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I am trying to replicate the circuit I found here:

schematic
(source: cogeco.ca)

This is a photograph of what I have built:

Update

In my photo I have labeled one diode in the bridge rectifier which has burn out twice now. Unfortunately my supply of diodes is running low and I can not see what the error is in my circuit when I compare it to the schematic.

Can some one please try to spot what I have done wrong?

Edited: Updated Picture with new Layout that the Poster has made. Still burns out the same bottom diode.

Original: https://i.sstatic.net/Oigw4.jpg

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    \$\begingroup\$ Aside from what other folks are telling you, you should solder your connection to the transformer instead of just looping solid wire and hoping it works. Also, it looks like you have a bad solder joint on the wiper of your pot (though you can never be sure just by looking). You'll find that there's enough to worry about when doing this kind of stuff that you don't need to be worrying about the status of each individual connection. \$\endgroup\$
    – Scott Seidman
    Commented Mar 25, 2013 at 20:47
  • \$\begingroup\$ This is some of the worst work I have ever seen. Bend the diode leads with a pair of piers, at right angles, near the entry to the body of the component, before you insert them. \$\endgroup\$
    – user207421
    Commented Mar 25, 2013 at 22:06
  • \$\begingroup\$ I would also add that this assembly is DANGEROUS! You have access to the main without any protections. The cables in front of the transformer seem to be accessible. I hope that you use a residual-current circuit breaker in front of your setup. I understand that this is a prototype. Anyway, If you do something wrong (nobody is perfect) you could die !!! My advice: take care and use appropriate security system. \$\endgroup\$
    – Blup1980
    Commented Mar 26, 2013 at 7:35

2 Answers 2

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The yellow wire from your transformer is connected to the wrong thing. You have it connected to the "Common" in the schematics, but instead it should be connected to two of the diodes.

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  • \$\begingroup\$ Especially considering that's highlighted in red on the picture OP posted :D \$\endgroup\$
    – Passerby
    Commented Mar 25, 2013 at 20:26
  • \$\begingroup\$ I have connected the yellow wire to output end of the diode that was burning out and connected the input of the burning out diode to the "Common". I tested with a new diode and the same diode burned out again. I took a picture of the changes I made: imgur.com/ZCG8XT5 \$\endgroup\$
    – codefi
    Commented Mar 25, 2013 at 20:32
  • \$\begingroup\$ One of the other diodes was fused, after replacing it the circuit works like a charm. Thank you for all your help. \$\endgroup\$
    – codefi
    Commented Mar 25, 2013 at 20:48
  • \$\begingroup\$ @Passerby No, it isn't. The red text advises the builder not to bridge the common of the secondary-side circuit with the ground on the primary side. \$\endgroup\$
    – Kaz
    Commented Mar 25, 2013 at 22:29
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    \$\begingroup\$ @Adamme The lesson there is that when some silicon blows, you cannot assume that only that silicon device is faulty. This is the case in other situations. E.g. if in a push pull amplifier one transistor blows, you can't assume the other one is okay. It could seem to be, but actually it could have internal damage that will prevent it from performing properly. \$\endgroup\$
    – Kaz
    Commented Mar 25, 2013 at 22:31
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Your wiring does not look right. It looks like you're using the horizontal strip as the common for the secondary-side circuit. This is obvious because the capacitor grounds there as well as the potentiometer.

But, oops, you have connected one of the wires from the transformer to this common ground.

Note that the schematic has no such junction. The bridge rectifier has four distinct nodes. The transformer connects to two of them, supplying AC, and the other two go to the circuit's voltage rails, supplying DC to it.

In other words, it looks like you may have done this:

enter image description here

And so note how the diode simply shorts out the transformer on every half-cycle, which means that it will be fried from a lot of current.

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