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Re: rectifying
Hi Malcom, all,
Original Poster: "Malcolm Watts" <malcolm.watts@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
Let�s 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 @ 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 @ 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 doesn�t 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. It�s
like paralleling equal DC sources. Most, if not all, tube rectifiers
work this way, btw. This dual diode & ct�d 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@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