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16th August 2019, 03:35 PM  #21 
diyAudio Member
Join Date: Oct 2013

Hi Guys,
Thank you for sharing your knowledge. The question is how low a DCR is appropriate for Choke input ps ? Using the formula , if the circuit only draws 25ma so it's 350/25 = 14H the DCR will surely be high, example Hammond 157L dcr is 429 ohms Thank you again 
16th August 2019, 04:44 PM  #22  
diyAudio Member
Join Date: Dec 2012
Location: SF Bay Area

Quote:
In terms of power supply droop under realworld loads, remembering that one's HV power supply Linput LC section will have good sized C followin the L, in an SET/SEP only varies 20% or so between quiescent and clipping, and for a PP power stage (depending heavily on how close to A, AB, AB2, pure B it runs), anywhere from the same 20% to well over 100%, you're still in a pretty small range of ΔV values. 20% of 25 mA = 5 mA … ×0.43 V/mA → ±(2.2 ÷ 2 = 1.1) VMinor indeed. I could be fullofbeans, but at least that's my take on it. Just saying, GoatGuy ✓
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John Curl's Golden Rule…: 100 kHz bandwidth, 3 μs risetime, 100 W mean output, 100 V/μs slew rate, 2 Ω dynamic load, 20 amp min current source/sink Last edited by GoatGuy; 16th August 2019 at 04:47 PM. 

16th August 2019, 05:28 PM  #23 
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16th August 2019, 07:27 PM  #24 
diyAudio Member
Join Date: Dec 2012
Location: SF Bay Area

I believe it to be a empirical ruleofthumb which encapsulates other design criteria into a simple constant.
{goes away… calculates … scratches beard … comes back 2 hr later} So, I was right. It is empirical, but also fairly magical. First, the '350' is the V_{RMS} of the transformer. The rectification drop is small compared to that at high tension, so it is ignored. BUT… that 350 = V_{RMS} is totally coincidental … to F = 50 Hz, and rectification being fullwave bridge. Completely. Change F to 60 Hz, and while the same inductor will work, the minimal one would actually 'heuristic out' as: L = 350 × 50 / ( mA • Hz )Obviously, 50 / Hz = 1 when Hz is 50. But at 60 Hz, the critical L is smaller. In the end, we also need to remember that the use of a larger value inductor than the one specified by the heuristic will work for the intended purpose; it'll also deliver the 90% of V_{RMS} as the filtered voltage. Only when L drops below the critical value does the LC “choke input” filter begin to deliver (ironically) more voltage, on average. More, up to when the DC resistance of the thing starts to suck up its share. The one big (for goat) takeaway from this is … it is really pretty hard to actually formulæcalculate what the minimum critical inductor size need be, given a particular current. Hard, because rectified sinusoids are anything but formulæeasy integratable. Hard. I ended up with a 30,000 line spreadsheet with 20 columns. It would have been easier to use R or C or PERL, I think. So… L = (50 ⋅ V_{RMS}) / (mA ⋅ Hz)That's the real heuristic. The right one. Just saying, GoatGuy ✓
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John Curl's Golden Rule…: 100 kHz bandwidth, 3 μs risetime, 100 W mean output, 100 V/μs slew rate, 2 Ω dynamic load, 20 amp min current source/sink Last edited by GoatGuy; 16th August 2019 at 07:33 PM. Reason: typos 
16th August 2019, 09:04 PM  #25  
diyAudio Member

Quote:
I have always gone by this formula that is given in the ARRL "The Radio Amateur's handbook" chapter 7 "power supplies". Under "ChokeInput Filters" ....Minimum choke inductance.... L(henries) = E(volts) / I(ma) I have also read of more elaborate versions of this equation but essentially it boils down to this same formula. Last edited by DAK808; 16th August 2019 at 09:05 PM. Reason: correction 

16th August 2019, 11:37 PM  #26 
diyAudio Member
Join Date: Jun 2016

Yes, those good old ARRL handbooks come through again. I like using simple formulas that work, that I can remember, and that I use.
Just remember, one manufacturer might tell you his choke is 7 Henrys @ 130mA DC. It may not be so (at 50Hz or at 60Hz). So a larger inductance and higher current rating gives a little margin to make the supply a true choke input supply. Or check the part yourself (for most of us, the only test equipment we have for this is to put it in an amplifier, check the input ACV, versus the output DCV). Do not forget to account for the voltage drop due to the choke DCR. If the DCR is large enough, the DCV will be less than 0.9 x the ACVrms, even though the inductance is far less than the critical inductance. When you are in the middle of the Pacific Ocean on a Destroyer, and the Captain wants you you to fix that broken piece of electronic gear, and there are no correct parts, you learn to figure a thing or two. The average, unfiltered DCV, of a full wave rectified sine is 0.6366.... times the ACVrms. It also is is 2/pi times the ACVrms. Knowing the why and the full calculations is one thing, knowing something you can remember and use is another. Find the form you most prefer. If you wonder what an application of the 0.6366... is, the old VOM with a d arsonval meter movement and a rectifier to measure ACV. It is calibrated to measure the rms volts of Sine Waves. Any other wave shape, and it responds to, and reads out, in Average volts, not rms volts. 
17th August 2019, 01:53 AM  #27 
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Join Date: Jun 2003
Location: Maine USA

Resistance in the *first* (beforecap) choke has no large effect on final DC value, but damps the turnon surge.
In this sim, V7 is 400V peak, so 283V RMS. With choke input we expect 254V DC. Choke DCR is varied from 1 Ohm to 600 Ohms. The final DCV value hardly changes. However the (unlikely!) 1r DCR peaks to 455V for a dozen mS; the 600r DCR peaks to 335V. While the peak is too quick to blowup an electrolytic cap, even brief 455V on a 300V cap would worry me, 335V for 30mS is no worry at all. 
17th August 2019, 02:40 AM  #28 
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Join Date: Jun 2005
Location: San Diego, CA

The "critical" inductance value is based on what is refereed to as the boundary between Continuous and DisContinuous for the inductor ripple current... The handbook equations are not always applicable.... If your going to use a handbook equation, you should understand where these numbers are derived from... In many cases the author will add in their own "fudge factor"... You ideally do not want to be at the critical inductance value..since that is a boundary region...but rather well into the CONTINUOUS region for some design margin... It all falls back to the basic equation E = di/dt * L

17th August 2019, 04:24 AM  #29 
diyAudio Member
Join Date: Jun 2016

PRR,
The transient voltage at the beginning also depends on the DCR of the secondary (or DCR of the half primary in the case of a full wave secondary w/center tap). I am not as sure of this, but I believe the DCR of the primary also is a factor, as it translates 'upward' due to the turns ratio. cerrem, As I said in some recent threads, I do not always trust manufacturer's choke ratings, either for current or inductance. That is why I recommended using more than critical inductance. Know your vendor. Also, what happens if later you decide to reduce the current of the output tubes, change tube types, change output transformers that have lower current rating. Changes have implications, especially if there was little or no margin there to begin with. There is good engineering, so so engineering, and the original Tacoma narrows bridge. 
17th August 2019, 05:29 AM  #30 
diyAudio Member
Join Date: May 2011
Location: Little Rock

As already pointed out, critical inductance is the bare minimum needed under optimum conditions. It's value depends on both applied voltage and current draw, so formulas with only a current term are incorrect, or only correct for a particular voltage.
A simple but correct number is usually stated as: critical inductance in Henrys equals the equivalent resistance of the load in K Ohms. Just be several times this under all load conditions and you're golden. All good fortune, Chris
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