EWorkshop1708 said:
Oh really? How long is 'eventually'? I gave up after about 72 hours, but maybe I should have left it longer. This is ALWAYS the reason I have to buy a car new battery - was that a waste and I could have restored the battery by charging long enough?
I keep meaning to build a low current charger (les than an amp) that I can leave on 24/7/365.
I'm in the process of trying this out on a completely dead battery that was in a car I've had sitting for a long time. I tried charging it with my car charger but it just keeps cycling on/off.
I was trying to get a 60w bulb hooked up to the circuit, but getting a wire attached to the steel base of the bulb proved to be an excesses in futility. I ended up using a 7.5uF motor-run cap instead.
When I first plugged it in the voltage was around ~17V and has slowly dropped to ~12.4V (and still dropping). Current draw measures around 270mA.
I will post back with my final results later.
I was trying to get a 60w bulb hooked up to the circuit, but getting a wire attached to the steel base of the bulb proved to be an excesses in futility. I ended up using a 7.5uF motor-run cap instead.
When I first plugged it in the voltage was around ~17V and has slowly dropped to ~12.4V (and still dropping). Current draw measures around 270mA.
I will post back with my final results later.
Success! 
I left the "desulfator" on the battery for around 5 hours at which point the voltage was ~12V. I felt this was high enough for my automatic charger to take over so I set it at 6A and left it over-night.
After the charger completed it's charge cycle the voltage was ~13V with the charger in trickle charge mode. I then connected the battery to a transformer out of a an old linear regulated 12V 10A supply (with a rectifier of course). After just a few minutes the voltage was at ~15.5V and the electrolyte was boiling rapidly. I think I need to put some resistance in series, the transformer is capable of too much current.
Anyway, it seems to have worked on this >5 year old battery. I will probably do a slow discharge and charge it back up again before putting it back in the car.
I left the "desulfator" on the battery for around 5 hours at which point the voltage was ~12V. I felt this was high enough for my automatic charger to take over so I set it at 6A and left it over-night.
After the charger completed it's charge cycle the voltage was ~13V with the charger in trickle charge mode. I then connected the battery to a transformer out of a an old linear regulated 12V 10A supply (with a rectifier of course). After just a few minutes the voltage was at ~15.5V and the electrolyte was boiling rapidly. I think I need to put some resistance in series, the transformer is capable of too much current.
Anyway, it seems to have worked on this >5 year old battery. I will probably do a slow discharge and charge it back up again before putting it back in the car.
theAnonymous1 said:Success!
I left the "desulfator" on the battery for around 5 hours at which point the voltage was ~12V. I felt this was high enough for my automatic charger to take over so I set it at 6A and left it over-night.
After the charger completed it's charge cycle the voltage was ~13V with the charger in trickle charge mode. I then connected the battery to a transformer out of a an old linear regulated 12V 10A supply (with a rectifier of course). After just a few minutes the voltage was at ~15.5V and the electrolyte was boiling rapidly. I think I need to put some resistance in series, the transformer is capable of too much current.
Anyway, it seems to have worked on this >5 year old battery. I will probably do a slow discharge and charge it back up again before putting it back in the car.
Sweet! Glad to see you got that old battery to take a charge.
7.5uf, pretty sweet. Is there any heating of the capacitor, or does the circuit remain cool?
EWorkshop1708 said:
The cap and rectifier remained completely cool.
I'm not too good at this stuff, but doesn't the cap form a first order high-pass filter? If so, according to the current draw I measured (~270mA), shouldn't the cap be dissipating around ~32W? It didn't feel like it was dissipating any power at all.
I would like to find another battery and try this again with a slightly larger capacitor.
EDIT: Oops, there was a flaw in my above thinking. I didn't measure the current draw on the AC side, I measured it on the DC side when the voltage was under 12V. That would make it more like 3W of dissipated power in the cap, correct?
Hope you understand this better now
and
Current certainly flows through capacitors; but only alternating current. Capacitors block DC.
When (AC) current flows though a capacitor, the current is phase shifted by 90° relative to the voltage.
Power dissipation is defined V.I.cos(phi) - i.e., the scaler product of current and voltage.
