can someone recommend a really TRUE RMS meter?

P

panfilero

Jan 1, 1970
0
In principle that will work, but if the waveform is complicated, so
will be the calculations.

ok... after reading some of the documents you guys have pointed me to,
what I'm gathering is that if I do Vrms*Irms I will just get the
apperant power, and in order to convert that to the real power I need
to multiply by the cosine of my phase shift, but that is only true for
sinusoids, since I have a messy pulse I think my situation has become
a lot more complicated.... I'm dealing with a pretty small fan and
feeding the windings from a microcontroller sending out pulses to each
winding, like a pwm to control the speed.... these watt meters look
like they are for big AC type circuits, are there watt meters for low
voltage non-sinusoidal applications?

appreciate the help
 
R

Richard Henry

Jan 1, 1970
0
ok... after reading some of the documents you guys have pointed me to,
what I'm gathering is that if I do Vrms*Irms I will just get the
apperant power, and in order to convert that to the real power I need
to multiply by the cosine of my phase shift, but that is only true for
sinusoids, since I have a messy pulse I think my situation has become
a lot more complicated.... I'm dealing with a pretty small fan and
feeding the windings from a microcontroller sending out pulses to each
winding, like a pwm to control the speed.... these watt meters look
like they are for big AC type circuits, are there watt meters for low
voltage non-sinusoidal applications?

appreciate the help

There are cheap in-line meters intended for use by radio-controlled
vehicle hobbyists made by companies like Medusa Research.

http://www.medusaproducts.com/Power-Analyzers/PA-Main.htm
 
P

panfilero

Jan 1, 1970
0
"Phantom"
I gave you the answer in an earlier post.

Go read this page:

http://en.wikipedia.org/wiki/Power_factor

Then read this page for some more of the same explanation:

http://www.yokogawa.com/tm/tr/tm-tr0605_01.htm

followed by this page to explain how to actually measure the power
delivered to your 3 windings, except that you will probably need to use a
scope with trace math, rather than a wattmeter. The scope will have to be
able to multiply the instantaneous voltage and current and integrate
(average) that product. You can use the two-wattmeter method to get the
3-phase power delivered to the motor windings, with the scope taking the
place of the wattmeters.:

http://www.yokogawa.com/tm/tr/tm-tr0605_02.htm

Those links were really helpful, thanks for sending them... from what
I'm gathering according to the yokogawa links what I'm trying to do,
find the power going to the windings, can be done by "Active power is
calculated by averaging the products of the instantaneous voltages and
currents" and "Ideal active power is expressed as the product of the
instantaneous voltages and currents averaged over one period of
voltage or current. "
but then it also says "We know that the active power from the voltage
and current of the distorted wave is the sum of the active powers
obtained from the products of the voltages, currents, and power
factors of the same harmonic component (frequency)"

so would you say I can just multipy the instantaneous values of V and
I to get the real power? I don't even need RMS anything? Weird... I
thought the point of RMS is cause you can't do that.......

thanks
 
M

Mark

Jan 1, 1970
0
so would you say I can just multipy the instantaneous values of V and
I to get the real power?  


Yes... but they have to be the V and I taken _at the same instnat in
time__, you multiply them and that gives you the instantaneous power
at that time....then repeat for many periods.. take all the power
results and average them and you will have the true average power...

but why do you want to measure the true average power of a small fan
motor?

lets get to the meat of the real problem you are trying to solve..

Mark
 
P

panfilero

Jan 1, 1970
0
Yes... but they have to be the V and I taken _at the same instnat in
time__, you multiply them and that gives you the instantaneous power
at that time....then repeat for many periods.. take all the power
results and average them and you will have the true average power...

but why do you want to measure the true average power of a small fan
motor?

lets get to the meat of the real problem you are trying to solve..

Mark

Thanks Mark,

well basically Im tryign to characterize a small fan, and I want to
know how much power this little blower is consuming... the blower
comes with a little controller that sends pulses out to each winding
in order to keep the fan spinning. That is the only power the fan
gets, the controller gets power from somewhere else but I'm just
interted in the fan... so I have set out to measure the power each
winding consumes, and that's how've I've ended up with all these
issues.
 
J

John G

Jan 1, 1970
0
Yes... but they have to be the V and I taken _at the same instnat in
time__, you multiply them and that gives you the instantaneous power
at that time....then repeat for many periods.. take all the power
results and average them and you will have the true average power...

but why do you want to measure the true average power of a small fan
motor?

lets get to the meat of the real problem you are trying to solve..

Mark
Thanks Mark,
well basically Im tryign to characterize a small fan, and I want to
know how much power this little blower is consuming... the blower
comes with a little controller that sends pulses out to each winding
in order to keep the fan spinning. That is the only power the fan
gets, the controller gets power from somewhere else but I'm just
interted in the fan... so I have set out to measure the power each
winding consumes, and that's how've I've ended up with all these
issues.

I don't want to criticize you at all, but as you do not seem to understand
the difference between real and apparent power you are Way way out of your
depth wanting to worry about the power used by a small fan.

