Re: Power Inverters

BillW50 wrote:
spamme9 wrote on Mon, 24 Nov 2008 23:48:35 GMT:
BillW50 wrote:
spamme9 wrote:
BillW50 wrote:
spamme9 wrote:
nospam@xxxxxxxxxxxxxx wrote:
We're looking for a power inverter to recharge our Compaq notebook
in the car. Is it better to buy one with a larger rating (400W) rather than
a smaller one? Black & Decker has a 100W model ($13) and it gets
good reviews:

http://tinyurl.com/5fstl9

But the notebook uses something like 90W, so we'll be very close to
this inverter's capacity. The alternative is something larger, e.g.

http://tinyurl.com/6nfbrp

This one's 400W, but we're wondering if the extra cost is worth it.

Thanks for any info.

The 100W one isn't big enough.

Here's the ugliness of all this.

Any inverter you can afford will be a Chinese knock-off
of a knock-off of another knock-off of a maybe-good original
design.
The "designer" copied the circuit (not to be confused with the design)
and used the cheapest available
parts and the cheapest labor. They had not a clue to the original
design tradeoffs. They probably spent more money on the design of
the attractive display box than the design of the inverter.

It's tested to run a light-bulb in very infrequent use for as long
as the warranty...assuming you can find the vendor.

But your laptop is NOT a light bulb. It's a diode and a capacitor.
When the fast rise of the modified sinewave slams into the cap, you
get MUCH higher peak currents than you would with a light bulb.
Just twice the current for half the time is still twice the heat in
the output bridge of the inverter. And that assumes the transistors
can reliably take twice or 4-times or more the peak current.

When using a square wave or modified sinewave (not so square wave)
converter to drive a switching power supply, you need much more
"rated watts" than you'd think.

Typing this got me thinking...
Newer power supplies have built-in power factor correction to make
life easier for the power companies. What happens when you try to
run one of those from a square wave???

Nice theory! Although when you run at 110VAC and the power supply is
rated at 110v to 240vac, the slightly higher voltage won't matter. Plus
computers use power switching supplies and they don't care if you feed
them AC, DC, square wave or a variation of therefore.


That wasn't part of the discussion.
I'm NOT talking about the ability
of the load to run from a square wave source.
I was discussing the original topic of the thread.
I'm talking about factors that affect the ability
of the inverter SUPPLY to DRIVE a switching supply load.

I re-read my post multiple times. The ONLY mention of "voltage"
was yours...

And I am saying your statement is hogwash. And when you talk about a square wave source, whether you know it or not, you *are* talking about voltage. And computers use power switching supplies. And they don't care if the input voltage is a bit higher, square wave, sinewave, or DC.

Wash your hog in whatever you want.
I NEVER mentioned voltage in the quoted text.
Voltage is IRRELEVANT to the ORIGINAL TOPIC being discussed here.

You either don't know what you are talking about or are playing us for fools. Okay I'll call you bluff. I'm an electronic engineer. Here we go...

I don't know where you got your engineering degree, but you might think
about asking for your money back.


spamme9 wrote on Sat, 22 Nov 2008 04:09:33 GMT:
> nospam@xxxxxxxxxxxxxx wrote:
>> We're looking for a power inverter to recharge our Compaq notebook
>> in the car. Is it better to buy one with a larger rating (400W)
>> rather than
>> a smaller one? Black & Decker has a 100W model ($13) and it gets
>> good reviews:
>>
>> http://tinyurl.com/5fstl9
>>
>> But the notebook uses something like 90W, so we'll be very close to
>> this inverter's capacity. The alternative is something larger, e.g.
>>
>> http://tinyurl.com/6nfbrp
>>
>> This one's 400W, but we're wondering if the extra cost is worth it.
>>
>> Thanks for any info.
>>
>>
>>
>>
> The 100W one isn't big enough.
>
> Here's the ugliness of all this.
>
> Any inverter you can afford will be a Chinese knock-off
> of a knock-off of another knock-off of a maybe-good original
> design.
> The "designer" copied the circuit (not to be confused with the design)
> and used the cheapest available
> parts and the cheapest labor. They had not a clue to the original
> design tradeoffs. They probably spent more money on the design of
> the attractive display box than the design of the inverter.

