Yes, Nicoh, I know. For low pw application EI Xfrm is my favourite, while split bobbins and R core ones follow closely. The problem is: if I must have it made, I can only get some toroids. Selectronic in France sells some nice R core made in China, but even if not cheap and not considering how expensive is to get them in Italy, they don't carry the V I would need.
Sorry, can't help you with that. See attached picture for examples. The more milli Henry you use, the lower the filter frequency will be. Anything from 10 to hundreds mH helps, mind the suitability for at least 500mA of current passing. Connection to ground at the last capacitor. Besides, you can roll one of your own.le´flu;1965084 said:
There's a lot of info on the forum already on Schottky, hexfred and other diodes. Making a PSU is not hard, be aware of the consequences of ringing and select your components with care. No need to make it expensive though.
Both EI-core and toroid have their strong points. For the sake of a small electromagnetical field I would opt for a toroid if placed nearby the amplifier circuit. For a separate box I'd take the EI-core for better filtering mains pollution.
Pity you guys don't have access to redundant equipment like an old Hifi amp. The larger units often contain a 2x30V tranny, some of Japanese origin are good quality. Forget anything under 2x25W.
Attachments
I told you five month ago ......
I remember that 
But in that time I was "focused" on low electro field because of using 6 tx inside the CDP box and I wanted to close one tx inside the riaa box.
At the time I did not get your idea... why are EI better in some cases ?
Now I am starting to get a picture:
" Both EI-core and toroid have their strong points. For the sake of a small electromagnetical field I would opt for a toroid if placed nearby the amplifier circuit. For a separate box I'd take the EI-core for better filtering mains pollution. " (Thank you disco)
But in that time I was "focused" on low electro field because of using 6 tx inside the CDP box and I wanted to close one tx inside the riaa box.
At the time I did not get your idea... why are EI better in some cases ?
Now I am starting to get a picture:
" Both EI-core and toroid have their strong points. For the sake of a small electromagnetical field I would opt for a toroid if placed nearby the amplifier circuit. For a separate box I'd take the EI-core for better filtering mains pollution. " (Thank you disco)
Just received the Block RK 100/30
It looks very good and the inscriptions in the TX reveal:
I have two independent secondary windings, each for 30vout and 1,66A so I got a 2x50VA TX.
It is smaller than spec but that is very good.
Hi Salas
I believe I will have between 42 and 45v output after rectification and smoothing.
I need to drop between 6 and 8 volts.... what resistor value should I use in series with the 15000uF cap ?
Ricardo
It looks very good and the inscriptions in the TX reveal:
I have two independent secondary windings, each for 30vout and 1,66A so I got a 2x50VA TX.
It is smaller than spec but that is very good.
Hi Salas
I believe I will have between 42 and 45v output after rectification and smoothing.
I need to drop between 6 and 8 volts.... what resistor value should I use in series with the 15000uF cap ?
Ricardo
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R ? = V (6 or 8 volts drop as you wrote) devided by I (you know how the CCS is set).
I must be in A, not mA, thus if you have set it for 400 mA, you use 0.4 in the formula
Then how big the resistor should be? Again, the Ohm's law says: P = V x I
(current always in A, not mA)
The result (in Watt), x 2,5 or better x 3 in order to get the right wattage for the required resistor.
P.S. if the Xfrm is smaller is not a very good sign..... high fluz maybe... BTW, how "square" its shape is? I mean, is the height the same at the ourter edge and at the inner edge of the "doughnut"?
I must be in A, not mA, thus if you have set it for 400 mA, you use 0.4 in the formula
Then how big the resistor should be? Again, the Ohm's law says: P = V x I
(current always in A, not mA)
The result (in Watt), x 2,5 or better x 3 in order to get the right wattage for the required resistor.
P.S. if the Xfrm is smaller is not a very good sign..... high fluz maybe... BTW, how "square" its shape is? I mean, is the height the same at the ourter edge and at the inner edge of the "doughnut"?
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R = 6 / 0.2 = 30 Ohm, correct.
With 33 ohm (practical value), you have: V = R x I or V = 33 x 0.2 = 6.6 V drop.
Now: P = V x I or P = 6.6 x 0.2 = 1.32 W.... then x 2.5 - x 3 means a 3 W or higher reisistor .
A nice idea is to "split" this resistor for the V + and V - side and thus use a couple of 68 Ohm/2W on each side of your p.s.
Connect the V- to the chassis gnd at the second caps (you're using a CRC, aren't you?)
With 33 ohm (practical value), you have: V = R x I or V = 33 x 0.2 = 6.6 V drop.
Now: P = V x I or P = 6.6 x 0.2 = 1.32 W.... then x 2.5 - x 3 means a 3 W or higher reisistor .
A nice idea is to "split" this resistor for the V + and V - side and thus use a couple of 68 Ohm/2W on each side of your p.s.
Connect the V- to the chassis gnd at the second caps (you're using a CRC, aren't you?)
Because its strong ripple current there it gets lower R than predicted in my experience. Has to do with peak charging. Maybe 15R 5-10W Ricardo. Just stick some 10R there and get an indication in conjuction with your CCS setting that governs how much the Tx will sag, the charge cycle with 15mF etc. Then you can move accordingly.
The windings must overlap at the centre more than at the perimeter.
That ensures that ALL toroids are higher in the centre.
Ricardo,
I guess you're using two power supplies to feed your regulators. Exchanging the load one regulator is presenting to your supply for a resistor, you can experiment to hearts content with resistors, caps, coils etc. Just measure the output voltage over the load.
If I remember correctly you're set for .2A and 28V. It would take a resistor of 28/.2= 140 ohm which produces .2 x .2 x 140= 5.6W of heat. You can make a serial or parallel connection with 10W resistors, if you have some.
Placing resistors between the bridge and the cap is possible and is sometimes (if small) preferred. You can have more sections of Resistor and Capacitor but always end your filter with a cap, the largest goes last.
I guess you're using two power supplies to feed your regulators. Exchanging the load one regulator is presenting to your supply for a resistor, you can experiment to hearts content with resistors, caps, coils etc. Just measure the output voltage over the load.
If I remember correctly you're set for .2A and 28V. It would take a resistor of 28/.2= 140 ohm which produces .2 x .2 x 140= 5.6W of heat. You can make a serial or parallel connection with 10W resistors, if you have some.
Placing resistors between the bridge and the cap is possible and is sometimes (if small) preferred. You can have more sections of Resistor and Capacitor but always end your filter with a cap, the largest goes last.
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This answers my question !
I will try to use 10r before the capacitor.
I am using dual mono psu and Jensen 4 pole as smoother... the resistors must be before the cap.
Before cap. Between bridge and cap. RC filter and drop. Think valve circuits. Don't manipulate the ground BTW.