You can ditch the zeners and just accept that you have no voltage ref., or go down the LED ref. path.
But really for a pre-amp, there is little to gain by going much further than a run of the mill regulator like the one seen in the BOZ article. I have added a CLC to the one I have in my test setup....I found no benefit (but also no drawback besides the price of the inductor).
Magura
But really for a pre-amp, there is little to gain by going much further than a run of the mill regulator like the one seen in the BOZ article. I have added a CLC to the one I have in my test setup....I found no benefit (but also no drawback besides the price of the inductor).
Magura
That is not the obvious way to do it! If you want to make a CLC supply, you simply choose a trafo that fits! The option to do both a CLC and a reg. supply will be like shooting a goose after its dead allreadyStefanoo said:
A CLC supply is very fast and responsive and the sound is just about the best that can be had!! If you dont like the big coils for whatever reason, make a regulated supply as suggested in my earlier reply.
Strictly speaking from practical experience!!Steen
Magura said:
Magura,
I'm not fussed with rolling them, it's just that finding the right wire (50m of 14 gauge) is posing a bit of a problem via the internet (and it's been raining here for a week). When it clears up I'll have to do some looking myself...steenoe said:
Steenoe,
I've been looking at the power supply in http://www.passdiy.com/pdf/zen-v5-hires.pdf, and it seems to be the same as the bipolar regulator in the pdf you mentioned (but with a CLC
). Which picture were you referring to, if I might ask?
Cloth Ears said:
Magura,
Hmm, may I suggest the cheap, easy and quick solution?
Simply buy a roll og 14avg and be done with it for any foreseable future ? I'm quite sure it's cheaper than any of the other solutions, and you will have enough for making a few experiments.
Magura
GRollins said:
But I've never breathed before (and some guy keeps turning me upside down and
I'm not trying to be difficult, but any solution (spools of wire, rolls of 14awg, 4 inductors off the rack) adds up to over $200 down here in Australia. My apologies if not being a spendthrift (OK, being cheap) offends...
i would like to have guys your opinion on this power supply design.
It's for the Aleph P1.7.
Since the preamplifier absorbes a constant current (correct me if i'm wrong) of about 69mA, i place the constant current source as load.
If i'm mistaken, plese correct me.
Here's the schematic
It's for the Aleph P1.7.
Since the preamplifier absorbes a constant current (correct me if i'm wrong) of about 69mA, i place the constant current source as load.
If i'm mistaken, plese correct me.
Here's the schematic
Attachments
here a screenshoot of the regulation capability of this circuit.
I simulated it with a good resolution (5usec as maximum step size).
The rail voltage for the P1.7 is supposed to be 60V.
As you can see with this specific it's 63V.
This is because i picked out the Plitron transformer attached on the link
http://www.plitron.com/shopping/shopexd.asp?id=170
I would you use two of them for the two channels. (i still have to figure out a way to get them without paying this high shipping expences).
Here's the simulation screenshoot
I simulated it with a good resolution (5usec as maximum step size).
The rail voltage for the P1.7 is supposed to be 60V.
As you can see with this specific it's 63V.
This is because i picked out the Plitron transformer attached on the link
http://www.plitron.com/shopping/shopexd.asp?id=170
I would you use two of them for the two channels. (i still have to figure out a way to get them without paying this high shipping expences).
Here's the simulation screenshoot
Attachments
Cloth Ears said:I'm not trying to be difficult, but any solution (spools of wire, rolls of 14awg, 4 inductors off the rack) adds up to over $200 down here in Australia. My apologies if not being a spendthrift (OK, being cheap) offends....
ditto.
Not sure though if you can order from overseas, but this place in Singapore has 1.4mm (just about 15awg) air core inductors for $32.20 singapore $ (about 21 USD).... would that do the trick? I guess with postage it'll come to about AU$120 -- plus the risk of customs tax...
Hi,
CLC is better than CRC at attenuating the higher frequencies.
The saw tooth output from CRC is simply mains waveform plus harmonics.
The output from the CLC has fewer harmonics and so looks more like the sinewave.
Due to the extra filtering inherent in the CLC one can choose a DCR slightly less than the R in the equivalent CRC. This will improve the output regulation.
Pi filters whether CRC or CLC or multiples are very good at removing mains ripple.
The last C supplies the majority of the peak current demand of the amplifier.
I suggest that the last C be scaled to supply all the peak current demand with the level of output induced ripple that you consider appropriate.
Here in the UK we can buy 500gms of enamelled copper for about £6 to £7 on a plastic reel. 1kg is only £11. That is becoming a big inductor. But, not nearly enough for a choke regulated supply unless you can accept the DCR of finer wire.
CLC is better than CRC at attenuating the higher frequencies.
The saw tooth output from CRC is simply mains waveform plus harmonics.
The output from the CLC has fewer harmonics and so looks more like the sinewave.
Due to the extra filtering inherent in the CLC one can choose a DCR slightly less than the R in the equivalent CRC. This will improve the output regulation.
Pi filters whether CRC or CLC or multiples are very good at removing mains ripple.
The last C supplies the majority of the peak current demand of the amplifier.
I suggest that the last C be scaled to supply all the peak current demand with the level of output induced ripple that you consider appropriate.
Here in the UK we can buy 500gms of enamelled copper for about £6 to £7 on a plastic reel. 1kg is only £11. That is becoming a big inductor. But, not nearly enough for a choke regulated supply unless you can accept the DCR of finer wire.
Using Wheeler's formula L=7.87*M^2*N^2/{3M+9B+10C} where
M=mean diameter
N=number of turns
C=spool width
Approximate inductance for air cored coil and DCR as follows:
wire - - - spool weight
diam - - - 500gm - - - - 1kg
1.2mm - - 3.2mH - - - - 10mH
- - - - - - - 0r7 - - - - - - 1r4
1.4mm - 1.7mH - - - - 5.6mH
- - - - - - 0r4 - - - - - 0r78
1.6mm - 1mH - - - - - 3.3mH
- - - - - - 0r22 - - - - - 0r46
M=mean diameter
N=number of turns
C=spool width
Approximate inductance for air cored coil and DCR as follows:
wire - - - spool weight
diam - - - 500gm - - - - 1kg
1.2mm - - 3.2mH - - - - 10mH
- - - - - - - 0r7 - - - - - - 1r4
1.4mm - 1.7mH - - - - 5.6mH
- - - - - - 0r4 - - - - - 0r78
1.6mm - 1mH - - - - - 3.3mH
- - - - - - 0r22 - - - - - 0r46
Cloth Ears said:
hitting me)!
I'm not trying to be difficult, but any solution (spools of wire, rolls of 14awg, 4 inductors off the rack) adds up to over $200 down here in Australia. My apologies if not being a spendthrift (OK, being cheap) offends....
Offended by frugality? Don't be silly. I have Scottish blood in my veins and was raised to squeeze pennies until they squealed.
You just need to be more creative in where you source your wire.
If you buy wire from electronics places, they will charge you more than if you go to a local motor repair shop and buy it off their spool. For a six pack of beer, they may even let you have some for free.
Think, lad, think!
Work smart, not hard.
Grey
GRollins said:
Grey, you have a way with words - I hope you're either a writer or a teacher.
You picked up on my Scottish heritage eh? I'm off to visit my grandmother tomorrow (who, herself, had only one grandparent from south of the border - ie. England) for her 96th birthday. Difficult to catch her at home, as she's always driving off to the golf club or to play bridge, but we've pinned her down to afternoon tea on Sunday.
"Work smart, not hard." You're right. That means not searching the WWW, but going and finding it myself.
...!!!
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.