Sciencemadness Discussion Board

Electric motor driven alternators for high amp 6V current..

evil_lurker - 21-7-2008 at 00:21

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?

len1 - 21-7-2008 at 00:30

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.

ShadowWarrior4444 - 21-7-2008 at 01:35

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.

Twospoons - 21-7-2008 at 14:42

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.

len1 - 21-7-2008 at 15:04

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]

Magpie - 21-7-2008 at 15:41

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?

len1 - 21-7-2008 at 15:56

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.

Magpie - 21-7-2008 at 17:29

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?

len1 - 21-7-2008 at 17:48

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

ShadowWarrior4444 - 21-7-2008 at 18:33

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]

Twospoons - 21-7-2008 at 18:36

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.

len1 - 21-7-2008 at 19:24

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]

Magpie - 21-7-2008 at 19:41

Quote:

Fully regulated for either constant voltage or constant current.


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?

Twospoons - 21-7-2008 at 20:07

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.



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.

len1 - 21-7-2008 at 22:17

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]

ShadowWarrior4444 - 21-7-2008 at 22:57

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.


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]

len1 - 22-7-2008 at 00:29

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

ShadowWarrior4444 - 22-7-2008 at 00:52

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.

len1 - 22-7-2008 at 11:12

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

Twospoons - 22-7-2008 at 14:24

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. .



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.

ShadowWarrior4444 - 22-7-2008 at 14:45

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]

len1 - 22-7-2008 at 14:54

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

Twospoons - 22-7-2008 at 15:09

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 :)

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]

ShadowWarrior4444 - 22-7-2008 at 15:30

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]

len1 - 22-7-2008 at 15:49

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

ShadowWarrior4444 - 22-7-2008 at 16:08

Quote:
Originally posted by len1
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


Since you obviously have not read the article, I will paraphrase it:

It clearly states that the stiffening capacitor supplies the instantaneous Supply Current that the amplifier requires and that the battery is too slow to supply. This goes to prove that a battery is not capable of instantaneously supplying high current, and therefore not capable of adequately smoothing a high-current signal.

You are trying to obfuscate a very simple concept of response time with irrelevant and inadequately delineated equations. The fact is that for a power supply to have a smooth voltage you need two components: a capacitor and a regulator. Neither of those are a lead acid battery, and any high current electrolysis supply, whether transformer driven or alternator driven, will utilize a capacitor and regulator over a battery.

Why you insist on recommending the more dangerous, more costly, and less effective solution is still unclear.

Ancillary: 3000 RPM is only 50Hz on a 2-pole alternator.

P.S. It is entirely possible to run the cell off a lead-acid battery, however I would only recommend doing this if you had a battery charging circuit. It would only be more economically efficient if you already had such devices, or if you wished to run the cell away from a mains source.

[Edited on 7-22-2008 by ShadowWarrior4444]

gregxy - 22-7-2008 at 16:43

The confusion here seems to be on current and frequency.

A battery can supply more charge (current over time) than
a capacitor, but the battery may have series inductance
that limits the rise or fall time for the current. It is therefore
useful to place a capacitor in parallel with the battery
to supply the current during the time that it takes for the
battery current to ramp up. Typically a 1uf cap should
be enough.

Large electrolytic caps have same problem of
series inductance, so often small
ceramic disk caps (which have very low inductance) are
place in parallel with large electrolytics.

Finally batteries are non linear, most can source much
larger currents than they can sink

tumadre - 22-7-2008 at 18:46

Car alternators are 6 pole, 2:1 off the engine so about 1200-1800 rpm at idle, where most cannot supply rated current, and full rated current is typically at >4000 RPM due to the exceptionally high losses.

BUT it is three phase, which inherently has 4% ripple in theory, more like twice that, but under 10%. Adding a battery as a filter for a three phase rectifier will only decrease power*, although it will drag the average voltage toward Vp.

Car alternators... the smoothing cap is relatively nonexistent except at >10Khz, as an RF and EMI filter, as others have stated it does a poor job at that!

If you want a smooth dc voltage on a single phase Tx good luck achieving >70% efficiency using transformers at or beyond their Va rating. It certainly could be done with a one farad cap but you will burn up at least 40% more copper loss in both the secondary and primary coils as opposed to no filter. I could be wrong, it might be 20 and 41.41% ..still..)

A much better choice is a large inductor, at least the current forms a square wave, at or near 120 degrees of conduction for both half waves, rather than a current spike decreasing in width and increasing in amplitude as the cap gets bigger.

