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evil_lurker
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Electric motor driven alternators for high amp 6V current..
I've seen a whole lot of discussions in the past on attemps at convert PC powers supplies to laboratory power supplies.
Unfortunately for me all those circuit diagrams might as well be printed in traditional chinese.
I was wondering if anyone had considered taking an AC Delco 10SI or 12SI alternator, convert it over to 6V, and drive it off a 5hp continuous duty air
compressor motor... then use a 6V battery to smooth out the current.
I already have the motor and the alternator and battery should only cost about another hundred bucks.
I'm thinking it should produce a steady 6V @ 35amps or so.. large enough for just about any lab power supply.
What do ya'll think?
Not all chemicals are bad. Without chemicals such as hydrogen and oxygen, for example, there would be no way to make water, a vital ingredient in
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len1
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Lurker - theres a huge energy imbalance here - a 5hp motor, at 750W/hp = 3850W
and you want to use this axe to crack a 6V x 35A = 210W nut. 1/2hp is more appropriate.
But overall such wattages dont justify mechanical contraptions. Ive converted many PC supplies. If you post both sides of a PC power supply board of
yours I most likely can tell you exactly what to replace.
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ShadowWarrior4444
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I should note that a battery will not smooth out current; you will need to use a few capacitors to do that. A couple electrolytic caps directly after
the rectifier bridge should do it.
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Twospoons
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It would be far easier just to use a plain old mains transformer. If you can, find one of the old filament transformers for valve amplifiers
(typically 6.2V output). You could also use the filament winding from a microwave oven transformer.
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len1
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I think it would be nice if posters would at least express some doubt in what they are saying if they dont really know.
A lead-acid battery is far better than almost any capacitor you can find for smoothing the rectified current. Its a voltage source in series with a
very low resistance, you can expect a ripple of 1/2 volt or so due mostly to overvoltage.
Now lets check what kind of capacitor would achieve this
I = C dV/dt -> C = I/(dV/dt) = 35/(0.5/.01) = 0.7
Its hard to get a 0.7 Farad capacitor - and one that will handle 35A of current - a motorcycle battery is much cheaper.
Valve heater currents are of order 1A or less, so a plain old valve transformer generally supplies no more than 4-8A of current - a waste of time. A
microwave transformer is not wound to supply 35A on the heater - though it is capable of suppling an equivaent wattage through its HV supply
[Edited on 21-7-2008 by len1]
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Magpie
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Please forgive my naivete but what are these "DC lab power supplies" used for? Are these primarily for electrolysis? If so, what are the desireable
characteristics of a power supply for electrolysis? Constant voltage with variable current, or what?
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len1
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Magpie, Ive been using these types of supplies for electrolysis, for both Na and H2/O2 generation. The most desirable characteristics Ive found are
1) High current 30-80A or so. To make a mole of substance - assuming its a one electron oxid/red you need 10^5 coulombs. 1A for 1hr is 3600 C. So
if you want a mole an hour at 50% efificiency thats about 60A.
2) Precisely variable max voltage in the range 3-8V at the aboce current
3) Reliability
In regard of 3) switchmodes are not nearly as good as 50Hz transformer approach, thats why I used the latter in my Na cell.
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Magpie
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Thank you Len. Are you saying that you take power off your mains (220VAC, 50Hz?) to a variable transformer, taking it down to whatever voltage you
want for electrolysis, then convert this to DC via electronics (diodes, etc)?
Is this cheaper than buying a commercial unit, or just more fun/satisfying?
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len1
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I use the variable voltage 50A supply I posted for the Na project generally for electrolysis. Its picture and circuit is also given. It basically
works as you stated. If anyone wants to make it I will be happy to help. There is no smoothing required in this type of electrolysis so capacitors
are not needed. It is much cheaper than an equivalent commercial supply , though probably about the same cost as a 300W switchmode. Switchmodes are
OK for the unform current loads of computers, but in electrolysis due to possible shorts, current spikes etc, they are not so good. The transformer
and fan in my unit is from a welder, cost about $80, 4 35A bridges about $20, shunt and meter about another $20, and about a day to make. Len
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ShadowWarrior4444
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Quote: | Originally posted by len1
I think it would be nice if posters would at least express some doubt in what they are saying if they dont really know.
