smoothing dc transformer ?

Discussion in 'General Electronics Chat' started by Mathematics!, Nov 21, 2009.

  1. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
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    I have a power transformer 273-1385B
    input 120 vac 60 Hz --> output 12.6 vac 60 hz
    max output current I believe can be 300mA
    typical is 80mA

    Primary is plugged into the house 120 vac 15 amp wall circuit .

    When I measure the secondary side with a digital multimeter I get approx 16 volts ac +- .5 volts or so. I have my multimeter set correctly to ac. So my question is I would think the value should be closer to 12.6 as the transformers ratings say on the side of the transformer. ( I am getting more voltage! Is this a typical tolerance kind of like a resistors tolorance or something???)

    Now when I put a full bridge rectifier on the secondary coils of the transformer rated at 4amp 50volts (which is the only rectifer I had but definitely well over the ratings of the transformer so I am ok to use it)
    I am just wondering when I measure the dc I get it dropping a approx 1 volt from the rms 16 volts ac before the rectifer?
    So it seems like a retifier has some resistance to it? Superized that it has so much resistance but this could be because the ac was measured in RMS where as the dc was measure normally?

    Either way my multimeter reads around the same value for the dc of the rectified ac? I would think my multimeter would flucuate between 0 and 15 volts ??? So does this full wave bridge rectifier also act as a regulator ???

    Could it be that this rectifer smooths dc as well as converts it from ac to dc? The part is 276-1146 radioshack.

    If it's still pulsating dc then why am I not seeing the dc changing on my multimeter?

    Read below if it is pulsating dc

    Ok , if it is pulsating dc I have a 12 volt voltage regulator but it says on the back it's
    input voltage 35V
    Operating temp 0 C - 70 C
    Max junction temp 150 C

    It is a voltage regulator 7812 from radioshack part 276-1771
    Can I stick this regulator in front of the rectifier or do I need a 35 V operation source <--(or will the 15 volt be ok to run it on )

    If that is not going to work then I have some high rated , high capacitance capacitors wondering how high I would need the capacitance to be to make the dc smooth ?
     
    Last edited: Nov 21, 2009
  2. SgtWookie

    Expert

    Jul 17, 2007
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    You are measuring the output of the transformer under no-load conditions.

    Try connecting a 158 Ohm 2 Watt resistor across the output as a load, and measure the voltage with the load in place.

    Then remove the resistor before you connect it back up to the bridge rectifier.
    Rectifiers will have a voltage drop across them, the actual voltage drop depends upon the individual rectifier and the current flowing through it.

    If you do not have a capacitor across the + and - terminals of the rectifier, you are reading rippled DC instead of DC. If you connect a capacitor across the + and - terminals, you will read about 1.4 times the voltage that you are reading now.

    No, it does not.

    No, it does not.

    You will need a capacitor on the output of the bridge rectifier to obtain filtered DC from rippled DC. 1,000uF would be a good starting place. Use a capacitor rated for at least twice the expected voltage.

    In order to get regulated DC, you will need a regulator circuit.
     
  3. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
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    Well I would but the rectifier is already solder into to the secondary transformer legs.
    If I had done this and measured the voltage a cross the resistor would I get RMS = approx 12.6 not the 16 that I am seeing with my multimeter before hand???

    Well then why is my digital multimeter not moving from 0 to 15 ,...etc it stays at one fix value ? If this where pulsating dc won't I see changes in voltage on my multimeter when I measure the voltage between the + and - legs of the rectifier? <-- Would the reason be because I am not measuring under a load? (or need to measure rippled dc with the multimeter set to ac ? )

    For the 1.4 times the voltage correct me if I am wrong but this would be the peak of the pulsating dc? So the capacitor would smooth dc out to it's pk dc voltage. If this is true you said 1000uF cap or greater would do provided the voltage rating is twice. So I have a 2200uf , rated at 50DCV I am going to use that.
    Anyway How do you compute what caps will work ? Like how did you compute 1000uf would do it?

    Thanks
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Your transformer is rated for 12.6VAC output when the input is 120VAC and the load current on the output is a certain value. If there is no load current, the output voltage will be higher. If the load current is higher than the rated current, the output voltage will be lower.

    It's rippled DC at 120Hz. The changes in voltage are much too rapid for a typical DMM to report.