Car audio caps are specifically for smoothing out sub 100Hz ripple, that 400 Amp peak drain is for those 3000Watt subs, not the 2000 hz midrange where inductance becomes an issue, keep in mind the first thing a car audio amp does is boost the 12 volts into 100 or more volts, it must have enough voltage to push 2000 watts into a 4 ohm resistor, and maintain linearity. A cheaper place to put those caps is on the 60-120 volt bus, depending on the wattage rating and model, but who wants to do that?

*Assuming power is limited by allowable temperature rise, not cell resistance. Remember, the voltage is changeable, if you include the restraint of a fixed voltage, then you definitely do not want an inductive input filter, as the output voltage will be (for single phase) 90% of Vrms, rather than the capacitive 1.2-1.3 Vrms, 1.41 being the peak

[Edited on 22-7-2008 by tumadre]

[Edited on 22-7-2008 by tumadre]

len1 - 22-7-2008 at 19:35

Quote:
Originally posted by tumadre

BUT it is three phase, which inherently has 4% ripple in theory, more like twice that, but under 10%. Adding a battery as a filter for a three phase rectifier will only decrease power*, although it will drag the average voltage toward Vp.

Car alternators... the smoothing cap is relatively nonexistent except at >10Khz, as an RF and EMI filter, as others have stated it does a poor job at that!


I think thats meant to be 14%

min( cos(x), cos(x+120) ) in the upper quadrant occurs
at x=30 degrees, cos(30)=0.866

it'll actually be worse due to diode switching.

A 3 stator winding alternator putting out rectified DC means to make any sense a capacitor must discharge considerably less than 14% over 1/6 rather than 1/2 period. The audio bypass capacitor is compeltely useless for that as you say.

I will also note that the cell resistance is highly non-linear. So you have to maxmimize useful I rather than the power. I absolutely agree that driving the generator into a battery (or any capacitive load) will only detriment P = V I, but it can in many cases increase cell efficiency. In a simple form an electrolysis cell can be represented as a voltage source in series with a (rather low) resistance. This means V drops very rapidly with voltage - especially so if the resitive power loss is small and cell efficiency high. On the other hand they have an optimum electrode current density which should not be exceeded.

Its an interesting observation that the car battery will supply much more current than itll sink. I wonder the chemical reason for this.


[Edited on 23-7-2008 by len1]

ShadowWarrior4444 - 22-7-2008 at 20:55

Quote:
Originally posted by len1

Its an interesting observation that the car battery will supply much more current than itll sink. I wonder the chemical reason for this.

[Edited on 23-7-2008 by len1]


This is most likely due to the difference in thermodynamic stability between PbO2 and PbSO4. The electrolytic conversion of PbSO4 to PbO2 may also have a lower quantum efficiency than the reverse reaction, as evidenced by the need to use 15v to effect a ~99% conversion.

evil_lurker - 22-7-2008 at 21:21

Well to add to this thread, I'm only after inorganic stuff, like manganese ammonium alum, hydrogen, (per)chlorates, and maybe even one day some zinc plating. Precision is not my goal, only something that half ass works and is above all simple.

Till I can soak it all up...


[Edited on 22-7-2008 by evil_lurker]

ShadowWarrior4444 - 22-7-2008 at 21:24

Schematics that may be of interest:

http://www.osti.gov/bridge/servlets/purl/10178414-Y8y6y8/101...

http://www.elecfree.com/electronic/lm7232n3055-power-supply-...

Simplest:
http://www.aaroncake.net/CIRCUITS/supply2.asp

They range from complex to simple high current supplies, usually around 12v, however they can most likely be modified for 6v.

Paralleling two computer power supplies may be useful for high currents as well. Modification/cannibalization may also be useful:
http://www.wikihow.com/Convert-a-Computer-ATX-Power-Supply-t...

(This is likely the most simple method for obtaining an electrolytic power supply [other than buying a lab one.])

