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Old 18th October 2004, 02:35 PM   #1
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Default Vero board under load

Ok ive got my bridges using MUR860 diodes built onto Vero board (The board with the strips). Now the power supply is fine under no load and measures approx 37 volts on each rail.

But once connected to the circuit the strip which connects two diodes and a cap and becomes the ground burns off the board literally.

Is this the circuit drawing too much current for the board or is there something else wrong?

Will get a schematic up tomorrow morning.


Thanks for your time,
Joel
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Old 18th October 2004, 05:02 PM   #2
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Joel, I have made a number of discrete rectifier bridges that are built on Vero board and never had a problem with them.

Do you have a circuit diagram of how you have connected everything up? The cap may be the problem!
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Old 18th October 2004, 05:35 PM   #3
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ive done some on strip board aswell and all fine if a little warm!!

try leaving the legs long, bending them along the strip board and soldering all the way along, similar to point to point. that can help with heavy current loading.

and im talking 15 amps, 40 volts. so it can be done. IF, your circuit is correct.

hope this helps, see you soon, steve.. ..
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Old 19th October 2004, 07:10 AM   #4
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Hope this works out

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Old 19th October 2004, 07:46 AM   #5
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Thanks Joel. That schematic looks OK but what we really need to help you, is to see what you have actually constructed.

Can you draw us a simple sketch of how you have connected up your transformer, rectifier bridges and caps?
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Old 19th October 2004, 08:04 AM   #6
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OK ill work on that now. But another question about the windings. The dot on the schematic denotes the start of the winding but does it matter what part of the bridge it is connected to?

Like does it matter if A and B where swapped around?

Sorry for my n00b questions but transformers are new for me.
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Old 19th October 2004, 08:52 AM   #7
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I can't see it well enough on your copy, but looking at the power supply drawing I think that for the lower rectifier section the + and - might be swapped.

In other words: The + out of the second/lower rectifier section should be connected to ground of the first bridge and the make the GND for the amp.

The negative output of the second rectifier section should be connected to -V

Also you should make sure that A and B are indeed not swapped. This can be verified when powering up without the rectifiers, connect B and C and measure the voltage between A and D. It should be double so around 44 Volts.

regards,

Maarten
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Old 19th October 2004, 10:06 AM   #8
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A rectifier bridge has two input points and two output.

On a ready-built bridge, the (AC) input points will probably be marked with the AC symbol liked this ~

The (DC) outputs will be marked + and -

On a discrete bridge, you need to understand the corresponding points. If you look at my
GC FAQ page you can see this diagram which will enable you to identify the correct connection points.

Click the image to open in full size.

Also look a the picture on the FAQ page just below that diagram. You can see the red and black DC connecting wires. The AC wires from the transformer go between the two pairs of diodes.
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Old 19th October 2004, 11:02 AM   #9
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Hi Scribble

one question, does your transformer have 4 wires on the primary or 2?

if i am guessing correctly, it is 2, which means you are shorting out the transformer when you are applying a load! eek!

you cannot use that transformer/bridge setup if you are using a locally available transformer of which all are single 240V primaries.

where did you get your transformer from?
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Old 19th October 2004, 11:08 AM   #10
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So this is my PSU that supplies two channels. Each channel takes a - and + from each bridge. The -'s meet up to form the star ground and the -v and +v go to there respective pins.

Upon measurment both rails give about 37volts under no load but once connnected to the circuit KA-BLAMMO!

There goes the strip that connects the -v, ground connections (ie: cap and two anodes)

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