A math Q

[artv]

Hi All,...If a wall charger for a 6 volt nicd battery , says 6VDC out at 500ma....and it takes an hour to charge the flat battery....
So if you  charge it in 30 mins. that's 6VDC out at 1 amp??.......15mins. 6VDC at 2amps??
Every time you halve the time ,...you double the amps???...........artv

[wdyasq]

By any chance, did you go to a government school?

Ron

[artv]

I thought all schools were run by the government......So if a single phase at max rpm is 3.6VAC........there is no way to charge the six volt nicd???........that is in 5mins ,so the baterry voltage dosen't drop???...........artv

[rossw]

So if a single phase at max rpm is 3.6VAC........there is no way to charge the six volt nicd???

OMG. Are you for real? Or are you just trolling for someone to bite?
I could explain it to you.... but I can't understand it for you.

"The lord helps those who help themselves.....   (but god help those *caught* helping themselves)"
(That's my way of saying "would you please go do some research - THEN come back with more appropriate questions")

[wdyasq]

There are private schools that actually teach rather than just process children from K-12.

Ron

[kurt]

i will post one word of warning a nicad charger is more than just a wallwort it has circuitry in it to keep from overcharging the battery and just hooking a nicad battery to any old power source without doing the research to know what you are doing could prove interesting for you. 

[rossw]

could prove interesting for you. 

I was in a room with someone who was charging a NiCad pack inadvertently at C instead of C/10.
When it went off, it took lumps out of two brick walls, and a smashed window where it exited.

interesting barely begins to describe it.

[Bruce S]

The word could be DANGEROUS or DEADLY!!
I'd like to say that I'm pretty good at making sure my NiCds and safely charged I have quiet a few larger banks than normal
Here's a pic of one that got away from me.
take a good close look at the lower right.


IF this can happen to someone who works with 100s of them just thing what can happen to someone who is not totally aware of the dangers involved.
BTW, This took 10 seconds to happen, and I'm still cleaning up. Luckily I was the only person around and had the correct stuff to keep everything else safe.

Bruce S

[artv]

Sorry for my ignorance .....If a single phase of a three phase altenator puts out 3.5 VAC at max rpm....
3.5VAC x 1.414(rms)
 = 4.9VDC,....4.9VDC - 1.4VDC (diode drop) =3.5VDC x1.73=6.14VDC???
Why is it when I test the alt that max rpm says 3.5VAC..........after going through the diode bridge the meter only reads 2.9VDC
I think I understand rms (an average of the sine wave?) .....the diode drop is obvious....where does this 1

[rossw]

Sorry for my ignorance .....If a single phase of a three phase altenator puts out 3.5 VAC at max rpm....
3.5VAC x 1.414(rms)
 = 4.9VDC,....4.9VDC - 1.4VDC (diode drop) =3.5VDC x1.73=6.14VDC???
Why is it when I test the alt that max rpm says 3.5VAC..........after going through the diode bridge the meter only reads 2.9VDC
I think I understand rms (an average of the sine wave?) .....the diode drop is obvious....where does this 1

RMS is not the average. It's the square root of the mean of the squared instantaneous values. For some waveforms, RMS = average, but not all.

Measuring the unfiltered DC value of the full-wave-rectified value won't tell you what you think it will. And a capacitor across it will tell you something between the same value, and the peak value, depending on the capacitance.

If *I* had a 3.5V AC source, and I *WANTED* to charge a (say) 6V nominal pack, I'd use a schottky bridge or a synchronous rectifier followed by a boost type DC-DC converter, into a suitable nicad charger.

If you don't understand what I just said, you should do some research to understand the implications and risks involved before proceeding.

