Date: Wed, 3 Mar 2010 03:48:40 EST
From: TexasRF@aol.com
Subject: Re: [Amps] soft start, how to determine if needed?
Hi Alex, no, I didn't forget what inrush protection is; that is why I
suggested the 25 ohm resistor and shorting relay.
You are the one suggesting installing a one ohm resistor and leaving it in
the circuit. Please go back and read your email. I was simply responding
to your suggestion and explaining why it is a bad idea.
## Gerald, what he is implying is that when T=0.. on turn on, the
ENTIRE 240 vac is dropped across the 1 ohm, leaving little or
no voltage across the pri. This is how any step start works.
Place one dvm across the main 240 input, and 2nd dvm across the
step start R... and a 3rd dvm across the PRI. On turn on, the entire
240 vac is across the step start R. Zero on the PRI. A few msecs
later, the vdrop across the pri goes up, while the V drop across the
step start R goes down. The 2 always total 240 vac.
Also, on the idea of the capacitors supplying current peaks during the
peak of the AC cycle and recharging them during the remainder of the cycle,
please consider this:
### what's he's saying is the caps supply current for the voice peaks,
and also all the RF. Meanwhile, the xfmr is still pumping out high level,
short duration pulses every 8.3 msecs. [120 per 1000 msec]. These days
they call that a ..'pulse train'.. and the reference is always to switch mode
supply's. About the only thing that has a 'pulse train' is switch mode
supplies and ham radio HIGH C B+ supply's. It also accounts for the
lousy power factor of high C, plus you have globs of harmonics on any
pulse train. You require a HP current probe and mating FFT, to see
the harmonics [3-5-7-9-11] If u just measure V x A, and arrive at
VA power, you will not know how much of the lousy power factor is
caused by the C filter, and how much is caused by harmonics. It's a real
eye opener.
73,
Gerald K5GW
In a message dated 2/28/2010 1:43:39 A.M. Central Standard Time,
alexeban@gmail.com writes:
you forget what inrush current protection is!
when switching on a power supply with empty capacitors, the zero impedance
of the capacitor bank is reflected across the primary so that at this time
the current can reach about 250 A worst case and gets worse the better the
transformer is (less internal resistance). at this time all you have as
protection is that resistor and wiring resistance. what the mains sees is an
R-C of 1 ohm and about 50uF network, with a time constant of about
50-100usec.
### = .00005 sec.
## with 50 ohms and 50 uf = .0025 sec. It takes a helluva lot longer
to charge a 50 uf cap with a 50 ohm resistor.. than .0025 sec.
## It takes 1-3 seconds to get the B+ way up, then it slowly keeps
climbing after that. I just tried that exact combination
[ 50 ohms + 48 uf]
after this time, even for a 20A average current you get a sag of 10% in
the voltage. actually it's less since the 20A is a peak current 220V x 20A
equal 4400 Watts! during the time of the peaks, which are quite short
duration , what comes into play is the dynamic regulation of the supply
whereby
the capacitors supply the peak current and get recharged during the rest of
the cycle. that's why they are there!
Alex 4Z5KS
### agreed... BUT... 240 vac / 1 ohm = 240Amps... which is a bunch.
And will burn up ur on /off switch's etc. Check out the german fellows
website. They recorded 190A surges in a L4B. with NO step start
protection. Same deal for another amp. [150A] MLA-2500
## In the unique case of the MLA-2500, it has the fil xfmr built into
the plate xfmr. The start up surge into the poor tubes was massive,
>65 amps. Check it out.
## I think the part ur missing is the RC. This is not ur normal RC network,
with a cap being charged up from an external B+.... via a resistor !!
In this case, the R is BEFORE the xfmr... and the C.. is AFTER the xfmr.
When u turn on the AC voltage, you have a huge 240 vac drop across the
1 ohm R. You also have no voltage across the pri of the xfmr.. that takes
time.. and a lot longer than .00005 sec.
## In any event, the 190-240A surge will fry components. Even with 1 ohm
inserted into the primary, the surge is still too great. 1 ohm is not high
enough for a R... and it can't be left in the circuit either.
Jim VE7RF
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