Tesla Switch Solid State (John Bedini)

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  • เผยแพร่เมื่อ 10 มิ.ย. 2024
  • Solid State version of John Bedini's "Telsa Switch".
    Circuit was created by Mr. Ronald Brandt who claimed he had a car that used the circuit to drive a DC motor that never needed recharging. John Bedini made a small version of it that ran off 5v NiCad batteries and demonstrated it at a 1984 convention. His circuit powered some small lights and he claimed the batteries never lost charge.
    The circuit sends power from a 24v battery bank to a 12v battery bank and a load is connected between the bank negatives. The banks then switch the other direction. Conventional measurements show only 50% of the power is recovered in the lower 12v bank, plus whatever power the load sees, and accordingly the unit should not be spectacular. However, many experimenters have tested it and claimed loads can be powered well beyond their standard capacities. This is my solid state version of the circuit.
    Unfortunately, I did not see any efficiencies beyond or even close to 100%. Batteries do not perform well when they charge/discharge in rapid succession. The circuit may work better if it pulsed power from the 24V to the 12V bank for several minutes and then reversed course.

ความคิดเห็น • 28

  • @newmonengineering
    @newmonengineering 25 วันที่ผ่านมา +2

    There was some videos i saw of a guy with 3 aa batteries in series and 3 in parallel and some led inbetween. He claims it runs for months. He just hard wired them and every few days he switched the battery cells between the banks. I have no idea if what he was doing worked or not but his setup was very simple compared to yours. It would be interesting to try but I haven't found the time yet recently. Keep up the experimenting!

  • @craigpierce7996
    @craigpierce7996 20 วันที่ผ่านมา

    I did the same experiment in 2005 with a (6) floating gate mosfet circuit. It had less than 50% efficiency. I thought about it and was so pissed at myself for not seeing something very obvious. The source current from the side in series enters the parallel side and has to branch. Therefore, the effective charging current then becomes half of the source current. There's no load that can be connected that can change this inherent relationship. Hard circuit to build! I should have seen this relationship up front and passed on the build. .. Don't let yourself get sucked into this one folks!

    • @energyhack3538
      @energyhack3538  20 วันที่ผ่านมา

      Correct - 50% recovery into lower batt bank is max. (+ whatever load gets). In any case, Not an OU circuit. Also, will destroy the batteries if you run it long enough. I ruined a few LIFePO3 batts letting it run for a few weeks.

  • @windpowerandmore6310
    @windpowerandmore6310 26 วันที่ผ่านมา +1

    That's some impressive work there! Very interesting results too

    • @energyhack3538
      @energyhack3538  26 วันที่ผ่านมา +1

      Thx buddy - does some strange things - have even seen all batteries rise in voltage for a period but eventually they die off. Anyhow, u suggested SRRs and I found they do work well!

    • @windpowerandmore6310
      @windpowerandmore6310 26 วันที่ผ่านมา

      @@energyhack3538 That's awesome!
      Are those LIFEPO4 batteries?

  • @sandysmagicmix
    @sandysmagicmix 26 วันที่ผ่านมา +1

    Bedini Bearden Psy Op. Simple to "replicate", extremely difficult or impossible to achieve significant OU. Impractical to scale up. How many tens of thousands of hours collectively have we all spent on this? Hope you all find what you seek!

    • @energyhack3538
      @energyhack3538  25 วันที่ผ่านมา +2

      LOL. Well put. Perhaps my video will save a few others the time.

    • @VentodiCampagna
      @VentodiCampagna 24 วันที่ผ่านมา

      Batteries cannot recover energy cause their resistive behavior. Only a reactive device can do it. That circuit is a fake! One more thing: the transformer introduces a loss of 30%, cause its magnetic circuit.

  • @Kangsteri
    @Kangsteri 26 วันที่ผ่านมา

    Thanks for sharing! Super capacitors might improve the results. Resistance of the battery chemistry also seems to be important thing. Lead acid doesn't like short cycles, specially starter batteries without added graphite and some bismuth trioxide.

