O2 amp charger
May 27, 2013 at 6:12 AM Thread Starter Post #1 of 11

worrblade

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I am not sure if this is the right place to post this! so sorry
I am currently in china and i need a charger for my O2. I was looking around online for the wau16-400 but its a pain for me to try to set up an account and try to order something online because i dont have a chinese bank account. 
I was looking online but i can not find any kind of data sheet for the O2. I looked around today and found a lot of 2.1mm chargers that would fit but i didn't dare try to use one because i dont know what my O2 needs.
if anyone knows what charger I could use other than the triad chargers please let me know! if you can tell me the specs I can go find one. 
 
May 27, 2013 at 8:29 AM Post #2 of 11
15VAC-20VAC with 500ma min. is all u need
 
You can prob find winstar variable vac transformers easily, dont need one with a fixed voltage imo.
 
May 27, 2013 at 9:03 AM Post #3 of 11
I found a link to the O2 power supplies and what will work. Could some one please explain to me why it NEEDS to be AC/AC and why AC/DC will not work?
i dont know why AC/DC would not charger the batteries
 
May 27, 2013 at 2:38 PM Post #4 of 11
Quote:
I found a link to the O2 power supplies and what will work. Could some one please explain to me why it NEEDS to be AC/AC and why AC/DC will not work?
i dont know why AC/DC would not charger the batteries

I'm pretty sure there's rectification and linear voltage regulation on the O2 PCB.  In that case, you'd completely waste the DC power supply and lose several volts in the rectification and regulation process.  There wouldn't be enough voltage left to overcome the charging impedance of the batteries. IOW, it wouldn't charge.
 
In the process of rectification, AC voltage gives you DC voltage that's 1.414 times the AC voltage.  You have to subtract the voltage drop in the rectifiers, though, that's about 0.7V per diode, so for a full-wave bridge, about 1.4V.  So for a 20VAC power supply, you'd have (1.414 x 20) - 1.4 = 26.9VDC.  I'm not familiar enough with the O2 to know if the charging circuit is ahead or behind the regulation, but it doesn't really matter - two 9V NiCad or NiMH batteries would total 16 - 18VDC, so you'd have plenty of voltage left to charge those batteries.
 
If you used DC - then you'd start with 20VDC, go through rectification which wouldn't gain anything - but you'd still have to subtract the diode voltage drop.  Or, 20VDC - 1.4VDC = 18.6VDC.  If you used higher-voltage NiMH batteries, chances are by the time you went through the charger circuit your power supply voltage would be less than the battery voltage.  That wouldn't work.  If you started with a 15VDC power supply, you'd never have a prayer.
 
That's probably confusing to you, but the thing to remember is that turning AC voltage into DC gains you more voltage, whereas starting out with DC limits your voltage to that initial amount and no more.  The other thing to remember is that you cannot charge batteries using a voltage that's less than the battery's rating.
 
May 27, 2013 at 5:20 PM Post #6 of 11
Quote:
How would AC to DC increase your voltage? How can it magically become more, I always tought that when you converse AC to DC it becomes less because it's a sine wave and you have to take the average.
 
This tool seems to agree with me: http://easycalculation.com/physics/electromagnetism/ACtoDC.php

 
That webtool calculates potential current, not voltage.
DC Voltage is sqrt2 *AC voltage, as Tomb posted above.
 
 
The O2 has 2 half-wave rectifiers in it, followed by positive and negative regulators.
The O2 PSU makes a two rail +/- 12volt supply from a single AC input. When selecting
transformers, 12v AC is enough for the PSU overhead in the O2, provided the transformer
has enough current capacity. There is no benefit in going with extra voltage or current
capacity in the transformer used. A 12vac 400mA wau12-400 is plenty of transformer.
More voltage just heats the regulators and more current goes unused.
 
The PSU design in the O2 is a little wonky IMO. It goes against common practice to use
half wave rectification for audio, because of the ripple generated compared to full wave. The
fact that the rails are powered by +/- 12 volts from the mains, or +/- 8.4 volts from the
batteries is a bit wonky as well because of the amps clipping behavior. If you get the gain
setting right for the rail voltage and source used, it obviously works well. The problem
comes from the lack of understanding among typical consumers, as evidenced by the
numbers of posts from owners hearing clipping and not understanding its source.
 
It would be nice if the designer would finish the design into a nice desktop format more
suitable for the general consumer.
 
May 27, 2013 at 5:22 PM Post #7 of 11
Charging circuit is past the rectifiers. Actually, you wouldn't charge the negative-side battery at all with DC input, the way it's configured. It's only a half bridge.
 
schematic
 
 
How would AC to DC increase your voltage? How can it magically become more, I always tought that when you converse AC to DC it becomes less because it's a sine wave and you have to take the average.
 
This tool seems to agree with me: http://easycalculation.com/physics/electromagnetism/ACtoDC.php

 
1 V rms input AC sine wave has 1.414 V peak. You get the same ratio for a higher AC input. A suitably sized capacitor on that power input may hold close to the peak value (minus diode drops), discharging because of power drawn by the rest of the circuit during most of the cycle but recharging during the part of the AC input that is higher.
 
 

 
May 27, 2013 at 8:26 PM Post #8 of 11
So the ac/dc power supply already has a rectifier that is changing ac to dc. so when it goes to the amp it is going into a 2nd rectifier and will have a lose. i understand the rectifier and why it would have a lose. 
I am not as familiar with voltage regulators and how they work.
 

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