A lead-acid battery is far better than almost any capacitor you can find for smoothing the rectified current. Its a voltage source in series with a
very low resistance, you can expect a ripple of 1/2 volt or so due mostly to overvoltage.
Now lets check what kind of capacitor would achieve this
I = C dV/dt -> C = I/(dV/dt) = 35/(0.5/.01) = 0.7
Its hard to get a 0.7 Farad capacitor - and one that will handle 35A of current - a motorcycle battery is much cheaper.
Valve heater currents are of order 1A or less, so a plain old valve transformer generally supplies no more than 4-8A of current - a waste of time. A
microwave transformer is not wound to supply 35A on the heater - though it is capable of suppling an equivaent wattage through its HV supply
[Edited on 21-7-2008 by len1] |
This is flatly not true; there is a reason that absolutely no manufacturer will use a rechargeable battery to smooth current--especially at 6V. (Not
the least of which would be the heat and gas concerns.) [Also, should the voltage of the battery not remain precisely the same as the input
voltage...]
"The capacitors act as a local reserve for the DC power source, and bypass AC currents from the power supply. This is used in car audio
applications, when a stiffening capacitor compensates for the inductance and resistance of the leads to the lead-acid car battery."
If he wishes to use a mains transformer, it is very possible to wind his own to supply anywhere for 35A to 1000A. 120-200 windings primary, 6
secondary using high gauge copper available from any home supply store would work nicely. He could even make it a variable core transformer--lowering
the core in and out of the windings to vary to current supplied. (As in an industrial arc welder.)
I might also point out that your calculations are wrong:
Smoothing capacitor for 10% ripple, C = (5 × Io)/(Vs × f)
C = smoothing capacitance in farads (F)
Io = output current from the supply in amps (A)
Vs = supply voltage in volts (V)
f = frequency of the AC supply in hertz (Hz)
(5 x 35)/(6 x 60)=(175)/(360)= .4861F = 486,100uF
If greater than 10% ripple (0.6V) is tolerable, the capacitor can be quite a bit lower.
[Edited on 7-21-2008 by ShadowWarrior4444]
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Twospoons
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The filament tfr and MOT suggestions were just for cheap quick solutions. If you want to get serious you could buy a 500VA toroid, and wind your own
secondary with fat wire. len1's welder tranny is also an excellent idea.
I tend to have somewhat more rigourous requirements for my DC supply, which is why I was so pleased to score a used HP lab supply for NZ$300.
0-60V,0-50A, max 2.5kW output. Fully regulated for either constant voltage or constant current. Its a switcher/linear hybrid, and I can only just
lift the bugger.
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len1
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But that is precisely how the battery is used in automobiles! To guard against overcharge you need a voltage governer circuit - as most alternators
have.
As for my formula being wrong the one you sued is exactly the same numerical differences being due to your 0.6V ripple, 60Hz freq, and approx. I like
the result quoted to 4 sig figs. - funny man.
[Edited on 22-7-2008 by len1]
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Magpie
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Quote: |
Fully regulated for either constant voltage or constant current.
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For electrolysis, production is directly proportional to current. So for this application it seems that it would be desireable to have a regulated
current rather than a regulated voltage. Is this true?
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Twospoons
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Quote: | Originally posted by len1
Its hard to get a 0.7 Farad capacitor - and one that will handle 35A of current - a motorcycle battery is much cheaper.
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68,000uf 10V electrolytic from Digikey = $5.50ea for 10 or more. Rated at 5 amp ripple current . 10 in parallel of those would be more than
adequate. So the capacitor solution is dead easy, and has the advantage of operating at any voltage up to the rated voltage - not stuck at 6V.
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len1
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Not a good way to go. An alternator delivering 50A at 6V nominal will normally spike at least 30% higher 8 - 9V no load, and youre close to blowing
the capacitors - these actually need overspecification. The batterys voltage increases much less with cycle charge than a capacitors, meaning current
can be sourced over alarger range of duty cycle, rather than be sourced in a narrow peak - for high current constant voltage supplies much better to
use battery rather than capacitive storage/ You car is everyday living proof of this.
Magpie, constant current is the way to go in many applications. In my electrolysis cell I found a small increase in current, leads to heating of the
solution, reduction in its resistance, and hence a further increase in current. This positive feedback can blow the supply if its a constant voltage
type.
