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ForthReich


Member


Joined: 02 May 2007
Posts: 228

Posted: 20 Apr 2009 05:07:45 pm    Post subject:

I'm working on a project that I have to have assembled by tomorrow. I have to take an electric motor and hook it up to 12V leads so that it can be powered. However, I want to reduce the voltage from 12V to 6V so that this particular motor runs cleaner and hopefully with 2x the amperage. Anyone know anything about voltage conversion? btw I can't really run to Radio Shack or anything lol.
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DigiTan
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Joined: 10 Nov 2003
Posts: 4468

Posted: 20 Apr 2009 06:18:47 pm    Post subject:

I well, I gotta warn ya: the only way you'll get x2 the current is with a buck converter or a transformer. What parts do you have?

Last edited by Guest on 20 Apr 2009 06:19:12 pm; edited 1 time in total
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ForthReich


Member


Joined: 02 May 2007
Posts: 228

Posted: 20 Apr 2009 06:31:30 pm    Post subject:

Yup the bad news is that i have no parts Sad
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DigiTan
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Joined: 10 Nov 2003
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Posted: 20 Apr 2009 06:48:45 pm    Post subject:

Well I guess since motors are generally more efficient at lower voltages, I'd stick to 6V battery. If the project absolutely *must* involve voltage conversion from 12V to 6V, you can make a voltage divider out of a number of materials. Like pencil graphite.
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NETWizz
Byte by bit


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Joined: 20 May 2003
Posts: 2369

Posted: 05 May 2009 01:24:54 am    Post subject:

It is not possible. I have had too many electronic courses to know what you said goes against the laws of Physics

Facts:
Wattage = Voltage * Amperage
Voltage = Amperage * Resistance

These are the two formulas and you can write them however you want; I am not going to argue with the letters you choose to use :)


The amount of energy a motor uses is referred to in Watts, which are Volts * Amps
Note: 746 watts is roughly 1 Horse Power



A DC motor will have a certain amount of resistance that you cannot change unless you swap it for a different motor. I.e. You can swap a 12v motor with a 6v motor for instance!


Say that resistance is 2 ohms and you have a 12 volt battery.


12 volts = 6 amps * 2 ohms

12 volts * 6 amps = 72 Watts (Amount of Energy Motor consumes)


Okay, so now you feed the motor a lower the voltage (6 Volts)... The amperage will NOT increase; in fact, here is what happens:


6 Volts = 3 amps * 2 ohms

6 Volts* 3 Amps = 18 Watts



Ohms Law:
Decreasing the voltage with an unchanging resistance results in lower amperage (current)


You can work the Algebra to solve for anything you want with either formula or a combination of the two.
Volts = Amps * Resistance
Amps = Volts / Resistance
Resistance = Volts / Amps


Honestly, chances are good that a DC motor operates somewhat like a transformer or inductor with collapsing magnetic fields. It likely has changing impedance instead of resistance; in other words, its resistance may change similar to how inductive reactance works in AC circuits.



Please, search google for out of phase impedance, inductive reactants, capacitive reactance, and you will know much more.


I do NOT have the time now to a project like this, but here is how I would have gone about it:

Get a regulated DC power supply and set the voltage at 6 Volts and measure the current with an ammeter (multimeter works great). It needs to be wired in SERIES to do this! Now, you have the voltage and the current, so calculating the resistance of the motor is not hard.

Now, set the regulated DC supply at 12 Volts and again measure the current then calculate the resistance. This is a method to determine if changing the voltage impacts the motor's internal resistance!

FYI, when light bulbs get hot, their internal resistance changes! I have only AC light bulbs, but for the purpose of lighting AC electricity is the Same as DC only the RMS AC voltage = the DC. Look up RMS

DO NOT do this at home unless you are experienced or careful working with dangerous voltages.

I have a 60 Watt incandescent light bulb, and I measured it with an Ohm meter (actually the ohm function of a multimeter) and I got roughly 20 ohms. (You can check the resistance without adult supervision... can't get shocked doing that). The Lighb bulb was totally removed from the lamp fixture to get a reading.

I measured the AC Line Voltage (DO NOT do this). It is 122 Volts AC (RMS Volts).

