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Old 28th August 2006, 09:15 PM   #11
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Location: Dallas, Texas
Default Pics of main and PS boards

Here is a pic of the main board and the small boards for the power supply circuit.
They are sitting on one of the heat sinks. The heat sinks for the main transistors
are 12"x19.5", but I think they need to be a little be bigger.

Although it is not shown, the main board is mounted in the center of the heatsink
with the transistors wired to the green terminal blocks. With about 2 inch leads,
the transistors are mounted on the heatsink with equally spacing. The front of the
board (with the POTS) then faces the top (or bottom) of the heatsink, so the
POTS are fairly easy to reach. Note that both POTS are on the same side of
the board so that I don't have to move very far to make adjustments.

You might be able to see that the caps on the main board are not an exotic brand.
Does anyone know if, say Black Gate caps would really make a difference? I
think the BG caps are about $2.50 each, but I haven't tried them because there
are minimum orders... Also, I have no idea what Nelson means when he mentions
bypass caps, so if they make a difference, and if anyone would like to enlighten
me, I need some schooling.

On the other hand, I have learned a lot from doing this project. And it has been
a lot of fun! Pics of the amps in boxes soon...

-- Robert
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Old 28th August 2006, 09:28 PM   #12
Mad_K is offline Mad_K  Norway
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Default Re: Pics of main and PS boards

Quote:
Originally posted by audiorob
H I
think the BG caps are about $2.50 each, but I haven't tried them because there
are minimum orders... Also, I have no idea what Nelson means when he mentions
bypass caps, so if they make a difference, and if anyone would like to enlighten
me, I need some schooling.
The reason we are using plastic film (polyester, polypropylen) bypass caps across electrolytic caps is that the electrolytic caps becomes more inductive with higher frequencies. Placing a film cap (1uF is a good value) across the electrolytic lets the higher frequencies bypass the electrolytic via the film cap.
Conclusion: Much better sound in the mid/high frequencies. Some people actually likes the more romantic sound you get from not bypassing the el.cap. Using a tantalum electrolytic makes the sound even more "romantic".
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Old 28th August 2006, 09:38 PM   #13
Zen Mod is offline Zen Mod  Serbia
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Default Re: Pics of main and PS boards

Quote:
Originally posted by audiorob
Here is a pic of the main board and the small boards for the power supply circuit.
They are sitting on one of the heat sinks. The heat sinks for the main transistors
are 12"x19.5", but I think they need to be a little be bigger.

Although it is not shown, the main board is mounted in the center of the heatsink
with the transistors wired to the green terminal blocks. With about 2 inch leads,
the transistors are mounted on the heatsink with equally spacing. The front of the
board (with the POTS) then faces the top (or bottom) of the heatsink, so the
POTS are fairly easy to reach. Note that both POTS are on the same side of
the board so that I don't have to move very far to make adjustments.

You might be able to see that the caps on the main board are not an exotic brand.
Does anyone know if, say Black Gate caps would really make a difference? I
think the BG caps are about $2.50 each, but I haven't tried them because there
are minimum orders... Also, I have no idea what Nelson means when he mentions
bypass caps, so if they make a difference, and if anyone would like to enlighten
me, I need some schooling.

On the other hand, I have learned a lot from doing this project. And it has been
a lot of fun! Pics of the amps in boxes soon...

-- Robert

first-AVOID terminal blocks for any semiconductor,especially output transistors;just solder them

second- bypass caps are small ones ,soldered in parallel to bigger ones;they are ,presumably, there to speed up things (ya know-mambo jumbo about parasitic inductivity in large caps etc) ;in some cases there is benefit in bypassing,in some other cases isn't ;

your ears will decide ;try with some 1uF bypass film caps,then decide what you like.
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Old 29th August 2006, 01:21 AM   #14
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Join Date: Apr 2006
Location: Dallas, Texas
Default Zen 5, in a box

First, a big thanks to Mad_K and Zen Mod for the explanation and information. I
will add experimenting with the caps to my "to do" list. Thank you again!!!

Zen Mod - initially, I didn't understand the jibes about the mess (this is my first
home built amp). Now, I understand. In my case, I have a mess in about
5 different places where I work (not enough room at a single work place!). I
still have a huge mess to clear. But I'm going to listen to this thing first.