    That is correct. The 1.4 is just shorthand. It's really closer to 1.414213.
    That's fine.
    It's a function of what's an acceptable amount of ripple voltage, the load current, and the frequency of the ripple. 1,000uF was a somewhat arbitrary value, since you didn't specify a load current or desired maximum ripple voltage.

    This page explains it in detail: http://www.tpub.com/neets/book6/22e.htm
     
  5. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
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    As for the DMM not reporting the voltage changes fast enough. Could you get around that by using the DDM set to ac? Either way if that is not possible is their any cheap device that can give indication that it is pulsating dc and what it's pk and min voltage are? I know an ossilscope but those are much to expensive for me right now.

    Is it just using fc = 1/2*pi * R*C where R is the load resistance and fc is the cut off frequency. If so I have my load resistance and I can compute the capacitor provided I have the cut off frequency. But what does the cut off frequency have to be to make this dc smooth? ( am using the formula for a high pass filter)

    I would think dc can never be fully smooth because that would imply infinity cut off frequency so you can block all the frequencies out? If this is true then what is an exceptable cut off frequency to make dc smooth enough to run a computer off of or create a 12 volt charger ,....etc?

    Thanks maybe the formula for high pass filters is the same formula used for something else total different and the fc is not the cut off frequency because using their example with fc = 120hz then I am all set. This would imply that freqencies higher then 120 would be passed thru and frequencies less then or equal to 120hz will be blocked.

    Thanks
     
    Last edited: Nov 21, 2009
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Not really. Cheap DMMs don't give true RMS readings, but RMS readings won't do it either.
    You could use something like Zelscope on your PC; it uses the line input on your sound card as an input device.
    Zelscope is available here:
    http://www.zelscope.com/
    14 day demo. About $10 to register.
    Note that if you supply more than about 1v P-P to your sound card's line input, you will probably destroy it.

    Did you even bother to read that page I provided a link to? :rolleyes:

    Where on earth did you come up with that?
    Why don't you just read and digest the page I linked to?
     
  7. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
    1,022
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    Ok , got you on this.
    I am wondering since now I got a smooth 12volt dc, current max rating 300mA
    I am wondering if I can use this to charge a car battery?

    Since a car battery is 12 volts dc I figure it should work.
    But the only thing I am wondering about is the fact that a car battery produces alot of current so if you hook up + to + on my 12 volt power supply and - to my - on my power supply I would think the current would be the only issue.

    In terms of power the car battery has more because P = V * I = 12v * I
    I is much greater in car battery then 300mA ,...etc?

    But then I am think resistance R = V/I = 12volts/I
    So the car battery would have less resistance because it can deliver more current.

    If you look at I from my supply the resistance would be greater.
    So I would think the battery would charge slowly. Because the current would flow in the path of least resistance into the battery?
    But maybe I am over looking something?

    I am also going both ways because if the car batteries current is to high and is being delivered to my power supply then it could fry some of the components because it will exceed their max current ratings. (Maybe a very strong rated diode would fix this issue ??? so as to allow current only to flow in the car batteries direction??? )

    Either way I am curious to know which way current will flow?

    Seems to me if transformers power is the issue how big a transformer would you need to make a car battery charger from a 120 vac 60hz on the primary?

    Thanks
     
    Last edited: Nov 21, 2009
  8. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
    69
    The battery would charge the smoothing capacitor, but no other current will flow back to the supply as it will be blocked by the rectifier.

    However, due to the possibilities of very high current from the battery if you mis-connect something, you should fit a fuse in line with one of the battery terminals before connecting anything else to that. Without a fuse, an accidental short could cause the wiring to catch fire while you are holding it!

    Also be aware that lead-acid batteries release hydrogen and oxygen while charging, which can accumulate in the casing. A small spark anywhere near the battery, such as removing a charging lead from a terminal, can cause the battery to explode - which considering the acid content is not pleasant - it has happened to me (I turned off the wrong switch..)

    The supply will only trickle charge the battery.
    A '12V' lead-acid battery needs around 14 - 15V to bring it to full charge.

    The charging time from dead flat is based on the capacity of the battery (in Amp-Hours, AH) and the charging current.

    A 40AH battery would charge in eight hours at 5A, or 40 hours at 1A.

    If the battery is not fully discharged, the time will be less.

    It's usual to extend the time by 20 to 40% to allow for losses, but be careful not to overcharge as this may damage the battery.

    A simple, safe charger has both current and voltage limiting.

