power that goes to the microphones, but the problem is that it goes through them like a kid in a candy shop. I get about 2.5 hours out of a single battery when powering two microphones. With the amount of sessions I do, this is becoming expensive. And if I forget to change the battery, it inevitably dies in the middle of the next session.

    As the converter does not have a plug for a power adapter, I was thinking about making one myself. This is the sanity check part as the device costs about $400 to import from the US (I'm in Australia) or almost $800 to buy here.

    I was looking for something that I could just buy outright, but I can't seem to find anything... Probably a bit too niche of a need. So I came up with the following plan instead:

    1) Get a 9v AC/DC wall adapter... they have one there that is 9v, 500mA output only with the exchangable tips that you can flip around + and - on...

    2) Get a pack of the little leads that you plug a 9v battery into, they come with the red and black wires already attached...

    3) Get a connector from one of the bins that you can unscrew to attach the 9v connector wires to... One that fits one of the exchangable tips in the package of course...

    4) Connect the 9v leads to the connector, keeping the polarity the right way around (I'm going with the guess that this does matter) and plug it into the 9v AC/DC adapter...

    5) Somehow get the 9v connector in the right position in the device (it goes into a little sliding tray), plug it into the wall and turn it on...

    6) No expensive smelling smoke...


    In my head this works. But I also don't know this stuff as well as you folks here do - so I'm probably missing something terribly obvious which would result in a bad day.

    Can anyone let me know if this will work, or if I have to adjust it in some way? I'm also interested in what I might be able to do to test it before I plug it into expensive equipment...

    Thanks much for your time.

    Tim
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Yes, that would work - HOWEVER!
    The 9v battery adapter w/leads, the one you said "pack of the little leads that you plug a 9v battery into, they come with the red and black wires already attached"

    You must connect that up to your 9v supply with negative to the red lead, and positive to the black lead.

    This isn't intuitive at all, but realize that you're going to be using the 9v battery adapter backwards. They're set up on the premise that you will be connecting them to a 9v battery, not another battery adapter.

    If you tried hooking the red wire to the + of your supply, the black wire to the - of your supply, and there were no internal protection diodes in the phantom mic power adapter, you would let all of the magic smoke out of your phantom mic power adapter.

    Testing it: Connect a 150 Ohm 1 Watt resistor across your powered 9v battery adapter. At the same time, connect your voltmeter across the powered battery adapter with the + lead to the smaller (male) clip. You should read in the vicinity of +8.5v to +9.5v.

    To make sure you're reading it correctly, try it on a regular 9v battery. You need to get the same polarity and roughly the same voltage from your adapter.

    The 150 Ohm 1 Watt resistor is a simulated load. A typical 9v rechargeable battery has a capacity of 150mAh, or 60mA for 2.5 hours.
    So, since R=E/I, and E=9v and I=0.06, then R = 9/0.06 = 150.
    Power is E x I. So, since E=9 and I=.06, power = 9 x .06 = 0.54 Watts. Normally, we double it for safety, but this is just for a short duration test.
     
    Last edited: Jun 15, 2008
  3. Audioguru

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    Dec 20, 2007
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    The battery does not have mains hum. The AC to DC adapter might produce lots of mains hum. You might need to use a "regulated" 9V AC to DC adapter to eliminate hum.
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    Back when I was building power supplies (wall warts are a blessing!) zeners knocked the hum dead, but the catch is it has to be a low power system.
     
  5. tsservo

    Thread Starter New Member

    Jun 15, 2008
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    Thanks for the information, especially on how to test it to make sure I don't cook anything when hooking it up...

    I wanted to clarify how I was going to connect the 9v connector and leads to the power adapter...

    Looking at my Duracell 9v, it indicates that the (+) terminal is the post terminal, not the "spring" socket terminal. I was going to arrange the 9v connector with the red and black leads in the sliding drawer so that it was the same way as the battery and connect the red lead to the positive from the power adapter...

    (inside the device it is just two metal plates which make contact with the 9v and cutouts on the sliding drawer so you don't turn it around the wrong way. It's not like a home smoke detector with the two leads and a connector that you have to snap the battery into).

    I think that's what you're referring to where I have to flip it around the other way... That the black lead goes to the positive from the adapter. In my mind I don't see how the power adapter is being used backwards - but that's probably the non-intuitive part that I'm not understanding correctly...