Not so! As others have testified. Please provide evidence of your claims.

> It's tested to run a light-bulb in very infrequent use for as long
> as the warranty...assuming you can find the vendor.

Also where are the facts? Others have tested telephones, TVs, stereos, answering machines, and yes light bulbs too. So were is your evidence that they only test them with light bulbs? I bet you made that up didn't you?

I SURRENDER. I hereby apologize to the Chinese vendors. They have fine
designers. They have the highest quality products available.

It's about how far removed the implementation is from the actual
design. Think of the game "telephone" where a verbal message
gets transferred through several people and comes out the end
significantly different from the input.

Incompetence is NOT limited to Chinese engineers, it's pervasive.
SHEESH, can we get back to technical issues?


> But your laptop is NOT a light bulb. It's a diode and a capacitor.
> When the fast rise of the modified sinewave slams into the cap, you
> get MUCH higher peak currents than you would with a light bulb.
> Just twice the current for half the time is still twice the heat in
> the output bridge of the inverter. And that assumes the transistors
> can reliably take twice or 4-times or more the peak current.

First of all a laptop isn't a diode and a cap. That is pure BS! It is a switching power supply with FETs, triacs, caps, resistors, coils, transistors, chips, etc. that sees the UPS, inverter, or wall outlet as its input.

Post a link to the schematic for a "current design" laptop off-line switching power supply. All we need to see is the input connector, the hash filter
network, the rectifier diode and the input cap that the charge is slamming into. The rest is not relevant to this discussion.
If you have a design that does not need an input rectifier and bulk
storage cap, I'd like the patent number.


When you speak of current slamming into a cap, you are talking about an applied voltage to cause a current flow. You may not understand this, but you are. It is impossible to have current flow without voltage being applied or created (i.e. magnetic field).

On the first day of engineering school, they should have taught you that
I = C * dV/dT.
The current through an ideal cap is related ONLY to the
derivative of the terminal voltage...for constant C. The actual VOLTAGE is irrelevant!

Start at any DC voltage you like. The instantaneous current
through the (ideal) cap
is related ONLY to the derivative of the terminal voltage.
This is not my opinion. It's the DEFINING equation for a capacitor.


And no, a modified sinewave doesn't cause anymore higher current to a cap then a regular sinewave does.
From the above equation, repeated here for those with short attention span:
I = C * dV/dT,

dV/dT for a sinewave (of amplitude 1) peaks at 1. In the steady state,
by the time the input sinewave voltage reaches the point that the rectifier diode turns on, the slope is much less than one.

dV/dT for a (perfect) square wave is INFINITE. For a typical inverter output, it's way less than infinite, but WAY more than 1.

Again, not my opinion...it's math.


Caps are rated in volts anyway and not
in current. Current doesn't not hurt a cap at all. It is the voltage that will hurt a cap. You can actually plug a cap straight into your outlet. And no, it won't blow (assuming you don't exceed the voltage rating of the cap). And no you won't even blow a fuse either.

Folks, please don't try this at home.

Seriously...get your college money back. They failed you miserably.

Take a look again at I = C * dV/dT.
It certainly is possible to pick a value of C such that the
current is within the current rating of the cap.
Caps across the line are routinely used for power factor
correction by people who know what they're doing and do the math.
In the general case, your statement is absurd.
I'm too lazy to do the math, but, for any ac voltage
and fuse rating, you can easily calculate the capacitance
value required to blow the fuse.

Be aware that REAL caps have parasitic losses and will explode
given sufficient current.

AGAIN...PEOPLE...DO NOT PLUG A CAP INTO YOUR WALL SOCKET.
DEPENDING ON THE CAP YOU PICK, YOU MAY GET DEAD.

