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Re: rectifying
Hi Malcom, all,
Original Poster: "Malcolm Watts" <malcolm.watts-at-wnp.ac.nz>
>Hi Reinhard,
>I'm afraid I'm going to have to agree to differ on this one:
More comments below ;o))
>No. Each diode in the centre tapped arrangement only
>delivers on alternate half cycles. There is no reason why
>you can't pull the same current from either arrangement.
>However, each winding in the centre-tapped supply is
>being used for only half the time.
Not gonna let you off the hook, here ;o)). Let me try my
hand at ASCII schematics, so that we can see, if we are
talking about the same circuit (use courier font to view):
X---------P||S------->|-----|
P||S D1 |
P||S |
P||S |
P||S-------O1 |---O2
P||S |
P||S |
P||S D2 |
X---------P||S------->|-----|
X,X are primary mains terminals
P is the primary winding
|| iron core
S is the secondary winding with a center tap
>| are the two diodes, with "|" being the cathode
O1 is the negative output terminal
O2 is the positive output terminal
Letīs stay with my example of 12-0-12V, with each winding
capable of 1A. I say, if you ignore the above schematic and
connect the outer terminals across a 4 way bridge, you get
24V -at- 1A. If you use a 4 way bridge and use the center tap
as one connection you get +12 from the top and -12V from
the bottom, but still at 1 amp each, so total VA is still 24VA.
If you use the above (schematic) dual diode full wave rectifier,
I say you get 12V -at- 2A between O1 & O2, still being 24VA.
And here is my explanation:
The diode D1 conducts on the first half cycle, while diode D2
is still reversed biased. On an alternate half wave, the situation
is reversed (D1 doesnīt conduct, D2 does). If you take an
O-scope and look at each diode section separately, you will
see a waveform (diode forward biased), then a "hole" (diode
reversed biased), then a waveform, etc. The other diode is
exactly the same, except that the "hole" is exchanged for a
"waveform". I.e., when D1 has a "waveform", D2 has a "hole"
and vice versa, so the fundamental frequency is 2x FMains
(be it 50Hz->100Hz or 60Hz->120Hz).
And now to the current ;o)) The center-tapped, dual diode
setup supplies current from BOTH windings (to the load), so
you get 2x the current possible per single winding. In our case
above, it would be 2 x 1A. Picture the current flowing though
D1 as I1 and the current flowing through D2 as I2. The load
you connect between O1 and O2 sees I1+I2 (or in our case,
as I1=I2= 2 x I1 or 2x I2) However, at only one half the
voltage across the two windings (24V/2), as you are paralleling
both windings (each having 12 volts) through the diodes. Itīs
like paralleling equal DC sources. Most, if not all, tube rectifiers
work this way, btw. This dual diode & ctīd setup is also used
in most DC welding rigs (except for TIG).
Of course, if you have two separate windings and 4 way rectify
each one and then parallel them, you also get 12V-at-2 x 1 amp
or 12V at 2 amps. So, no matter how you rectify a center
tapped xformer (be it dual diode or 4 way bridge), the total VA
remains the same, BUT the voltage and current DO differ ;o))
Okay, your shot ;o))
Coiler greets from Germany,
Reinhard