[Edited on 22-7-2008 by len1]
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ShadowWarrior4444
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Quote: | Originally posted by len1
But that is precisely how the battery is used in automobiles! To guard against overcharge you need a voltage governer circuit - as most alternators
have. |
Indeed, there is a rather complex device used to prevent overcharging of the battery; and, as I have previously mentioned, he would *still* require a
capacitor to smooth the current coming from the battery.
Quote: | I like the result quoted to 4 sig figs. - funny man.
[Edited on 22-7-2008 by len1] |
Significant figures don't exist for theoretical computations. Restrictions regarding sig figures only apply when the calculations are based on actual
measurements.
Quote: | Magpie, constant current is the way to go in many applications. Consider electrolysis. A small increase in current, leads to heating of the solution,
reduction in its resistance, and hence a further increase in current. This positive feedback can blow the supply if its a constant voltage type.
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Blasphemy! Whether you need constant voltage or constant current is entirely dependant on what you are electrolyzing. If you are for instance making
acetaldehyde electrolytically, you need a very rigorously stable voltage at 1.43V--lower and nothing happens, higher (1.6v) and you get acetic acid.
Quote: | Not a good way to go. An alternator delivering 50A at 6V nominal will normally go at least 30% higher 8 - 9V no load, and youre close to blowing the
capacitors - these actually need overspecifrication. The batterys voltage increases much less with cycle charge than a capacitors, meaning current can
be sourced over alarger range of duty cycle, rather than be sourced in a narrow peak - for high current constant voltage supplies much better to use
battery rather than capacitive storage/ You car is everyday living proof of this. |
Cars, as mentioned previously, have capacitive reservoirs coming off of the battery because the battery doesn’t smooth the current well enough. You
seem to forget that a battery does not react instantaneously to changes in voltage, whereas a capacitor does. Also, were he to use a battery, he would
need additional control circuitry to prevent overcharging. The best way by far is quite obviously to skip the battery and use capacitors and a voltage
regulator. The regulator prevents any overvoltage on the capacitors.
In addition, if he is using a 3-phase alternator like most automotives have, it renders the battery even more pointless. A 3-phase system rectified
will put out pulsed power at 120Hz or more, thereby reducing the required size of the capacitor.
Ancillary: A 6V lead acid battery, in addition to its sluggish response time (compared to a cap) [see energy density vs power density] cannot even
provide 35A instantaneous current at the low points in the cycle. A car electrical system is only 25A at 13.8V. A motorcycle's is vastly less. While
35A 12V batteries are sold, they are around $77 US, and utterly pointless.
[Edited on 7-22-2008 by ShadowWarrior4444]
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len1
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Why do you keep writing utter rubbish?
Quote: | Cars, as mentioned previously, have capacitive reservoirs coming off of the battery because the battery doesn’t smooth the current well enough. You
seem to forget that a battery does not react instantaneously to changes in voltage, whereas a capacitor does. Also, were he to use a battery, he would
need additional control circuitry to prevent overcharging. |
I suggest people put their car lights on full - so the alternator kicks in and youre drawing current (youll see a voltage jump when the engine is on)
and a scope on the battery terminals - if what the chap has written above is true - youll see the chopped half of a sinusoid at large voltages, if not
you wont. I remember having to switch to AC and upping the gain before the otherwise flat waveform showed variation of the order of tens of
millivolts - the measurements must be made direct at the battery terminals.
Common cars with capacitor banks and 3 phase alternators ..
an optimum electrolysis voltage is heavily dependent on concentration and temperature as anyone who studied Nerst eqn will know.. I wont bother
anymore with this chap, Ive no time to waste
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ShadowWarrior4444
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Quote: | Originally posted by len1
Why do you keep writing utter rubbish?
Common cars with capacitor banks and 3 phase alternators ..
an optimum electrolysis voltage is heavily dependent on concentration and temperature as anyone who studied Nerst eqn will know.. I wont bother
anymore with this chap, Ive no time to waste |
http://www.allaboutcircuits.com/vol_6/chpt_4/8.html
http://stason.org/TULARC/entertainment/audio/car/2-9-What-is...
http://www.sciencemadness.org/library/books/electrochemistry...
Do not resort to insults and anecdote when you are incorrect.