    Thanks again.

    Tim
     
  6. tsservo

    Thread Starter New Member

    Jun 15, 2008
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    So is there anything specific that I'm looking for when I get the power adapter (it was going to just be you standard wall wart...)? Is it going to say "regulated" somewhere on the package or will it be written in some other way?

    Or is there something else I can do or a specific adapter I can get that will minimise or eliminate the possibility of the hum?
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    If you did so, you would have a high probability of letting the magic smoke out. :eek:

    Take another look at your Duracell 9v battery, and that small male "post" terminal that's marked "+".
    When you plug a 9v battery adapter/clip into it, you plug the adapter/clips' larger female side over the batterys' smaller male post, and the batterys' larger female post gets the adapter/clips' smaller male post plugged into it.

    The battery adapter/clip RED lead is connected to the battery adapter/clips' FEMALE terminal which snaps over the batterys' MALE terminal which is positive. But you're going to be using the battery adapter/clip to REPLACE the battery. So to match the terminals up, you have to swap the red and black leads around, too.

    Well, normally these clips are used to accept power from a 9v battery, or perhaps to charge up a battery, not as a replacement source for a battery.
    It would be very easy to get confused on this.

    Have you ever opened the unit?

    Were it mine, I would be sorely tempted to add an external power jack somewhere convenient on the chassis, and a small regulator and filtering circuit on the inside, protected against reverse polarity on the input with a Schottky diode. If there were room, a compact battery charging circuit could be added to top up a rechargeable NiMH 9v battery.
     
  8. tsservo

    Thread Starter New Member

    Jun 15, 2008
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    Ah, yes, this makes sense... I got it now. I also sketched it out on some paper to make sure that it continues to make sense for me... :)

    Yes, you're absolutely correct, I would have likely released all of the magic smoke... Hopefully in a $400 piece of equipment they've got something more failsafe than some cutouts on a battery tray to minimise damage if someone puts the battery in wrong way around.

    That sounds like a fantastic idea, especially the Schottky diode, because I've got no idea what on earth that is and neither would anyone I told about this... :) But it's well over my level of expertise, and I'd be a bit apprehensive to experiment and learn on the one (and difficult to replace) piece I've got if it involves soldering anything that's inside the magic containment case.

    Regarding the mains hum that was mentioned earlier, is there anything that I can add to this picture that would filter some of that? Say, build the regulator and filtering circuit in an external box? Or is that down to the quality of the wall wart I get?

    Thanks again.

    Tim
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    Wall warts don't typically regulate well, if at all.

    I'd get a 12VDC wall wart capable of at least 200mA output, and use an LM7809A regulator, with a 220uF capacitor on the input, and a 10uF and a 0.1uF capacitor on the output. That should keep things very quiet.

    You need a 12v wall wart because the old-technology LM7809A regulator drops a minimum of 2v across itself.

    Here's a link to LM78xx regulators on Fairchild's site:
    http://www.fairchildsemi.com/pf/LM/LM7809.html

    Yes, there are more modern regulators out there, and some exotic buck/boost converters, but I'm just trying to keep this simple and inexpensive.
     
  10. tsservo

    Thread Starter New Member

    Jun 15, 2008
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    If you could see me I'd have a polite smile and nod look on my face... I suppose that I could make something like that - but if I find a regulated power supply somewhere (I found one online so I can try to find one here in Oz) would that effectively accomplish the same thing?

    I suppose that if I've got hum I could pop back in for some instructions on making the thing that you just described...

    I'll get my parts together in the next few days and let you know how it goes. Thanks for all the help.

    Tim
     
  11. Wendy

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    Mar 24, 2008
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    Don't let them intimidate you, they are easy to build, but I still like zeners. How much current does this take you suppose?
     
  12. SgtWookie

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    Jul 17, 2007
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    Bill,
    He mentioned before that he was using rechargeable 9v batteries.
    I think these usually have a capacity of 150mAh, and his die after 2.5 hours of use.
    So, that would be around 60mA. 9v/60mA=150 Ohms, which is what I was using for a load resistor when fiddling with some circuits simulations.

    Now that I re-read it, he said "change", not "charge" - so my calculations are off.

    If he's using an alkaline cell, that's about 500mAh, or 200mA current. That's quite a bit.

