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charging batteries without a coil

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  • charging batteries without a coil

    Yesterday afternoon I mutually charged AA NiMH batteries with only
    1 transistor,1 cap and one resistor, by connecting the base with a 10K resistor
    to the + of the source and the - of the source with a wima capacitor polypropylene
    0.1/250V MK8 4. and connecting the collector tthrough a 1N4007 diode with the +
    of the charge battery and both - of source and charge connected with the emitter
    of the TIP31C transistor.After 4 hours
    source from 1.17V to 2.30V (2 AA NiMH)
    charge from 1.23V to 8.48V (8 AA NiMH)
    So , no expensive wire spools.

    PatDet

    ps: can someone replicate and confirm this!

  • #2
    trigger

    Somebody lent me a scope and it produces sharp needle spikes. So it's loading
    with voltage spikes. I'll use it as trigger in an amplification unit.

    patdet

    Comment


    • #3
      Originally posted by patdet View Post
      Yesterday afternoon I mutually charged AA NiMH batteries with only
      1 transistor,1 cap and one resistor, by connecting the base with a 10K resistor
      to the + of the source and the - of the source with a wima capacitor polypropylene
      0.1/250V MK8 4. and connecting the collector tthrough a 1N4007 diode with the +
      of the charge battery and both - of source and charge connected with the emitter
      of the TIP31C transistor.After 4 hours
      source from 1.17V to 2.30V (2 AA NiMH)
      charge from 1.23V to 8.48V (8 AA NiMH)
      So , no expensive wire spools.

      PatDet

      ps: can someone replicate and confirm this!
      Can you post a exact schematic of your exact setup?

      Thanks,
      -Dave Wing

      Comment


      • #4
        Originally posted by citfta
        Is this what you are describing? I don't see any connection from the source battery to the charge battery.
        simple charger 001.jpg

        @citfta - very close, but he called for a TIP31C High power NPN transistors.
        in your drawing you have shown a PNP transistor, also you have your batteries polarities reversed, reference : Battery (electricity) - Wikipedia, the free encyclopedia

        I ran the circuit in a simulator and did not see the spikes that patdet speaks of.

        perhaps this is a case of - build it and the spikes will come . . .

        we should just wait for an exact schematic to be posted by patdet
        Last edited by prembold; 03-27-2015, 07:08 PM. Reason: test

        Comment


        • #5
          Yes you are correct. I just jotted that down real quick. I should have paid more attention to what I was drawing. I still don't see any connection between the source battery + and the charge battery. I think there is still something missing. Thanks for catching my mistakes. I'll redraw that but I wish he would show what his schematic looks like.

          Carroll
          Just because someone disagrees with you does NOT make them your enemy. We can disagree without attacking someone.

          Comment


          • #6
            I couldn't find a way to change that schematic to a corrected one so I just deleted the post. Here is a corrected schematic as best I can understand from the description.

            Carroll

            PS: I went ahead and posted this thinking he may not have a way to post schematics so maybe he can tell us what is wrong with this one.
            Last edited by citfta; 10-06-2015, 09:14 PM.
            Just because someone disagrees with you does NOT make them your enemy. We can disagree without attacking someone.

            Comment


            • #7
              Spikes

              Your drawing is correct. I'll intend to play around with it varying capacitor and
              resistor values and redirecting it to an optocoupler.

              Patdet

              Comment


              • #8
                The diode is spiking. Ultra fast diodes can spike as high as 20 volt if the current drops rapidly. Normal diodes 3-10 volt. 1n4007 is a normal diode.
                You can look at it like a light switch hooked up on the supply side, Current on the supply side is suddenly stopped from flowing and you get small arc which creates a spike on the line towards the load. Over time this will either burn the switch or the load.
                In your case the power that is left is between the transistor and the diode.

                Its something you have to be very aware of when using diodes in line with a gate on a mosfet. Over time the continued spiking of the diode will eat away at the gate and wreak havoc with your turn off time.

                Matt

                Comment


                • #9
                  Diode spike

                  Matt, could you explain a bit more about the dynamic in this circuit and elaborate a bit on your explanation? Thanks for helping.
                  Last edited by wayne.ct; 03-30-2015, 01:21 AM. Reason: My bad
                  There is a reason why science has been successful and technology is widespread. Don't be afraid to do the math and apply the laws of physics.

                  Comment


                  • #10
                    " both - of source and charge connected with the emitter
                    of the TIP31C transistor"
                    The updated drawing only shows the one battery set connected to the emitter.
                    I'm always missing something, but would like to try this one and I've already missunderstood something.

                    Andrew

                    Oh, one more thing, is the charging bank in series or parrellel ? 1.5 volts (parrellel) or 12 volts (series)
                    Last edited by BobBrown; 03-30-2015, 02:09 AM.

                    Comment


                    • #11
                      Originally posted by wayne.ct View Post
                      Matt, could you explain a bit more about the dynamic in this circuit and elaborate a bit on your explanation? Thanks for helping.
                      So if you have a power source, then a switch, then a load, then ground, and you turn the switch on get power flowing, then cut the switch off You will get a small surge of voltage via an Arc. Any time power jumps across air its a transformation of power from current to voltage. It steps up. It also has the physical property of something in motion wants to stay in motion like fly wheel. This is actually a pretty well known effect.

                      So the same thing happens in a diode. The diode is in front of the load, you turn the current off with the transistor. The diode has capacitance between the anode and cathode so it stores current. When the diode substrate is turning off that bit of current left then basically arcs into the load causing a boost of voltage.

                      You can maximize this effect by following the data sheet and the "Turn Off" time of a given diode and pulsing the diode at 30% duty cycle.

                      Remember though this is not free energy in fact its probably a loss of about 30-40% of total power. But the closer you end up to the turn off time and capacitance the more efficient you get.

                      Hope that helps
                      Matt
                      Last edited by Matthew Jones; 03-30-2015, 04:31 AM.

                      Comment


                      • #12
                        That drawing is just my idea of what I thought patdet was describing in his first post. He has posted it is correct. However, I see no way for that circuit to charge the batteries as there is no connection from the source batteries to the charge batteries on the positive side.

                        To answer your questions. The negative of both charge and source batteries ARE connected to the emitter. And in his first post he states tne battery voltages after charging that would indicate the source batteries are in series. He also indicates the charge batteries are in series. Hope this helps.

                        Originally posted by BobBrown View Post
                        The updated drawing only shows the one battery set connected to the emitter.
                        I'm always missing something, but would like to try this one and I've already missunderstood something.

                        Andrew

                        Oh, one more thing, is the charging bank in series or parrellel ? 1.5 volts (parrellel) or 12 volts (series)
                        Carroll
                        Just because someone disagrees with you does NOT make them your enemy. We can disagree without attacking someone.

                        Comment

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