# What characterizes a powerFET for audio use?

Discussion in 'Electronic Repair' started by N_Cook, Jan 1, 2014.

1. ### N_CookGuest

Other than p channel in this case , same for BUZ901P nch
eg BUZ906P 200V, 8A ,datasheet says
"POWER MOSFETS FOR
AUDIO APPLICATIONS"
but also
"FEATURES ... (for use in)
HIGH SPEED SWITCHING ... "

Would a powerFET designed solely for high speed switching use and 125W
rating be derated in power handling terms to only 50W say for linear 10
Hz use. Or secondary oscillation liability if paralleled up devices? or
some other operational failing in audio use not found with smps say ?

2. ### Phil AllisonGuest

"Nutcase Kook" is a pig ignorant pommy ****

** The Semelab app note makes it pretty clear there is a HUGE difference
between "switching" and audio ( ie lateral) power mosfets.

http://products.semelab-tt.com/pdf/ApplicationNoteAlfet.pdf

** Yawnnnnnnnnnnnnnnn....

More brainless, fucking TROLLING !!!!!!!

..... Phil

3. ### William SommerwerckGuest

"dave" wrote in message
Any (???) device can produce more peak power in pulsed mode than continuous,
because it spends less time in the region where it dissipates the most power.

There is no such thing (other than in a mathematical sense) as RMS power.
Voltage and current can have RMS (root-mean-square) values. Squaring voltage
and dividing by resistance, or squaring current and multiplying by resistance
produces the //average// power.

4. ### Maynard A. Philbrook Jr.Guest

Oh really? have you found some over unity energy somewhere?

Jamie

5. ### William SommerwerckGuest

"Jeff Liebermann" wrote in message
The heating value of a waveform is the average value of its power, NOT the RMS
value of its power.

Many years ago I sat down and calculated this for a sine wave. I assure you, a
sine wave's average power and its RMS power are not at all the same. RMS
applies ONLY to voltage and current.

6. ### Phil AllisonGuest

** Yaaawnnnnnnn - more stupid fucking bullshit !!!!!!!!!!!!!!

One colossal fool asks an idiotic question and even BIGGER fools reply.

..... Phil

7. ### Phil AllisonGuest

"William Sommerwerck"
** But even a complete fool like Somerwanker should know that the term "RMS
power" is not meant literally.

It is NOT the RMS value of the instantaneous power level.

In the world of audio, it normally refers to a power measurement done using
the RMS value of a sine wave (or other wave if specified ) voltage delivered

A simple bit of calculus shows that the RMS value of a sine wave to be half
the square root of two ( 0.7071) times its peak value.

..... Phil

8. ### Maynard A. Philbrook Jr.Guest

I got into his long ago, and all the proclaimed experts came out of the
wood work to say I was wrong. RMS power ratings have been stuck on
equipment as long as I can remember.

Jamie

9. ### William SommerwerckGuest

Many years ago I sat down and calculated this for a sine wave. I assure
I can't be a complete fool, because on rare occasions you agree with me.
He knows nothing of the sort. Just because people say "pressure" when they
mean "force", doesn't mean pressure and force are the same thing.

I see ads proclaiming that an amplifier delivers "125 watts RMS power". No, it
doesn't. There is a perfectly good term for that... "continuous average
power", or perhaps "continuous average sinewave power". This term has a
specific meaning, and it IS NOT the same as RMS power.

10. ### Phil AllisonGuest

"William Sommerwanker = IDIOT"
** Yes you are and about to prove it yet again.

** Proof given

** " ... the term "RMS power" is not meant literally."

Cos it is not what the writer meant PLUS there is simply no such quantity
if you do.

FUCKWIT !!!!!!!!!!!!

..... Phil

11. ### Phil AllisonGuest

"Jeff Liebermann"
** There is no such quantity as " RMS power " if you are so stupid as to
interpret the term literally.
** To get " true RMS " volts of course.

** The RMS value of a voltage or current could hardly be more useful.

My god, you are SOOOOOO illiterate and SOOOOO retarded.