[Edited on 7-23-2008 by ShadowWarrior4444]

gregxy - 23-7-2008 at 07:46

For my chlorate cell, I used a variac, a transformer from a
dead UPS and a large fullwave bridge to give a variable
voltage from 0 to 12V at about 10 amps. No filter at all.
It worked great for the application.

dann2 - 23-7-2008 at 07:55

Hello evil_lurker (that name always makes me laugh),

A good (relatively crude) power supply can be made from rewound microwave oven transformers. Put lots of taps on the rewound secondary to give yourself different voltage tap off points. You can even use bare Copper tubing for the secondary if it is self supporting and you insulate it from the core with some pieces of plastic.
A variable voltage transformer on the input would be even better. Sometimes you can get lucky at your local industrial brakers yard and pick up variacs for the price of the scrap Copper in them.
If you use a center tap on the output can use a two diode rectifier (instead of four).
Some pics here:
http://www.geocities.com/CapeCanaveral/Campus/5361/chlorate/...

You may need no smoothing depending on what you are making/doing.

Regarding exact voltages for making a particular substance in an electrolythic cell.
The quoted voltages for to obtain a substance is always measured using a third electrode in the cell (saturated calomel electrode, aka SCE, Hydrogen electrode etc etc). You won't be using any of these. The cell voltage (measured accross cell) is not a suitable measurement for an exacting reading if you are attempting to make something that requires a very accurate voltage to form. Most stuff we make does not require a very exact voltage to form so the 'humpy' unsmoothed voltage will be OK.
If you are making some stuff that has a narrow fixed formation voltage go buy a Saturated Calomel Electrode + a well regulated (smooth voltage and current out) power supply and go knock yourself out!!! (and done forget to tell us how you got on!!!)

$$$$$$$$$$
No doubt is is nice to have a very large output, well regulated, constant voltage/constant current (whichever suits the particular job in hand) DC supply with Current and voltage readouts and an inbuilt Coulumb meter.
$$$$$$$$$$$$

Going off topic:
A question I have always been unable to answer it this:
A variac has a brush which contacts two wires at a time (one winding) as you turn the dial. Why is it that you do not get a hugh (one winding shorted, low voltage) current flowing in this winding?

Dann2

[Edited on 23-7-2008 by dann2]

tentacles - 23-7-2008 at 08:56

I've been curious if it's possible to modify one of those inverter type DC welders to supply variable voltage (they are already variable current). Or, if they will supply voltages low enough for electrolysis (chlorate, perchlorate) as is. My gut feeling is that they won't.

I'm talking about the cheap 80A units (50A @ 100% duty). Maybe all they need is a sturdy voltage reg. circuit on the output to prevent going over voltage when the anode goes titsup?

They're like $89 on sale at harbor fright - http://www.harborfreight.com/cpi/ctaf/Displayitem.taf?itemnu...

[Edited on 23-7-2008 by tentacles]

dann2 - 23-7-2008 at 11:20

or you could put a more crude (easier to do) affair which would simple sense output voltage and switch the whole unit off with a relay on the input socket when the output voltage rose to (say) 7 volts.
Easy do do with a comparator or some such.
They are definitely a cheap high current power supply, and you will be getting a welder thrown in for free!!
Dann2

tumadre - 23-7-2008 at 12:58

Quote:
Originally posted by tentacles
I've been curious if it's possible to modify one of those inverter type DC welders to supply variable voltage (they are already variable current). Or, if they will supply voltages low enough for electrolysis (chlorate, perchlorate) as is. My gut feeling is that they won't.

I'm talking about the cheap 80A units (50A @ 100% duty). Maybe all they need is a sturdy voltage reg. circuit on the output to prevent going over voltage when the anode goes titsup?

They're like $89 on sale at harbor fright - http://www.harborfreight.com/cpi/ctaf/Displayitem.taf?itemnu...

[Edited on 23-7-2008 by tentacles]


i would be interested to know what is in those things. if its an IGBT and a ferrite core, yes, you could rewind it. I don't think it would be worth the trouble, and you might blow it up the first few times unless you redesign the entire thing.

@len1 you are right, its 14%, i was thinking of the 12 pulse (cos 15)

Diode overlap is a much bigger issue for the 'star' arrangements, because the leakage inductance is drained through the conduction of the other diode, rather than coupled through the other two coils to simply be a lagging power factor.

Below say 6-10 volts, the single diode drop can cut losses enough to justify the increased Tx loss. and for single phase, it certainly cuts diode cost in half;)

tentacles - 23-7-2008 at 19:36

tumadre: There's a review of a very similar (but 125A) unit here: http://phelum.net/info/Mini142/

I suspect they are all very similar, IGBTs and ferrite cores.

len1 - 24-7-2008 at 23:17

Quote:
Going off topic:
A question I have always been unable to answer it this:
A variac has a brush which contacts two wires at a time (one winding) as you turn the dial. Why is it that you do not get a hugh (one winding shorted, low voltage) current flowing in this winding?


