The critical inductance(minimum) for the choke to work is:
Henries= (Volts/Current in milliamps) * 0.88(for 60Hz)
Grollins is correct. I should have said it's the much higher voltage of tube circuits that require larger chokes in Henries. As you increase voltage in the formula more henries is needed for the choke to work.
Henries= (Volts/Current in milliamps) * 0.88(for 60Hz)
Grollins is correct. I should have said it's the much higher voltage of tube circuits that require larger chokes in Henries. As you increase voltage in the formula more henries is needed for the choke to work.
This thread has been inactive for a while and the issues of the best LC scheme remain unanswered.
How reliable are the manufacturer saturation curves for predicting the transformers behavior under load?
Is there anything about using an LC power supply configuration that will distort this curve and if so, what are the determining factors?
When accounting for diode losses, do we subract these from the VAC (before rectification) or VDC (after rectification)?
In an LC power supply we will have a DCR loss from the wire of the coil which reduces the VDC but where do we apply the induction loss of the LC configuration? Do we reduce the VAC or the VDC? I've seen several conflicting formulas which calculate the induction loss on voltage from an LC configuration but which is the most accurate for this Class A amp power supply?
How reliable are the manufacturer saturation curves for predicting the transformers behavior under load?
Is there anything about using an LC power supply configuration that will distort this curve and if so, what are the determining factors?
When accounting for diode losses, do we subract these from the VAC (before rectification) or VDC (after rectification)?
In an LC power supply we will have a DCR loss from the wire of the coil which reduces the VDC but where do we apply the induction loss of the LC configuration? Do we reduce the VAC or the VDC? I've seen several conflicting formulas which calculate the induction loss on voltage from an LC configuration but which is the most accurate for this Class A amp power supply?
Depends...if you go with an L filter (inductor followed by a cap), a large enough inductor will actually reduce your rail voltage somewhat, though more current will be available. If you go for a PI (cap, inductor, cap), your rail will approach the full peak of the incoming AC, less a volt or two for rectifier losses.
All things considered, I'd say to go for the PI. You'll get better filtering that way. Think of it as an 18 dB/oct filter as opposed to the L being a 12 dB/oct, maybe that will help.
Grey
All things considered, I'd say to go for the PI. You'll get better filtering that way. Think of it as an 18 dB/oct filter as opposed to the L being a 12 dB/oct, maybe that will help.
Grey
GRollins
What formula do you use to determine the voltage loss caused by the LC configuration? In the post provided by jonk, the formula indicates that with 20VDC rails and 8A of throughput, the inductor should have a value of around 5mH. I am suspicious of these formulas because they seem to indicate that capacitance in an LC configuration is not as critical to reducing noise ripple as the inductance. Have a look at the page and play around with some numbers to see if they jibe with your experience.
Alain Dupont
Nice to have you on this thread. As was mentioned earlier in this thread, I want to use the LC configuration because the three box design better suits that configuration. I realize that a CLC configuration allows the use of a smaller coil but I prefer simplicity. What do you think of the formula page provided by jonk is the post above?
What formula do you use to determine the voltage loss caused by the LC configuration? In the post provided by jonk, the formula indicates that with 20VDC rails and 8A of throughput, the inductor should have a value of around 5mH. I am suspicious of these formulas because they seem to indicate that capacitance in an LC configuration is not as critical to reducing noise ripple as the inductance. Have a look at the page and play around with some numbers to see if they jibe with your experience.
Alain Dupont
Nice to have you on this thread. As was mentioned earlier in this thread, I want to use the LC configuration because the three box design better suits that configuration. I realize that a CLC configuration allows the use of a smaller coil but I prefer simplicity. What do you think of the formula page provided by jonk is the post above?
GRollins
Thanks for the explanation the Pi 18 dB/oct filter as opposed to the L being a 12 dB/oct is for me most interesting.
