24 May
2009
24 May
'09
1:51 a.m.
Taken from Stancor Transformer Design Guidelines Page 7
The formula for the relation between secondary RMS current (IAC)
which the transformer has to deliver and the D.C. output current taken
from the rectifier (IDC) is: IAC = KFF x IDC where KFF
is the form factor.
The factor for each circuit and type of load is as follows:
Capacitor Load
Form Factor Rectifier Circuit
1.2 FWCT ( Full Wave Center Tap )
1.8 FWB ( Full Wave Bridge )
How can you say the rectifier topology doesn?t matter ?
Best regards, Steven
----- Original Message -----
From: "dwight elvey" <dkelvey at hotmail.com>
To: "General Discussion: On-Topic and Off-Topic Posts"
<cctalk at classiccmp.org>
Sent: Wednesday, May 20, 2009 6:52 AM
Subject: RE: S-100 power supply transformer
Hi
Wow, it just goes to show that you can't believe everything
you read. What a bunch of bull. While I'm sure most of the
book is quite useful, this part if flat wrong.
Ever wonder why every high power switcher I've ever seen
used a center taped transformer? I guess those engineers
never read this book.
Lets take things one at a time. First, it is correct that
the voltage is half but that is just a matter of doubling
the turns on each winding.
Now the next part. Each winding carries current for only
half the time( so far ok ).
Now for the point of deviation. Why would the current
be double? It takes two pulses of current during each
cycle to charge the capacitor. Does it make any difference
were that current came from? Even a more important question
is where did double the current go? Seems like there is
a rule for current that says what goes in must come out
( Nortons rule as I recall ).
For the bridge circuit, there are two pulse of current
per cycle. Each one of these pulse is split into the
two halves of the center tapped. Each pulse is exactly
the same current except one is on each leg of the center
tapped transformer ( again Norton rule demands it ).
This means that each half of the winding has half the power
of the single winding ( assuming that the windings are the
same resistance ).
The number he states as 1.4 is completely bogus.
Please do think it through. To have the same loss in the
secondary, the winding have to be exactly 2 times as much
wire but being the low voltage windings in our case, they are just
not that much bigger on the transformer.
All I can say is WOW, what a blupper!!!
Dwight
----------------------------------------
From: steven.alan.canning at verizon.net
To: cctalk at classiccmp.org
Subject: Re: S-100 power supply transformer
Date: Tue, 19 May 2009 14:34:12 -0700
I respectfully disagree. ? In a center-tapped full-wave rectifier circuit
the output voltage is half what you get if you use a bridge rectifier. It
is
not the most efficient circuit in terms of transformer
design, because
each
half of the secondary is used only half the time. Thus
the current through
the winding during that time is twice what it would be for a true
full-wave
circuit. Heating in the windings, using Ohm?s law, is
I^2R, so you have
four
times the heating half the time, or twice the average
heating of an
equivalent full-wave bridge rectifier circuit. You would have to choose a
transformer with a current rating 1.4 times as large as compared with the
( better ) bridge circuit; besides costing more, the resulting supply
would
be bulkier and heavier ?. from the good book ? ? The
art of Electronics ?
The disadvantage of a bridge is two diode voltage drops. Diodes are a lot
cheaper than transformers though.
Best regards, Steven
-------------------------------------------------------------------------
Hi
This is the result of when the current actually flows.
It only flows when the voltage is near the peak. This
changes the average power dissipated in the secondary winding
( the VA rating ).
Altough, the 1.414 is the worst value and the real value
is somewhere between 1 and 1.414.
It make no difference what type of rectification is used,
fullwave bridge or center tap. Watts are watts and there is
no way around it. It is true that the load current rating
is halved for the secondary turns for the center tapped
fullwave rectification but the VA rating of the transformer
remains the same.
I'm not sure what the book you reference says. I do
know how power is calculated and it is always the vectored
volts times amps. The current flows out of the transformer
when the output voltage is close to the peak. This means
that the rms current of the secondary needs to be compensated
to account for when the current actually flows.
If your transformer has a secondary rated at 10 amps that
can only feed a rectifier and capacitor filter of 7.07 amps.
That would change if feeding a inductive filter because of
when the current flows.
If it was a center tapped with two diodes, it would be
14.14 amps instead of 20 amps. Still, the VA rating of the
transformer does change.
Dwight
----------------------------------------
From: steven.alan.canning at verizon.net
To: cctech at classiccmp.org; cctalk at classiccmp.org
Subject: Re: S-100 power supply transformer
Date: Mon, 18 May 2009 23:36:52 -0700
Dwight,
I think a little explanation is in order, otherwise your info borders on
incorrect. You need to increase the CURRENT rating of the transformer by
1.4, but this only applies to a full wave rectifier topology and not a
full
> wave BRIDGE topology.
>
> reference: The holy book " The art of electronics " page 47
>
> Best regards, Steven
>
>
>> ----------------------------------------
>>> From: lynchaj at yahoo.com
>>> To: cctalk at classiccmp.org
>>> Subject: S-100 power supply transformer
>>> Date: Sun, 17 May 2009 16:51:44 -0400
>>>
>>> Hi! Does anyone know of an available transformer suitable for making an
>>> S-100 power supply? I need 120VAC input and 10VAC output and +/- 18VAC
>>> outputs.
>>>
>>> The best I solution I can find is two separate 80VA transformers; one
>>> parallel 10VAC output and another 36VCT output which can be configured
> for
>>> +/-18VAC outputs.
>>>
>> ---snip---
>>
>> Hi
>> What you want is an 8 volt and a 13.5 volt AC. You forget
>> that the DC voltage is created from the peaks not the RMS
>> average. You have to account for diode drop as well.
>> That is how I got the numbers. I assumed full wave rectification
>> and .6v per diode.
>> Dwight
>>
>
> Hi
> I forgot to mention. You need to derate the transformer
> as well. The problem is that the current flows when the
> voltage is the highest ( or close to it ). The math is a little
> complicated but it means you need to increase the rating of
> the transformer by at least 1.414.
> If you can't find a transformer with the right windings,
> I recommend looking at toroidal transformers. These can
> easily have turns added to increase or buck the voltage
> of the secondary.
> Dwight