This is just a pic of me biasing one monoblock. The "bonnet" (heatsink) is up
so I can better reach the POTS and take temps.

-- Robert
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Old 30th August 2006, 11:10 PM   #15
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Join Date: Apr 2006
Location: Dallas, Texas
Default 20KHz square wave trace

This iis a trace of a 20KHz square wave through my Zen 5 amp. I
admit that I thought that the traces would be more square, but again,
this is not my field of expertise.

Does anyone know if this is good, bad, or normal? Is there a
formalized or generally accepted method of performing square wave
tests? I may not have performed this test correctly. I only have the
vaguest idea of what I'm doing. For example, this trace has no load
except for the O-scope. Traces into an 8 ohm load look far worse,
but that is true when applying the signal through the resistor alone
(no amp, just the resistor).

Assuming I have done this correctly, I wonder if this would improve with
the bypass caps. Does anyone know?

BTW, this trace is from the USB/PC based scope mentioned in previous
posts.

Thanks,
Robert
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Old 30th August 2006, 11:34 PM   #16
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Location: Dallas, Texas
Default measuring wattage

Here is another trace, this time using 1KHz triangle waves, just before the
amp clips. The yellow trace is the input. The magenta trace is the amplifier
output connected to an 8 ohm resistor load.

So, do we call this a 6V input or a 3V input?

The amp is swinging +/- 17V. If wattage is VA, then we need to know
A. From the equation V=IR, we know V and R. I'm guessing that we use
V=17 and not V=34. So:

17V = I * 8 ohms
17V/8 ohms = I
2.125 Amps = I

Then using W=VA we have:
W = 17V * 2.125 Amps
W = 36.125

Does this sound about right to everyone? (I have to ask because *I* would
have said that V should have been 34. But using that value in the equations
yeilds 4.25A and 144.5 Watts, which seems unreasonable.)

Does that look right?

thanks,
Robert
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Old 30th August 2006, 11:55 PM   #17
Zen Mod is offline Zen Mod  Serbia
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looks good enough to me,taking in account that you have pretty crappy 'scope
find attached one pdf ,author can be known from file properties-it's bergerons ;
look for caps named C102 ,C104,C105

102 and 105 are there for stability reasons,and 104 is there to shape your rectangle waves.........

btw-I like that future case for your amp
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File Type: pdf aleph30.pdf (29.9 KB, 373 views)
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Old 31st August 2006, 06:26 AM   #18
Mad_K is offline Mad_K  Norway
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Default Re: measuring wattage

Quote:
Originally posted by audiorob

Assuming I have done this correctly, I wonder if this would improve with
the bypass caps. Does anyone know?
Not likely.


Quote:
Originally posted by audiorob

The amp is swinging +/- 17V. If wattage is VA, then we need to know
A. From the equation V=IR, we know V and R. I'm guessing that we use
V=17 and not V=34. So:

17V = I * 8 ohms
17V/8 ohms = I
2.125 Amps = I

Then using W=VA we have:
W = 17V * 2.125 Amps
W = 36.125

Does this sound about right to everyone? (I have to ask because *I* would
have said that V should have been 34. But using that value in the equations
yeilds 4.25A and 144.5 Watts, which seems unreasonable.)

Does that look right?

thanks,
Robert
To get W RMS you need to divide your answer (36W) by 2. So yes, it is correct (18W RMS)
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Old 31st August 2006, 07:37 AM   #19
AndrewT is offline AndrewT  Scotland
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Location: Scottish Borders
Hi,
when defining your triangle wave you can use peak to peak (Vpp) or just the ref to peak voltage (Vpk). If it was asymetrical you would define the DC offset as well.

Peak power into a load using sinewave signal is P = Vpk^2 / R = Ipk^2 * R = Vpk * I. All three give the same answer. Use the version for which you know the operands.

The power is the peak power divided by two, P = Ppk / 2 = Vpk^2 / 2R.

Your DMM reads rms (AC average) when doing these measurements.

In this case the power using rms measurements is P = Vac^2 / R, omitting the divide by two.
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Old 1st September 2006, 09:50 PM   #20
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Zen Mod said:
> btw-I like that future case for your amp

The cases are completed, and the amps are installed. I have been listening
to it for a couple of days.

-- Robert
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