I think thats an interesting problem - I did some calc on this sometime - this might be of interest.

We have an autotransformer with turns N, voltage V, coil resistance R, Power P, so current at full power I = P/V

If you short out one turn the power developed is

(V/N)^2/(R/N) = 1/N V^2/R

If transformer efficiency is E, then assuming most of efficiency loss occurs as resistive loss in the core (rather than in the hysteresis curve of the iron) we get

I^2R= =(P/V)^2 R = P (1-E)

so R = (1-E) V^2 / P

and the power loss shorted turn is

P(1 turn) = P/N (1-E)

Transformer iron for small-medium power transformers is wound at about 10 turns per volt, so a 230V primary will have as a representnative figure 2300 turns (less on larger transformers - as the maximum flux is about proportional to Xsecional area). Efficiency at full power is about 80%, take power = 500W

P(1 turn) = 500 / (2300 0.2) about 1W

This a 500W transformer will dissipate without notice.

It is further reduced by the unpredicatble value of contact resistance.



[Edited on 25-7-2008 by len1]

12AX7 - 25-7-2008 at 01:52

What, you calculated R as estimated iron resistivity? It's obviously the wire resistance, which depends on guage and length per turn.

Without knowing DCR, a better estimate would be assuming half resistive losses in the wire. A 90% efficient transformer (you're hard pressed to find one less efficient than 90%, especially in a large size) will have 5% loss in the iron (hysteresis and eddy currents) and 5% in the wire. Therefore, a 500VA (not watt), 230V autoformer might dissipate 50W under full load, 25W of which is resistive. Although current varies along the winding (depending on where the tap is at), if that's assumed to be full length, then the full load of 2.17A is generating a voltage drop of 11.5V across the resistance (11.5 * 2.17 = 25W). Therefore, the total DC resistance is 5.3 ohms.

Alternately, one can assume 5% voltage sag under full load for each winding of the transformer (this obviously works out differently for an autoformer), which is also around 11.5V. The similarity of these two figures shouldn't come as a surprise, as they come from the same estimates, merely expressed differently.

That said, an estimation of turns is needed. For a 500VA model, I would realistically guess closer to 2 turns per volt. This means 460T full winding, so one turn has about 1/2V on it, and a resistance of 12 miliohms, for a current of 43A and 22W dissipation, which would quickly burn it up.

The brush is usually made of graphite, which has appreciable resistivity. A mere fraction of an ohm will dissipate this voltage effectively while having no effect on output voltage, because output current is much lower. The caveat is, all that power is dissipated by the brush, which being smaller than the wires, could get very hot indeed.

Tim

ShadowWarrior4444 - 25-7-2008 at 13:40

The question was why there isn't a large current in a single winding when the brush contacts two windings?

Well, the current through any primary winding is dependant on the load and the saturation of the core. Bridging two windings will simply reduce the current through the first one in sequence and increase the current in the second one.

The brush will not heat up appreciably unless large amounts of current are being drawn through the variac in the first place.

12AX7 - 25-7-2008 at 20:49

No, primary current is the sum of reactive current, idle losses and load current. Saturation has nothing to do with it, because power transformers are not usually designed to operate in saturation (examples that do include MOTs, ferroresonant transformers and magnetic amplifiers).

Each turn in the variac corresponds to a tap on the winding, therefore each tap corresponds to a reasonably low impedance voltage source. The impedance is especially low between two adjecent turns. The series (primary and load) current has very little effect and the shorted turn can be modeled in isolation as a seperate winding with little error, at least for low contact resistances. For high contact resistances, load and short-circuit current have to be added to calculate the correct current and dissipation in the brush.

My point, to answer the question, was that the brush must have high contact resistance specifically in order to avoid this dangerously high short-circuit current.

Tim

len1 - 25-7-2008 at 22:23

What youve done is use different numbers applicable to more powerful transformers

E = 0.95 instead of 0.8, N=460 instead of 2300. Plugging this into the formula I gave

P(1 turn) = P/ N (1-E)

theres a factor of 5 increase due to N, and factor of 4 due to 1-E. So you get 22W instead of my 1W. But these are all just approximate numbers.

Having said that I have seen autotransformers with a metal brush and far more turns than 460 on the primary hence my result. I dont have an auto with a carbon brush.

