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  #1  
Old 01-12-2009, 02:50 PM
Mathematics! Mathematics! is offline
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Default turn AC into DC?

Ok, I have been reading about diodes. I was wondering if this would work.

Say I cut one end of a wire and say that one end of a wire can be pluged into the wall. If I but a diode that is strong enough to hold 120volts back.
Can I charge a DC battery.

For Example I want to plug a plug into the wall and run it to a car battery.

Is it like

-
|----------------diode----- aligator clamps --- dc car battery
-

When I split the wire can I use any of the 2 wires to connect the diode ?

120 volts is the voltage of my house at 60 hz so when I convert it to DC
what is the voltage ? I would think it would vary from 0 to 120 over the 1/60 of a second?
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Old 01-12-2009, 03:04 PM
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You must use a transformer.

If you tried what you are proposing, the battery would explode, the diodes would vaporize, the wires would melt, or you would trip the breaker on the circuit. Please do not try your idea.

Household voltage is given in VAC(RMS) - the RMS part is Root Means Squared.

A quick way to get the peak voltage is to multiply the RMS voltage by 1.414 or divide it by 0.707; either way you'll wind up with 169.7V.

If you really want to build your own charger, a good start would be to get a transformer that has a 36V center tapped secondary rated for 4A.
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Old 01-12-2009, 03:40 PM
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Ok I just want to understand why it will melt and do bad things.

A diode would make the flow of current only one way.
So why would you get more voltage ?
And why would this method melt the wire ...etc etc???

Quote:
You must use a transformer.
Is this the only way to convert ac into dc???

And how about converting dc into ac ?
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Old 01-12-2009, 04:00 PM
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You don't get more voltage, you get the voltage already present on the outlet. However, when you say the outlet is 120V you are talking about the RMS value of the voltage. The peak value of it is Vrms*sqrt(2). For 120V rms the peak value is 169.7V. The diode melt because a lot of current flowed from the outlet to the battery. Diodes are rated for some current and voltage. Also, this is not the only way to convert AC to DC but its the safest for newbies in this area. Always take care when working with the mains voltage, it can be lethal.
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Old 01-12-2009, 04:08 PM
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Quote:
Originally Posted by Mathematics! View Post
Ok I just want to understand why it will melt and do bad things.

A diode would make the flow of current only one way.
So why would you get more voltage ?
And why would this method melt the wire ...etc etc???
An automotive battery is a low-voltage high-current device.
Household mains power is high-voltage, medium current.

To keep the explanation brief, it's normal to charge automotive batteries by using a constant voltage source of around 13.8v-15v until they are nearly charged. A safe charging current for auto batteries is 5A.

When you use a diode to rectify voltage, you still wind up with nearly the peak voltage output (less about 1 volt drop across the diode itself).

I=E/R, or Current = Voltage / Resistance (Ohm's Law)
Let's say we have a battery that we're charging with a commercially-built battery charger; it's putting out 14V at 5A, so the battery is getting charged up. Let's experiment a bit with Ohm's Law. What's the equivalent resistance of the battery?
R = E/I, or Resistance = Voltage / Current
R = 14/5 = 2.8 Ohms
OK, now let's see what would happen if we tried charging it from the mains with your diode in the line:
I = E/R = 169/2.8 = 60 Amperes! That is a LOT of current, or 3x what most typical household circuits are rated for (20A).

re:transformer:
Quote:
Is this the only way to convert ac into dc???

And how about converting dc into ac ?
It's not about the only way to convert AC to DC; it's about exchanging high voltage, medium current for low voltage, high current.

Converting DC into AC in such items like uninterruptible power supplies, inverters, etc is typically performed using transformers or inductors.
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Old 01-12-2009, 04:10 PM
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Hey you can convert AC to DC only using diodes.. You can not reduce the volatge.. According to your proposal, you will end up firing a bomb to charge a 12v car battery with 120volts ac just using diodes??

-
|----------------diode----- aligator clamps --- dc car battery = EXplosion
-

Its is simple. Transformer is used to step down the supply to the desired voltage. And Diodes convert AC to DC.. Ac always varies between 0 to 120 but that variation occurs 60 times per second !! You cannot catch it at the point it becomes 12 v!!
And Diodes have a property known as reverse breakdown voltage.. That is the maximum voltage that can pass through it safely when connected in reverse.. No normal Diode has a reverse voltage of 120volts ... sorry 169 volts
Current will flow only in one way but voltage is voltage.. You need to use a transformer to reduce it..

