Voltage Converter

Hi,

I have Netgear Wireless Router (from U.S.A) that needs a input of 120v
AC ~ 60Hz 16W

Here in India, I bought a Voltage Converter which convert from
220/240v AC ~ 50/60Hz 20W to 110/120V AC 20W

* Converter doesn't mention what is the converted value of 60Hz.

Question: Is it safe to use this Converter with the Netgear Router.
Will 20W input to 16W requirement ruin the router?

Thanks!

 

Be *careful* !!!
A number of cheap "converters" is nothing more than a diode and is
"good" only for heaters and incandescent lights (the life of those will
be severely decreased).
There are some converters that are transformers, and so safe to use.
Frequency will not be changed, so if your power is 50Hz, the output
will also be 50Hz - which for most equipment is acceptable.

 

Will 20W input to a 16W requirement damage the router?

* I got the converter checked using a multimeter and it converts to
110v

 

"Robert Baer"

** No such silly thing exists - you pathetic half wit.






........ Phil

 

No. That would give you double the wattage. There's a bit more than a
simple rectifier in there (think 'dimmer').

If it's rated for only 20W and heavy it should be fine.

Right kind:
http://cgi.ebay.com/DE-LUXE-50-W-AC...ryZ88759QQssPageNameZWDVWQQrdZ1QQcmdZViewItem
(also a splendid example of the capabilities of HTML)

Wrong kind: http://www.samstores.com/_images/products/148.jpg


Best regards,
Spehro Pefhany

 

No. 20W is a maximum, but in practice with something electronic like
that 16W is about at the maximum due to the way it draws current. It
might get warm (the converter) but both converter and router should be
okay.

Good. If it was the "wrong" kind of converter it would read
considerably lower than that unless the multimeter was a
fancy-schmancy type.


Best regards,
Spehro Pefhany

 

Not relevant; one could use a 200 KW generator to run a 5W shaver.
What kind of multimeter?
If analog, the reading is not meaningful as it is an average.
If a DVM, i cannot say if that reading is relevant or not.
A transformer-type converter has some weight to it - about a pound or
so; a diode type is usually smaller and is very light.
The transformer types are billed as being useable for a shaver.

 

A diode does not "double" the wattage.

 

Use your head for more than a hat rack; 1000W "converters" are sold
and "plainly" marked "for heaters and irons only".

 

"Robert Baer = FUCKING MORON "

** Know them well.

They use triac dimmers circuits set to 66 degrees ( out of 180) conduction
angle per half cycle.

NOT BLOODY DIODES - you FUCKWIT





........ Phil

 

"Robert Baer " = a STEAMING GREAT FUCKWIT

** It would supply DOUBLE the wattage the ( heating) appliance is rated to
handle.

YOU STEAMING GREAT FUCKWIT !!





........ Phil

 

In this case, a diode would "halve" the average wattage by giving you
"half" of the cycles.

Now, let's do a little Gedankenexperiment-- what would the power
dissipation of a "100W 120V" heater be if operated on 240VAC?



Best regards,
Spehro Pefhany

 

Well, if one had a square wave source, when the diode conducts, one
has 240V during that time, for P=E*E/R or four times the power then; the
other half of the time one has zero power; heating power is double as
claimed.
For a rectified/half sine wave, i do not know what the "RMS" would
be, but i would think it would be less than 240V.

 

The shape of the waveform doesn't matter, so long as the positive and
negative halves are symmetric.

If you apply 240V RMS to a 100W, 120V RMS resistive load, you get 400W. If
you halve the duty cycle with a diode, you get 200W.

Whether a square wave, sine wave, triangle wave, etc, halving the duty
cycle halves the power and thus reduces the RMS voltage by sqrt(2).

 

Okay, assuming ideal diodes:

If you have a FULL wave 240V RMS sinusoidal source, the power is
quadruple the 100W rating, or 400W.

If you have FULL wave rectified 240V RMS sinusoidal source, the power
is STILL 400W because resistors don't know for polarity.

If you have a HALF wave rectified 240V sinusoidal source, the power is
400W average for the half-cycles that the diode conducts, and zero for
the half-cycles that the diode blocks. So the average power is 200W,
or double the rating.

No calculus required, although it's not difficult to do the definite
integral of voltage*current over a half cycle for a sinusoid, which
leads to the same conclusion.


Best regards,
Spehro Pefhany

 

Gotcha; now understand...

 

Spehro Pefhany posted to
sci.electronics.design:

That is an ok 0th cut, too bad the heating element is a strongly
temperature dependant device. It increases in resistance strongly
with increased temperture, this helps it self regulate its
temperature and consequently its power.

 

Yes, the simplifying assumption is made that the resistance of our
imaginary "120V 100W" heater is constant.

Nichrome actually has quite a low TCR.. about 1/20 that of most pure
metals or around +0.01-0.02%/K, which is not what I'd call "strongly
temperature depend..nt". Maybe your imaginary heater is made of some
other material or runs much hotter than my imaginary heater. Or
perhaps you're thinking of an incandescent light bulb.



Best regards,
Spehro Pefhany

 

Spehro Pefhany posted to
sci.electronics.design:

Haven't looked at it for a long time, but i have had issues with
wirewound resistors changing resistance with temperature. All of the
resistance versus temperature curves i have seen look like a pure
exponential curve, perhaps not paying enough attention to the
vertical scale has tripped me up. I do remember that by the time it
is glowing the resistance and power gets to acting pretty strange
versus added voltage.