"John Robertson" wrote in message
...

klem kedidelhopper wrote:
On Feb 28, 7:21 pm, Jamie
t wrote:
klem kedidelhopper wrote:
On Feb 27, 6:21 pm, Jamie
t wrote:
klem kedidelhopper wrote:
On Feb 26, 6:39 pm, Jamie
t wrote:
wrote:
On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:
"klem kedidelhopper" wrote in
message
...
On Feb 20, 6:06 pm, amdx wrote:
On 2/19/2013 5:01 PM, klem kedidelhopper wrote:
On Feb 19, 12:51 am, "Phil Allison" wrote:

With a full-wave bridge rectifier, if you use an 8300 uF
capacitor,
the
volts of ripple will equal the amps of load current. (Don
Lancaster
taught me this.)
** The correct value is 6300uF.
Such a cap will drop 1 volt in 6.3mS, the typical discharge
time in a
full
wave rectifier at 60Hz.
For 50Hz full wave supplies, the value is 7500uF for 1V ripple
at 1
amp.
.... Phil
This is some great advice. Thank you everyone for all your
input.. I
especially like the diode and Zener ideas. They/re cheap and
should
work well. I've got lots of diodes around here and I'm going to
experiment with them. However after all this I remembered that
the
transformer is center tapped! At the time, I simply wire nutted
the
tap, tucked it down under the transformer and forgot about it.
How
this simple fact and the possibilities it presents slipped my
mind is
anyone's guess. I've been building this thing in the evenings
after my
usual work and perhaps I've been tired. The transformer measures
25.2V
at 120V line unloaded. Loading the transformer in this fashion
will
probably unbalance it somewhat and drop the 24V a bit as well
but I
don't see it as a real problem. I should be able to use one side
of
the secondary and the tap now get at least 15 -17 VDC out of the
bridge and filter, and that will provide a healthier input to my
regulator. Lenny
No need to unbalance the transformer, just use 2 diodes and use
the
center tap as negative. See Here,
http://metroamp.com/wiki/index.php/F..._Tap_Rectifier
Mikek
But won't that still give me 36VDC into my regulator? Lenny
No, it won't
Arfa
I looked over what I had previously written and perhaps I didn’t
explain this properly. I have a 24VCT transformer. The transformer
needs to power 24VAC equipment as well as 12VDC equipment. For the
DC circuit I first tried using a bridge directly off the 24V
winding. As soon as I connected up the filter cap the DC output went
to 35V, which is probably too high to feed my 12V regulator.
Several people came up with some great ideas to address this, and I
then realized that I had a center tap that was not being used. So I
connected my bridge across the center tap and one side of the
secondary. This time the 13VAC when FW rectified using the bridge
went to about 19VDC, which is a safe input to the regulator.
Although this worked, I wasn’t happy about unbalancing the
transformer this way so I posted my results.
Unless I seriously misunderstood It was suggested here that I come
off the full secondary output with two diodes, cathodes tied
together, (typical FW rectifier), and use the center tap as my
negative return. I didn't think that would make any difference in
the output voltage from using a bridge without the center tap
however I tried it anyway. As I suspected it would, the DC output
again went to 35V when I connected the filter up.
I've been playing with power supplies all my life and maybe my age
is catching up with me. I guess I just don't understand the
explanation of how to get a lower voltage to my DC regulator by what
I thought sounded like just configuring the transformer properly. So
at the risk of sounding like a complete idiot here can someone
please explain this further? Thanks, Lenny
Your problem is simple
If the 24V AC devices does not come in contact with the common or
ground of this supply you have, you can use a 2 diode full wave
config.
The center tap would be your common for the 12 volt supply, one
diode
from each outter leg joined together to form a full wave. The diode
output alogn with using the center tap as your common will give you
12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess
you
could do that. Use a LDO type.
As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd
or
DC out..
o o
| |
| | to your 24V AC device
| |
| |
| |
| |
| |
| |
+ | |
--------------+. ,-o---------|-----|-+--+
)|( | | 18DC
)|( | +------++
Line Voltage -. ,-+----+| | | ===
)|( === | | /-\
)|( GND | | |
+-------------+-' '+----------o-----|+----+ ===
GND
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)
Unless I missed something, I don't see a problem?
Jamie
I'm not sure if my last reply worked so I'll try it again. I know that
it's very difficult when all we can use here is text but I'm afraid
that I just can't understand, (read) your schematic. From you
description however it appears to be the FW arrangement that I've
already tried and got 35 V out of after I connected up the filter. Is
there any way to make your diagram any more readable in this medium?
Thanks, Lenny
THe FW you tried was a 4 diode bridge, this is a 2 diode full wave
config.
THe difference being is, the CT (Center tap) is your common, the
negative terminal and where the Cathode of the 2 diodes come together
are the (+) terminal which will be around 18VDC with Cap attached
If you truly do have a 24 volt transformer with CT, this means you'll
get 12 Volts from each outer leg with respect to the CT. Because they
are out of phase with each other (180), this is a split phase
configuration, like seen in residential pole pigs in the USA.
If you have done this and gotten 36 volts, then you actually have a
48V transformer with CT
Jamie
What you're referring to is a 12VCT transformer. Connecting either a
bridge and not using a center tap or just two diodes across the full
secondary with cathodes tied together using CT as common connected to
an output filter will yield 18VDC. A 24 V transformer under the same
circumstances will produce 36VDC. I've tried it and it does. Lenny.
absolutely not, you are greatly mistaken or you don't know what you have
for a transformer.

THe transformer is spec'ed out via its outer legs, the highest
voltage. The CT is the voltage divider. when you use the CT as your
common, the supply will yield only half of the rated voltage of that
transformer and is when you use only 2 diodes from the outer legs to
form a full wave into a single node where this becomes the (+) terminal.

With a real 24V Transformer with CT and using the CT as your common
(-) terminal, you will get 12*1.414 = 16.968-0.650 = 16.318VDC at the
peak of a marginal ripple.

You have your logic mixed up

Jamie

What I decided to do is use the full secondary and employ the two 6.8V
zeners in series with the cap. That arrangement gives me 21V out of my
power supply to feed the 12V regulator. As far as I can tell the other
methods would be using the center tap and one side of the transformer,
a method which I didn't want to use. I tried this full secondary
arrangement with a 1.2A load and the 21V across the zeners holds
pretty steady. I have a nice heat sink for the two diodes, (they will
be insulated) that will be going into the box as well. Thanks everyone
who advised and helped me on this project. Lenny

Until one of your zener diodes short out (a zener will eventually short
out due to inrush current issues is my guess) and you then punch around
19V to the circuit that wants 12....not a good design as a result, more
of a hack.

John :-#(#



Well then you want another protection Zener shunted on the 12v line.



Gareth.