The Aleph-X

[BrianGT]

So, by this:

Step 1:
target: 120W into 4 ohms

Step 2:
P=V^2/R, so V=sqrt(P*R) = sqrt(480) = 21.91v
P=I*V, so I = P/V = 120W/21.91v = 5.48A

Step 3b:
V * .8 = 17.5v

Step 4b:
I=5.48A

My question for this, is that for step 1, don't you have to account for the fact that each side of the Aleph-X is only seeing half the load impedence, and I should be using 2 ohms instead of 4 ohms?

Then it would be: I = sqrt(P/R) = sqrt(120/2) = 7.75A of total bias current.

Which bias current is correct?

--
Brian

 

[Fred Dieckmann]

and my girlfriend says that the chassis should dissipate 300W pretty easily

Sounds like a pretty hot mama. Just out of curiosity, how did you wind up discussing this with her? I used to eat breakfast in the cafeteria with a pig farmers daughter, so she could help me with my physics homework. This was at the University of Arkansas and she was not a hot mamma but was a very sweet and intelligent young lady who wore overalls, pigtails (her hair,not actual pig tails), and thick glasses . I am still married to a classical music major I met at U of A. She is practicing the clarinet in the next room for a Dallas Wind Symphony rehearsal.

http://www.dws.org/main.htm

Go Hogs Go,
Fred

 

[BrianGT]

Re: and my girlfriend says that the chassis should dissipate 300W pretty easily

Sounds like a pretty hot mama. Just out of curiosity, how did you wind up discussing this with her?

My girlfriend is a chemical engineering major and when I was putting my chassis together, I noticed that her homework for her thermodynamics class involved problems with calculating heat dissipation for heatsinks. I had her run some numbers from my heatsinks (gave her the base thickness, number of fins, fin thickness, total area). I forget what the exact numbers came out to. She said she would run the calculation again in a couple of days, once she finishes her other work, if anyone wants to see the calculations.

--
Brian

 

[MEnsing]

Which bias current is correct?

I think that the calculated bias current of 5.48A is correct. Keep in mind that the original XA-200 runs at about +/-31V and a bias current of 8A.

 

[jh6you]

Brian

I am confused a bit about the power dissipation.
The original circuit on page 1 shows Q1, Q2, Q10 and Q11 with the source resistors of 0.22R. It means that each Q is biased at about 3A. If the rails are +/- 15V, the total power dissipation (Q1, Q2, Q10 and Q11) per channel would be 4 x 15 x 3 = 180W. Go ahead at full speed and finish the project with this. Then, later if you like to increase the bias to 4.5A, you could do it by adding 0.47R parallel with the 0.22R (i.e. 0.47R//0.22R = 0.15R). If so, you would see the total power dissipation of 270W per channel.

Good Luck!

JH

 

[BrianGT]

I think that the calculated bias current of 5.48A is correct. Keep in mind that the original XA-200 runs at about +/-31V and a bias current of 8A.

Alright, if the second one is correct, which follows Grey's formulas, here are other dissipations calculated with that:

16v
5A bias current
~100W into 4 ohms
~160W poser dissipation

18v
5.625A bias current
~125W into 4 ohms
~200W poser dissipation

20v
6.25A bias current
~150W into 4 ohms
~250W power dissipation

25v
7.8A bias current
~240W into 4 ohms
~390W power dissipation

32v
10A bias current
~400W into 4 ohms
~640W power dissipation

The XA would fall into the last one, putting out 200W into 8 ohms. These figures seem fairly accurate, with most likely some overhead in the bias current figures, if the XA200 really has just 8A bias current. It is quite a beast.

I am thinking of aiming for around 20v with 6.25A bias current, then I can raise the current later, depending on my chassis temp.

With my the pair of parallel output devices, I will set the output current for each to: 6.25/8 = .78A per device and that leads to the resistor values: .5^2/.78A = 0.32ohms (will use 0.33ohm).

--
Brian

 

[BrianGT]

Brian

I am confused a bit about the power dissipation.
The original circuit on page 1 shows Q1, Q2, Q10 and Q11 with the source resistors of 0.22R. It means that each Q is biased at about 3A. If the rails are +/- 15V, the total power dissipation (Q1, Q2, Q10 and Q11) per channel would be 4 x 15 x 3 = 180W. Go ahead at full speed and finish the project with this. Then, later if you like to increase the bias to 4.5A, you could do it by adding 0.47R parallel with the 0.22R (i.e. 0.47R//0.22R = 0.15R). If so, you would see the total power dissipation of 270W per channel.

Good Luck!

JH

Each Q is biased by 0.5v^2/0.22ohms= (v^2/r, remember ohm's law). This brings the total bias current per channel to 4.55A.

The total dissipation would be 15v x 2 x 4.55A = 136.5W for Grey's model. The 15v rails give ~88W into 4 ohms ((15v/.8)^2/4ohms).

--
Brian

 

[Peter Daniel]

Well, I have finished my pcb layout and am sending the design out tommorrow after I look over it one more time.

Brian,
I was thinking about your layout and it seems like making it perfectly symetric on a PCB, will complicate things outside the board.

One side of the FETs will have to be mounted under the board. Let's say you'd decide to mount devices directly to the board, without wires. How will you align them, mount to heatsink and solder to the board? I do it like this: mount them loose to the sink first, put the board over the leads (this step aligns the fets on a sink), then tighten the screws on the fets to the heatsink and solder the leads. If you mount them under the board you won't be able to tight the screws on devices unless you'll use a template.

