klem kedidelhopper wrote:
On Feb 27, 6:21 pm, Jamie
t wrote:
klem kedidelhopper wrote:
On Feb 26, 6:39 pm, Jamie
et 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