[Edited on 26-7-2008 by len1]

len1 - 25-7-2008 at 23:10

Its interesting to calculate for the carbon brush

The bulk resistivity rho of graphite shows quite some variability but on average about 1000 that of copper. The resistance of a piece of size w x h x l is

rho x l / (w x h)

Lets assume the carbon brush makes a w length contact along the winding, with the wire being l x l in cross section, and neglect the width of copper insulation. The wire turn length is l2

The carbon resistance is approx (assuming contact along entire width of wire!!)

rho x l / (l xw) = rho/w (current in poissons eqn penetartes ~ length of contact)

the copper turn resistance is

0.001 rho x l2 / (l x l)

so the ratio of carbon brush resistance to the copper turn is

1000 l^2/ (w x l2).

If l=1mm, w = 5mm, l2=200mm the brush resistance is about same as that of copper turn so the power dissipation is decreased only by half, worse, the brush heats up much more than the wire.

So this problem is not so simple, the contact of the graphite must be far smaller than turn diameter - which it would be anyway with wire being circular cross section. The contact is probably along only 1/4 the diameter, decreasing dissipation by factor of ~ 5.

ShadowWarrior4444 - 26-7-2008 at 00:14

Quote:
Originally posted by 12AX7
No, primary current is the sum of reactive current, idle losses and load current. Saturation has nothing to do with it, because power transformers are not usually designed to operate in saturation (examples that do include MOTs, ferroresonant transformers and magnetic amplifiers).

Each turn in the variac corresponds to a tap on the winding, therefore each tap corresponds to a reasonably low impedance voltage source. The impedance is especially low between two adjecent turns. The series (primary and load) current has very little effect and the shorted turn can be modeled in isolation as a seperate winding with little error, at least for low contact resistances. For high contact resistances, load and short-circuit current have to be added to calculate the correct current and dissipation in the brush.

My point, to answer the question, was that the brush must have high contact resistance specifically in order to avoid this dangerously high short-circuit current.

Tim


*Raises an eyebrow* I don't actually see a short-circuit in the problem. Nothing seems to be shorted to ground.

I believe the reason that there is no high-current short lies in the inductive nature of the autotranformer. It is simply a large tapped inductor after all, and inductances resist the flow of AC current.

Empirically, I have not noticed any appreciable heating of the brush when bridging two windings. (K-type thermocouple)

The solution should lie in the magnetic properties of the variac, perhaps saturation wasn’t the correct term--if I can only manage to remember what the applicable theory is.

*One must note that current does not flow through the copper winding without a load when it is connected to an AC source, even though it is one continuous conductor.

12AX7 - 26-7-2008 at 00:43

Quote:
Originally posted by ShadowWarrior4444
*Raises an eyebrow* I don't actually see a short-circuit in the problem. Nothing seems to be shorted to ground.


A short doesn't have to be to ground. If you must, move the "ground" to one of the wires and then look at it. Please read up more on electronics before taking an authoritative stand on the subject!
Quote:
I believe the reason that there is no high-current short lies in the inductive nature of the autotranformer. It is simply a large tapped inductor after all, and inductances resist the flow of AC current.


If that were the case, then the load would see the exact same effect, and it would be a very shitty "transformer", in fact not being a "transformer" but an inductor! Obviously this is not the case. Voltage is fixed by turns ratio.
Quote:

The solution should lie in the magnetic properties of the variac, perhaps saturation wasn’t the correct term--if I can only manage to remember what the applicable theory is.


The primary winding has self-inductance which causes a small idle current (inductive phase), nothing else. It does not flow in the output or taps.
Quote:

*One must note that current does not flow through the copper winding without a load when it is connected to an AC source, even though it is one continuous conductor.


Except it does, due to self inductance. Again, please study electronics some more. A lot more. Until then, I recommend you stick to chemistry.

Tim

len1 - 26-7-2008 at 01:55

Tim I wouldnt bother arguing, this guy is either mad, or hes taking the piss

ShadowWarrior4444 - 26-7-2008 at 02:30

Quote:
Originally posted by 12AX7

A short doesn't have to be to ground. If you must, move the "ground" to one of the wires and then look at it. Please read up more on electronics before taking an authoritative stand on the subject!


Of course it doesn't have to be to ground, it has to be an unintended low resistance connection causing excessive electrical current. Again, that is obviously not the case here.
Quote:
If that were the case, then the load would see the exact same effect, and it would be a very shitty "transformer", in fact not being a "transformer" but an inductor! Obviously this is not the case. Voltage is fixed by turns ratio.