So I will put the inductor in the Pi config, between the cap devided in two.
with 40VDC and 2*56000 uF what would be the
average inductor value ?
yldouright
Thanks for the welcome ;
and for the link, very valuable info.
Thanks for the explanation the Pi 18 dB/oct filter as opposed to the L being a 12 dB/oct is for me most interesting.
So I will put the inductor in the Pi config, between the cap devided in two.
with 40VDC and 2*56000 uF what would be the
average inductor value ?
yldouright
Thanks for the welcome ;
and for the link, very valuable info.
Long ago when tubes ruled the world. Capacitors where very expensive. Chokes where a lot cheaper. Huge chokes were used because it allowed the use of fewer capacitors. Remember chokes store energy in a magnetic field. Capacitors store energy in an electric field. The choke is sort of adding to the amount of energy in your power supply. When capacitors got cheap chokes went out of style because of cost and size. Later regulator chips came along too.
To get an LC to work. What you need is the working circuit, capacitors, choke(s), one similar VA transformer with a higher voltage output than you need. And a variac(an adjustable A/C voltage transformer). Simulations work ok to get close. But, you need a real world test rig to get the voltage and ripple you are looking for. The variac allows you to adjust the a/c in a real circuit. Now your free to try different chokes and capacitor combinations. LC or PI. Once you find those. You will have a very real idea on what the correct voltage is for the transformer. LC is more work to get right. R and D is expensive. Hence the reason products cost a lot more than their parts add up to.
To get an LC to work. What you need is the working circuit, capacitors, choke(s), one similar VA transformer with a higher voltage output than you need. And a variac(an adjustable A/C voltage transformer). Simulations work ok to get close. But, you need a real world test rig to get the voltage and ripple you are looking for. The variac allows you to adjust the a/c in a real circuit. Now your free to try different chokes and capacitor combinations. LC or PI. Once you find those. You will have a very real idea on what the correct voltage is for the transformer. LC is more work to get right. R and D is expensive. Hence the reason products cost a lot more than their parts add up to.
I've been doing some calculations on winding my own air coils. An 8mH air coil will require around 50 turns on a 12" former and require nearly 200 feet of wire. That is an impractical size to work with and prohibitively expensive since I would need 8 for my application. A 6" former with the same number of turns would bring us down to 2mH and use half the wire which is workable but the value doesn't really help quiet much over an RC configuration. I think I'm going to need a core to increase the permeability. Nelson Pass had stated earlier that he doesn't know of any coils that won't saturate when you push over 6A through them so that leaves me with two choices. Find a core that won't saturate at 8A and wind around it or lower my bias and up my voltage. I really thought that I would need 8A (2A per FET) to completely drive a 2.5 ohm impedence dip but I may have to live with a loosened grip in order to build thiis amp. Can anyone on this forum advide me on the best material and shape to build a coil that won't saturate with 8A through it.
You can get any size iron core choke you want. Your problem will be the massive size and cost. Hammond 195J10 10 millihenreis 10 amps .07 dcr 8 pounds. Just do not drop it on your toe.
look at www.hammondmfg.com heavy current reactors. You can get a price quote from Mouser.com.
look at www.hammondmfg.com heavy current reactors. You can get a price quote from Mouser.com.
Digging from stored knowledge:
XL=2*Pi*f*L; where XL=inductive reactance(ohm), Pi=3.1416, f=frequency(Hz), L=inductance(Henry)
Note that f=120Hz after the diode bridge. The xformer secondary voltage will increase by a factor of 1.4 after rectification. The value of XL determines the voltage drop (loss) in the inductor.
Somebody correct me if I'm wrong.
XL=2*Pi*f*L; where XL=inductive reactance(ohm), Pi=3.1416, f=frequency(Hz), L=inductance(Henry)
Note that f=120Hz after the diode bridge. The xformer secondary voltage will increase by a factor of 1.4 after rectification. The value of XL determines the voltage drop (loss) in the inductor.
Somebody correct me if I'm wrong.
