YOU THINK YOU KNOW, BUT DO YOU?

[Dave]

12V vehicle Inverter

200watt AC device @ 120v (do vehicle inverters push 110 or 120v?).

Is there an equation that will convert what an AC device draws on the
inverter to what in turn the inverter will draw on the car? Using the
above example..?

 

[Peter Bennett]

On 10 Mar 2005 11:42:58 -0800, "Dave" <dmcomm_ads@yahoo.com> wrote:

12V vehicle Inverter

200watt AC device @ 120v (do vehicle inverters push 110 or 120v?).

Is there an equation that will convert what an AC device draws on the
inverter to what in turn the inverter will draw on the car? Using the
above example..?

Power = Voltage times current.

Your 200 watt device will draw 1.7 amps at 120V, and the inverter will
draw 17 amps at 12 volts (roughly - the inverter will only be 80 - 90%
efficient, so will draw a little more power than it produces, but the
battery voltage would be over 13 if the engine is running, so it would
draw a little less current than the 12 volt calculation shows.)



--
Peter Bennett VE7CEI
email: peterbb4 (at) interchange.ubc.ca
GPS and NMEA info and programs: http://vancouver-webpages.com/peter/index.html
Newsgroup new user info: http://vancouver-webpages.com/nnq

 

[Dave]

Getting the 17 amps is what I was asking. What is this equation?

 

[Rich Grise]

On Thu, 10 Mar 2005 13:24:34 -0800, Dave wrote:

Getting the 17 amps is what I was asking. What is this equation?

Go back and reread your answer.

This should also be an object lesson in quoting context.

Good Luck!
Rich

 

[JeffM]

This should also be an object lesson in quoting context.
Rich Grise

As a clueless Google Groups poster, Dave still won't get it.
All he has to do is scroll up the page.
He doesn't understand that the majority of folks use newsreaders
and don't visualize Usenet as he does.

 

[Dave]

One more time for the ubers...

Power= Voltage x current. Yeah, we learned that in grade school.

What was the equation used to arrive at the inverter drawing 17 amps?

There must be a correlation between what amps an AC device draws
compared to what amps this makes the inverter draw from the car.????

 

[Robert Monsen]

Dave wrote:

Getting the 17 amps is what I was asking. What is this equation?

The forumula is Power = DC current x DC voltage.

So,

200W/12V = 16.66A

However, that assumes your converter is 100% efficient, which it is not.
So, say it's 70% efficient, then

200W / 0.70 = 286W, so 286/12V = 23.8 A

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Dave]

How did you arrive at 17 amps?

 

[Dave]

This should be an object lesson contributing something meaningful to a
discussion instead of an obviously nerdy, smartass, socially oblivious,
human contact lacking, left brain comment.

 

[Peter Bennett]

On 10 Mar 2005 13:24:34 -0800, "Dave" <dmcomm_ads@yahoo.com> wrote:

Getting the 17 amps is what I was asking. What is this equation?

For a rough approximation, the power is constant through the inverter,
so VI at the inverter input equals VI at the inverter output.



--
Peter Bennett VE7CEI
email: peterbb4 (at) interchange.ubc.ca
GPS and NMEA info and programs: http://vancouver-webpages.com/peter/index.html
Newsgroup new user info: http://vancouver-webpages.com/nnq

 

[Dave]

Ah, true Usenet nerd police. I was waiting for your arrogant comments;
they always come, here goes the volley.

Clearly a case of a nerd who's little (although I'm sure intelligent)
one track mind puts meaning on what he thinks is fact.

What is actually so, is that I use a newsreader at home, but I can't at
work, so I use Google. I did scroll up son, but the equation was not
the one I was seeking. I always get what I need here.

 

[Dave]

Is the 70% used to factor in what the AC device draws on the inverter?
The 200W device is an AC device drawing on the inverter.

Pardon the novice.

 

[Don Bruder]

In article <1110483778.410642.278160@l41g2000cwc.googlegroups.com>,
"Dave" <dmcomm_ads@yahoo.com> wrote:

12V vehicle Inverter

200watt AC device @ 120v (do vehicle inverters push 110 or 120v?).

Is there an equation that will convert what an AC device draws on the
inverter to what in turn the inverter will draw on the car? Using the
above example..?