Volts = Amps * Resistance
122 = Amps * 20

Amps = 122/20
Amps = 6.1

Watts = 6.1 * 120


At the instantaneous point you turn on a 60 Watt lightbulb, it draws 732 watts when cold. Obviously, that amount of power instantly heats the filament and the resistance goes up A LOT causing the current (amperage) to drop A LOT given relatively constant, unchanging voltage being delivered to the light bulb.

Please, don't ask how fast this happens, unless you want me to respond with a page full of Calculus since in reality it happens about as quickly as electricity flows though the filament takes a bit longer than that to heat. We are talking about small fractions of a second that are not even close enough to cause any heat or pop circuit breakers. It is generally in the interest of forum members to stick to Algebra with explanation because almost everyone can follow along. Seriously, you can take it further with derivatives if you would like to show the relationship of the instantaneous rate of change of resistance with respect to time dR/dt. Quite simply, you would plug in some info and differentiate. It is not that Calculus is tough or that you will not get it because in reality most of our members on this forum are very intelligent and would find Calculus easy. However, we feel that given their age it is unlikely many have had it or the background knowledge required to contemplate that way of thinking. I.e. Limits come first since by its very definition a derivative is actually a formula describe in terms of limits :-)



Does any of this make sense?

Great Also Consider?
It sounds totally silly since you would think the motor is the only item in the circuit, but I tell you it is not! In fact, a cheap power supply or battery WILL drop some of the voltage across its own internal resistance. I.e. Perhaps you get 5.9 volts across the motor and could calculate that .1 is across the power supply internally.


Another one for you:

What are High-Discharge batteries, and why are they preferable in high-drain devices?


Hypothetical:
Cheap Battery 10 ohms internal resistance
Better Battery 1 ohm internal resistance

Let's say each batter registers 10 volts on a volt meter.

We then wire a 10 ohm load to each battery.

Well, with the 10 ohm battery and 10 ohm load, 5 volts is dropped across the battery, and the output voltage drops to only 5 volts on a 10 ohm load. Hence, under stress, the lower discharge battery drops more voltage across itself and turns into an internal heater.

In contrast:

The 1 ohm battery with a 10 ohm load will drop 10 * (10/11) or about 9 volts across the load meaning it is actually delivering 9 volts. The other 1 volt gets dropped across the battery internally.



Make sense?


Last edited by Guest on 05 May 2009 01:41:24 am; edited 1 time in total
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benryves


Active Member


Joined: 23 Feb 2006
Posts: 564

Posted: 05 May 2009 06:31:57 am    Post subject:

A complete aside, but do Americans also say "Ohmage" for resistance? If not, why "Wattage" for power and "Amperage" for current?


How can you expect me to know what MOST Americans say? The only thing typical of Americans is that we are each completely different people. Most Americans leave this stuff for electricians and electrical engineers. They are not familiar with these terms or what they mean. In our schools and colleges, students are taught to connect a light bulb, two wires, and a battery. For anything more complicated, we seek outsourced help. Smile


Last edited by Guest on 06 May 2009 09:04:50 pm; edited 1 time in total
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DigiTan
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Joined: 10 Nov 2003
Posts: 4468

Posted: 05 May 2009 01:11:56 pm    Post subject:

I think ForthReich was referring to having a 2:1 transformer double the current. You'll get maximum power transfer when the battery and motor match each other in resistance. At normal rpm's the motor's R will be much higher than the battery's in most cases, so even though things are moving, you may not be getting the most out of your battery.


With a 2:1 transformer in between, you double the current, halve the voltage, and halve the resistance of the motor (in the form of reflected resistance from the batteries point of view). But since transformers are DC-blocking tools, this is all moot unless you can turn back the clock and gather the parts to use it.


Last edited by Guest on 05 May 2009 01:17:29 pm; edited 1 time in total
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ForthReich


Member


Joined: 02 May 2007
Posts: 228

Posted: 05 May 2009 07:23:48 pm    Post subject:

thanks Netwizz...that was pretty comprehensive
and yeah, it did make sense
the project is done...Rcfreak we won right? hahaha yeah we pwned.
and now I know loads more about electrical....loads....yeah...lame joke Neutral
thanks!
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rcfreak0


Advanced Member


Joined: 27 Mar 2007
Posts: 354

Posted: 05 May 2009 08:21:25 pm    Post subject:

yes, we did win, by far...just lost in the towing part due to a bad prop....
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