Wiring a Dust Buster to a car's power...questions about when amps are drawn and how.

[Thomas G. Marshall]

I have a dust buster (hand held vac) that is considered the 7ish volt
version and whose internal battery is charged with a 11 volt (and some small
number of milliamps) AC->DC converter.

Ok, two fundamental questions:

1. If I wire the vac (complete with its battery) to the car's battery (via
the aux "cig lighter" port), is it possible for the car to somehow overwhelm
the vac's battery and fry it? I know that frying is possible when charging
up a car battery with a charger left on manual for too long, so I figure
that maybe batteries are considered shorts in the circuit somehow and will
draw every amp you give it (???) That would mean that the vac battery would
be fried?

2. If I remove the vac battery and just connect the 12V car line to the
motor, will the motor draw just the number of amps it needs? (Ignoring the
difference in voltage, which I'm assuming will only speed up a DC motor).

Thanks

 

[Homer J Simpson]

1. If I wire the vac (complete with its battery) to the car's battery (via
the aux "cig lighter" port), is it possible for the car to somehow
overwhelm the vac's battery and fry it?

Yes. You need at least a series resistance.

2. If I remove the vac battery and just connect the 12V car line to the
motor, will the motor draw just the number of amps it needs? (Ignoring
the difference in voltage, which I'm assuming will only speed up a DC
motor).

Incorrect. You will smoke the motor.

 

[[email protected]]

I have a dust buster (hand held vac) that is considered the 7ish volt
version and whose internal battery is charged with a 11 volt (and some small
number of milliamps) AC->DC converter.

Ok, two fundamental questions:

1. If I wire the vac (complete with its battery) to the car's battery (via
the aux "cig lighter" port), is it possible for the car to somehow overwhelm
the vac's battery and fry it? I know that frying is possible when charging
up a car battery with a charger left on manual for too long, so I figure
that maybe batteries are considered shorts in the circuit somehow and will
draw every amp you give it (???) That would mean that the vac battery would
be fried?

2. If I remove the vac battery and just connect the 12V car line to the
motor, will the motor draw just the number of amps it needs? (Ignoring the
difference in voltage, which I'm assuming will only speed up a DC motor).

Thanks

Or 3.)
You could get some cell phone cigarette plug charger to replace the
wall-wart charger and use it to just charge the duster buster
normally.

 

[Thomas G. Marshall]

Yes. You need at least a series resistance.

Wouldn't a resister just heat up and waste power? I'm missing some
fundamentals here...see below.

Incorrect. You will smoke the motor.

Hmmmm........I better ask this a different way.

When does an item draw "just what it needs" and when is current force fed to
it? A lightbulb in an a/c circuit if rated for 100W will draw 100W and no
more regardless of how much current is available . Why? And yet a motor in
my car will fry up?

 

[Gareth]

Wouldn't a resister just heat up and waste power? I'm missing some
fundamentals here...see below.



Hmmmm........I better ask this a different way.

When does an item draw "just what it needs" and when is current force fed to
it? A lightbulb in an a/c circuit if rated for 100W will draw 100W and no
more regardless of how much current is available . Why? And yet a motor in
my car will fry up?

Things usually draw the correct current when connected to a power supply
of the correct voltage. You said in your original post that this hand
held vac was "7ish volt". A car battery is 12ish volts, which is nearly
twice as much.

It may work if the manufacturers have been very cautious with the motor
rating, or if you only run it for very short periods of time and allow
the motor to cool down before running it again, but I would NOT
recommend it.

--

 

[Homer J Simpson]

Wouldn't a resister just heat up and waste power? I'm missing some
fundamentals here...see below.

No. The resistor should limit the current while not operating to a trickle
charge to the battery.

Hmmmm........I better ask this a different way.

When does an item draw "just what it needs" and when is current force fed
to it? A lightbulb in an a/c circuit if rated for 100W will draw 100W and
no more regardless of how much current is available . Why? And yet a
motor in my car will fry up?

If the voltages match you are fine. If you apply 11,000 volts to your TV it
will never work again.

Voltage is pressure. Current is flow.

If the water pressure is too high the tap will blow off the wall. If the
possible water flow is unlimited, the tap will allow what it needs.

 

[ehsjr]

Wouldn't a resister just heat up and waste power?

Yes - which is exactly what you would need it to do.

I'm missing some
fundamentals here...see below.





Hmmmm........I better ask this a different way.

When does an item draw "just what it needs" and when is current force fed to
it? A lightbulb in an a/c circuit if rated for 100W will draw 100W and no
more regardless of how much current is available . Why? And yet a motor in
my car will fry up?