Yes, you must not exceed the voltage rating of the cap,
but I didn't mention VOLTAGE anywhere in my original analysis
because it's IRRELEVANT...as long as you don't design with
caps that have insufficient voltage rating.
I DON'T UNDERSTAND WHY YOU GUYS INSIST ON WORRYING ABOUT
VOLTAGE. IT DON'T MATTER TO THE ORIGINAL THREAD STATEMENT
except to the extent that it affects dV/dT.

And no the transistors don't handle twice or more current either. As the current remains the same in any case. Why? Because it gets switched on and off as much as it needs like you turning the water on and off in your sink. You only turn the knob enough as much as you need. The switching power supply does the very same thing with voltage.

The only analog I can think of on short notice is "water hammer"
Turn on your faucet, then turn it off instantly. The pipes will bang
against the rafters as the energy of the flowing water is dissipated.
Turn it off slowly and there's no water hammer, because the rate of change of velocity is much lower. The energy of the flowing water
is dissipated over a longer time and the peak stress on the plumbing
is greatly reduced. The initial velocity of the water
does not matter. Only the rate of change of velocity matters.

> When using a square wave or modified sinewave (not so square wave)
> converter to drive a switching power supply, you need much more
> "rated watts" than you'd think.

That makes no sense whatsoever. Switching power supplies don't care if it is sinewave, square wave, modified sinewave or DC.

You've still got it backwards. The thread started with the question,
"What do I need to DRIVE my laptop?" The discussion is not about
what the laptop can handle. It's about what the INVERTER can supply.
If you want to discuss the issues surrounding what the laptop supply
can tolerate on its input, related issues apply. But that's not the
topic of this thread. Start your own thread for that.

And when you talk
about such, what do you think you are talking about? Yes the sinewave in the AC outlet is talking about the wave of the voltage. Get it?

No, you don't...

And when
you talk about the output of an inverter as square wave, modified sinewave, or whatever, you are talking about the shape of the voltage once again.

You say the words, but don't comprehend.
The SHAPE of the voltage is EXACTLY what we're talking about.
dV/dT describes the shape. And that's ALL that's relevant to the cap...as long as you don't design with a cap that can't handle
the peak voltage.

And when you are talking about watts... you are talking about current times voltage. Or voltage times current. The result is the same. Look it up!
DC voltage times DC current equals DC watts.
If it ain't DC, you have more math to consider.
That was taught on the second day of engineering school.

> Typing this got me thinking...
> Newer power supplies have built-in power factor correction to make
> life easier for the power companies. What happens when you try to
> run one of those from a square wave???

Not isn't true either. It has nothing to do with making anything easier for the power company.

Just because you don't know something, don't accuse me of making it up.
The patents hanging on my wall suggest that I AM very good at making
stuff up, but that's not the case here.

Plug "power factor correction" into your favorite search engine.
From the first page you'll get:
http://en.wikipedia.org/wiki/Power_factor
yes, it's wikipedia, so may not be 100% correct, but it's a place
to get you started.

What am I thinking...if I don't quote it, another firestorm
will ensue over some other irrelevant detail...

Quoting:
A typical switched-mode power supply first makes a DC bus, using a bridge rectifier or similar circuit. The output voltage is then derived from this DC bus. The problem with this is that the rectifier is a non-linear device, so the input current is highly non-linear. That means that the input current has energy at harmonics of the frequency of the voltage.

This presents a particular problem for the power companies, because they cannot compensate for the harmonic current by adding simple capacitors or inductors, as they could for the reactive power drawn by a linear load. Many jurisdictions are beginning to legally require power factor correction for all power supplies above a certain power level.
END quote

You sure make a lot up in your stories I must
say. It has to do with switching power supplies can adjust to what is fed to them automatically. All they need is enough voltage 50 to 60 times a second (anything between 110 to 240 volts usually works just fine).

There you go with the VOLTAGE again and at the WRONG END of the problem.

If you don't believe anything I say,

I decided that long ago.
But I AM entertained...up to the point where you tell people to
plug a cap into their wall socket...please don't do that.

then look it up. I dare you! ;)

I'm not intimidated unless you "double-dog" dare me.


.

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