Blues
If you play around with the formula posted above, you'll note that it doesn't work out this way in practice. In an LC configuration the relationship of AC secondaries to final rail voltage is closer to a 1:1 ratio. Much is dependent on the resistance in the coil.
Gnomus
Thanks for the link but 8 of these coils would blow up my budget for this amp at the price quoted.
If you play around with the formula posted above, you'll note that it doesn't work out this way in practice. In an LC configuration the relationship of AC secondaries to final rail voltage is closer to a 1:1 ratio. Much is dependent on the resistance in the coil.
Gnomus
Thanks for the link but 8 of these coils would blow up my budget for this amp at the price quoted.
Blues
Even with a coil that has a miniscule 100 milliohm resistance the 1.414 AC/DC conversion factor does not apply according to the formulas used on that page. The best that you are likely to get out of the taps according to that page is 0.9VAC. I don't know if the formulas are peculiar to an LC configuration. It may very well be a different situation using an RC layout. Nelson Pass uses thermistors in his AX line of amps so the DC voltage rises as the amp warms up but so does the ripple. Try the formula on that page and see what I mean.
Even with a coil that has a miniscule 100 milliohm resistance the 1.414 AC/DC conversion factor does not apply according to the formulas used on that page. The best that you are likely to get out of the taps according to that page is 0.9VAC. I don't know if the formulas are peculiar to an LC configuration. It may very well be a different situation using an RC layout. Nelson Pass uses thermistors in his AX line of amps so the DC voltage rises as the amp warms up but so does the ripple. Try the formula on that page and see what I mean.
yldouright,
I played around with the formulas on the weblink and they look okay to me. I keyed in your requirements of 20Vdc and 8A and came out with the following:
inductor=2.212 mH
actual inductor value=2.2 mH, ESR=0.52 ohm
actual cap value to determine ripple V=100 mF
ripple V=75 mV
Transformer sec V=26 Vrms; after the diode bridge this voltage ramps up to about 36V @ 120Hz (26V*1.4 and 60Hz*2). At 120 Hz and 2.2mH the XL on the formula I provided is about 1.66 ohms and 1.66*8A=13.3V=the voltage drop on the coil. The final rail V will be about 36V-13.3V=22.7Vdc=pretty close to your 20Vdc requirement. If in your final power supply you measure more than 20Vdc, you can always series a resistor (as in the figure on the webpage) to drop some voltage to suit your desire. If you will connect this supply to a regulator, you can always choose your zener values to your heart's desire.
You can key in your actual values to get actual ripple V. Actual coil values are from internet. In the formula ESR and additional series resistance lowers minimum cap value. anyway I also tried the Zen v4 supply requirements and they were pretty close.
I played around with the formulas on the weblink and they look okay to me. I keyed in your requirements of 20Vdc and 8A and came out with the following:
inductor=2.212 mH
actual inductor value=2.2 mH, ESR=0.52 ohm
actual cap value to determine ripple V=100 mF
ripple V=75 mV
Transformer sec V=26 Vrms; after the diode bridge this voltage ramps up to about 36V @ 120Hz (26V*1.4 and 60Hz*2). At 120 Hz and 2.2mH the XL on the formula I provided is about 1.66 ohms and 1.66*8A=13.3V=the voltage drop on the coil. The final rail V will be about 36V-13.3V=22.7Vdc=pretty close to your 20Vdc requirement. If in your final power supply you measure more than 20Vdc, you can always series a resistor (as in the figure on the webpage) to drop some voltage to suit your desire. If you will connect this supply to a regulator, you can always choose your zener values to your heart's desire.
You can key in your actual values to get actual ripple V. Actual coil values are from internet. In the formula ESR and additional series resistance lowers minimum cap value. anyway I also tried the Zen v4 supply requirements and they were pretty close.
Perhaps ist's a good moment to throw in this wonderful program again: http://www.duncanamps.com/psud2/download.html.
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