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len1
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What that article is referring to is that for audio applications one should bypass the battery with a small cap at audio frequencies, a common
practice because the battery inductance is significant at HF audio. A Neanderthal translation of that is that a car battery can not smooth the
voltage from an alternator - which is utter rubbish as the simple experiment I have described above shows
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Twospoons
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Quote: | Originally posted by len1
I suggest people put their car lights on full - so the alternator kicks in and youre drawing current (youll see a voltage jump when the engine is on)
and a scope on the battery terminals - if what the chap has written above is true - youll see the chopped half of a sinusoid at large voltages, if not
you wont. .
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I have done this. I saw 13VDC with about 1V of noisy half-sine ripple on top. Obviously the results will vary depending on the condition of the
battery. I was looking because I wanted to run a 400kHz signalling carrier over the car wiring, which turned out to be impossible due to the amount of
crap coming from the alternator.
And if you are not happy with 10V caps (chosen for the mains transformer solution), then buy 16V ones at $7 each! My point was that achieving 0.7F
would be quite easy.
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ShadowWarrior4444
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Quote: | Originally posted by len1
What that article is referring to is that for audio applications one should bypass the battery with a small cap at audio frequencies, a common
practice because the battery inductance is significant at HF audio. A Neanderthal translation of that is that a car battery can not smooth the
voltage from an alternator - which is utter rubbish as the simple experiment I have described above shows |
http://stason.org/TULARC/entertainment/audio/car/2-9-What-is...
"refers to a large capacitor (several thousand microfarads or greater) placed in parallel with an amplifier"
"The electrical theory is that when the amplifier attempts to draw a large amount of current, not only will the battery be relatively slow to
respond..."
"a capacitor in parallel with a load acts as a low pass filter (see Section 3.10), and the voltage level dropping at the amplifier will appear as an
AC waveform superimposed upon a DC "wave". The capacitor, then, will try to filter out this AC wave, leaving the pure DC which the amplifier
requires."
I do not know why you continually insist on using an electrochemical battery for smoothing a rectified 3-phase signal. It is completely ridiculous.
Not only will the battery overcharge (car alternators switch off intermittently to prevent this), it will still be unable to provide 35A
instantaneously at the low points in the cycle. Running at 120Hz, the minimum capacitance he would need is .243F.
[Edited on 7-22-2008 by ShadowWarrior4444]
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len1
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And if you do it with a dead battery you can get almost the full half-sinewave. My battery is 2 years old and gives tens of millivolts variation.
Even with a near-dead battery an alternator should put nothing out that would be an obstacle to a 400kHz signal - a simple CR filter will eliminate LF
noise
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Twospoons
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The LF noise was not the problem. It was the harmonics and diode switching noise extending way up the spectrum that were causing trouble. Given that
this was meant to become a commercial product, and so had to work with any car/battery, I shifted to 303MHz ![:)](./images/smilies/smile.gif)
Shadowwarrior, have you looked at the CCA rating of a car battery? Even a small one can deliver 300A.
[Edited on 23-7-2008 by Twospoons]
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ShadowWarrior4444
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Quote: | Originally posted by Twospoons
Shadowwarrior, have you looked at the CCA rating of a car battery? Even a small one can deliver 300A.
[Edited on 23-7-2008 by Twospoons] |
The above was mistyped and has been edited.
My comment referred to the, as mentioned constantly, power density and response time differences between a battery and a capacitor.
http://en.wikipedia.org/wiki/Image:Supercapacitors_chart.svg
Ancillary:
A useful page on basic power supply construction:
http://www.kpsec.freeuk.com/powersup.htm
[Edited on 7-22-2008 by ShadowWarrior4444]
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len1
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I know arguing with f..ls is a waste of time - but just this once more ..
The guy above is continuing to insist that the article quoting a 'large' capacitor of several thousand microfarads across the audio amplifier is not
meant for bypassing the supply at HF audio as I said, but is actually supporting his bizarre notion that the battery cant smooth the alternator
voltage and a capacitor must do the job
Lets calculate the voltage drop across this capacitor during a single cycle engine revolution with the lights on - drawing 10A at 3000rpm = 50Hz, and
see how well it does - assume its 4700uF = .0047F.
dv = I dt/C = 10 * .005/.005 = 10V ripple at alternator frequency.
Of course no audio amp will operate well with the voltage varying 2-12V, and its the battery whcih is charged with the task of smoothing alternator
voltage, the cap is too small for this, its effect is to bypass the supply at HF audio
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