    The best source of data for actual figures is a battery manufacturors web site, as that should show details of charging requirements.
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    Your 12.6vac transformer connected to a bridge rectifier with a filter capacitor should have roughly (12.6 -1) * 1.414 = 16.4 volts across the capacitor under it's typical load (you had specified 80mA).

    The 12.6 -1 is a rough estimate of the RMS output less the drop across the bridge rectifier.

    The filtered DC supply would try to charge the battery, but it would take a great many hours to charge the battery. If a completely discharged battery were rated for 40AH, at the very least (300mA output) it would take 40/300mA = 133 hours (5.56 days) to charge.

    Much of the time, your transformer would be operating in overload unless the current output were limited.

    It might be able to output enough current to maintain a fully charged 12.6v lead-acid battery at a float voltage (somewhere around 13.3v-13.5v, depending on battery chemistry and internal temperature).

    A standard charger for a typical lead-acid battery might output a maximum of 5A-6A during charging up to 14.5v, and then maintain the battery at a float voltage.
     
  10. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
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    Ok , got you guys on this.
    I do realize that the battery will take some time to charge.

    However their is a couple questions I have

    So initially current will flow from the car battery to the filter capacator Why? they are both pushing at the same voltage and the resistance of the car battery is less. I would think because my supply has more resistance current will flow in the car battery direction path of least resistance?

    If I am wrong and current initially flows to the filter cap / rectifier will it be damaged or will the capacitor explode? Because
    the cap has capacitance = 2200uF and Vots = 50 volts so
    Q = C V => Q = (2200 uF ) (50 volts) = .11 coulumbs.
    How fast does a car batteries current Amps = coulumbs/sec so going by is amp-hours would that exceeded .11 Coul ?

    If 40amp-hours => 2400amp-seconds if battery is giving 4amps =>
    2400amp-sec /4amps = 600seconds time
    => 4amps * 600sec = 2400 coulumbs much more then .11coulumbs maybe I am using the formulas incorrectly though?

    If the cap is not going to be a problem from the car batteries current then what about the bridge rectifer it is rated at 400volt max 4amps max. So if current from the car battery is greater then the 4 amps it could damage the rectifer I would think?

    If this is true then I would think the 5/6 amps would damage the rectifier?

    Would this hurt or damage/weaken the transformer in anyway over time?

    And also if I put a diode after the rectifier/filter cap this should prevent any current from the car battery from getting to any of my 12 volt supply components. Just wonder what the diodes ratings would have to be to totally protect the components from current. (i.e reverse/forward max current ,...etc etc)

    Also what fuse for safety would you suggest i.e what voltage/current ratings?
     
  11. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
    69
    You should only connect or disconnect the charger from the battery whilst the mains supply is off, so I am assuming the capacitor would be discharged when you connect the battery.

    There will be a fairly high current pulse as the capacitor charges, but it is so short that the current rating of the wiring etc. is irrelevant.

    The current through the rectifier will only ever be that supplied by the transformer, as long as you don't connect the battery backwards.

    The fuse should be lower rating than the wire you use, but higher than any sustaned current you may be working with. 5A may be reasonable to start with?
    On a 12V supply, the voltage rating of the fuse does not really matter, it's only a maximum not a requirement.

    As far as possibly overloading the transformer, the simple thing to do is try it and see if the transformer gets exessively hot. As long as it does not get too hot to hold, it is OK. Try it in short bursts to start with to allow heat in the windings to percolate to the casing.
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    Manufacturers generally rate their batteries in AH capacity for discharge over a 20-hour period. So, a 12v battery with a 40AH capacity was tested by giving it a constant 2A load for a 20 hour period.

    Auto batteries are generally also rated for CCA's, or Cold Cranking Amperes. A typical auto battery might put out 400A or to 700A for a short period of time.

    The battery current won't flow through the rectifier, unless you connect it backwards (reverse polarity). If you connect it backwards, the capacitor and rectifier bridge will be destroyed immediately.

    Your transformer is not likely capable of supplying more than perhaps 1A even if the output were shorted. Of course, this would be very bad for the transformer.

    Yes, you can do that. A 1N540x series 3A rectifier diode would work just fine.
    I suggest a 3A diode over a 1A diode, because the 3A diode's Vf will be less than a 1A diode like a 1N400x series when you are running so little current through it.
     
  13. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
    1,022
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    Ok , Is their a diode I can use after the components to prevent this?
    Just in case somebody accidentally connects negative to positive and positive to negative of the car battery with my power supply?

