Aleph 2 Power resistors = V2/R

Has anyone tried or using 2 watt metal film power resistors in their Aleph 2 project?

I am thinking about this and tossing up whether to use plain 5 watt ceramic wire wound or go for 2 watt power film resistors.

My sums suggest the power disipation will be 0.50 watts per resistor where V 2sq / R where V=voltage = 0.50 volts R = 1.00 ohm (A=0.50 amps) ie 6 x 0.50 amps = 3 amps bias current.

Alternatively if more power is needed 2.2 ohm and 1.8 ohm (0.99 ohms) could be placed in parrellel for 4 watts due to temperature drift.

Please forgive me if my sums are wrong.

If this correct then I could use obtainable Philips metal film resistors in 2 watt ratings from Farnel!

This assumes of course metal film resistors be preferable over standard ceramic cased wire wound versions?

All comments and opinions are most welcome.

regards

macka
 

UrSv

Member
2002-01-31 5:48 pm
Sweden
As far as I can tell calculation would be that you have 1*0.5*0.5 W (R*I*I) per resistor which is 0.25 W which then gives some room for using 2 W metal oxide at 1R and not think more about it. I would go for it. If you do not feel comfortable with that 2 0R47 in series works as well giving 4 W maximum pwer handling for the resistors.

/UrSv
 
Opinions on Vishay RS5 / RS2B / Mills MRB

I guess a lot of people use Caddock power resistors for the 0.47 and 1.0 ohm resistors, but has anyone tried either Vishay RS5, RS2B or Mills ‘MRB’ Series?

The Mills MRB series are non inductive, but does it matter?
How much do the power resistors actually influence sound quality?

Thx and regards

Axel
 

BrianGT

Retired diyAudio Moderator
2002-01-21 2:42 pm
near Atlanta, GA
www.chipamp.com
If you look at the Aleph 2 in the PassDIY gallery by Wayne Sankey. instead of using 6 - .47 ohm resistors, he used 7 .5 ohm RS2B Vishay resistors.

He seems to have no problems, and I have planned to go the same route. The RS2B resistors are 3W and work fine for the .47. He used the RS5 for the 1 ohm resistors, also with no problems. If you wanted, you could use 2 RS2B 2 ohm resistors in place of the 1 ohm resistors, as suggested by some people in earlier discussions.

--
Brian
[email protected]
 
Power resistors

Yes the 0.47 power resistors are in the signal path and act as degeneration for the output fets and have an influence on the open loop linearity on the amp. Using a good resistor here will have sonic benefits. The Mills MRA-5 non inductive wire wounds have an excellent reputation for this type application and would be on my short list. I use two Dale 1 ohm 1 watt wire wounds SMTs in parallel in my modified Aleph 3 and they sound great. Surplus shoppers look for 1% wire wounds. CaddockMP-820s would also be excellent. There are some real sonic differences in small value wire wounds.

H.H.
 
Thanks for your replies Brian, HPotter and Harry

Great info for a novice :)

So I guess that the absolute value for each of the power resistor can be 0.5 ohm and that the difference (0.03 ohm) is not critical.
Please correct me if I am wrong.

But matching of the total parallell resistance pr channel might be a good thing, right? I.e making sure that R22-27 for both channels have the same Rtotal?

Well, I am going to try Vishay RS2B for the 1.0 ohm (R40-R51) and RS5 for the 0.5 ohm (R22-R27) power resistors. Time will show wether it works or not.

Thanks guys

Axel
 
Why 7 x 0.5 ohm Resistors and not 6?

Could someone explain why one would use 7 x 0.5 ohm resistors instead of 6 x 0.5 ohm?

With 6 x 0.47 ohm resistor I get:
Rt=1/(1/0.47+1/0.47+1/v+1/0.47+1/0.47+1/0.47)
I get Rtotal = 0.0940 ohm for 6 x 0.47 ohm resistors.

With 6 x 0.50 ohm resistor I get:
And if I use Rt=1/(1/0.5+1/0.5+1/v+1/0.5+1/0.5+1/0.5)
I get Rtotal = 0.0833 ohm for 7 x 0.5 ohm resistors.

With 7 x 0.50 ohm resistor I get:
If I use Rt=1/(1/0.5+1/0.5+1/v+1/0.5+1/0.5+1/0.5+1/0.5)
I get Rtotal = 0.0714 ohm for 7 x 0.5 ohm resistors.

