Re: Aw: Re: AW: Re: AW: Re: Re: Re: Re: Re: AW: [EVGRAY] Re: "Impressive"!

Database ID: 106790
2018-03-03T20:14:15+00:00
werner held <[email protected]>

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YES SVEN correct the diode stopped the motor from drawing from the battery&allowed the battery to charge when the RV was running!I am not very up on electronics so I don't know most of what the group talks about(diode plug/thyrisistors/etc) .Up above someone said capacitor charging a lead acid battery does NOT WORK .WLL TO THE CONTRARY I SAY IT DOES WORK .I have several sizes of  capacitor chargers that I use very regular .One I use on my 36 VOLT GOLF CART  THE BATTERYS ARE GEL CELLS APPROXIMATELY 15 YEARS OLD &STILL WORKING .
 

   

 On Saturday, March 3, 2018 11:31 AM, "'Sven Friedrich' [email protected] [EVGRAY]" <[email protected]> wrote:
 

     Hi Ole,

with a load resistor I waste the energy yes, I would like to push this into the battery.

If I switch 2x 230V / 24V transformers primarily in parallel I reduce the impedance since the energy is distributed to both transformers.

You mean to cache the energy after the push-pull transformer and then discharge it. Help that gets more and more complicated.

There are also people who have loaded the capacitors directly into the battery, apparently the lead-acid batteries keep that out. They are only sharp pulses. It would be nice if you could install a kind of electrical sink, so that the energy of the transformer falls almost without resistance, perhaps with large capacitors and then discharged in the battery with a separate circuit or the capacitors swallow the energy peaks and give them with lower voltage from.

I'm a bit overwhelmed, and working on a new constellation with shorter and equal connections between the diodes. I also wonder if I should twist the lines back and forth all together.

regards

Sven Gesendet: Samstag, 03. März 2018 um 19:57 Uhr
Von: "[email protected] [EVGRAY]" <[email protected]>
An: [email protected]
Betreff: Re: AW: Re: AW: Re: Aw: Re: Re: Re: Re: AW: [EVGRAY] Re: "Impressive"! Hi Sven,

For transformation between output to load impedance the ratio of the transformer has to fit the impedance ratio to be able to do its job. Else the output voltage and impedance must be chosen to fit the load voltage and impedance or vice versa.

Or the load has to be disconnected from the capacitors even when the capacitors haven't fully discharged which gives the problem of switching off the SCRs. A load that can absorb the energy before the next charging cycle requires the proper resistance for the capacitance value chosen. This resistance can be calculated when the other parameters of the circuit are known. The other circuit parameters can also be calculated for a given load and worked backwards if the load is known as everything affects each other.

"I have already considered whether to connect the secondary coil of the push-pull transformer with a FWBR and conducts the energy in a mains choke and then connect in the battery with flyback diode as a step-down converter with only 50 Hz or 2x 230V / 24V transformers in parallel to the secondary winding to lower the impedance."

Or use a power resistor to dissipate the excess power as heat. This can transfer the energy of the capacitors into heat at almost any rate when correct chosen. The series inductance will be what limits the current and thus must be small enough to be able to discharge the capacitors fast enough to turn off the SCRs before the next capacitor charging. This power resistor can be added in parallel to make the load impedance less or in series to make it greater.

Regards
Ole

---In [email protected], <s.friedrich@...> wrote :
 Hi Ole,

you just have to create a low impedance at the push-pull transformer. The problem is that this transformer is just a isolating transformer so the voltage is not down-converted to 24 volts.

The 230/24 volt transformers I have to connect to the secondary side of the pushpull generates a higher impedance which ensures longer switching times at the SCR.

I have already considered whether to connect the secondary coil of the push-pull transformer with a FWBR and conducts the energy in a mains choke and then connect in the battery with flyback diode as a step-down converter with only 50 Hz or 2x 230V / 24V transformers in parallel to the secondary winding to lower the impedance.

The only question is when the 24V winding is connected with a FWBR to the battery, the battery will act against it with its own voltage and create additional impedance in the push-pull transformer.

The easiest way would be to connect the secondary winding directly to the battery and to charge it with the energy peaks.

I'm just not comfortable because the batteries are in excellent condition and I do not want to destroy them.

regards

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