He connected the diodes in series, between the positive of the supply and the positive of the Zeus. The cathodes on all are pointing to the Zeus. Each one drops about 0.7V. Just remember they must be able to handle the full current of the strips and will dissipate 0.7V x the current in power, so they may need a heatsink. That's why he used the two diodes in series in a bridge rectifier module; it's easy to get high current and to mount to a heatsink.
This is better than just using a series resistor since the drop across a diode is (mostly) constant, while the drop across a resistor depends on the current (Ohm's law). Of course, the switching buck converter is a better solution, but the diodes are an easy thing to try.
/mike
Bingo, give the man a cigar he is completely right.
A SILCON diode will drop about .7 volts in the forward direction it is called the forward voltage drop. Next problem is how to do this.
1) single diodes are actually a problem on where to mount them and cost almost what a full wave bridge does. A full wave bridge comes in nice little package that can be mounted on the side of the power supply. You CAN NOT USE ALL 4 DIODES IN THE BRIDGE. You can only use 2 of the diodes so to get a reasonable voltage drop you can put 4 diodes in series and get a little over 3 volts dropped.
So I decided to BUTCHER (you will have to butcher you cables to do this) my power supplies and put these diodes in. I've attached a pic.
Now for power dissipation I have something like 10 amps average going thru these and it'll be 1.4 volt drop across each bridge. So each bridge is going to have to dissipate about 14 watts and a 2nd one is right next door to it so in that little area It'll be 28 watts. I might get away with it BUT I REALLY WOULDN'T RECOMMEND DOING THIS.
I believe Mike talked about a heat sink and I believe one should be used but right now I'm just trying to get every thing working.
My connections are as follows for the bridge.
I bring in the 12 volts to one of the AC sides of the bridge, Then I jumper the + side to the - side and get my output from the other AC side. I then BUTCHER my cables and ran the 12 volts into this mess and pulled the 9 volts out from it.
NOTE: The big 24 pin connector also has 2 12 volt lines going into it so you have to cut those and cut the other 12 volt lines and run your 9 volt lines into them. You also have watch the amount of current you are running on these wires.
MY SETUP IS FOR TEST!!!!!!! AND ONLY RUNS 1 STRING OF LIGHTS. So do not think that little wire will work for 20 amps ( you will melt the wire ).
I got my bridge at radio shack simply because I needed them for testing. They are rated at 25 amps and have a forward voltage drop of 1.7 volts. I am seeing about 3.2 volts across the pair of bridges. NOTE: You only use 2 diodes at any time in a full wave bridge so its forward voltage drop is what you can expect so when it states a 1.7 volt drop that is for both diodes and like I saw 3.2 volts across 2 bridges gives you 1.6 volts across each one. So you can expect to see what the forward voltage drop is.
NOW FOR THE PUREST
A diode does have resistance and this resistance will cause a change in voltage drop with a change in current. I've seem a couple tens of volt change in my testing with only one string. My option is there is a regulator on the chip I'll let it deal with these variation, besides my lights are junk any fluctuation in brightness is a huge improvement over not working. So I DON'T CARE and I don't think it will matter.
Hope this helps and I wouldn't recommend it unless you have some tech background to understand the variables not addressed above.
Joe