[Ubel]

Here is a link to the schematic/diagram http://www.5mv.com/ebay/banzaivcm-1306042668-36729.jpg
 
Basically as you can see it runs on 2x9v batteries. I wish to run it off a DC transformer as well..
Short of using two transformers which seems dumb, can't I just use an 18v transformer somehow? Would I be required to build a simple splitter as done in Tangent's cmoy?
 
I'm not sure. I'm not that great at reading schematics so I'm a bit confused. I know that one of the negative and one of the positive terminals are connected.. like it was wired in series, but I don't believe so? Also it has a bias jumper, what exactly does that do? I know it has to do with keeping the output going under load or not and means it's class A or B?
 
Thanks for any help, I'd really appreciate.. not looking forward to burning through batteries on this amp as it's used 98% of the time at my desk .

 

[deltaydeltax]

I myself might go with an AC transformer, create a full-wave bridge rectifier circuit, use two larger electrolytic caps, two smaller value ceramics, and add two voltage indicating LEDs.

 

[Ubel]

That sounds a bit in depth for me, if not overkill for such a simple amp? I don't have much experience when working with electronics on a scale like this so I'd like things to be as simple as possible.. here's what I used for my last cmoy and the way it accomplishes its power supply: http://tangentsoft.net/audio/cmoy-tutorial/misc/cmoy-tangent-sch.pdf
 
I assume something like the power supply section in that schematic would suffice but I'm not sure..... honestly I'm about to go find two wallwarts with the same outputs and rig this thing up. I don't really want to be buying parts either but I will if needed. I just want it to work.
 
I appreciate your advice though and I might go that route if need be but again it sounds like overkill for a cmoy.

 

[TheLaw]

It's actually not complicated at all. You should use a center tapped transformer which is noted by having 3 wires on the secondary side of the transformer.
 
I made a diagram for you to show you how to do it. There's really not a more simple way to do it. In fact, it only gets harder. So essentially you have the two red wires with voltage running into the rectifying bridge. For the positive rail (top), the rectifier blocks out the negative AC waves and for the negative rail, the rectifier blocks out the positive half of the AC wave. The end result is 1 rail of positive DC, and 1 rail of negative DC, which is just what your cMoy wants.
 
The problem is: The waveform is still really wavy, and not suitable for use yet. So to get rid of this "ripple", you put one high capacitance electrolytic capacitor in parallel with 0V (black wire of the transformer "center tap"). You put in one for each rail. You are simple putting one leg of the capacitor on the +7V and one leg on the 0V. And then you do that on the other rail.
 
The next thing you should add, although not strictly mandatory, but highly recommended, is a small value ceramic capacitor, or film capacitor. Using a 0.1uF ceramic is a safe bet, though you can't really "do it wrong". The reason you put this ceramic capacitor in is because electrolytic capacitors (the 470uF) are really slow. If there is a spike or a quick change in power draw, they cannot react very fast and thus you get crappy waveform. The 0.1uF ceramic makes up for the electrolytics downfall.
 
(You might also ask...why is it 5VAC on the input and then 7VDC on the output. This is because after rectification, the voltage in DC is roughly 1.41 x Voltage AC.)
 
Next step is to add a resistor and a LED to tell you whether it's on or not. However, most likely, if you just put an LED in this circuit without a resistor, it would burn out, so you need a resistor to limit the amount of current available to the LED, and thus preventing it from burning out.You can use an LED calculator like this: http://led.linear1.org/1led.wiz
 
Choose the LED you are going to use, and make sure you know it's forward voltage and operating current. Also know the voltage of the circuit it's going to be in. In this case, it is 14V (the difference between +7V and -7V = 14V). Attach one leg of the resistor to the positive rail (+7V), and the other leg to the positive leg of your LED. Then attach the negative leg of the LED to the negative rail (-7V), (not 0V!)
 
The 0V is not attached to your LED.
 
It may seem complicated because I wrote a lot, but trust me, it's not hard at all. When you layout the circuit, you can pretty much copy the schematic for the layout. You don't need anything fancy. I was in your position a while ago too, when it all looks daunting. Just follow the schematic and you'll be fine.
 
If you are wondering what kind of transformer to get, something like this is fine: http://cgi.ebay.com/12V-Transformer-6V-0-6V-CT-1A-110Vac-12Vac-Chassis-/120736134976?pt=LH_DefaultDomain_0&hash=item1c1c6f3740
 
I actually own that one. It works well.

 

[Ubel]

Thanks again for the help, it does look pretty simple. I wouldn't be surprised if I have a 12vac transformer just like that around here somewhere. However I'm confused about the rectifier part, is that just diodes to stop the current flow? Also I'm confused about how it would connect to the cmoy. I see one of each negative and positive terminals, do the others connect to the 0v or what?

 

[Flops]

The diodes make DC out of AC and therefor are called rectifier.

 

[TheLaw]

This is a diode rectifier bridge:
 
It's very simple. You hook up the AC to the middle two legs, and then you have a negative and a positive. That's all there is to it. You COULD make your own diode bridge, but it takes more time and thinking. These are much easier and fault proof, and they are quite cheap as well.

 
 
To hook up the power supply to the circuit you linked to, all you need to do is hook up the 0V to the places that would get the black wires (ground/return), and the +V to the place that gets the positive voltage and the -V to the hole that gets the negative voltage.
 
Simple!

 

[Ubel]

Quote:Originally Posted by TheLaw /img/forum/go_quote.gif

This is a diode rectifier bridge:
 
It's very simple. You hook up the AC to the middle two legs, and then you have a negative and a positive. That's all there is to it. You COULD make your own diode bridge, but it takes more time and thinking. These are much easier and fault proof, and they are quite cheap as well.
 
