Re: How do laptops get their power?

M.I.5¾ wrote:
"mike" <spamme9@xxxxxxxxx> wrote in message news:VK9Di.1316$s06.514@xxxxxxxxxxx
M.I.5¾ wrote:
"GlowingBlueMist" <nobody@xxxxxxxxxxx> wrote in message news:46dce159$0$47155$892e7fe2@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
"Barry Watzman" <WatzmanNOSPAM@xxxxxxxxxx> wrote in message news:46dcadc8$0$18915$4c368faf@xxxxxxxxxxxxxxxxx
The battery normally does not serve as a power conditioner.

Laptops have an internal switching power supply. Both the battery and the external AC adapter are just inputs to this switching power supply, which also charges the battery when on AC.

Don't worry about the generator. Both the AC adapter and the power supply in the laptop are switching power supplies, and switching power supplies are extremely tolerant of abberations in their own input power.


grizdog@xxxxxxxxx wrote:
Seems like this ought to be a faq, I'd like to know where to look.

The question is, when you plug a laptop into the wall, does the
battery serve as a power conditioner for the laptop? I may have to
use an old generator to power my laptop, and want to know if I will be
better off never running it while it is plugged in to the generator,
and rather only use the generator to charge the battery and then run
the laptop on battery only.

If it matters, this is a 5 yr. old thinkpad.

Check to find out what kind of power the generator will be producing. Some generate a sine wave like the power company provides while others produce a modified power output using some sort of square wave or other simulated sine wave output.

If the output is a true sine wave then you will have no problem but if it is one of the square or modified wave shapes be very careful when attaching computers to it. Many power supplies, computer or other electronic equipment, do a melt down when they are plugged into a square or modified sine wave output.
Why? The first thing encountered by the mains current is a filter circuit (which filters nasty spikes and things from either direction). The second thing encountered is a full wave rectifier to convert the AC to DC. Rectifiers actually work better from square waves than sine waves
The current in the filter cap
Ipk = c x dv/dt. The dv/dt of a sine wave is pretty low near the peak
where the diodes turn on. The square wave/modified sine wave has a very high dv/dt where the diodes turn on. Higher peak diode current, higher
peak cap current, higher fuse current...all increase the failure rate of the input components. This is mitigated somewhat by the resistance in the input filter.


Can you smell something? Gas? Drains? No, It's Bull***!

You are talking out of your arse hole.

First the impressive looking differential formula that you give is completely invalid (and it certainly didn't even look right). If you perform a dimensional analysis on the c x dv/dt part, it gives a result in coulomb seconds (dimensions [IT^2] - which does not correspond to any recognised unit) and *not* amps (dimensions [I]). In any event peak current in diodes is also largely completely irrelevant. It is the average diode current that matters, failure rate being determined largely by power dissipation (i.e. heat). The only other thing you have to worry about beyond that is that the I^2t rating for any current pulses does not exceed the I^2t rating of the diode (essentially its ability to emulate a fuse).

For an ideal square wave the output voltage of an unloaded full wave rectifier is contstant, so you arguement fails completely as there are no perceptible current pulses anyway. The only current flowing through the diodes is the load current (on each half cycle). The current flowing through the reservoir capacitor should be zero.



In any event the dv/dt in
your flawed formuala would be infinity for an ideal square wave and *very* large for a practical one.

THAT IS EXACTLY MY POINT!!!
It's true that for a perfect square wave, the infinite current only
happens at turn on. But for a perfect (typical shape) modified
sine wave, it happens twice every cycle.


LOL
You never cease to amaze me. You should write comedy.

So, you're saying that you slept thru math AND circuit analysis class...
Please post YOUR formula for the relationship between
time-variant voltage and current flow in a capacitor.

I really, REALLY enjoy reading your many musings on various subjects.
The depth of your "understanding" is...well...bottomless.
The concern I have is that someone may take your input as relevant
and get hurt.
These news articles hang around forever.

So, keep up the good work entertaining us. I'm sure there's no stopping you anyway.
I'll chime in when your advice seems harmful. When it's just
strange or misleading,...well...I just don't have enough time for that.

I can't stop laughing....
mike

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