|[AMRadio] Mod transformer turns ratio|
km1h at jeremy.mv.com
Sat Nov 17 17:26:51 EST 2007
Don, it is rated at 7.34KW and weighs in around 250-300#. Came from a 10 KW
Martin BC TX; PP 4CX5000A's modulating another pair. No other markings on it
other than the mfg (Avdin Energy Corp) part # and date code of 7752.
Its in oil and about 17" square and maybe 22" high.
Serious overkill but the price is right and Ive been looking over a year for
something. The size doesnt bother me, just as long as I can use it. Maybe I
should dust off the 304TL's and forget about the 810's?
OK on the choke requirement. I'll start looking once I determine exactly
what the heck Im doing.
----- Original Message -----
From: "D. Chester" <k4kyv at charter.net>
To: <amradio at mailman.qth.net>
Sent: Saturday, November 17, 2007 4:18 PM
Subject: [AMRadio] Mod transformer turns ratio
> Carl KM1H wrote:
> "Looking at a modulation xfmr that is marked 4596VCT RMS Primary and 3747V
> RMS secondary. This appears to be a 0.82:1 voltage step down or a .64
> impedance step down if I did the math right. Question is what sort of
> room is there in both impedance ratios and actual voltages used?
> The power rating is way more than sufficient for my needs and it is rated
> down to 50 Hz. I can use modulator tubes to supply sufficient power to
> overcome a reasonable mismatch but what is reasonable and how does the
> mismatch affect distortion? I havent been able to find any formulas or
> That would be more normally expressed as 1.23:1 turns ratio, or almost
> exactly 1.5:1 impedance stepdown.
> Does it give a power rating or current rating? Or load impedance rating?
> Without that information it is difficult to determine what the optimum
> impedances are, but that usually isn't critical. That's the way the
> popular multi-match modulation transformers like the UTC VM series works.
> With a given turns ratio, a wide variety of impedances can be used. The
> only thing that cannot be changed is the impedance ratio. So you might
> use a pair of modulator tubes working into a 15000 ohm plate-to-plate load
> to modulate a final amplifier at 10000 ohm modulating impedance. Or your
> modulator tubes may operate into a p-p load of 9000 ohms, so the
> modulating impedance would be 6000 ohms. A 4000 ohm modulating impedance
> would yield a 6000 ohm p-p load.
> That is a good ratio for using a common power supply for modulator and
> final, for achieving good positive peak capability with plenty of
> headroom beyond 100% in the positive direction.
> How big is the transformer and how much does it weigh? That might give a
> clue to its power rating if that data isn't given. Given the voltage
> ratings you listed, if the power rating or current rating is known, the
> optimum modulating impedance can be calculated using Ohm's law, and thus
> the power rating if that isn't given. It sounds like broadcast iron if
> it's rated down to 50 Hz frequency response.
> That brings up another issue. Is it desined to carry the DC to the final
> through the secondary? If not, you will need a modulation reactor and
> blocking capacitor to go with it. A good way to find out, if it open
> frame where you can see the core laminations, is to see if there is a gap
> in the core. If the laminations are stacked in such a way that there is a
> gap, filled with paper or some other kind of insulation, then it is
> designed to carry the DC. If it is cross-laminated like a power
> transformer, with adjacent sets of laminations reversed so that there is
> no gap in the iron core, then most likely it is not designed to carry the
> DC. Most broadcast modulation transformers are NOT designed to carry the
> DC to the final.
> Don k4kyv
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