 
 
To hook up the power supply to the circuit you linked to, all you need to do is hook up the 0V to the places that would get the black wires (ground/return), and the +V to the place that gets the positive voltage and the -V to the hole that gets the negative voltage.
 
Simple!

 
Okay, either I'm dumb or none of you guys are understanding me. This cmoy needs DUAL power supply. TWO separate 9v rails. I don't see how I can hook them both up to the power supply we've been discussing. If I could just hook up both rails to one source, why the heck can't I just use a wallwart? Either I'm missing something huge or all you guys have shown me is a power supply for one out of two "rails". There is 2x of each negative and positive terminal and as far as I can tell they are mutually exclusive in the circuit. This cmoy needs a separate supply for each channel. I do not know where the ground/return is on this amp. It does not have a virtual ground circuit that I know of either, like Tangent's cmoy does.
 
Edit: As I mentioned earlier, one of the negative and one of the positive terminals ARE linked to each other according to my multimeter. I tried hooking up a power source to the OTHER positive and negative terminals(while doing nothing with the ones that are linked, seeing as they're linked) but nothing happened. Are these "linked" pos/neg terminals the ground you're referring to? Because this amp design has me confused and so do all of your comments. I do not understand how I can power my cmoy, which needs two sources of energy, when all the schematics you've shown me appear to only have one source.
 
 

 

[Avro_Arrow]

In a dual rail power supply, there are two voltages and a common center voltage.
For example, a 9 volt, dual rail power supply would be +9 volts, 0 volts and -9 volts.
0 volts is often referred to as ground. The Banzai uses two 9 volt batteries to form
a dual rail power supply.
To form an AC dual rail power supply, a center tap transformer is often used.
The "ends" of the transformer go to the bridge rectifier and the center tap
of the transformer is the 0 volts (ground) connection. That is what post #4
is trying to show.

 

[Ubel]

Quote:

In a dual rail power supply, there are two voltages and a common center voltage.
For example, a 9 volt, dual rail power supply would be +9 volts, 0 volts and -9 volts.
0 volts is often referred to as ground. The Banzai uses two 9 volt batteries to form
a dual rail power supply.
To form an AC dual rail power supply, a center tap transformer is often used.
The "ends" of the transformer go to the bridge rectifier and the center tap
of the transformer is the 0 volts (ground) connection. That is what post #4
is trying to show.
 
Yeah.. that's what I asked. Still doesn't help me because I still don't understand HOW to hook it up. I said this like three times: WHICH wire goes WHERE. That is what I need to know. No one has answered me. I see a v+, v- and ground. That's 3 connections, I need 4. I'm assuming at least one of the battery hook up terminals goes to the ground but I don't know which. Your post explains the power supply just fine, but not how to use it. I understand the whole +9, 0 and -9 thing. I just don't understand how to hook up two positives and two negatives to three connections.
 
This is about as clear as I can make it even though I'm pretty sure you already understand me, there is:
 
Pair A:
9v+
9v-
 
Pair B:
9v+
9v-
 
All of which need to be hooked up to three wires, your aforementioned 9v+, 9v- and 0(ground). How do I accomplish this?

 

 

[TheLaw]

Chill my dude. We're trying our best to help you.
 
 
 
It's NOT:
 
+9V
-9V
 
+9V
-9V
 
It IS:
 
+9V
0V
 
-9V
0V
 
The 0V is a common ground.
 
In the power supply, 0V is only one wire, but as you can see, you need 2 wires for your board. So all you do is split the 0V somewhere (into two wires) and put each of the 0V wires into holes that have that are marked for ground. On your picture, they are the bottom holes.Then you put one of the +9V in one of the pairs and one -9V in the other pair.
 
It's as easy as that.
 
 

 

[Pars]

In your Cmoy, one of the 9V battery - terminals (say Pair A, above)r is hooked to the other battery's + terminal (Pair B), forming the "ground" or 0V point. It is probably connected on the PCB, hence you don't see it. The + of the first battery becomes the V+, or +9V, and the - of the second battery becomes the V-, or -9V.
 
From this, i trust you should be able to figure out how to connect the AC supply to your amp...

 

[TheLaw]

This is all you are doing. Now you have your dual supply

 

[Ubel]

Thanks.. I already mentioned how one of the positive and negative terminals were linked and that I assumed they were the ground, like ... three times, but no one seemed to notice or reply to that so I didn't know. That's seriously all I needed to know, sorry if I seemed angry, I wasn't. Just kinda ironic how I had to make my point like three times before someone understood me lol
 
So basically you're saying in the diagram I linked, the bottom power poles are the ground..? That's not entirely correct because when doing a continuity test, these are the ones(circled in neon green) that are linked and thus must be the ground: http://i.imgur.com/vdUAX.jpg
 
Either way I understand now. Just wanted to make sure before I accidentally connect the wrong terminals and blow something.
Thanks again, I appreciate it everyone.

 

[TheLaw]

Could be. You can look where the traces go and then look at the datasheet for the op-amp you are using to see which are the supply pins and see if you have it hooked up right. It's fairly easy to kill an op-amp so just make sure you've got that right.
 
The wiring diagram for the Banzai is kind screwing with my mind but yeah you're right. The 2x9Vs form a "center tap" and one of the grounds goes to the top of the pair and the other goes to the bottom off the pair.
 
I was just going off of the basis that red was + and black/grey was -. But yes, you're right.