Does`nt 6 x 0.50 ohm bring one closer to what is specified in the schematics? Am I missing something here?

Thanks

Axel
 

BrianGT

Retired diyAudio Moderator
2002-01-21 2:42 pm
near Atlanta, GA
www.chipamp.com
You have the calculation for the 6 .47 ohm resistors wrong. It is .07833 ohms. For same element parallel, it simplifies to R/(number of elements). 0.094 is the value for 5 .47 in parallel.

I am not sure why 7 was chosen instead of 6. I would assume that a lower resistance value would be better in the circuit.

I am still a little confused at the exact purpose of the resistors. I showed the circuit to my professor here at Georgia Tech, Prof. Leach, who designed the Leach amp, and it confused him for a while, then gave me some explanation, which I don't fully understand.

--
Brian
[email protected]
 
Am I missing something here?

I think you are missing the big picture. The values for those resistors are not critical at all. For R40-51 you can use any value close to 1 ohm. Your bias will change slightly which you might have to adjust in the end anyway depending on your hitsinks. Just trim R19. For better description check the thread:
http://www.diyaudio.com/forums/showthread.php?threadid=783
From what I understand resistors R22-27 act as a sort of a fuse. In case of a shorted output the voltage drop appears across them and is detected by the circuit around Q5 which cuts the current to the output. The number of the resistors and their values are also not important as long as the total value is close to 0.1 ohm. And that's what I'm using: one Caddock resistor with value of 0.1 ohm rated 30W.
Or am I missing something?
 
Brian

Quote:

"I am still a little confused at the exact purpose of the resistors. I showed the circuit to my professor here at Georgia Tech, Prof. Leach, who designed the Leach amp, and it confused him for a while, then gave me some explanation, which I don't fully understand."

Have a look at Nelson's description of the Aleph Current Source on page 3 of 'The Zen Variations - Part 2'. This should give you a clearer idea of the purpose of the resistors. If you want more detail, then read his Patent No. 5,710,522.

Geoff
 
I would think that the patent should be required reading before attempting the circuit, but perhaps I'm just feeling unusually grumpy this morning.
R22-27 a fuse?
Gracious me.
Q4 senses the current across R46 (just the first one in line, used as a proxy for the entire set) and drops the drive if there's a short. It doesn't read from R22-27.
R22-27 are a current sensing array for the output. The signal derived from those resistors is seen by Q5 and is used to modulate the current sources (Q6-11), increasing or decreasing the current, hence enabling the designer (aka Nelson) to achieve higher efficiency from the amp. It's also a feedback loop reducing distortion.
Configured properly, it will also scratch your back and whisper sweet nothings in your ear.
Those who wish to use voltage feedback can eliminate the output array and sense directly from the output.
There will be a test Friday. If you don't pass, ya can't build your Aleph...

Grey

PS: I used the ordinary blue Panasonic metal film resistors from Digikey. Sounds fine. I have nothing against Caddock, et. al. and for all I know the amp would sound better with them, but I can't afford such extravagances at the moment. Maybe the Aleph-X will get the full treatment, then again, maybe not.
 

BrianGT

Retired diyAudio Moderator
2002-01-21 2:42 pm
near Atlanta, GA
www.chipamp.com
The patent page is very helpful in understanding the circuit:

Patent page

"The load-current sensing resistor R4 (R22-R27) senses the direction and magnitude of the current passing through the output node 110. This serves as a signal that is fed back 120 to the current source 26' to control the amount of current it produces. Specifically, the output current of the current source 26' decreases in proportion to the magnitude of current flowing from the load Z to the input gain device 10. Since the input gain device sinks current from both the current source 26' and the load Z, the effect of the operation of the current source is to decrease the amount of current that the input gain device must sink during the positive going portions of the input signal. Conversely, the output current of the current source 26' increases in proportion to the magnitude of current flowing to the load Z. Thus, during the negative going portions of the input signal, the current source provides proportionately increasing amounts of drive current to the load Z. In this manner, the peak-to-peak current variation experienced by the input gain device is only a fraction of the current flowing through the output node 110 of the amplifier. "

If you look at the images, in tif format, they help even more.

--
Brian
[email protected]