Couple of meta42 questions. . .
Jul 28, 2002 at 3:20 AM Thread Starter Post #1 of 13

andrzejpw

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I'm about to start building the meta42(probably wed or so), and I've got a few questions.

1) At first, I'll be operating on battery power only. Just 2 9V. I ordered the parts to go into class A(the transistors), but should I do it at first? I may add a current source later.

2) Is there a place on the board to connect a switch? (wait, its the thing marked SW, isn't it?)

3) I have an old wallwart that I've ripped from an old set of non working labtec speakers. Here are the specs:

15VDC 1000mA.

Polarity: + 0 - (trying to draw that picture.
smily_headphones1.gif
)

Is it worth it to bother with this one? I mean, I have it.
smily_headphones1.gif
And I may upgrade to a better one soon. How hard would it be to change polarity if I need to, inside the meta, I mean? Just flip the wires on the jack?
 
Jul 28, 2002 at 5:59 PM Post #2 of 13
Do you want to build it all once and never mess with it again? Or you could build it as a basic unit then add the current source and actually hear what it adds to the quality of the sound? Then add the double buffers and try to hear if there is a difference? But this would entail removing the board and working on it again several times. Heck you could probably socket the buffers then be able to do a pretty quick A/B test. And add a switch to the current source and switch that live to see if it does what you want? Personal goals would dictate this.

The wall wart should be OK unless it’s noisy? You can use diodes or a switch so it can be powered by either batteries or the wall wart without opening it. To change the polarity in side just swap the wires.

Good Luck!
 
Jul 29, 2002 at 10:04 PM Post #3 of 13
Quote:

should I do [the current source] at first?


Unless you know for a fact that adding it will help your op-amp, I say "no". Even if you know it will help, you may want to hear the before vs. after. (Why would the current source not help, you ask? Because some chips are already running in class A at the low current levels that exist between the op-amp and the buffer in the META42 design.)

Quote:

Is there a place on the board to connect a switch?


There are two sections in the META42 documentation that explain the passive parts and all the wirepads. You should read these sections carefully before starting so you know which sections of the board have which parts and which parts you should ignore. The META42 shouldn't be blindly "stuffed" with parts -- you need to look at each one, understand what it's for, and then decide on a value for that part, or decide to leave it out/jumper it.

Quote:

I have an old wallwart that I've ripped from an old set of non working labtec speakers.


This will be a pretty low-quality power supply -- itp probably only cost Labtec $3 in quantity. Chances are excellent that it's either an unregulated supply or a switching power supply. You can do better.

If you insist on using this, first measure it to find out a) what voltage it's _really_ giving, and b) what the polarity is. Don't try to figure polarity out from the diagram -- it's much more reliable to just measure it. You'll also need a DC power jack for it; probably it's a 5.5/2.5 or 5.5/2.1mm barrel connector. You can pick up good jacks for this at Radio Shack. Avoid the "closed circuit" jack if you don't have to use it. The "heavy duty" jack is a better idea.

Quote:

How hard would it be to change polarity if I need to, inside the meta, I mean? Just flip the wires on the jack?


Given the polarity of the power supply, this will tell you how to wire the DC power jack. As long as the wall wart's positive terminal connects to the META42's V+ pad and the negative connects to V-, you're golden.
 
Jul 29, 2002 at 10:54 PM Post #4 of 13
In your experiance, does the 8610 already run in Class A?

Oh, and I also ordered transistors for the cascade circuit. Could I put these in at once? Or should I wait for a current power supply?
 
Jul 29, 2002 at 11:53 PM Post #5 of 13
ppl says the AD8610 benefits from a current source. And again, I would put the cascode in immediately only if you don't want to hear what the "before" sounds like.

Another reason to hold off on adding that is that it's prudent to build the amp in testable stages. If you try and stuff the entire board and then flip the power switch, you risk blowing everything up if you did something wrong in assembly. Or at least, something fails and now you end up suspecting everything on the board because you don't know for sure which parts are working and which aren't. If you build it a stage at a time and test at each step, you will have fewer things to blame if something goes wrong.

Personally, I add the resistors, C1 C2 and C3, D1, TLE2426, ground buffer and sockets, then apply power. I then test that the right voltages are on the appropriate socket pins, and that the other pins have 0V on them. Then I add the pot, LED and other buffers and try again. No smoke, no heat, right voltages? Good. Finally I add the op-amp and wire up a single input and output pair (this is if I have dual inputs and outputs). By now I should be hearing good music. Only at that point do I start with optional stuff like current sources, C4, C5, R8, R9, extra inputs and outputs, crossfeed, etc.

Take it from one who's spent too many hours debugging when too many things have changed to make diagnosis easy: it doesn't have to be that way.
 
Jul 30, 2002 at 5:06 PM Post #9 of 13
Yes indeed, you are confused: R10 and the JFETs are the current source. (R10 isn't strictly part of the current source, but it's a buffer for the current source, so it's closely associated.)

It may be helpful if you know what's going on here rather than blindly following instructions. A "voltage source" is EE-speak for a perfect source of a fixed voltage that does not waver under any current demand. Such a thing does not really exist, but a battery is reasonably close: it stays steady unless you put a large current demand on it.

Contrariwise, a "current source" is a perfect source of a particular amount of current which doesn't waver no matter the surrounding voltage difference. You've read through that section in the META42 documentation, so you know of several kinds of current sources that approach the ideal: a bare resistor, a CRD, a JFET with the drain pin tied to the gate pin, and a pair of JFETs arranged into a cascode.

EDIT: That should be "source pin tied to the gate pin".

All having batteries has to do with the current source in the META42 is that adding a current source to drive the op-amp into class A draws more current. In battery-powered amplifiers, a higher current draw means less battery life. If you insist on using a current source in a battery-powered amp, you should take care to make its current draw as small as you can. Shoot for 0.5 mA or so per channel.
 
Jul 30, 2002 at 6:29 PM Post #10 of 13
Also, some opamps benefit more from the current source than others. Class A opamps such as the AD843 probably won't benefit at all. The AD825 is vastly improved. The AD8610 experiences only modest improvement. All this courtesy of reading back posts. I would not have a problem sparing 1-2mA for a battery powered amp, but your mileage may vary.
 
Jul 30, 2002 at 8:11 PM Post #11 of 13
wow. I feel stupid. All this time I thought a current source meant something that plugs into the wall.
eek.gif
frown.gif


Tangent, you said that worst case, your meta42 drew 40 mah? So, 2 9V batteries, say, around 600 mah, so that'll be 15 hours of time, worst case, which is ok. . . right?
 
Jul 30, 2002 at 8:29 PM Post #12 of 13
You could call the wall wart a "voltage source" -- a good regulated power supply is a pretty good approximation of an ideal voltage source.

As for battery life, that's a complicated issue. I recommend searching back posts. But here's just one monkey wrench I can throw into your ideal pen-and-paper world: to last 600 hours with a 1 mA current draw, the battery has to be completely drained. For a typical 9V battery, this means getting it down to 5 or 6V. But what if your op-amp starts sounding bad at 7V? or 8V? There are lots of good back posts describing why that happens despite guarantees on the datasheet of operation down to 3V or 5V or whatever the claim is. There's a section at the end of my CMoy assembly tutorial that touches on this as well.
 

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