[artv]

I don't know what just happened........where does the 1.73 come from
does that have to do with all the phases together
None of this math makes any sense .........3.5VAC rectified should be what??.........I get 2.9VDC.........
If this question causes you grief or aggrevates you just ignore it ............not trying to cause anybody any grief just looking for answers
Thanks for your time........artv

[Watt]

I don't know what just happened........where does the 1.73 come from
does that have to do with all the phases together
None of this math makes any sense .........3.5VAC rectified should be what??.........I get 2.9VDC.........
If this question causes you grief or aggrevates you just ignore it ............not trying to cause anybody any grief just looking for answers
Thanks for your time........artv

It's part of three phase power calculation and 1.73(2) is a constant.  http://www.ehow.com/how_6521700_calculate-3-phase-power.html

* Just caught the rest of your question.  I'm pretty sure the 1.73 must have been used incorrectly on the example above.  I don't believe the 3.5vdc will be increased by 1.73.  The power available with factor will be increased by 1.73 from all three phases.  *

 3.5vac { (rms) measured with an rms meter } = ~4.9vac peak to peak ( this is what is rectified to dc ) as ross pointed out, this will not be a smooth dc voltage.  

I think if I was going to charge batteries with this idea, I would build a step up transformer from the 4.9vac and then rectify and regulate the 8 or 10vdc to 6vdc to charge my batteries.  But, * I'd * most likely had just built a huge contraption to heat my feet under my desk as the wattage out would be too limited to be practical for battery charging.

[Flux]

Sorry for my ignorance .....If a single phase of a three phase alternator puts out 3.5 VAC at max rpm....
3.5VAC x 1.414(rms)
 = 4.9VDC,....4.9VDC - 1.4VDC (diode drop) =3.5VDC x1.73=6.14VDC???
Why is it when I test the alt that max rpm says 3.5VAC..........after going through the diode bridge the meter only reads 2.9VDC
I think I understand rms (an average of the sine wave?) .....the diode drop is obvious....where does this 1

I think this will be pointless but I will try to explain a bit of it.

If one phase of a 3 phase alternator puts out 3.5vac, then that will be the voltage developed across one third of the winding. What I expect you to have measured is the voltage between a pair of the output terminals. This would be equal to phase voltage for a delta connected alternator, but if it is star connected the phase volts will be referenced to the star point and will in fact only be 3.5/1.73 volts.

Assuming what you have in fact measured to be line volts and you are only dealing with one phase then ( assuming the waveform is a sine wave and that is not likely to be exactly correct).  The mean dc of 3.5v rms will be  3.5/1.11  or 3.15v. You now loose two diode drops from the bridge. Unloaded the diode drop is not very well defined and even loaded the claimed 1.4v drop is much of a compromise figure.

If you manage to read 2.9v then that seems to be well within the realms of reality although probably a bit on the high side.

All the maths work if you take things in context.

The equation you are thinking of here is for a 3 phase star connected machine feeding a 3 phase rectifier into a battery.

The phase voltage here will be from one line to star point. The line voltage between a pair of output leads will be phase volts x 1.73 ( root 3).

When rectified in a 3 phase bridge the mean dc will be close to 1.4 times the line voltage.  Subtracting diode drops is not mathematically very realistic but the intrinsic drop of a silicon diode is near 0.6v depends on temperature and diode construction. Adding the forward slope resistance for a reasonable load current brings you up to something near 0.8 to 1v depending on the diode.

In your example I assumed that you were rectifying the single phase that you mentioned, with a single phase bridge. If you were actually using all 3 phases with a 3 phase bridge then your figure would be 3.5 x 1.4 then less any diode drops. From your 2.9v figure I suspect you were dealing with the single phase case.

Flux

[dualsport54]

Sorry for my ignorance .....If a single phase of a three phase altenator puts out 3.5 VAC at max rpm....
3.5VAC x 1.414(rms)
 = 4.9VDC,....4.9VDC - 1.4VDC (diode drop) =3.5VDC x1.73=6.14VDC???
Why is it when I test the alt that max rpm says 3.5VAC..........after going through the diode bridge the meter only reads 2.9VDC
I think I understand rms (an average of the sine wave?) .....the diode drop is obvious....where does this 1

Hi Art,

Some of the error you're seeing may be due to your generator not producing a "true sine" waveform. The 1.414 AC to DC conversion factor only applies if you're dealing with a pure sine waveform. Physical details in your generator's construction will change the shape of your output wave. If you could get access to an oscilloscope you'd be able to look at what you're producing. The more angular time (degrees of rotation) your waveform spends above you're minimum voltage (diode drops, battery minimum charge voltage, etc.) the more power you'll be able to deliver to your batteries.