The variation of incremental inductance versus DC magnetisation was presented back in 1927:
https://www.dalmura.com.au/static/THE%20MEASUREMENT%20OF%20CHOKE%20COIL%20INDUCTANCE.pdf
That paper also indicates the variation of incremental inductance with AC current level.
For a CLC filter, the AC current and DC current reduce significantly if power supply loading decrease (such as with just a bleed resistor).
For a LC filter, the impedance of C and load R are typically very low compared to the filter L, and so all the rectified AC voltage drops across the inductance and the resulting AC current changes little with loading variation.
https://www.dalmura.com.au/static/THE%20MEASUREMENT%20OF%20CHOKE%20COIL%20INDUCTANCE.pdf
That paper also indicates the variation of incremental inductance with AC current level.
For a CLC filter, the AC current and DC current reduce significantly if power supply loading decrease (such as with just a bleed resistor).
For a LC filter, the impedance of C and load R are typically very low compared to the filter L, and so all the rectified AC voltage drops across the inductance and the resulting AC current changes little with loading variation.
The change of permeability in a gapped core is largely swamped-out by the gap. A 6H 60mA choke, at 8mA, may be 7H or 5H, but not real-low H.
Only if the voltage is thousands of Volts. (Which probably exceeds the choke insulation rating.)
The formula in reply #11 is general. Simplified it means the L in Henries must be more than the R/I of the DC load (in k-ohms). A 6H choke can soothe an approximately 6,000 Ohm load. An 8mA, 0.008A, load at say 400V is 50,000 Ohms. We need a 50H choke for full choke action.
It will give "some" smoothing and probably less than full 1.414 peak voltage, but more than the 0.9 of proper choke-input rectification. Rectifier current will NOT be continuous. (I don't know if I care at a mere 8mA.)
PSUD says we need, for 300V and 8mA (37.5k) at least 40H to get "continuous" rectifier/choke current. Any less, the choke sucks pulses, the output voltage rises to something between 0.9 and 1.414.
Only if the voltage is thousands of Volts. (Which probably exceeds the choke insulation rating.)
The formula in reply #11 is general. Simplified it means the L in Henries must be more than the R/I of the DC load (in k-ohms). A 6H choke can soothe an approximately 6,000 Ohm load. An 8mA, 0.008A, load at say 400V is 50,000 Ohms. We need a 50H choke for full choke action.
It will give "some" smoothing and probably less than full 1.414 peak voltage, but more than the 0.9 of proper choke-input rectification. Rectifier current will NOT be continuous. (I don't know if I care at a mere 8mA.)
PSUD says we need, for 300V and 8mA (37.5k) at least 40H to get "continuous" rectifier/choke current. Any less, the choke sucks pulses, the output voltage rises to something between 0.9 and 1.414.
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To me that is a fairly clear question about the variation of choke inductance with current. The initial responses assumed the same. Then there was a side-discussion about rectifiers. It was only in post 10 that the opposite assumption was first made about the OP's original question.abj1 said:
Anyway, he could clear up the confusion by telling us what he meant in post 1. It is clear to me that he meant one thing, and equally clear to some others that he meant something different.
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I see the ambiguity in the question, but I'm putting words in his mouth! - of course a choke input filter is still a filter but for chokes used downstream of the PSU after the dc has been created already, I don't think current variable inductance should be an issue.
I wouldn't wait for the OP. His question is nearly 15 years old and he hasn't participated on the forum for a decade.
I wouldn't wait for the OP. His question is nearly 15 years old and he hasn't participated on the forum for a decade.
Blame Bigun. He popped-into a 14yo thread talking about his kids hooking-up (apparently "connecting" to an old gripe about the forum dirty-word censor).
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There is a formula to calculate critical inductance. Which is the minimum inductance required for a specific filter task. L(crit)=E/I Where E is voltage from the supply, and I is current drawn through the filter (in ma). Case in point; My HF amplifier, 250ma at 2000v. so we divide 2000 by 250ma.=8h is the minimum inductance required for input filter choke. For my 6L6 stereo amp, 350v divided by 60ma = 5.8333h (round up to 6h to be safe)
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WHY are we still beating this dead horse?
From *you*.
reply #11, this thread
It is an approximation: 1,130 is not 1k, and 60Hz is not used everywhere. But it gives you a quick answer to a question which obviously confuses many people (including me: thank you trobbins for the correction).
From *you*.
reply #11, this thread
It is an approximation: 1,130 is not 1k, and 60Hz is not used everywhere. But it gives you a quick answer to a question which obviously confuses many people (including me: thank you trobbins for the correction).
If you can see correctly my post from Seely, there is a parameter w = 2πf (as o-mega) that has a frequency term, so I'm not confused nor in error. It appears that you need a new pair of glasses. Again, E/I gives R, or Volts/Amper = Ohms (Reactive or resistive depending if AC with reactive components; or DC)
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You are being personally rude for no good reason. Yes, I had eye trouble, but $4k of surgery made me 20/25; PC glasses let me read the monitor excellently.
Let us go through the math together.
Lc = 2*Rl / (3*2*pi*F)
or
Lc = Rl * (2 / (3*2*pi*F))
What is the value of (2 / (3*2*pi*F)) for the common power frequencies?
(2 / (3*2*3.14*50)) = 942
(2 / (3*2*3.14*60)) = 1131
At 50Hz: every 942 Ohms of DC load needs 1H minimum inductance.
At 60Hz: every 1131 Ohms of DC load needs 1H minimum inductance.
It is *coincidence* that both values are "1k Ohm" for practical purpose. (We don't buy 5% tolerance chokes, we always round-up to the nearest available value.)
Therefore you figure the DC load like a resistor, divide by "about 1k" (942 or 1131), that is the choke value that you need to exceed.
Dear PRR: Osvaldo CAN NOT be personally rude for the very good reason that he does not know you (besides these postings in DIYAudio) and he has NO WAY to know or even guess that you had an eye operation or whatever.
Besides, "you need glasses" is a common expression to mean "you did not understand it" or "not read it well" and applies to anybody, don´t read any bad meaning into that.
Resistance- R=V/I, where R= resistance, V= volts, I= current in amps.
So, in the case of my RF amp, R=2000v/0.25a, R= 8000ohms.
Critical inductance is L=V/Ima. Where, L is inductance, V is volts, and current is miliamps, so L=2000/250ma= 8h. See?
The formula comes from ARRL handbook.
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