Inverters can (theoretically - you'll almost certainly hit a wall due to
the limits of real-world materials) push any voltage the maker wants
them to, depending on how they're designed. You want 240 volts out of
it? Wire it right, and you'll get 240 out of it. You want 120 volts out
of it? Wire it right, it'll spit 120 volts. And so on for any desired
voltage from 0 to infinity. It's all about how it was designed.

To get the answer you want, you turn the 200 watts and 120 volts your
device wants into a "PIE" - the equation P = I * E (where P is power in
watts, I is current in amps, and E is voltage in - wait for it! - volts)
will give you that answer - and solve for I, like so:

Plug in the values we know:
200 (P, watts) = I (amps) * 120 (E, volts)
Transform the equation to make solving it give us current in amps by
applying basic algebraic principles:
200 / 120 = I
And do the math:
I = 1.666... = 1 and 2/3rds amps
Inverter output: 1 and 2/3rds amps at 120 volts.

But you want to know how much of a load that places on the car...
How do you do that?

You "make another PIE":

We already know that 1 and 2/3rds amps at 120 volts is 200 watts, so we
also know that the inverter is in turn drawing 200 watts (Note 1) from
the battery. We do exactly the same equation as when we calculated how
many amps the device draws from the inverter: Solve P = I * E. Only this
time, we're working with how much the inverter draws from the car's
power supply - 200 watts at 12 volts.

Plug in our known values:
200 = I * 12
Transform to make current (I) the unknown:
200 / 12 = I
And do the math:
I = 16.666... = 16 and 2/3rds amps
Your inverter needs to draw 16 and 2/3rds amps to power your 200 watt,
120 volt device when it's running from a 12 volt source (Note 2). Plug
it into a circuit with a 20 amp fuse, and you're good to go.

Note 1: This assumes the inverter is running at 100% efficiency - a
HIGHLY unrealistic assumption, but we'll ignore that factor for now
since there are too many unknown variables to say anything
that'saccurate.

Note 2: This ignores the fact that when the engine is spinning, the
voltage available to the inverter from the car's system is going to be
closer to 13.5 or 14 volts than 12 volts, which will change the actual
current being drawn from the car slightly - finding out exactly how
"slightly" is left as practice for you.

--
Don Bruder - dakidd@sonic.net - New Email policy in effect as of Feb. 21, 2004.
Short form: I'm trashing EVERY E-mail that doesn't contain a password in the
subject unless it comes from a "whitelisted" (pre-approved by me) address.
See <http://www.sonic.net/~dakidd/main/contact.html> for full details.

 

[Robert Monsen]

Dave wrote:

Is the 70% used to factor in what the AC device draws on the inverter?
The 200W device is an AC device drawing on the inverter.

Pardon the novice.

No, it's losses in the conversion process, and it's just an estimate.

200W is what your plugged in AC device requires. I'm assuming that it
takes a bit more power to manufacture that 200W AC. Thus, I used a 70%
fudge factor, and used that to compute the power required of the battery.

From that, and the assumption that the battery is 12V, I calculated the
current draw of from the battery while a max load is plugged in.

Your converter will probably tell you how much current it needs in it's
product literature.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Kitchen Man]

On 10 Mar 2005 14:42:24 -0800 in sci.electronics.basics, "Dave"
<dmcomm_ads@yahoo.com> wrote msg
<1110494544.568179.118660@f14g2000cwb.googlegroups.com>:

One more time for the ubers...

Power= Voltage x current. Yeah, we learned that in grade school.

What was the equation used to arrive at the inverter drawing 17 amps?

Power = voltage x current. Or, if you prefer, P = I * E. Yummy pie.
Two pieces of pie, one for each side of the converter. Both pieces of
pie are ideally the same volume, with different dimensions. P1 = P2.
You already know E1 and E2. The rest is a piece of cake.

There must be a correlation between what amps an AC device draws
compared to what amps this makes the inverter draw from the car.????

As others have mentioned, the efficiency of the inverter must be
factored, and you want to deal in rms units for the ac calculations.
Are you worried about power factor? Unless your load is heavily biased
capacitively or inductively, power factor shouldn't be an issue.

--
Al Brennan

 

[mike]

Dave wrote:

12V vehicle Inverter

200watt AC device @ 120v (do vehicle inverters push 110 or 120v?).

