Re: The things you see when ya go lookin'......
- From: Jim Lesurf <jcgl@xxxxxxxxxxxxxxxxxx>
- Date: Fri, 24 Feb 2006 17:20:07 +0000 (GMT)
In article <43ff1d72$1_1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx>, Serge Auckland
<serge.auckland@xxxxxxxxxxxxx> wrote:
"Jim Lesurf" <jcgl@xxxxxxxxxxxxxxxxxx> wrote in message
news:4dfe236609jcgl@xxxxxxxxxxxxxxxxxxxxx
Nominal impedance 10k. 48dB attentuation in 2dB steps. Inductance 80H.
Since I am unfamiliar with the use of such transformers for audio,
I'll risk asking a few simple questions. :-)
Can you explain what you mean by "10k bridging" in the above? Does it
refer to the input load/arrangement?
For many years now, pro-audio has abandoned the old 600 ohms terminating
(that is, 600 ohms sending impedance, 600 ohms receiving) for the more
modern very low sending impedance (typically less than 50 ohms) into a
high receiving impedance (typically 10kohms) The 10k impedance is
referred to as "bridging" as it can be put across (bridge) a 600 ohm
load without materially affecting the level. The use of the terms is
now something of an anachronism, but is still retained to indicate a
high load impedance, and to distinguish it from the now rarely used 600
ohm terminating imedance.
So is "bridging" in this context used to mean "mismatched, but using a high
load impedance to avoid over-loading the source"?
[snip]
Under what conditions of use does the system present (?) 10kOhms?
If a transformer is designed for 10k use, it needs to be presented with
a high load impedance on the secondary. A 10k transformer presented
with a 600ohm load will distort at a much lower level than with a 10k
load, as the core saturates.
In that case I am puzzled by what relevant the specific value "10k" has in
this context. Is the following incorrect for some reason?
The input impedance presented by the transformer to a source connected to
its primary will essentially be the transformed value of the impedance
connected as a load to its secondary.
Thus if the load on the secondary is 'high', the impedance presented by the
primary will be even higher - by the relevant transformation ratio. Given a
load of > 600 Ohms on the secondary this implies the load the primary
presents to the source will always be well above 600 Ohms, and vary with
the chosen 'attenuation' setting.
If so, is it the case that the effective load isn't likely to be "10k"?
So the "10k bridging" simply means "a large value that varies with the
chosen attenuation"? If so, why on earth call it "10k"? :-)
Conversely, if a transformer is designed for 600 ohm use is used with a
10k load, it won't saturate or distort, but it could have an
undesirable peak in the treble response, although with a well-designed
transformer, this won't be too serious a problem. Nevertheless, the
cousel of perfection is to terminate transforers with their design
impedance.
I can see that being straightforward with a fixed transformer ratio.
Indeed, I've done that more than once. :-) However how do you do that with
a changing tap of a multi-tap transformer used as an attenuator? Does
anyone do so?
>
What would be the levels of series resistances, shunt capacitances,
etc, for the above device? You quote '80H' for an inductance, but
don't distinguish the coupled (mutual) value from the uncoupled values.
This will depend on the specific transformer, so I can't answer this....
Iain?
I was wondering if Iain would know the specific values...
Slainte,
Jim
--
Electronics http://www.st-and.ac.uk/~www_pa/Scots_Guide/intro/electron.htm
Audio Misc http://www.st-and.demon.co.uk/AudioMisc/index.html
Armstrong Audio http://www.st-and.demon.co.uk/Audio/armstrong.html
Barbirolli Soc. http://www.st-and.demon.co.uk/JBSoc/JBSoc.html
.
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