Aleph-X builder's thread

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#822
Nelson Pass said:


The capacitance seems a bit high on them, but at the same time
they dissipate more and they have higher current capacity.
I would try to get away with using as few as possible - I think
you could idle them at say 50-60 watts if you give them enough
sinking.

😎
Click to expand...


I have a question about the capacitance and how that affects the circuit.

I noticed that the

IRFP240 has;
Input capacitance=1300
Output capacitance=400
Reverse Transfer Capacitance=130

IRFP044 has;
input capacitace = 2500
Output capacitance= 1200
Reverse Transfer Capacitance= 200

MTW32N20E has;
Input capacitance=3600
Output capacitance=130
Reverse Transfer Capacitance=690

Nelson said that because the capacitance is higher on the MTW's that I should use less of them. Would this also be the case with the IRFP044's as well since they have higher capacitance than the IRFP240's?

I had wanted to use 12 outputs per channel to keep down the individual temp per device. My logic tells me that having three devices mounted across a large heatsink would get better heat transfer than having only one device mounted on the same heatsink and dissipating the same heat as the three combined. Is this faulty logic?

When Nelson says to use a few as possible does this mean 4 outputs rather than 12? Is 8 too many? Is 12 too many?

I have 40 of the MTW30N20E's. I had hoped to get 4 sets of 6 matched device from them. I'm still trying to undertand Nelson's circuit for testing them. As soon as I get that figured out I plan to start matching them. Have any of you built that circuit? Could you take a couple of pics of it so I could get a better Idea of how to build it?

Thanks, Terry
 
#823
As far as I can tell, the die for the 240 and 044 are similar,
but they have been doped for different voltages, and the
higher voltage types have less transconductance and less
capacitance. (This is speculation on my part)

The MTW's clearly have greater dissipation to go with the
capacitance, and clearly you can use fewer of them, and I
would recommend that. If it were me, I would use 1/2 as
many if I could. How many watts were you planning for
output?
 
#824
Nelson Pass said:
As far as I can tell, the die for the 240 and 044 are similar,
but they have been doped for different voltages, and the
higher voltage types have less transconductance and less
capacitance. (This is speculation on my part)

The MTW's clearly have greater dissipation to go with the
capacitance, and clearly you can use fewer of them, and I
would recommend that. If it were me, I would use 1/2 as
many if I could. How many watts were you planning for
output?
Click to expand...


Hi Nelson,

How kind of you to help me.

I was hoping for at least 40-50watts per channel into 8 ohms but I would like to be able to run 4ohm. The spread sheet calculator says that 6A is best for 4ohm with 4 output devices but that puts the dissipation per device at about 48W. Is this OK?

The speakers that I wanted to use for this are 4ohm.

If I can get by with only the 4 outputs that mount on the board that would of course be the best. Everyone has scared me that the heat will be too much.
 
#826
If I can get by with only the 4 outputs that mount on the board that would of course be the best. Everyone has scared me that the heat will be too much.
Click to expand...

Terry,
I fan cooled mine and its almost necessary unless you have water cooling or really large sinks. It certainly runs as hot as the KSA-50 does. Either lots of sinking or a small amount with a fan is whats required.

I didn't have very much trouble setting mine up and getting them to work correctly, I did have to change some parts values to get it to balance out but it was not a real big deal.... I think you really should build them... its a good expereince. Me..... I'm holding out for the real X circuit to be released someday by the Master, then I will attack building it again.

Mark
 
#828
Hi Guys,

I didn't get the parts order so I spent time matching the outputs.

I used my PSU for the amp, which puts out 17.7VDC unloaded. I wired 1R36 10w worth of resistors in series with the devices. I hooked up the +rail to the gate and drain and hooked the source to ground through a switch. When measuring the rail voltage would drop to 14.73vdc. Is this normal?

These are the voltage readings I got from the 30 devices I have so far. They are the MTW32N20E FETs from On Semi.

4.75
4.78
4.80 x 2
4.81 x 2
4.82 x 2
4.84 x 2
4.94 x 2
4.96 x4
4.98
4.99 x 2
5.00
5.01 x 3
5.02
5.05
5.06 x 2
5.07
5.08
5.10 x 2

Is this test sufficient for matching these FETs? If I were to use 8 outputs rather than 4, which range would you suggest I draw them from.

Thanks, Terry
 
#829
heatsink height

Terry,

The bloke who makes heatsinks in Australia, Conrad is the company, agrees with Nelson P. Essentially, as you double the height of the heatsink, the heat dissipation only increases by 1.4: it is a square root thing.

Hope this helps.

(Note to self: I must build one of these rather than spend all my time reading about it.)

Regards,
George.
 
#830
While reviewing the ax wiki regarding the ac current source adjustment I think I may have found an error.
I may be wrong about all of this but the given equation of adjustment is (1- currentR5/current R2/R3) X100=percentage of current source.
Now the way I understand the diagram is that R5 is the source resistor of the current source.The more ac current we measure there the more the current source is contributing to the total current output and vice versa.
If the current source was not variable then there would be no AC current over R5 only DC current.
However substituting 0 for current over R5 in the equation gives us (1- 0)X100=100%!! So the current source is contributing 100% instead of 0!!
To me this means everything is the other way round.
It should be current R5/current R2/R3 X100= percentage of c.s.
Of course those that adjusted for 50% will not have a problem but somebody that adjusted for 65% will actually have 35%!
Or have I missed something?
 
#832
It was me who wrote that piece of the WIKI with Nelson's ZenV4 and Edwin Dorre's theory in mind. Knowing that my math is awful, it might be there is a mistake. I'll have a look ASAP and hope brighter minds will also come to the rescue. 😀

/Hugo 🙂
 
#833
Hi Protos,

I always prefer to call the things by name and not by number.
So ac-current-gain is ac current through the active source-source resistors devided by the total ac current through the output resistors.
Or in words the percentage of current the active-current-source delivers related to the total output current.

William (who will have a look at the WIKI now)
 
#835
It looks like the theory is correct but the outcome is wrong.
The only thing I can find is:
(1 – ( 0.425 / 0.91)) * 100 = 53.3%
Should read:
(1 – ( 0.425 / 0.91)) * 100 = 45.7%
Close to 46.7%
Please correct me if I’m wrong again.

/Hugo 🙂
 
#836
Hugo,

now you´re confusing me a bit:

(1-(0,425/0,91))*100=53,3%

This is correct (but not the ac-current-gain)


if you leave out the 1- part you get:

0,425/0,91=0,467=46,7%

wich is the ac-current-gain

The upper calculation would be ok if you look at the source resistors connected to the negative rail.


William
 
#840
wuffwaff said:
Hugo,

now you´re confusing me a bit:

(1-(0,425/0,91))*100=53,3%

This is correct (but not the ac-current-gain)


if you leave out the 1- part you get:

0,425/0,91=0,467=46,7%

wich is the ac-current-gain

The upper calculation would be ok if you look at the source resistors connected to the negative rail.


William
Click to expand...

Hugo,

To expand on what William is saying above:

The calculation without the "1-" is the percent of current contribution of the Aleph current source or what we call the AC Current Gain;

and the calculation with the "1-" is the percent of current contribution of the amplifier output itself.
 
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