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13th January 2002, 11:27 PM  #1 
diyAudio Member
Join Date: Jan 2002
Location: New York City

SOZ Choke input filter?
Want to experiment with a choke input filter(inductor after rectifier bridge) compared to PI filter. For 60Hz mains power I found the formula L(henries) = E(volts) / I(milli amps) x .88. This gives the absolute minimum choke size.
My question is how many amps can I expect from a 30 Volt DC supplied SOZ? Has anyone tried a Choke input SOZ? Or can share the general difference one can expect betwween choke input VS PI? Thanks in advance for any who respond. For a 50 Hz supply the formula is L=Volts / milliamps X 1.06
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Paul 
13th January 2002, 11:36 PM  #2 
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Join Date: Jun 2001
Location: Calgary

THe article on the SOZ gives the total power required for a given supply (actually viceversa, if I remember). Just look there and divide power by voltage to get amperage. Times 2 for stereo, maybe.

15th January 2002, 08:04 PM  #3 
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Join Date: Jan 2002
Location: Colorado

gnomus,
Since you found the formula for critical inductance, I suppose you know that a choke input filter gives a DC voltage of 0.9 times the transformer's RMS voltage while the capacitor input (pi) filter gives a DC voltage of 1.4 times the RMS voltage. In both cases remember that there will be a DC drop across the rectifiers and IR loss in the choke. If your load "goes away" the DC voltage of the choke input filter will go from 0.9 to 1.4 times the transformer voltage, which can cause not nice things to happen. I have not built SOZ (I have something similar, tho). If by a 30 volt supply you mean +/ 15 volts then I guess you will have about 2.5 amperes per amplifier. (amps per amp :) ) As paulb points out, multiply this number by the number of amplifiers you are running. I use choke input filters on all of my audio amps and recommend them whole heartedly. They are much quieter electrically and there is less audible noise from the power transformer. Inrush current is greatly reduced also. Another plus. Happy listening!
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FEThead 
4th February 2002, 08:09 PM  #4 
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Hi,
I'm making the SOZ, and am using a pi filter as shown on the web site. The pi configuration has caps going from each leg to the common ground an inductor, and then more caps from each leg to the common. The thing that seems wierd to me is that on the +Vs side the cap negative is toward the common ground. On the Vs side, the cap positive is towards the common ground. Is this right? Also, I think something Grey Rollins mentioned awhile back is becoming clear: I had a transformer with too high voltage. He suggested a choke supply. I guess he meant to use the .9 configuration vs. the 1.4 pi filter, which would make me end up with a lower output voltage, right? In the choke (.9) filter, I then follow the inductor with a bunch of caps? Mark
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"The geek shall inherit the earth" 2017 Burning Amp is November 12 in S.F. Thread here on diyAudio: http://www.diyaudio.com/forums/clubs...francisco.html 
5th February 2002, 01:19 AM  #5 
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Join Date: Feb 2001
Location: Columbia, SC

Ground, positive, and negative are all relative terms. Ground is more negative than the positive rail, yet more positive than the negative rail. An electrolytic cap only cares that the voltage at its positive terminal is more positive than the voltage at its negative terminal. Kind of an electrical analog of the old expression about one man's floor being another man's ceiling.
To use an inductor to lower the voltage coming out of a power supply, it must be first in line after the rectifier. Then come the caps. The voltage rating on the caps should still be high enough to withstand the 1.414 multiplier, though, since if the power supply is ever operated without a load, the voltage will rise up to the same value as a cap input filter. Grey 
5th February 2002, 06:00 AM  #6 
diyAudio Moderator Emeritus
Join Date: Jan 2002
Location: Michigan

Always a Load in a Class A?
In a Class A amp (specifically, the Alephs and Zens), wouldn't the constant bias current constitute a continueous load even if the speaker is disconnected? If that is the case, one would have to disconnect + or  rail from the amp, or one of the FET's would have to fail in order to unload the supply, in which case, you would have bigger problems to contend with.
I guess I wonder if I run a properly sized inductor on the input to the PSU, would the bias current draw of an Aleph be enough to hold the rail voltage to +/ .9 * Vac. How much will the rail voltage depend on the amount of capacitance following the inductor? Rodd Yamashita 
5th February 2002, 08:28 AM  #7 
diyAudio Editor

FEThead was the one that spelled it out so I finally grocked it.
Grey, thanks for the clarification. You can really communicate, which I 'm sure you will agree is difficult. I think it helps that the concepts are quite clearly worked out in your mind. I think for a lot of us, our knowlege seems really clear to us until we try to explain it to someone elsethen the gaps start to turn up. I had assumed that L filter was the same as the pi as far as voltage drop. How amazin' that the order of the components makes such a difference. OK, I found some 5.5mh inductors. They are about .6 DCR (DC resistance?) vs. about .3 for 2mh. They are 14 Ga. So... will they heat up too much with 20 amps at 23 volts? That is about the max I would ever use. Sound like time for Ohms law. Too bad I don't have a clue! Mark
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"The geek shall inherit the earth" 2017 Burning Amp is November 12 in S.F. Thread here on diyAudio: http://www.diyaudio.com/forums/clubs...francisco.html 
6th February 2002, 05:02 PM  #8 
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Join Date: Jan 2002
Location: Colorado

