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Bedini Solid State Oscillators

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  • selamatg
    replied
    Battery getting warm

    Hi All,


    I have 2 battery always getting warm after about 4 hours on charging with my SS.
    The voltage can raise up to 13.6 V (on charging). but when I try keep charge the voltage will drop slowly. I have try to stop the charging for few hours then charge again but seem the battery condition still a same (won't hold the charging for long period). both battery already try more than 10 cycle with same result.

    This only happen with this 2 battery, the other can charge up to 14.9 V (on charging) and the condition getting better.

    Anyone have same experience with this? or maybe any advice for me?

    Thanks
    Selamatg

    Leave a comment:


  • Shamus
    replied
    Originally posted by theremart View Post
    I have been looking for others who have done what that guide says to make sets of 4 batteries and rotate them with excess power.
    Me too. It seems that I have yet to hear of a success story from that angle (other than Kevin's ) and lots of failures.

    I can't help but think of the TUV tests on John's old pages that showed that very thing and if I remember correctly, the SSG he used wasn't that big. Need to look into this deeper.

    Leave a comment:


  • theremart
    replied
    Originally posted by Burned_NE2 View Post
    Hi EgmQC

    Oops I get what you were saying now, looks like I didn't correctly interpret the formula you were using (my math was a little dyslexic)

    Theremart thanks for the link to the SSG_STARTERS_GUIDE-Two.pdf it's very informative.

    Regards,
    Paul
    Actually Sephiroth was the one that gave you the starters guide ( the one that I have converted to audio and have listen to about 40 times

    I have been looking for others who have done what that guide says to make sets of 4 batteries and rotate them with excess power.

    Leave a comment:


  • Burned_NE2
    replied
    Hi EgmQC

    Oops I get what you were saying now, looks like I didn't correctly interpret the formula you were using (my math was a little dyslexic)

    Theremart thanks for the link to the SSG_STARTERS_GUIDE-Two.pdf it's very informative.

    Regards,
    Paul

    Leave a comment:


  • EgmQC
    replied
    Originally posted by Burned_NE2 View Post
    Hi EgmQC

    Sorry to butt in but according to JB the C20 discharge is NOT "80% of the amp rated" but 20%. If you discharge the battery the "80% of the amp rated" you will kill it in no time.

    To see the example that JB talked about see "18)The C/20 discharge rate." in the John Bedini: SG Collected Posts (PDF) | MERLib.org that Theremart posted earlier.

    Regards,
    Paul
    Hi Paul,

    When i wrote : "we should use 80% of the amp rated to make the C20 rules" , i mean 80% of the amp rated should be use to calculate the C/20, for exemple , a 10amp/h rated battery ,80% of 10amp is 8amp , so 8/20 = the C20 rate, 0.4amp max discharge.

    Best Regards,
    Eric

    Leave a comment:


  • Sephiroth
    replied
    I think for smaller batteries he recommended use 80% of the c20 rate.

    There is some good info on batteries in this pdf



    I found this to be a very good write up of the basics.

    Leave a comment:


  • Burned_NE2
    replied
    C20 Discharge rule

    Originally posted by EgmQC View Post
    Hi theremart,

    Ive got realy bad result when i was trying to do the test load with a inverter and a 5W Light. After like 30 unsuccessfull load test ive put a amp meter. The Inverter alone in idle mode took 300 mAmp , with the light that jumped to about 780 mAmp on a 8amp/h battery. As soon as i used the correct load my result got good as it should be, more power out than power in. Its why i asked you the question about your C20 and discharge. Ive read somewhere(i think its on Bedini_SG3) that we should use 80% of the amp rated to make the C20 rules, in my case (8/100)*80 = 6.4Amp/h, C20 = 6.4/20 = 320 mAmp.

    Best Regards,
    Eric
    Hi EgmQC

    Sorry to butt in but according to JB the C20 discharge is NOT "80% of the amp rated" but 20%. If you discharge the battery the "80% of the amp rated" you will kill it in no time.