    If I remember right, phantom power for mics is generally around 48v. (ETA: just confirmed that by looking at the site referenced in the first post.) Based on that wild guess, I'm going to assume for the moment that his phantom power supply has some sort of chopper/boost circuit in it. A 556 running at full blast driving alternate sides of a center-tapped primary boost transformer would neatly account for that kind of current draw. But that's all in the realm of wild guesses, of course. Without opening the box, it's hard to say precisely what's inside.

    There are a number of LEDs used as Vu meters. They'd suck up a fair amount of current. Supply impedance would need to be pretty low, so I'm afraid a resistor/Zener combination is a no-go. I have a feeling that this thing won't be overly sensitive to noise.
     
    Last edited: Jun 17, 2008
  13. Audioguru

    New Member

    Dec 20, 2007
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    The chopper probably operates at an ultrasonic frequency so that it is easily filtered and any ripple will not be amplified by the mic preamp. It probably does not have enough filtering to remove mains hum from a cheap wall-wart power supply.
     
  14. Wendy

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    Mar 24, 2008
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    So a 7808 with a diode on the base would do it. Might not even need capacitors.

    [​IMG]

    I think that is about correct.
     
  15. tsservo

    Thread Starter New Member

    Jun 15, 2008
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    That doesn't look too bad at all, actually. It might have to give that a go... I'll pull together the basic adapter (as I'll need that in the next few days) very soon and find where to get the other parts not terribly long after that, hopefully.

    Probably a pretty basic question, but the diode hanging off the 7808, is that just simply a diode? Or are there specific flavours of diodes?

    If anyone would like to see the inside of this thing, I'd be happy to open it up and take some photos for you...
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    As far as diodes go, rectifier-types like the venerable 1N4001-1N4007 1A diodes drop roughly 0.65v across them when forward biased at moderate current levels. Exactly how much depends upon the individual diode and how much current is passing through it.

    I just subjected a 1N4002 to a torture test, since it was a hapless victim waiting near my bench supply and meters. :D See the attached spreadsheet and data graph. The odd "hump" in the graph is because I didn't stick with the 1,2,5,10 progression from 10mA thru 50mA. Without the odd values, it's nearly a straight line.

    Of particular interest to you is the 5mA to 8mA range, because that is the quiescent current the LM78xx regulator IC's draw, which is sent through the ground terminal. This diode tested right around 649mV Vf @ 5mA, and 684mV @ 10mA.

    So, if you used a 7808 regulator with a forward-biased 1N400x diode in the ground path, you'd wind up with about 8.65v out.

    But it would be easier if you just bought a 7809, which is already a 9v regulator.

    I still would recommend having a filter cap on either side of the regulator, just to help ensure stability.
     
  17. tsservo

    Thread Starter New Member

    Jun 15, 2008
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    I'm guessing that if I use the 7809 I won't need a diode then?

    I've started to think about how to get this into a nice little package so that I don't have random little regulators hanging off a line somewhere (and where someone will inevitably break it off). I'm thinking that if I can find a little box about the size of a 9v battery (or hollow out an actual 9v) I could make this adaptor like device sit right in the existing tray of the converter... That might be pretty neat I think. Worst case, I could stuff it in a protective little external plastic box. Unless you know that to be a bad idea, like that regulator will put out too much heat or something...

    Any specific filter cap I should be looking for?

    If that's all I need, I might just go for it right out of the gate...
     
  18. SgtWookie

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    Jul 17, 2007
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    Yes, I already suggested that you use a 220uF cap on the input, and a 10uF and 0.1uF on the output. The 220uF and 10uF can be electrolytic, but should be at least double the voltage you're filtering for long life. So, use caps rated 25v. The 0.1uF cap should be ceramic, tantalum or poly. The small cap filters out the high frequency noise. The larger caps smooth out the larger bumps.

    You should be able to fit everything into a pretty small size package. Make sure that you insulate the leads using shrink tubing.
     
  19. tsservo

    Thread Starter New Member

    Jun 15, 2008
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    Ah, that's what that was... It's all making more sense now.

    Time to track down some parts.

    Thanks again.

    Tim
     
  20. Wendy

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    Mar 24, 2008
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    Shows how out of date I am, I didn't know about the 7809.
     
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