    Would this be ok in the event they connected + - ?

    I would think 700amps ,...etc would over power most diode ratings?
    When does a car battery produce this ?
    I guess what I am asking is under what load does the battery supply this current? Techanically I always thought the current was controlled by the load and the voltage V/R . So if their is no load you theortically get no resistance and infinity current at any nonzero voltage (provided no resistance of the wire which can never truely be,...etc but in theory infinity). So the 12 volt car battery would only produce 700 amps to a load that
    has R = V/I = 12volts/700amps = 0.01714..etc ohms?

    If this is true then what component in the car is 0.01714 ohms not the starter or alterator I would think this would be much more resistance?

    So then what is the crank amps stuff mean? No 12 volt dc source can every produce 700 amps unless it was connected to a load of 0.01714ohms or less. This is just ohms law.

    Also we established the current would go into the car battery from my power supply (path of least resistance ) but if I new the internal resistance of the car battery I could figure out how fast the current would be moving into the car battery.

    Is their a typical internal resistance for the standard car battery or could you calculate the internal resistance by the amp-hours in some way?

    Also I would think if the internal resistance of the battery was greater then the internal resistance of my power supply (components ,...etc) then current would flow in the oppsite direction into my power supply. In theory but because the internal resistance of the components are much more resistant then the internal battery resistance we will never have this occuring correct me if I am wrong? So that is why if I know the internal resistance I can determine how much current is going to flow into the car battery and from their no how many days it will take to charge approx.

    Maybe I will just use an amp meter.
    Either way if my reasoning is correct this would be a way to determine what direction the current will flow when 2 same voltage sources are hooked together + to + - to - the current would flow in the direction of least resistance which would mean it would flow in the direction of of which ever had the least internal resistance. If I am correct then if they had both same internal resistance and volatage then no current would flow.

    Thanks
     
    Last edited: Nov 22, 2009
  14. gootee

    Senior Member

    Apr 24, 2007
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    I am joining this thread late. But here are a couple of URLs with very good information about the DC Power Supply portions of your questions:

    http://www.zen22142.zen.co.uk/Design/dcpsu.htm

    http://sound.westhost.com/power-supplies.htm#rectifiers

    If you use a regulator, make sure that you calculate the worst-case regulator input voltage, i.e. mains AC at 10% below nominal and use the minimum value of the DC that occurs, i.e. during the 'trough' of the ripple. That minimum regulator input voltage should be greater than the desired output voltage plus the regulator's specified 'dropout voltage' (at your operating conditions) plus some safety margin. (Don't forget to factor in the 'transformer regulation'.)

    P.S. Current does not follow the path of least resistance. Current follows ALL paths, in inverse proportion to their resistances.

    - Tom Gootee
     
  15. SgtWookie

    Expert

    Jul 17, 2007
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    That's why I suggested using a 1N540x series diode after the cap. It will prevent damage if connected in reverse.

    That kind of current only occurs when the battery terminals "see" a near-short circuit.
    When the electric starter motor turns the gasoline or diesel reciprocating engine crankshaft. That high amount of current is usually needed for only short durations; perhaps 30 seconds at most. If it won't start within 30 seconds, it probably won't start at all.

    The starter motor is the item which requires the highest current in a standard gasoline or diesel powered vehicle.

    Correct. Starter motors have very thick wires for field and armature windings, and very large commutator brushes.

    With a current supply limited to 300mA, an automotive battery would seem like an enormous sponge; it would take in that current without so much as a burp.

    It would vary between battery chemistry, battery structure, AH rating, etc. etc. ad nauseum. However, even large lead-acid batteries have limits as to how much current they can supply. You would have to consult the manufacturer of the battery in question to get a definitive answer.

    Where do you come up with these ideas? :confused:

    The charging and discharging of lead-acid batteries is certainly not 100% efficient. Some of the energy is converted to heat, and some of it may result in the generation of hydrogen and oxygen gases. Basically, the more quickly you charge or discharge a lead-acid battery, the more heat/gases will result.

    You'll need an ammeter and a voltmeter.

    If you have a protection diode on the output of your "charger", then current flow can only be in one direction.
     
  16. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
    1,022
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    Ok , got you.

    If intead I substituted a transformer that was rated at 120vac 60hz to 12 vac or 24 vac at 60hz with max current rating of 4 amps or 8 amps would this be better to charge a car battery in a short or reasonable amount of time?