A transformer *is* an inductor. Do you know just how the voltage is fixed by turn ratio? Might it have something to do with a magnetic flux induced in the core?
Quote:

Except it does, due to self inductance. Again, please study electronics some more. A lot more. Until then, I recommend you stick to chemistry.

Tim


I was obviously not talking about the exciting current of the transformer. Nor was I referring to any self-inductance from the copper wires. (*mumbles something about incorrect terms*)

I was attempting to coax readers into thinking about the properties and underlying EM theory that go into a transformer. However, since there seems to be a bit of pig-headedness going around, I will give you the answer:

An autotransformer does not 'transform' 100% of the power delivered to the secondary. For example, a center tapped auto-transformer will transfer 50% power by conduction to the secondary, and the other 50% will be 'transformed' (extracted from the magnetic field.) If you have two windings bridged by the brush, all you are in essence doing is reducing resistance to the conducted component by removing the bridged coil from the circuit. Another thing this does is reduce the flux density in the core by a corresponding amount, since power is conserved.

So then, to recap: by bridging two windings, you increase conduction, and decrease transformation, thereby increasing the voltage across the load. (Exept for bridging the final two windings, of course.)

There may be a more simple, concise way to explain it, but honestly you should be able to see the solution now.

[Edited on 7-26-2008 by ShadowWarrior4444]

12AX7 - 26-7-2008 at 02:37

You seem to forget that the magnetic field is constant regardless of load current. Primary amps cancel with secondary amps in exact proportion, leaving magnetizing current. Which is why saturation doesn't enter in.

You've got a point Len, I'll take the high road now.

TIm

ShadowWarrior4444 - 26-7-2008 at 03:08

Quote:
Originally posted by 12AX7
You seem to forget that the magnetic field is constant regardless of load current. Primary amps cancel with secondary amps in exact proportion, leaving magnetizing current. Which is why saturation doesn't enter in.

You've got a point Len, I'll take the high road now.

TIm


The magnetic field is not constant. The 'flux' is constant only when the tap point is constant. Think about an autotransformer whose brush is sitting on the same tap as the input, forming a 1:1 connection. Most of the current is then transferred to the secondary by conduction, and the flux through the core is very low.

I in no way mentioned load current. I am talking about tap ("brush") location.

The phrase "power is conserved" means exactly what you just restated: "Primary amps cancel with secondary amps in exact proportion."

It is becoming clear that either of you have not worked on electrical circuitry professionally. Frequent erudition and unnecessary use of calculation in a theory discussion are both signs. And even when I attempt to provoke a bit of thought by stating hints and terms deliberately vague, you violently rebut it with your own version of misused jargon.

It’s simply mystifying how I can be insulted by people who suggested feeding an unregulated alternator into a battery and believe that a transformer's quadranture current is actually called "self-inductance."

That said, I do not hold anything personal against you; a great many people get caught up in a stance, and lose the ability to see different perspectives. I simply provide unbiased logic and research, what readers do with it is entirely up to them.

[Edited on 7-26-2008 by ShadowWarrior4444]

12AX7 - 26-7-2008 at 06:49

Quote:
Originally posted by ShadowWarrior4444
It’s simply mystifying how I can be insulted by people who suggested feeding an unregulated alternator into a battery and believe that a transformer's quadranture current is actually called "self-inductance."


You seem to be confusing me with someone else. You should check out my website, perhaps it will clear up some of your confusion. In particular for this thread,
http://webpages.charter.net/dawill/tmoranwms/Electronics.htm...

Bye.

Tim

len1 - 28-7-2008 at 19:17

Quote:
the magnetic field is constant regardless of load current. Primary amps cancel with secondary amps in exact proportion, leaving magnetizing current


Thats an interesting observation. Interesting because it would tend to suggest that since the flux in the core is - to a first approximation - independent of the current, whether the transformer is being run at full power or at no load, the size of the core to this approximation has no bearing on the power of the transformer.

To rephrase how does the power-handling of the transformer manifest itself in the amount of iron that needs to be used if the iron carries almost the same flux throught its regime of operation.

The key word here is carries. To a first (torroidal) approximation the amount of iron must be sufficient to store the entire energy transformed from primary to secondary in a half-period

E = P / 2 f

where P is the power rating of the transformer. This explains why switchmode transformers (or flyback) are so much smaller than their 50Hz equivalents.

