Diy Marshall power brake...Reactance expert consult needed!

[sajy_ho]

So I guess I better to stick with the current arrangement; but can someone tell me which type of core is suitable for the inductors to increase their saturation current? I mean I know that the best idea would be air core for the 1mH inductor, but what about the other one?

Thanks

[sajy_ho]

i wanted to know about the inductor with many taps used as attenuator

You mean the tapped inductor in the right side of the schematic? 
It's an autotransformer, you can use any type of tapped autotransformer. For that I used a 100V single winding speaker matching transformer that I found in my parts.

[Rob Strand]

Ipeak= Psquare/Vpeak = 2.12A

I calculated the current through the 12.5mH inductor alone using a circuit simulator and got the same sort of value at the worst-case frequency.

EDIT: I calculated the magnetic permeability(u) of these cores to be around 3850. So by increasing the core aread(a) by means of putting two or three cores in parallel, 'Al' becomes larger and subsequently 'N' gets around 177. According to the saturation current formula this won't increase the current capability very much, am I right?

I think the problem is the formulas say to use the magnetic permeability of the core (ur)
but I seem to remember you should use the "rod permeability" urod.  FYI rods often use low ur material.
The rod permeability is a function of the rod length to rod diameter ratio and is usually given as a graph:

http://www.magneticsgroup.com/pdf/erods.pdf

Again, from what I remember the AL factor doesn't change much and probably gets a little larger.

When you use n cores:
use the total area A total = n times A for one core, and,
for the diameter in the urod calculation you will need to use an effective value
d_effective = sqrt(n) * d for one core.

You might scrape by with two rods but three is safer from a max current perspective and is easier to bundle.

Edit:  If AL is about the same N is about the same and since A is n times larger the maximum current should be n times larger.

[Rob Strand]

Watch out playing around with inductor losses.  What seems better overall to the eye might not sound better.    The mismatch at low frequencies that occurs changes the "presence".  Hard to know what is better without doing AB tests.

[sajy_ho]

Ipeak= Psquare/Vpeak = 2.12A

I calculated the current through the 12.5mH inductor alone using a circuit simulator and got the same sort of value at the worst-case frequency.

EDIT: I calculated the magnetic permeability(u) of these cores to be around 3850. So by increasing the core aread(a) by means of putting two or three cores in parallel, 'Al' becomes larger and subsequently 'N' gets around 177. According to the saturation current formula this won't increase the current capability very much, am I right?

I think the problem is the formulas say to use the magnetic permeability of the core (ur)
but I seem to remember you should use the "rod permeability" urod.  FYI rods often use low ur material.
The rod permeability is a function of the rod length to rod diameter ratio and is usually given as a graph:

http://www.magneticsgroup.com/pdf/erods.pdf

Again, from what I remember the AL factor doesn't change much and probably gets a little larger.

When you use n cores:
use the total area A total = n times A for one core, and,
for the diameter in the urod calculation you will need to use an effective value
d_effective = sqrt(n) * d for one core.

You might scrape by with two rods but three is safer from a max current perspective and is easier to bundle.

Edit:  If AL is about the same N is about the same and since A is n times larger the maximum current should be n times larger.

You're right, I think I misunderstood the core permeabilty with the MTERIAL permeability. So I guess I need to go to store and buy 3 more cores.

Thanks again for all your help...
Sajad

[Rob Strand]

No worries - this stuff can get confusing. 

FYI, here's another spin of the inductance calculations:

http://agata.pd.infn.it/LLP_Carrier/New_ATCA_Carrier_web/DataSheets/New_Symbols/15th_edition_Fair-Rite_catalog.pdf\

see page 98.

[sajy_ho]

No worries - this stuff can get confusing. 

FYI, here's another spin of the inductance calculations:

http://agata.pd.infn.it/LLP_Carrier/New_ATCA_Carrier_web/DataSheets/New_Symbols/15th_edition_Fair-Rite_catalog.pdf\

see page 98.

Thanks again Rob.
I wound  the 1mH inductor again just for experiment, andThis time I went air core, it was my first air core inductor and I guess it'll be the last one

The inductance value turned out very good and the DC resistance is only 0.5R; the only problem is that some windings are so messy; so I guess the leakage inductance will be great. What would be the possible consequences?

[PRR]

> the DC resistance is only 0.5R

Use smaller wire.

The treble inductance is in-series with the ~~8 Ohm "voice coil resistance". You could use wire 1/4 diameter (1/16 area) and get the DCR up to 8 Ohms; then you do not need the resistor! However heat will be an issue (the wire can't run as hot as a resistor, the coil may have less surface area to throw heat). But it would be quite valid to wind for 1 or 2 Ohms in the coil, take the difference out of the "8 Ohm" resistor.

> some windings are so messy; so I guess the leakage inductance will be great.

Leakage inductance is mostly about transformers, not single-winding coils.

Also where leakage inductance "might" be a problem, the impedances are transitioning to the 20 Ohm resistor across the coil.

An air-core coil is best wound on a form. Old plastic solder-spool. Or a wood stick and two heavy cardboard flanges. For this purpose, you don't have to wind in neat layers, just don't get so bunched-up that wire pinches itself and breaks the enamel. (Even then, this would be "OK" because it approximates the losses of a real voice-coil in a conductive magnet gap; but we don't know how much is OK, and "pinch" is not predictable or consistent over time.)

[sajy_ho]

Leakage inductance is mostly about transformers, not single-winding coils.

Also where leakage inductance "might" be a problem, the impedances are transitioning to the 20 Ohm resistor across the coil.

Thanks again Paul, here's the mess that I'm talking about:

So if there isn't any problem with being messy, this inductor would be good in series with 8.2R resistor.
The only thing remaining is the resonance inductor; that I probably follow Rob's idea and stack 3 or 4 rod cores together and wind another one.

Thank you all guys...

[KMG]

My up to 15W loadbox

Very interesting idea about creating transformerless attenuator with equal In/out FR

Only limitation to keep In/out FR:
P(out)=K*P(in) @ K<= 0.16.
Original article (unfortunately in Russian language)
http://forum.amtelectronics.com/index.php?topic=33.0

[PRR]

> attenuator with equal In/out FR

Neat.

Note that output impedance is not emulated (that is very difficult).

A basic SS amp is zero Ohms out.

Classic Fender (5F6a) is (for 8 Ohm connection) 15-20 Ohms output impedance.

This plan is near 4 Ohms (on any amp).

The reaction of driving impedance on the loudspeaker in use affects its response and damping.

However this plan does give good power-loss and response quite simply.