    Is it typical to charge a car battery with 4 amps of current flowing thru it?

    Ofcourse I would get stronger components to match the stroner transformer,...etc

    Also for the smoothing of DC we used one capacitor. What happens if I wanted to use different loads with different resistances to charge or supply voltage to . I would think that I would keep needing to compute the new capacitance values for the capacitor to smooth things out.

    Is their any way other then a variable capacitor to keep the dc smooth other then a regulator.

    Like could I put a fixed resistor in parrell with the filter capacitor and then connect any load not worrying about the pulsating dc <-(their won't be any because the resistor in front is at a fix value ?).

    I just don't want to have to put a new capacitor in for the 2200uF one everytime I change the load resistance 120 = 1/2*pi*(2200uF) * R .
    The whole formula is based on the capacitance .

    Maybe just using a resistor in parrell with the filter cap (for the RC filter) instead of using the load as the resistor would remedy me having to very the capacitance value for different loads ?


    Another thing, would it be bad to charge a car battery with pulsating dc instead of just smoothing the dc (provided I was only going to use my power supply for charging a car battery) I would think it would take longer but still charge it???

    I also have a 12volt regulator but it say's input voltage is 35volts on the package. Does this mean I can only use it with a 35 volt supply voltage because if that is the case I won't beable to use this in my application. If ofcourse it is just a max rating then maybe I could use this for the smoothing of dc instead ? Radioshack part 276-1771 voltage regulator 7812.

    Thanks for clearing up these last minute questions .
     
    Last edited: Nov 22, 2009
  17. SgtWookie

    Expert

    Jul 17, 2007
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    I'm sorry, but you have exceeded your answers limit for today.

    Your next answer may be forthcoming tomorrow.
     
  18. gootee

    Senior Member

    Apr 24, 2007
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    The 7812's 35v spec is just the maximum input voltage (actually, it's just the maximum difference between the input and output voltages). Go to national.com and look up the LM7812. Also look closely at the related Application Notes. (National.com also has a web application that will design power supplies, for you, automatically. It might be useful to play with that.)

    Note that the 7812 can only handle about 1 Amp, maximum. And note that any regulator needs an input voltage that is greater than its output voltage by at least the 'dropout voltage' spec of the regulator. Also note that the greater the difference between the input and output voltages, the hotter the regulator will get, requiring a larger/better heatsink.

    My automotive battery charger has a 10-Amp "fast" mode and a 2-Amp "slow" mode.

    I think you might do better by charging a typical "12 Volt" automotive battery at something like 13.6 Volts.

    Why don't you do a google.com search for something like 'car battery charger schematic'?
     
  19. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
    1,022
    4
    Ok , I guess then the only question is do you need to smooth the dc for chargeing the car battery or can you just use pulsating dc?
    Or will this damage the battery in anyway?

    Provided you can charge with pulsating dc the car battery then
    Also how much longer would charging a battery with pulsating dc as opposed to charging with constant dc source take?

    Also if I used a 120 to 24 vac transformer would this be ok?
    I am just wondering how much more voltage you can use in charging a car battery? I would think if you used to much voltage (electrical pressure )
    the battery could bursted or if you used to much current to charge the car battery would start on fire ,...etc?

    Maybe 24vac is ok but their must be max voltage/current ratings for charging a car battery? (I know alot of batteries are probably different though in the max charge ratings)

    Thanks for your help
     
  20. SgtWookie

    Expert

    Jul 17, 2007
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    You can use rippled DC to charge lead-acid batteries. It was very common to do it that way in the "good old days". There is a project in "The Projects Collection" that uses rippled DC controlled by an opamp and SCR to charge a battery.

    Why don't you try it and find out?

    No. The output voltage from the rectifier (about 32v) would be far too high. The battery would look like a very heavy load to the transformer; and the transformer would very quickly overheat. If the transformer were large enough, the battery would quickly overheat.

    Yes. Charging voltage for lead-acid batteries varies by construction, but a typical lead-acid auto battery should be charged at 14.5v and charge current limited to 5A until it's nearly charged. AGM batteries can be charged at a higher current and voltage.

    Typical lead-acid construction auto batteries should not be charged faster than about 5A, or the battery will have excessive internal heating. The heatiing increases chemical activity, which shortens the battery life. Once a battery becomes heated internally, it takes a very long time for the temperature to come down; as it represents a large thermal mass.
     
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