And a transformer is not used to convert ac to dc.. It will produce ac itself but a different voltage depending upon the rating of it.. Diodes are used after it for that purpose..

For converting DC to AC, you need to produce a small ac current first using circuit and feed it to an inverted transformer i.e step up. Its the transformer we used before but its ouput becomes input and viv..
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Old 01-12-2009, 04:37 PM
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O, Ok I didn't realize.

So the diode only lets current flow one way. But then you still have the 120 volts. The diode equation only drops about 1 volt at most and the battery can only be charged with something in the range of 13 - 15 volts at 5 amps.

Ok so then cann't I use a load of resistors in series. Because then I could create a voltage drop so that the car battery is only supplied in the range of 13 - 15 volts. But maybe this won't work because the current won't be in the range of 5 Amps?

How did you know car batteries are charged with 13-15 volts and 5 amps?
And the current in a house is usually 20 amps with 120 volts? Did you measure it or is this a standard?
Their must be some respectable variation. And how do you determine what is a suitable range for current and voltage for a car battery?

Also last question when you read VAC this means RMS (root mean square).
I thought it meant alternate voltage current. My bad if I am wrong.

Thanks for your help.
I think if I use the transformer equation V1/V2 = N1/N2 , I1/I2 = N2/N1. I can work out the math. So do I put a diode then the transformer. Or can I just get a step down transformer that will take the 120 volts and produce 13 - 15 volts , 5amps?

Last edited by beenthere; 01-12-2009 at 04:43 PM. Reason: removed unacceptable language
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Old 01-12-2009, 05:05 PM
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Older electronics magazines (70s) show a really dirt cheap and simple circuit: One side of a 110vAC house-socket wiring line is securely connected to one side of a standard house-bulb socket (which will hold a 60-100w house-standard lightbulb), then the other side of this bulb socket connects to the Cathode end of a single 5-10watt silicon diode, and finally the free end of the diode (Anode) connects with a Battery Clamp or Spring-type Gator-thingy (Like those used on Jumper cables) to the + side (positive pole) of a car battery. The other side of the 110VAC line goes directly to the - side of the battery using the same type connection.

The inrush current, voltage surge and voltage-dropping function is performed by the 110/120VAC standard house bulb.

This will 'trickle-charge' any Lead-Acid battery, 6 to 24v, keep it warm on really cold wintery nights.

I have used this trick many times... One just has to connect battery FIRST, then plug it in. If the bulb burns out, there is NO connection to the wall AC.

Again... The house 110/120AC voltage and current are dropped from 110/120 (170RMS) to much lower level by the 110/120 standard screw-in house bulb(s), which is wired in series, and the remaining or leftover AC is run thru a large diode (Rectifier) which produces the pulsating-DC to charge the car battery.

Builders are urged NOT to plug it in without connecting a battery first, and never while the battery is STILL CONNECTED to the car ;-).

Think maybe this is what he was referring to?

Last edited by IT_Guru; 01-12-2009 at 05:09 PM. Reason: added clarity
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Old 01-12-2009, 05:27 PM
Mathematics! Mathematics! is offline
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How about my previous question about just using a diode and a whole bunch of resistors in series to drop the 120 down to a suitable 13 - 15 volt range?

Either way if I have to use a transformer what one do I buy?
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Old 01-12-2009, 05:44 PM
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Quote:
Originally Posted by Mathematics! View Post
How about my previous question about just using a diode and a whole bunch of resistors in series to drop the 120 down to a suitable 13 - 15 volt range?
That would be terribly inefficient; most of the power would be expended heating up the resistors instead of charging the battery.

Quote:
Either way if I have to use a transformer what one do I buy?
Your least expensive approach would be to purchase an off-the-shelf intelligent battery charger; these are designed to charge a battery safely, reasonably quickly, and prevent overcharging.

If you are determined to build your own charger, it will probably cost around $50-$60 in parts, and unless it is carefully designed and tested, will not likely perform as well as a commercially available charger.
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