    To see the example that JB talked about see "18)The C/20 discharge rate." in the John Bedini: SG Collected Posts (PDF) | MERLib.org that Theremart posted earlier.

    Regards,
    Paul

    Leave a comment:


  • theremart
    replied
    Originally posted by EgmQC View Post
    Hi theremart,

    Ive got realy bad result when i was trying to do the test load with a inverter and a 5W Light. After like 30 unsuccessfull load test ive put a amp meter. The Inverter alone in idle mode took 300 mAmp , with the light that jumped to about 780 mAmp on a 8amp/h battery. As soon as i used the correct load my result got good as it should be, more power out than power in. Its why i asked you the question about your C20 and discharge. Ive read somewhere(i think its on Bedini_SG3) that we should use 80% of the amp rated to make the C20 rules, in my case (8/100)*80 = 6.4Amp/h, C20 = 6.4/20 = 320 mAmp.

    Best Regards,
    Eric

    I also found that using a digital meter to measure the amp draw on my inverter gave me bogus readings. You know when you watch Bedini in his videos he does the exact opposite he tells you to do, he swaps the batteries from front to back without an inverter.

    Leave a comment:


  • EgmQC
    replied
    Originally posted by EgmQC View Post
    Hi All,

    I have a nice question, When the SSG send directly the radiant to the battery, the battery will transform it to usuable energy(by getting it from the vacuum) . With a capacitor setup, the radiant spike is sent to the cap and then the battery.

    My question is, What is inside the capacitor ?

    if we see for exemple 16 volt on a 30000uf cap before it dump it to the battery, that mean there 3.84 joules sent to the battery.

    There's radiant in the cap or its transformed already in usuable energy by the capacitor? if he is already transfomed that mean we only have to feed a cap at 16v and dump at X Frequency to get the same effect?

    Best Regards,
    Eric
    I finaly got my anwser from JB site: "Radiant currents are just like Tesla said a gas in nature , the transformation converter is the Capacitor." The Energy Pages Have

    So there no Radiant dumped to the battery, just ordinary volt/current because it already transformed by the capacitor but what is realy wierd, Mr Bearden say the radiant need some impedance to suck some energy from the vacuum , i understand that for a battery but for a capacitor, its exactly the inverse of a battery for the impedance point of view, low impedance at start and high at the end. Anybody know (or have a theory) on how the transformation take place in the capacitor ?

    Best Regards,
    Eric

    Leave a comment:


  • EgmQC
    replied
    Originally posted by theremart View Post
    Opps I missed this from Bedini, I should be using only .6 even on discharge..

    from :
    John Bedini: SG Collected Posts (PDF) | MERLib.org

    ---------------------------------------------------------
    I run everything on the C20 rate even on discharge, this is very important if you are to succeed.
    If Sterling with a .6 watt motor is charging his batteries and also running the wheel where is the energy coming from? Think about this, a .6 watt motor running a big wheel with rocks in the bearings plus the wind drag and anything else that could be wrong, but it is charging it's batteries and doing work. But Sterlings motor is only a .6 watt motor, how could that be, a .6 watts could not even drive a fly up the wall let alone keep that wheel turning and charge the back batteries. As we have said all along the machine is a 4 to 1 machine, one battery in four batteries out charged. If Sterling inputs one 12 volt 4.2 amp hour battery and he charges four on the back and then connects all four together he would have 16.8 amp hour battery that he could take .84 amps for twenty hours, if this goes beyond the twenty hours, where did the extra energy come from? You can not hurt your batteries if you follow the C20 rate of discharge.
    John

    -------------------------
    Hi theremart,

    Ive got realy bad result when i was trying to do the test load with a inverter and a 5W Light. After like 30 unsuccessfull load test ive put a amp meter. The Inverter alone in idle mode took 300 mAmp , with the light that jumped to about 780 mAmp on a 8amp/h battery. As soon as i used the correct load my result got good as it should be, more power out than power in. Its why i asked you the question about your C20 and discharge. Ive read somewhere(i think its on Bedini_SG3) that we should use 80% of the amp rated to make the C20 rules, in my case (8/100)*80 = 6.4Amp/h, C20 = 6.4/20 = 320 mAmp.