Is there an equation that will convert what an AC device draws on the
inverter to what in turn the inverter will draw on the car? Using the
above example..?

I haven't started a firestorm of controversy in a couple of days, so
here goes...

There ain't no such thing as a 200W AC device. There ain't no such
thing as a vehicle inverter.

What you can find is a device that happens to draw 200W under a very
specific set of conditions...120V RMS 60 Hz. sine wave is typical.
But unless the power factor happens to be 1, rms volts x rms amps won't
equal watts.

You can also find a device that converts 12VDC to 120VAC under a VERY
specific set of conditions that probably aren't anywhere near the
conditions specified for the hypothetical 200W AC device.

While you can find sinewave or squarewave inverters, the state of the
market is modified sinewave. That means alternate rectangular pulses
who's amplitude is the peak value of the desired sinewave being
approximated.
That's typically 120V X 1.4. The pulse width is adjusted so that the
RMS value of the waveform matches a sinewave.

If you're driving a resistor, or a light bulb, you can get 90%
efficiency give or take.
So, your output power is 90% of the input power. And you can do the
simple RMS voltage ratios to figger that out.

But what if you're not driving a light bulb? Then it depends on what
you're driving. Power factor can go all to hell...way further toward
hell than if you were driving the same device with a sinewave.

If you need a very accurate number, you have to model everything.

Or you could just use input current = 1/.9 X 120/12 X output current and
wing it. But a vehicle electrical system can vary quite far from 12V.

And don't get me started on the fact that the 200W number on the device
nameplate may or may not bear much relationship to the actual wattage
consumed in your particular application.

My personal favorite is to use an ammeter and measure it. (almost)Never
calculate anything you can easily measure.

mike



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[Robert Monsen]

mike wrote:

There ain't no such thing as a 200W AC device. There ain't no such
thing as a vehicle inverter.

And there ain't no such thing as a solid object. Turns out that they are
mostly empty space. What a shock that was when I found out.

What you can find is a device that happens to draw 200W under a very
specific set of conditions...120V RMS 60 Hz. sine wave is typical.
But unless the power factor happens to be 1, rms volts x rms amps won't
equal watts.

I always thought that for those (mythical) inverters, the 200W tag was a
limit, which meant they didn't want to supply more than 200W out the
device, and that if you did, they wouldn't be responsible if your
voltage sagged.

And I'm still waiting for Don Lancaster to use his magic sinewaves to
come up with a perfect sine wave inverter. What about it, Don?

http://www.tinaja.com/magsn01.asp

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Fred Abse]

On Thu, 10 Mar 2005 14:49:33 -0800, Dave wrote:

This should be an object lesson contributing something meaningful to a
discussion instead of an obviously nerdy, smartass, socially oblivious,
human contact lacking, left brain comment.

Hey, who's the one asking for help in this thread?

--
Then there's duct tape ...
(Garrison Keillor)

 

[Rich Grise]

On Thu, 10 Mar 2005 15:07:47 -0800, Dave wrote:

Ah, true Usenet nerd police. I was waiting for your arrogant comments;
they always come, here goes the volley.

Clearly a case of a nerd who's little (although I'm sure intelligent)
one track mind puts meaning on what he thinks is fact.

What is actually so, is that I use a newsreader at home, but I can't at
work, so I use Google. I did scroll up son, but the equation was not
the one I was seeking. I always get what I need here.

Yeah. Too bad that's so far from what you want.

'bye.

 

[Sean Leistico]

I learned it in Stagecraft class decades ago.

Ol' Mr. Peerbolte called it the West Virginia formula

The abbreviation for West Virginia (before two-letter postal codes) was W.Va

W VA

Watts = Volts x Amps

Algebra will give you the corrolaries for finding Volts or Amps

I know this formula is accurate for AC, and I'm assuming DC--can anyone give
confirmation on that?


"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:QeKdnbP6o7uFU63fRVn-sA@comcast.com...

Dave wrote:
Getting the 17 amps is what I was asking. What is this equation?


The forumula is Power = DC current x DC voltage.

So,

200W/12V = 16.66A

However, that assumes your converter is 100% efficient, which it is not.
So, say it's 70% efficient, then

200W / 0.70 = 286W, so 286/12V = 23.8 A

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.