.... Phil

12. ### Phil AllisonGuest

"Jeff Liebermann"

** The term " RMS power " of course -

you fucking BLIND nut case ASSHOLE !!!!!!!!!!!!
----------------------------------------------------------

** Quite wrong - there is a very close connection.
** RMS voltage or current value = the DC equivalent value.

So the DC power formulas: " I squared R " and " V squared / R "
still apply in power calculations using AC voltages & currents.
---------------------------------------------------------------------------

** The RMS value of a voltage or current could hardly be more useful.

My god, you are SOOOOOO illiterate and SOOOOO retarded.

** It's completely true.

You are the stupidest, most retarded, constantly lying fuckwit on the NG.

Get cancer and fucking DIE !

..... Phil

13. ### William SommerwerckGuest

"Jeff Liebermann" wrote in message
Unfortunately, no.

A fair question. Apparently, there was a time when AC power had a significant
percentage of harmonics (perhaps it still does), and engineers wanted to know
its "true" heating effect.

14. ### William SommerwerckGuest

"Jeff Liebermann" wrote in message

<http://www.eznec.com/Amateur/RMS_Power.pdf>
On Pg 6 it says:
"The RMS value of power is not the equivalent heating power and,
in fact, it doesn't represent any useful physical quantity."
and:
"The RMS power is different than the average power, and therefore
isnâ€™t the equivalent heating power. In fact, the RMS value of the
power doesnâ€™t represent anything useful."
Jeff, you're misreading this. The author is saying //exactly// what I said.
(It's unfortunate he doesn't use a sine wave for his example, but the math
would then require integral calculus.)

Average power /is/ the heating power. (The rationale is that a resistor's
temperature is determined by the average power applied to it.) When we know
the RMS value of any repetitive waveform, we can compute its heating power by
squaring that voltage (current) and dividing (multiplying) by the resistor's
value.

Voltmeters (of all sorts) are usually calibrated to show the RMS value //of a
sine wave//. If the waveform differs, the value shown is wrong. One of the
advantages of a true-RMS meter is that all readings are "equivalent" in an
easily-comprehended way, regardless of the waveform -- even if that
"equivalency" has little practical usefulness. (If you want to know the
details of a waveform, you use a 'scope.)

I remember a Popular Electronics quiz with questions about how a voltmeter (at
that time, a moving-coil device) would read, depending on the waveform
supplied. A key point was that the deflection was proportional to the average
current flowing through the coil, but the meter was usually calibrated for the
RMS value of a sinewave.

15. ### WondGuest

(snipped)
ISTR it used to be a DC equivalent thing, ie, you need 6Vrms ac to
operate a 6VDC light bulb at spec.
I think Leak confused the issue when he branded those wonderful
audio amplifiers "RMS", thus putting the term in the realm of marketting.

16. ### William SommerwerckGuest

"dave" wrote in message
Who is Llewelen?

RMS and average are the same only for square waves. True RMS can be computed
by analog circuits, one of which made its way in dbx devices.

17. ### William SommerwerckGuest

"dave" wrote in message
Exactly.

18. ### Phil AllisonGuest

"William Sommerwanker is a bullshitting ass."

** Wrong - it's a fuckwit question.
** Nonsense.

Any "moving iron" amp meter inherently reads "true RMS" values and they are
as old as the hills.

True RMS meters are mostly used to measure irregular signals like noise and
distorted current waveforms that exist with AC to DC supply conversion.

There is no way to compute these values ( from the peak or average rectified
value) as there is with sine or other regular waveforms.

..... Phil

19. ### Maynard A. Philbrook Jr.Guest

I'll be more than happy to put a Obama chia pet on your grave stone
for you.

I see Amazon has a deal for 28 bucks..

Jamie

20. ### Maynard A. Philbrook Jr.Guest

AVG and RMS is not the same with sine waves..

Avg = 2/pi * peak Voltage.

And RMS is simply sqr( VP^2 / 2).
you'll notice the root his to look for the square not the
average of total voltage for example.

One can shorten that to say RMS = Vp * .707

Just think of a half round circle an draw and find the
area where you can evenly fit a square box in that half circle,
the value will be .707 times the Peak value of that circle.

Averaging ends up to be 2/pi = 0.637 * the peak voltage which
obviously gives you a different number..

And of course, no matter how you slice it, square waves at 50%
duty give you the same all around. AVG and RMS = peak.. cause there
is no slope in time for the avg and the nice square box I stated
fits perfectly in the already square wave for RMS. etc..

Jamie