    Best Regards,
    Eric

    Leave a comment:


  • theremart
    replied
    RE: c20 discharge rate..

    Opps I missed this from Bedini, I should be using only .6 even on discharge..

    from :
    John Bedini: SG Collected Posts (PDF) | MERLib.org

    ---------------------------------------------------------
    I run everything on the C20 rate even on discharge, this is very important if you are to succeed.
    If Sterling with a .6 watt motor is charging his batteries and also running the wheel where is the energy coming from? Think about this, a .6 watt motor running a big wheel with rocks in the bearings plus the wind drag and anything else that could be wrong, but it is charging it's batteries and doing work. But Sterlings motor is only a .6 watt motor, how could that be, a .6 watts could not even drive a fly up the wall let alone keep that wheel turning and charge the back batteries. As we have said all along the machine is a 4 to 1 machine, one battery in four batteries out charged. If Sterling inputs one 12 volt 4.2 amp hour battery and he charges four on the back and then connects all four together he would have 16.8 amp hour battery that he could take .84 amps for twenty hours, if this goes beyond the twenty hours, where did the extra energy come from? You can not hurt your batteries if you follow the C20 rate of discharge.
    John

    -------------------------

    Leave a comment:


  • theremart
    replied
    Re: battery..

    Originally posted by EgmQC View Post
    Hi theremart,
    What's the C20 of your battery and your load amp during the discharge test ?

    Eric
    The battery is 12 amp hours.

    So... 12/20 = .6 amps would be the c20 rating

    I am pulling .75 amp from the battery on the load tests.

    Now, as i understand it the c20 is for discharge for being used at the primary driving battery on the SSG or SS.

    Thanks for your input Eric.

    Mart

    Leave a comment:


  • EgmQC
    replied
    Originally posted by theremart View Post
    Not a good day for testing results..





    I charged the battery with the Neos, and the battery put out only 1.34 amp hour as compared to the 1 amp hour it did with the Solid state. It seems this battery has gotten worse over time.

    I am going to charge another battery with the Solid state, and see if it does any better.
    Hi theremart,
    What's the C20 of your battery and your load amp during the discharge test ?

    Eric

    Leave a comment:


  • theremart
    replied
    More testing.

    Not a good day for testing results..





    I charged the battery with the Neos, and the battery put out only 1.34 amp hour as compared to the 1 amp hour it did with the Solid state. It seems this battery has gotten worse over time.

    I am going to charge another battery with the Solid state, and see if it does any better.

    Leave a comment:


  • theremart
    replied
    RE: questions.

    Originally posted by sinergicus View Post
    Hi to all...

    Because You have much more experiences and Knowledge than me ,I want to ask your opinion ...
    I see all around ,everybody trying to replicate some versions of Bedini motors,but all are just toys proving,more or less,some principles...

    With this toys cannot power a house....Or ,I suppose this is your goal in fact;to make some practical device at a big scale to power a house ;like Jim Watson unit for example.

    Nobody tried to replicate first Bedini motor?I understand is not so complicated;Jim Watson has built it.John Bedini has mentioned this can be build without electronics;I understand ,electronics cannot handle big power;
    Mr Bedini said in his book free energy generation:
    With three brushes and a commutator you can do away with all the electronics and handle 100 times the power back to the battery-no transistor to fail,no relays to stick,no resistor to burn up...



    What you think is worthy to give attention and time to this kind of Bedini motor?
    Can somebody help with information about how can be built at a bigger scale?


    Thanks
    Mike
    Hi Mike,

    You should start another thread as you are starting a topic that is a good question, but off track for this thread. Good question, but good to stay on topic in threads..

    Cheers.

    Leave a comment:

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