Your motor is rated at ~7 volts. If you give it the
voltage it was designed for, it will take "just what
it needs". You are going to ~double the voltage when
you run it from the car's electrical system. That is
why it will burn up - unless you reduce the power
to it with a resistor (which will waste power as
you noted) or by some other means.

A light bulb rated for 100 watts on a 120 volt AC
circuit would blow if you fed it with 240 volts -
it would not take "just what it needs" to produce
100 watts.

Ed

 

[Rich Grise]

I have a dust buster (hand held vac) that is considered the 7ish volt
version and whose internal battery is charged with a 11 volt (and some small
number of milliamps) AC->DC converter.

Ok, two fundamental questions:

1. If I wire the vac (complete with its battery) to the car's battery (via
the aux "cig lighter" port), is it possible for the car to somehow overwhelm
the vac's battery and fry it? I know that frying is possible when charging
up a car battery with a charger left on manual for too long, so I figure
that maybe batteries are considered shorts in the circuit somehow and will
draw every amp you give it (???) That would mean that the vac battery would
be fried?

2. If I remove the vac battery and just connect the 12V car line to the
motor, will the motor draw just the number of amps it needs? (Ignoring the
difference in voltage, which I'm assuming will only speed up a DC motor).

Is this a learning project? If all you want is a 12V car-vac, you could
get something like this:
http://www.amazon.com/Black-Decker-DustBuster-Vacuum-Cleaner/dp/B000BFWJNK

To modify your existing one, we'd have to know a lot more about the
specific unit that you have.

Good Luck!
Rich

 

[Thomas G. Marshall]

Your motor is rated at ~7 volts. If you give it the
voltage it was designed for, it will take "just what
it needs". You are going to ~double the voltage when
you run it from the car's electrical system. That is
why it will burn up - unless you reduce the power
to it with a resistor (which will waste power as
you noted) or by some other means.

A light bulb rated for 100 watts on a 120 volt AC
circuit would blow if you fed it with 240 volts -
it would not take "just what it needs" to produce
100 watts.

What I was talking about was the current: It will take "just what it needs"
if the voltage is the correctly matched, right? So there is no way to match
the voltage but push too much current into it, is there?

The analogy of pressure and flow are helping me here.

 

[Homer J Simpson]

What I was talking about was the current: It will take "just what it
needs" if the voltage is the correctly matched, right? So there is no way
to match the voltage but push too much current into it, is there?

Yes. If the voltage is correct, it will draw as much as it needs.

 

[ehsjr]

What I was talking about was the current: It will take "just what it needs"
if the voltage is the correctly matched, right? So there is no way to match
the voltage but push too much current into it, is there?

Right. When you talk about it taking just what it
needs that automatically brings current to mind.
Give it the right voltage, and there is no way
to push too much current into it.

Ed

 

[Thomas G. Marshall]

Right. When you talk about it taking just what it
needs that automatically brings current to mind.
Give it the right voltage, and there is no way
to push too much current into it.

Ah ok, I'm re-reading your reply before mine. You were pointing out that if
you, say, double to the voltage to a 100 watt device, that it wouldn't
simply draw half the current it needs to keep the total consumption at 100W.

 

[Homer J Simpson]

Ah ok, I'm re-reading your reply before mine. You were pointing out that
if you, say, double to the voltage to a 100 watt device, that it wouldn't
simply draw half the current it needs to keep the total consumption at
100W.

Unless it was designed to, no, it would draw double the current and four
times the power. Bye bye gadget.

 

[ehsjr]

Ah ok, I'm re-reading your reply before mine. You were pointing out that if
you, say, double to the voltage to a 100 watt device, that it wouldn't
simply draw half the current it needs to keep the total consumption at 100W.

Exactly. In fact, the power consumption would go
way up. (Well, briefly, because the device would likely
burn out and then power consumption would go to zero)

It is time for math, and ohms law.
E = IR E is voltage, I is current and R is resistance.

Consider a circuit with a ~12 volt battery (your car) and
a ~6 volt load device - your dustbuster. We don't have
more specs, so for the sake of the example, let's
assume that the dusbuster takes 1 amp when it is connected
to the proper voltage. E=IR, so 6 = 1*R, meaning R = 6.
Now, when you plug the dustbuster into the cigarette
lighter E changes from 6 volts to 12 volts. R - the
resistance of the dustbuster, does not change. So on
a 12 volt supply, 12 = I*6, meaning I = 2.
So doubling the voltage means you doubled the current.

Power (watts) = E*I. The dustbuster in the example
drew 1 amp at 6 volts, meaning it consumed 6 watts.
But at 12 volts, it draws 2 amps, meaning it consumes
24 watts - *four* times as much! That will cook
the dustbuster.

Ed