[AMRadio] coupling cap on the moduation transformer secondary


Bob Bruhns bbruhns at erols.com
Wed Aug 17 13:35:30 EDT 2005


Hi Geoff,

The inductances add arithmetically, because there is no
mutual coupling between them.  So you have 10 + 10 + 10
+ 16 = 46 Henries with the four of them in series.

I used 30 uF with 50 Henries back in the day.  I had an
810 (triode) modulator, a VM-5 mod transformer, and the
exact same as you have, 1500V at 300 mA on the RF final
(5000 ohm load impedance).  A 30uF coupling capacitor
and a 50 henry modulation inductor worked fine.  I had
30 Hz furnace rumble like you wouldn't believe.  I
could pass 15 Hz through the thing. (All I had was a
test record with tones on it.  I played it at half
speed to get 15 Hz.)

46 Henries will work just about the same as 50.  I had
the mod transformer secondary at B+, and the coupling
capacitor was just three 10uF 600V oil caps in
parallel.  But if you use the high-pass filter peaking
technique, you need a higher voltage capacitor, because
the resonant effects will put a lot of low frequency
audio voltage on the cap, even if you keep the DC off
of it.

It's a high-pass filter issue.  You have Z(source)
after the mod transformer - Inductance of the mod
transformer to audio ground - Coupling Capacitor -
Inductance of the modulation inductor to audio ground -
5K load to ground.  The inductance of the mod
transformer would be the inductance you would measure
on the secondary, with no load on the primary.

If you can model the high pass filter, you can play
with part values and see what will happen.  I thin the
plate resistance will transform to four times the plate
resistance of one tube, transformed by the impedance
ration of the mod transformer.  So if the tubes have
10K plate resistance and the mod transformer has a 2:1
impedance ratio, then the source impedance will be
40K/2 = 20K.  Somebody check me if I'm wrong on this.

  Bacon, WA3WDR



----- Original Message ----- 
From: "Geoff" <w5omr at satx.rr.com>
To: "Discussion of AM Radio" <amradio at mailman.qth.net>
Sent: Wednesday, August 17, 2005 12:30 PM
Subject: Re: [AMRadio] coupling cap on the moduation
transformer secondary


> Bob Bruhns wrote:
>
> >Do whatever works for you.
> >
> >You may actually get better low-end response with a
> >smaller cap, because the coupling capacitor and
> >modulation inductor make a high-pass L-C network
that
> >can have a peak before rolloff, depending on
impedance
> >and part values.  This peak can supplement a sagging
> >low end, although rolloff below the peak will be
> >faster.  Hopefully it will be low enough in
frequency
> >that it does not matter.
> >
> >A peaky high-pass filter will certainly mess up
phase
> >response, which would affect low-level negative peak
> >clipping and such.  And a high-pass filter will also
> >leave the modulator unloaded at the bottom end.
This
> >loading issue should not matter much if the
modulator
> >tubes are triodes.  If your modulator tubes are
> >tetrodes or pentodes, you could apply some negative
> >feedback from the plates or the transformer output,
to
> >lower the effective plate resistance and control the
> >unloaded output.
> >
> >Mostly, it's the sound that matters.  See what works
> >best and sounds best.  If it doesn't screw things
up,
> >the right capacitor value could add a few dB of
> >response a little bit above low-end rolloff, and
that
> >could work perfectly well for your voice.
> >
> >More capacitance will extend the low end response,
the
> >low-end phase response will be better.  The slight
> >low-end peak will be lower in frequency, and it may
be
> >gone altogether.  Less capacitance will make low-end
> >response worse, and the peak would rise in frequency
> >too.  If you want furnace rumble, you might find a
> >combination of capacitor and inductor values that
> >produces a resonant peak for you down around 20 or
30
> >Hz.
> >
>
> Well... only because I'm using what I have available
I wonder how the LC
> ratio changes, if I'm using multiple inductors in
series?  Of course,
> I'd prefer one 50HY choke at 500mA, but I don't have
one of those.
> Instead, I've got (3) 10HY at 500mA, and a 16Hy at 450mA
all strung in
> series.  Mutual Inductance calculations?  Two of the
chokes are potted
> and the other two are open-frame.  Unfortunatly, the
two open framed
> chokes are -not- identical.
>
> The total DC resistance is ~300ohms for all chokes in
series, and a
> total inductance of 46Hy.
>
> The calculations are for 1500VDC @ 300mA in the
final.
>
> Final impedeance is therefore 5000ohms.  The general
rule-of-thumb is
> 8Hy per 1000ohms Z which is 40Hy.  I'm covered there.
>
> Coupling capacitance?  This is where I'm stumped,
because of the
> multiple inductors in series.
>
>
> ---
> 73 = Best Regards,
> -Geoff/W5OMR
>
>
>
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