Rodd,
Yes, the constant current of the class A amp will keep the required load on the power supply. If the speakers become disconnected, no problem. And yes again, it would require a disconnect from the power supply or an FET failing open (short is OK but may over heat the resistors in series with the FET) to cause the power supply voltage to rise. You are correct: if the inductor is the proper size, the class A amp will keep the DC voltage at 0.9 * the VAC of the transformer. (Remember to include about 1 volt drop per diode and the IR drop (resistance) of the inductor when determining the output voltage. See my comments to Mark below.) The amount of output capacitance (after the inductor) will not affect the DC voltage. But the more capacitance the less the ripple voltage. Mark, Ohms law is: volts = amps * resistance. In the case you mention 0.6 ohms (DCR is DC resistance  or  ohms) times 20 amps is 12 volts! You will lose 12 volt across the inductor! Then, power is amps * volts. 12 volts * 20 amps = 240 watts! That will get hot!! For the 0.3 ohm inductor: 0.3 ohms * 20 amps = 6 volt drop. 6 volts * 20 amps = 120 watts. So, #14 is a little too small for your application. There are your clues, Mark. Sorry they did not turn out better for you!
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FEThead 
7th February 2002, 03:27 AM  #9 
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Join Date: Jan 2002
Location: Michigan

Thanks for the replies.
OK, it's starting to become clearer. If my goal is to eventually build an Aleph ~5/60, then I'll have to (or my power supply will) contend with both the + and  rail currents varying (sometime substantially) around the bias current level. Now since this changing current will be drawn directly from the power supply, if I were measure the dynamic voltage drop across the PSU inductor, it would changing with the current. Which in turn would cause my rail voltage to rise and fall with the voltage loss of the inductor, unless... If I have a large enough cap between the inductor and the supply rails of the amp, won't that smooth out the current draw across the PSU input inductor. This would be using a power supply configuration as follows: Bridge => Inductor => Large Cap in => Voltage Rail or Bridge => Inductor => PI filter w/Large Cap Last => Voltage Rail Would this be sufficient enough to smooth the rail voltage fluxuations enough to fix the voltage drop (hence the rail voltage) to a steady state (almost) value. I will be taking delivery on a Plitron 1000VA +/33V w/CT and would like to run about 28 or 29 Volt rails at relatively high current. Thanks all, Rodd Yamashita 
7th February 2002, 07:32 PM  #10 
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Join Date: Jan 2002
Location: Colorado

choking on chokes
Hi Rodd,
I am not sure what you mean when you said the transformer would have "+/ 33V W/CT". I think that means a total of 66 VAC with a CT. Based on that lets say your PSU looks like:    inductor    cap    + rail [ to gnd ] XFMR BRIDGE CT to ground amp(s) [ ]    inductor    cap     rail to gnd I hope that comes out OK and makes sense. (It did not! See below.) For an inductor I am going to use as example a Stancor C2688 simply because I have a couple of them in my power supply and do not have to do any research. These inductors are rated at .010 Hy, 12.5 amps, and .11 ohms each. The formula for critical (or minimum) inductance is: inductance in HY = (output voltage / output current in milliamps). (Or close to this. See earlier posts.) With a choke input filter your outputs will be about +/ 30 volts. (We will treat these as two seperate supplies.) Doing a little math we find that with these chokes you will need to have a minimum of 3 amps flowing to keep the chokes "working". (If you go below 3 amps the voltage will slowly rise as Grey pointed out in another post.) Note that the bridge will have 2 diodes conducting at any one time, and each diode will drop about one volt. So, you are down to about 28 VDC. Now you have 3 amps flowing thru the inductor. Three amps times .11 ohms gives a .33 volt drop across the choke. This brings you down to 27.67 VDC for each rail. Now suppose you pull 12 amps from each rail: 12 * .11 = 1.32 volts drop across the choke. Your rails have now dropped to 26.68 VDC. One volt variation, or about 5% for a four times increase in current. All of this assumes that your transformer voltage remains at exactly "+/ 33V" or 66 VAC total. I do not know anything about the transformer you mention, but a voltage variation of 5% from no load to full load is typical for many transformers. If this is the case, your VAC goes from 66 to about 63, and now your rails will drop to about 26.6 VDC each. And this assumes your line voltage stays the same as you go from 180 watts (3 amps * 60 volts total) to 720 watts. . . . but lets not go there! (Big smile icon here. I can't figure out how to work those things.) Hope that helps. BTW, this is a little off topic, but I suggest you put a small (1 uF, not critical) capacitor across the secondary of the transformer. When you turn off the PSU there might be a voltage spike from the collapsing magnetic field in the power transformer, and the small cap will absorb the spike so it does not kill your bridge. Upon posting this I saw that my attempt at a schematic got mangled big time. (Big frown icon here.) Let me try to explain it: CT goes straight to ground. + from the bridge goes to the inductor. The other side of the inductor goes to a cap to ground and the amp(s).  from the bridge goes to one side of the other inductor. The other side of this inductor goes to a cap to ground and the amp(s). (Big I Hate Computers icon goes here!)
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FEThead 
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