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Is there a Diode List some where with ratings & part numbers?

gary350

Well-Known Member
I drew my circuit for the 71 carbon rods that I have it turns out to be only 26 watts. I need to order titanium plates so I can make full use of a 1400 watt transformer. Circuit is 3.5 volts DC. Amps on the cathode plates .1 amp per sq in. I need to buy enough titanium to make the math work out. Diodes should be 10 volts 400 amps. My industrial chemistry book says 3KW for every lb of salt in water and water is 30% salt at 25 degree C. Before I do more math for surface area of titanium plates and tank size I need to know what diodes are available?

How much voltage drop is there on a diode? Is that on the datasheet?


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Diver300

Well-Known Member
Most Helpful Member
I'm not sure what your electrolysing salt water will do, but I'm fairly sure there'll be a lot of chlorine gas given off.

You are going to get around 1 V drop per diode. About half your power will go to heating the diodes.

You might be better to have active rectifiers to reduce the rectifier losses.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
You are going to get around 1 V drop per diode. About half your power will go to heating the diodes.
Agreed.

Also, titanium is not a good electrode material for electrolysis, unless the aim is to destroy it?
Titanium forms a fairly inert surface oxide layer in a similar way to aluminium, as far as I am aware?

Stainless steel electrodes are often used for simple electrolysis.
 

gary350

Well-Known Member
I think I have it figured out. 10 stainless steel positive plates each 6" x 12" in 5 gallons of liquid. Plates will be 14.4 amps each = 144 amps. I already have several 200 amp diodes. 144a x 3.89v = 560 watts.

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Diver300

Well-Known Member
Most Helpful Member
You are going to need a huge capacitor for it to be of any use. However the circuit will probably work fine without it.

I'm still not sure what you are trying to achieve.

I think you mean 5.5 V peak not 5.5 V dc.

You are going to get a lot of heat in the diodes. Have you got a photo of the setup?
 

gary350

Well-Known Member
You are going to need a huge capacitor for it to be of any use. However the circuit will probably work fine without it.

I'm still not sure what you are trying to achieve.

I think you mean 5.5 V peak not 5.5 V dc.

You are going to get a lot of heat in the diodes. Have you got a photo of the setup?
If transformer secondary is 3.9 volts ac that will really be 5.5 volts peak to peak. With a 1 volt voltage drop each of 2 diodes that will leave 3.5 volts peak to peak. I have several 74,000. capacitors 200 vdc that should help to smooth out the peak to peak. I have a 900, 1200 & 1400 watt E I transformer core I can wind a continious duty primary and secondary coil on. 900 w will be fine for this but wait and see how well wire fits it might be best to use only 500 watts of the 1400 watt EI transformer to have room for high amp wire. I have not done math on primary winding yet but from past transformers I have wound they all fall in the same area for 120 vac primary plus or minus a few turns. My memory is not very good these day but i think primary will be about 200 turns. Turn ratio is 21.818 makes the secondary 5.500 turns. 5.5 turns of 140 amp wire but not sure if 00 wire will fit or bend easy for the secondary winding. I might need to make 2 or 3 secondary windings in parallel to get 140 amps I have done that type thing before. I might just have to use what every wire fits in the space available if it turns out to be 75 or 100 amps than i will use what works. I have a 3KW E I transformer laminations I have been saving I think that is over kill for this project.
 

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