  • @SolarSeeker45
    @SolarSeeker45 26 วันที่ผ่านมา

    Tesla switches should be run with way slower cycling times, every few minutes or even better with a voltage sensor to detect when the drive batteries fall below a certain voltage. They should also be paired with a trickle charger such as a small solar panel. It's best to use lead acid batteries or a battery chemistry which can handle an overcharge. If you use super capacitors instead of batteries you'll see it's essentially an electricity pump or boost converter which can take a low voltage trickle charge and step it up to make usable power from very weak energy sources.

    • @energyhack3538
      @energyhack3538  26 วันที่ผ่านมา +1

      Agree..Switching at 60hz is tuff on a battery - it will eventually destroy them. But that’s the essence of the switch as developed by its creators - rapid back and forth switching and placing a load between the banks.

  • @Rssident
    @Rssident 26 วันที่ผ่านมา

    What's your switching speed?

    • @energyhack3538
      @energyhack3538  26 วันที่ผ่านมา

      Tried everything from 10 to 100 hz.. past 100 the amount of current drops off and at 150hz almost no current flow so the load doesn’t work.

  • @user-uk5jq4mp7v
    @user-uk5jq4mp7v 24 วันที่ผ่านมา

    Ok.... What's your charging ampere ???

    • @energyhack3538
      @energyhack3538  23 วันที่ผ่านมา +1

      Not much.. 350mA.. about 4 watts passing back and forth between batteries

  • @VentodiCampagna
    @VentodiCampagna 26 วันที่ผ่านมา

    Two batteries, 2:1 ratio, an inductor/resistor (load) in series with a capacitor. Two steps: first, charge the capacitor through the inductor/resistor; second, discharge it into the other battery. Almost 95% energy back, COP 5 - 8 depending on the solid state switches, inductors, capacitors, battery quality. There is not wasted energy in the load, otherwise you cannot get back the 95% of the energy from the source... ;) Things work when are simple.

    • @VentodiCampagna
      @VentodiCampagna 25 วันที่ผ่านมา +1

      You can prove what I said charging a supercap instead of a battery. Calculate the energy in the supercap...no matter if inductive or resistive load, the result does not change. Remember, 100 charges leave the source, 100 charges reach the charge battery/supercap through the load. No one less than this. That is the secret. Energy is potential and that is related to charge numbers. A load works as a break: 100 charges will flow through the load in 1 second or 10 seconds, but remain the same 100...

    • @VentodiCampagna
      @VentodiCampagna 23 วันที่ผ่านมา

      This is a practical guide to understand capacitor and inductor behavior... th-cam.com/video/qtPbejIEscs/w-d-xo.html

    • @craigpierce7996
      @craigpierce7996 20 วันที่ผ่านมา

      Have done this circuit as well. Basically, it's a ZVS circuit switching the capacitor via an inductor at it's min/max voltage points. At the right frequency, it is a series resonant circuit with voltage/current buildup between the two batteries. What keeps it from being OU is the phase shift between voltage/current and each battery experiences the same current. It draws down hard on the sourcing battery and I had very short run intervals tuning it at the resonant frequency. Best was just a bit slower than the resonant frequency. Interesting circuit, a great learning experience! Great for learning floating mosfet circuits...

    • @VentodiCampagna
      @VentodiCampagna 18 วันที่ผ่านมา

      @@craigpierce7996 Unfortunately current and voltage in a series resonant circuit are in phase because of resonance, of course...for this reason the circuit behaves as a resistive one. The inductive reactance and capacitive reactance cancel each other: max current takes place only limited by the resistance.

    • @craigpierce7996
      @craigpierce7996 18 วันที่ผ่านมา

      @@VentodiCampagna That isn't correct. Voltage and current are always phase shifted in a resonant circuit, series or parallel. It is the inductive and capacitive reactions that cancel, leaving the impedance to be just wire resistance. You can only shift the impedance in a resonant circuit by adding resistance, but that directly affects the Q of the circuit and thus the voltage rise in a series resonant circuit. If the current was in phase, then it would no longer be an L/C circuit.

  • @jensbuchholz5766
    @jensbuchholz5766 27 วันที่ผ่านมา