Also, when you place devices on one side outside the board and on the other side under the board, your symmetry is lost. And I think that it makes bigger difference than switching G and S leads for one side.😉

What do you think?

 

[BrianGT]

Brian,
I was thinking about your layout and it seems like making it perfectly symetric on a PCB, will complicate things outside the board.

One side of the FETs will have to be mounted under the board. Let's say you'd decide to mount devices directly to the board, without wires. How will you align them, mount to heatsink and solder to the board? I do it like this: mount them loose to the sink first, put the board over the leads (this step aligns the fets on a sink), then tighten the screws on the fets to the heatsink and solder the leads. If you mount them under the board you won't be able to tight the screws on devices unless you'll use a template.

Also, when you place devices on one side outside the board and on the other side under the board, your symmetry is lost. And I think that it makes bigger difference than switching G and S leads for one side.😉

What do you think?

My plan is to attach short wires to the output devices, since my chassis is 8" wide, and my boards are 6" wide, they won't fit on the heatsinks directly anyway. The devices will be mounted one upsidedown on one side and rightsideup on the other. The wires will be kept to a minimum length. The centers of the fets will be equal on both sides relative to the board.

I really don't think that this will mess up the symmetry too much.

I am open to any other suggestions for a future revision of this board. I sent my files off to get the boards made last night, so hopefully I should get them back this week and we will see how it works out 😀

--
Brian

 

[jh6you]

Each Q is biased by 0.5v^2/0.22ohms= (v^2/r, remember ohm's law).

Thanks for the ohm's law.

I am a very simple minded engineer, and therefore, P = VI is not ohm's law...?!?!
The designer put the 0.22R to get the maximum DC current of 3A (bias current). No? How do you explain then? For what MPSA18 working???

Probably, there must be something I do not know. 😉

JH

 

[BrianGT]

Thanks for the ohm's law.

I am a very simple minded engineer, and therefore, P = VI is not ohm's law...?!?!
The designer put the 0.22R to get the maximum DC current of 3A (bias current). No? How do you explain then? For what MPSA18 working???

Probably, there must be something I do not know. 😉

JH

You are quite right... V^2/R is a combination of P=I*V (power law) and V=I*R (ohm's law). I meant to write ohm's law in there. I made a mistake in the calculation. It should be v/r=i, which gives 2.27*2 = ~4.55A, with the current being the same on both sides. Sorry for the confusion.

Where are you getting this figure of 3A of current? Look at the first page of the thread:

Grey wrote:
The circuit, as drawn, will pull a bit over 4.5 amps

--
Brian

 

[eLarson]

If the "v^2" is intended to mean "v-squared", then
v^2/R would be the power dissipated by "R" when there are "v" volts across it.

<Gettin' all pedantic... now>
Ohms law is V=I*R.
Power equals Voltage (V) * current (I).

Restating in terms of current, Ohm's Law is
I = V/R

Therefore P = V * V/R (or V^2/R as was written above.)
</pedantic>

Erik
(Is this my first post to the Aleph-X thread? I can't really remember anymore...)

Edit (Of course I was a hair slow typing it all in. 🙂 )

 

[jh6you]

Brian

We need to understand for what MPSA18 is playing.
MPSA18 is playing to maintain about 0.66V across the 0.22R.
That gives us about 3A (I = V/R) bias.

JH

 

[mrfeedback]

Question Of Aleph-Xperts....

I am keen now to build a low power version for my nearfield monitors at my workstation.
I figure modern pc power supplies will supply 5v at 20 and more ameres, are self contained, regulated and cheap, real cheap.
Also that lower power devices are cheaper, and smaller, and less heatsinking required.
So can the ouput stage be powered from 5v and the ccs from 12v, or can the whole stage be powered from 5v successfully,
and what sort of output powercan be expected into 4 or 8 ohm ?.

Eric.

 

[jh6you]

The circuit, as drawn, will pull a bit over 4.5 amps.

Brian

Yes, I see it on the first page.

You know? 4.5A = 3A + 1.5A.
The 3A is DC bias current and 1.5A (50% of 3A) is maximum peak AC current gained by the active current source. He meant the 4.5A of the static+dynamic current. Unless Fred Dieckmann comes in and gives me a hammer punch with his “NO! NO! NO! a-shole…”, I am right.

JH

 

[Fred Dieckmann]

The circuit, as drawn, will pull a bit over 4.5 amps.

There are two halves the balanced circuit so each half is pulling about 2,25 amps or about 0.56volts across the 0.25 ohm source resistor. Some addtional base current to transistor is supplied by a 100K pot and 47k resitor to the transistor's base from the postive supply to allow adjustment of the bias current through the output mosfets. Check the Zen variations articles out for better description of the circuit function.

http://www.passdiy.com/pdf/zen-v4.pdf
http://www.passdiy.com/pdf/zen-ver2.pdf

No name calling required,
Fred

 

[jh6you]

No name calling required,

Sorry for my overaction.

JH

 

[AudioFreak]

JH,

It's been said before, the Aleph current source produces double the bias current on peaks when set to 50%.

2.25amps idle current per side.

 

[jh6you]

I agree on 2.25amps idle current only if the 0.56volts are across the 0.25R.

I said 3amps, considering the 0.66volts and the 0.22R. Then, I made two mistakes: (1) with the 0.66volts as in Zen and (2) with the 0.22R as shown on the original circuit. 😉

Thanks.

JH

 

[jh6you]

By the way,

Fairchild data sheet says that MPSA18 has Vbe(on) of 0.7V. How could we get the 0.56V across the source resistor? Due to +/- 20% tolerance? I am really confused. Your explanation would much appreciate.

JH