John G.
 
P

Phil Allison

Jan 1, 1970
0
"Phil Hobbs"
Phil said:
I didn't say I bought it off ebay.


* Quote from YOU here, in this thread on May 3.

" I dunno--I have an HP 400A with true-RMS, bandwidth ~ 10 MHz, cost on
Ebay ~ $75. "


** ????????????????????????????????

I bought it for work, new.


** If and when you ever arrive back on planet earth ........

Do try to get your FUCKING facts straight !!

IDIOT !!!!!!!!!!



.... Phil
 
P

Phil Allison

Jan 1, 1970
0
"Phil Hobbs"
Phil said:
You know, Phil, I can be wrong sometimes. Perhaps I'm wrong about the
specs on the meter in my lab--lots weirder things have happened.


** You are WRONG about the damn model number - IDIOT !!!

However, the fact that I bought my work meter new doesn't contradict the
statement that such things (analogue meters with no digital interface) are
going cheap on eBay at the moment.

** Yep a ** HP 3400A ** might go for such a low figure.

Rusty, bent needle and all.

Assuming that this discussion stays vaguely civil, I'll check the model
number, look up the specs, and post the results when I get back to the
lab.


** I won't hold my breath expecting an apology out of you.



.... Phil
 
W

whit3rd

Jan 1, 1970
0
A whirling magnet/coil gizmo can be a motor or generator.

Motors have a net power input. Generators have a net power output.
The exact nature of a thing spinning around is NOT
determined by averages of current and voltage, because the
power could be positive or negative. That 'Irms' and 'Vrms'
can only give a limit of the magnitude of the power, which
means the number you get can be correct, or 200% too high.
The factor that the averages miss can be anywhere in
the (-1, +1) range and (for simple sinewave case)
is the sine of a phase angle.

Multiply I times V, and average THAT, and you have
power. It is possible to divide power by Irms and Vrms
to get sine(phase)
 
P

Phil Allison

Jan 1, 1970
0
"whitless 3rd wit "
panfilero
"Phil Allison"
A whirling magnet/coil gizmo can be a motor or generator.


** The OP has a " brushless DC " fan motor - you ass.

" What I'm trying to do: measure the current and voltage going into each
winding of a brushless fan motor, each winding is recieving some ugly
looking pulses in order to run the motor, I'm intersted in find the
power per each winding. "
Motors have a net power input. Generators have a net power output.
The exact nature of a thing spinning around is NOT
determined by averages of current and voltage, because the
power could be positive or negative. That 'Irms' and 'Vrms'
can only give a limit of the magnitude of the power, which
means the number you get can be correct, or 200% too high.
The factor that the averages miss can be anywhere in
the (-1, +1) range and (for simple sinewave case)
is the sine of a phase angle.


** There are no sine wave voltages or currents involved.

Dickhead.


....... Phil
 
K

krw

Jan 1, 1970
0
[email protected]>, [email protected]
says...
ok... after reading some of the documents you guys have pointed me to,
what I'm gathering is that if I do Vrms*Irms I will just get the
apperant power, and in order to convert that to the real power I need
to multiply by the cosine of my phase shift, but that is only true for
sinusoids, since I have a messy pulse I think my situation has become
a lot more complicated.... I'm dealing with a pretty small fan and
feeding the windings from a microcontroller sending out pulses to each
winding, like a pwm to control the speed.... these watt meters look
like they are for big AC type circuits, are there watt meters for low
voltage non-sinusoidal applications?

You don't need to do all that work. Sample the currend and voltage
them multiply the instantaneous values, average over a cycle, or
hundred. No need to convert to RMS anything, though that might be
interesting (VA measurement) too.
 
K

krw

Jan 1, 1970
0
[email protected]>, [email protected]
says...
ok... after reading some of the documents you guys have pointed me to,
what I'm gathering is that if I do Vrms*Irms I will just get the
apperant power, and in order to convert that to the real power I need
to multiply by the cosine of my phase shift, but that is only true for
sinusoids, since I have a messy pulse I think my situation has become
a lot more complicated.... I'm dealing with a pretty small fan and
feeding the windings from a microcontroller sending out pulses to each
winding, like a pwm to control the speed.... these watt meters look
like they are for big AC type circuits, are there watt meters for low
voltage non-sinusoidal applications?

If you have access to the voltage and current waveforms, simply
sample them at the same time and multiply to get the instantaneous
power, then average over some time (a cycle or hundred is good).
There is no need to do any ugly math, though VA might be interesting
too... BTW, a current transformer may be better than a resistor.
 
L

legg

Jan 1, 1970
0
The most popular hand-held meter that gave true rms voltage used to be
the Fluke 87, which could cover the audio bandwidth, using an analog
rms converter (100KHz limit). It produced rms current measurements bt
measuring a shunt resistor voltage, which is the most common current
measurement method.Other meters are available that use the same kind
of hardware.

The old Fluke 8922a bench powered meter used to be usefull with higher
voltage/power circuits. It used a thermal converter, followed by
conventional voltage metering and is highly sensitive to calibration.
This has a 1MHz or 11MHz bandwidth, depending on input voltage range.
There are other similar meters.