The other feature is the maximum sustainable field in the core Bm. If V is the volume of iron required then the power formula becomes

Hm x Bm V > P / f

and power is proportional to volume.

What then fails in the first analysis? It will break down, and the secondary flux will cease to be linked to the primary when the energetically most favourable path for it ceases to be linked with the secondary winding - which will occur roughly when the inequality fails.

[Edited on 29-7-2008 by len1]

12AX7 - 28-7-2008 at 22:51

Quote:
Originally posted by len1
To rephrase how does the power-handling of the transformer manifest itself in the amount of iron that needs to be used if the iron carries almost the same flux throught its regime of operation.

The key word here is carries. To a first (torroidal) approximation the amount of iron must be sufficient to store the entire energy transformed from primary to secondary in a half-period


No, it doesn't. A flyback transformer does (commonly used in small switching supplies), but power transformers do not operate this way. In fact, these are only reasonably efficient at low power levels and high frequencies (typically 50-200kHz), where price and physical size aren't too bad.

Power is transferred essentially instantaneously. No magnetic middle-man. It goes directly from primary to secondary. The core just hangs around to increase inductance.

The basic idea is, since you must have a core to build a reasonable 50/60Hz transformer, you have to figure out how much you can get before reaching its limits. Limits are defined by inductivity and saturation field strength. And inductivity determines how many turns you need to avoid saturation. All those turns take up space, a well-defined amount of space because the winding has to handle load current. So you need a bigger core to handle more VAs.

Tim

len1 - 28-7-2008 at 23:20

Quote:
Originally posted by 12AX7
Quote:
Originally posted by len1
To rephrase how does the power-handling of the transformer manifest itself in the amount of iron that needs to be used if the iron carries almost the same flux throught its regime of operation.

The key word here is carries. To a first (torroidal) approximation the amount of iron must be sufficient to store the entire energy transformed from primary to secondary in a half-period


No, it doesn't. A flyback transformer does (commonly used in small switching supplies), but power transformers do not operate this way. In fact, these are only reasonably efficient at low power levels and high frequencies (typically 50-200kHz), where price and physical size aren't too bad.

Power is transferred essentially instantaneously. No magnetic middle-man. It goes directly from primary to secondary. The core just hangs around to increase inductance.

The basic idea is, since you must have a core to build a reasonable 50/60Hz transformer, you have to figure out how much you can get before reaching its limits. Limits are defined by inductivity and saturation field strength. And inductivity determines how many turns you need to avoid saturation. All those turns take up space, a well-defined amount of space because the winding has to handle load current. So you need a bigger core to handle more VAs.

Tim


Tim I think this is something you wrote someone
Quote:
Please read up more on electronics before taking an authoritative stand on the subject!


It would do you well to listen to your own advice. If not sure - ask, you might be dealing with someone who knows more than you, and you should always take that eventuallity into account.

This
Quote:
Power is transferred essentially instantaneously. No magnetic middle-man. It goes directly from primary to secondary. The core just hangs around to increase inductance.


if you pardon my French is absolute rubbish. I have lectured electronics in unversities for many years and you would get a very low mark for this - in fact repeat the course.

ShadowWarrior4444 - 28-7-2008 at 23:56

Quote:
Originally posted by len1
Quote:
Power is transferred essentially instantaneously. No magnetic middle-man. It goes directly from primary to secondary. The core just hangs around to increase inductance.


if you pardon my French is absolute rubbish. I have lectured electronics in unversities for many years and you would get a very low mark for this - in fact repeat the course.


Len is correct in this case, though particularly insulting.

A flyback operates by storing energy in the magnetic field which then rapidly collapses to produce high voltage. The core has an air-gap to increase magnetic reluctance.

Ancillary: Explaining this in plain terms instead of posturing would make you a better professor.

[Edited on 7-29-2008 by ShadowWarrior4444]

12AX7 - 29-7-2008 at 06:31

Quote:
Originally posted by len1
Quote:
Please read up more on electronics before taking an authoritative stand on the subject!


It would do you well to listen to your own advice. If not sure - ask, you might be dealing with someone who knows more than you, and you should always take that eventuallity into account.


So if you have a problem, would you be so kind as to show specifically what is wrong? As an academic, I'm sure you must understand my preference for fact over heresay.

I had prepared a more in-depth explanation of the phenomenon, but I've decided to hold off until we resolve these issues.

Tim