With a resistor, current and voltage are in phase, so phi=0, and
power =V.I
With the capacitor, phi=90°, so Power dissipated = zero!
Inductive loads shift the phase the 'other way' by 90°, a pure inductor will also dissipate zero power when carrying alternating current, though real inductors have a largish resistance too, so there are resistive heat losses involved as well, which is why inductors get warm.
The current flowing that dissipates no real (heat) power is called reactive power, and is a big issue for power distribution companies, because they have to install high current cabling, but get no return in revenue because the reactive power doesn't show up on the energy consumption meter. So industrial consumers are additionally charged for their reactive power usage.
So, anyone care to explain to a laymen like myself the function of power being transfered here?
It seems with the 7.5uF capacitor the max current flow after the rectifier can be no more than 288mA (measured into a complete short). With a 100R resistor in series the voltage is 28.5V @ 280mA. With a 5R resistor it's 1.45V @ 287mA. The measurements were taken without any capacitance after the rectifier.
It seems with the 7.5uF capacitor the max current flow after the rectifier can be no more than 288mA (measured into a complete short). With a 100R resistor in series the voltage is 28.5V @ 280mA. With a 5R resistor it's 1.45V @ 287mA. The measurements were taken without any capacitance after the rectifier.
The reactance of the capacitor at 60Hz is
1/(120.pi.C) = 353.6 ohms
The battery you say starts charging at around 17V DC. There is going to be a bit of an anomoly in the current measurements because the voltage difference between 17V DC and 120v AC means the waveform isn't a true sinusoid anymore. Most meters are only calibrated / accurate for true sinusoidal current.
But 17 is a fairly small fraction of 120, so the error won't be enormous.
120v - 17v gives 103v dropped over the capacitor reactance of 356 Ohm.
103/356 = 290mA.
Taking into account the error of the meter, the fact that the capacitor is only accurate to within about 10% of its marked value, your measured 270mA is within the expected range.
With a dead short, you get a measured 288mA. Here there is no DC error. 120V is dropped directly across the capacitor:
120/353.6 = 339mA.
So, either your AC line voltage is lower than 120, or the capacitor is out by 15% (6.37uF or 416 ohms at 60Hz).
120v/416ohm = 288mA - exactly as you measure it.
I don't understand where the 100R resistor or 5R resistor are put and where those voltages are measured, so I can't offer any input on those figures.
1/(120.pi.C) = 353.6 ohms
The battery you say starts charging at around 17V DC. There is going to be a bit of an anomoly in the current measurements because the voltage difference between 17V DC and 120v AC means the waveform isn't a true sinusoid anymore. Most meters are only calibrated / accurate for true sinusoidal current.
But 17 is a fairly small fraction of 120, so the error won't be enormous.
120v - 17v gives 103v dropped over the capacitor reactance of 356 Ohm.
103/356 = 290mA.
Taking into account the error of the meter, the fact that the capacitor is only accurate to within about 10% of its marked value, your measured 270mA is within the expected range.
With a dead short, you get a measured 288mA. Here there is no DC error. 120V is dropped directly across the capacitor:
120/353.6 = 339mA.
So, either your AC line voltage is lower than 120, or the capacitor is out by 15% (6.37uF or 416 ohms at 60Hz).
120v/416ohm = 288mA - exactly as you measure it.
I don't understand where the 100R resistor or 5R resistor are put and where those voltages are measured, so I can't offer any input on those figures.
So what is the theoretical and real world efficiency of such a capacitor-based charger? If it's very efficient, it could be a good charger for homemade EVs, provided the battery is isolated and a circuit used to verify isolation. Add some contactors and electronics to switch in and out different value capacitors and it can actually be a good smart charger.
Certainly the capacitor wastes no energy. But safety isolation would require a transformer, and if you are putting in a transformer I think you can do better than a simple resistance charger (which is what the capacitor effectively is, as seen by the battery)
Nickel-cadmium and Nickel-metal-hydride like constant current.
Lead-acid batteries like to be charged with constant voltage; constant current charging them over a long time period shortens their life.
Constant voltage can be achieved very efficiently by means of a switched mode power supply - several ICs exist to do this with minimal support components.
An old AT computer power supply should work - with a small modification to set the voltage.
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