More recently, given the availability of digital scopes with good
bandwidth and math capability, mathematical derivation is more common.

You can't just multiply an RMS current and RMS voltage measurement to
get power. Their phase relationship is important. As an extreme
example, an amplifier driving an inductor with 1Vrms will dissipate
internal losses equivalent to driving a short circuit, while the load
power is minimal.

A digital scope math calculation uses time-coherent multiplication to
produce a fairly reliable power indication for related waveforms in
its display/memory.

DC-offset will also affect measurements, the ability of the meter to
produce meaningful readings and the operator's ability to understand
what the reading means.

RL
 
P

panfilero

Jan 1, 1970
0
The most popular hand-held meter that gave true rms voltage used to be
the Fluke 87, which could cover the audio bandwidth, using an analog
rms converter (100KHz limit). It produced rms current measurements bt
measuring a shunt resistor voltage, which is the most common current
measurement method.Other meters are available that use the same kind
of hardware.

The old Fluke 8922a bench powered meter used to be usefull with higher
voltage/power circuits. It used a thermal converter, followed by
conventional voltage metering and is highly sensitive to calibration.
This has a 1MHz or 11MHz bandwidth, depending on input voltage range.
There are other similar meters.

More recently, given the availability of digital scopes with good
bandwidth and math capability, mathematical derivation is more common.

You can't just multiply an RMS current and RMS voltage measurement to
get power. Their phase relationship is important. As an extreme
example, an amplifier driving an inductor with 1Vrms will dissipate
internal losses equivalent to driving a short circuit, while the load
power is minimal.

A digital scope math calculation uses time-coherent multiplication to
produce a fairly reliable power indication for related waveforms in
its display/memory.

DC-offset will also affect measurements, the ability of the meter to
produce meaningful readings and the operator's ability to understand
what the reading means.

RL


"You can't just multiply an RMS current and RMS voltage measurement to
get power." - RL

So.... I've read that Vrms*Irms = Pavg..... but..... you're saying I
need to include the phase shift?! Isn't the phase shift implied in the
waveform? I have two meters hooked up, and I'm reading a Vrms and an
Irms, if I multiply these together don't I get the Apparent Power?
Aren't RMS values scalars? No phaeses involved.... and isn't
Vrms*Irms = P apparent?
 
P

panfilero

Jan 1, 1970
0
You really do have to worry about the phase shift. When the voltage
and current are in phase, the power is going in the same direction as
the current. When they're out of phase, the power is going in the
opposite direction. From the wall socket's perspective, that's the
difference between a motor and a generator.

If you imagine a load that drew some enormous current right when the
voltage was near 0, and much less current near the peaks, you could be
off by a factor of 100 or more if you just multiplied RMS voltage times
RMS current. The same is true of a purely reactive load, in which the
voltage is 1/4 cycle out of phase with the current. For instance, this
is what it would look like with a capacitor across the line.

I___ ___
/ X \ V
/ / \ \
/---/---\---\---/----------------------
/ \ \ /
/ \ X
-- --- --
| | | | |
| | | | |
1 2 3 4

In quarter-cycle 1, I >0 and V < 0, so the VI < 0 : power is going from
the capacitor into the wall socket.
In QC 2, I>0 and V>0, so VI > 0: the power is going from the wall socket
into the capacitor.
In QC 3, I<0 and V>0, so VI < 0 again, and
In QC 4, I<0 and V<0, so VI > 0.

Since the waveforms are symmetric, the net power transferred over a
cycle is zero.

Cheers,

Phil Hobbs

Hmmmmm..... ok, well my voltages will definitely be out of phase since
there's some reactance involved.... and I agree with everything you
are saying here, but this is looking at just one cycle of each.... I'm
assuming a meter is constantly sampling the signal and so the RMS
value is some sort of average that is constantly being updated, and
the average is maybe taken over a hundred cycles or somethign like
that not just one cycle, so do you think that since the meter is
taking an rms average of many cycles that somehow the phase shift
would then be included in there when I multiply the values together?
I'm fine with the fact that I'm getting the reacitve power (the
negative power going back to the source) in my readings, I'm fine with
the fact that I'm getting apparent power instead of true/active
power... but I only have meters and am trying to avoid the phase
issue.... I'm using the two-wattmeter method to do this (thanks
Phantom)


J.
 
P

panfilero

Jan 1, 1970
0
No, since every cycle is just like every other, if the total is zero
over one cycle, it's zero, period. Sorry about that.

Cheers,

Phil Hobbs

yes, completely agree, that a purely reactive circuit has no real/true
power component to it. I don't mind that, I know that I am measuring
Apparent Power which is not True Power.... but my question is, Do I
Have To Worry About The Phase Shift For Making Apparent Power
Measurements By Taking RMS Voltage and Current Readings From A DMM...
at this point, I do not think that I need to know the phase shift to
make a measurement of the apparent power... Vrms*Irms=apparent power,
RMS values are scalars and contain no phase information...
 
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