AC Voltmeter
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Posts: 46,084
I am attempting to build a Watt Meter. I have some basic questions
regarding the principles of the hardware design.
I was planning to measure the voltage across a resistor in series with a
load in order to figure out the current flowing to the load. Knowing the
current, I can figure out the power. The voltage across the resistor would
be fed to an A/D converter which would be hooked up to a BSII. Problem is
the voltage across the resistor will be AC. I believe I need to get the rms
voltage across the resistor in order to get a true measure of the power. If
this is so, how will I be able to get an rms voltage?
I need to be able to measure 0 - 25 A. I'd like to get as much resolution
as possible. Also, the measurement doesn't have to be real precise; I'd
like to keep the design as simple as possible. I can work with some fudge
factors here - but not too much [noparse]:)[/noparse]
Any ideas would be appreciated.
--Craig
regarding the principles of the hardware design.
I was planning to measure the voltage across a resistor in series with a
load in order to figure out the current flowing to the load. Knowing the
current, I can figure out the power. The voltage across the resistor would
be fed to an A/D converter which would be hooked up to a BSII. Problem is
the voltage across the resistor will be AC. I believe I need to get the rms
voltage across the resistor in order to get a true measure of the power. If
this is so, how will I be able to get an rms voltage?
I need to be able to measure 0 - 25 A. I'd like to get as much resolution
as possible. Also, the measurement doesn't have to be real precise; I'd
like to keep the design as simple as possible. I can work with some fudge
factors here - but not too much [noparse]:)[/noparse]
Any ideas would be appreciated.
--Craig
Comments
If the waveform is sinusoidal, it could be peak detector-rectified. Then
the rms value, (which is what you need for power), is peak/sqrt(2), or
0.7071*peak. This is DC that the A/D could read. Problem is that normally
a small value resistor is inserted to measure current, so the voltage may
only be 1 volt or less. The diodes used for rectification would introduce
~0.7 volt drop, (or slightly less for a schottky diode), and also
distortion, which makes them useless for low voltage measurement. There are
opamp diode feedback circuits that make low level peak detection possible,
which is probably the approach I would use. Maybe someone on this list
knows of ICs for the job.
Ray McArthur
> I am attempting to build a Watt Meter. I have some basic questions
> regarding the principles of the hardware design.
>
> I was planning to measure the voltage across a resistor in series with a
> load in order to figure out the current flowing to the load. Knowing the
> current, I can figure out the power. The voltage across the resistor
would
> be fed to an A/D converter which would be hooked up to a BSII. Problem is
> the voltage across the resistor will be AC. I believe I need to get the
rms
> voltage across the resistor in order to get a true measure of the power.
If
> this is so, how will I be able to get an rms voltage?
>
> I need to be able to measure 0 - 25 A. I'd like to get as much resolution
> as possible. Also, the measurement doesn't have to be real precise; I'd
> like to keep the design as simple as possible. I can work with some fudge
> factors here - but not too much [noparse]:)[/noparse]
> as possible. Also, the measurement doesn't have to be real precise; I'd
> like to keep the design as simple as possible. I can work with some fudge
> factors here - but not too much [noparse]:)[/noparse]
Don't know if it is an option for you but you could use a current transducer
with voltage output. For most of them you will need an additional power
supply of +-15V and they'll set you back $30-$40 but you can connect the
output directly to your A/D convertor.
Just an option...
It is not too difficult to do what you want to do. The resistor idea is a
good start. Next rectify the AC voltage with a diode. Use a signal diode,
1N4148 or better. This will give the peak voltage at the diode output less
the junction drop of the diode. The diode junction drop will always be the
same, regardless of the voltage. You will have to add this value back in
after you obtain the result from the A/D converter. You can measure the
junction drop with a Fluke DMM or the like in the diode test mode. Once you
have the peak value, multiply it by 0.707 to obtain the RMS value. You will
have to write a fixed point math routine as the stamp can only do integer
math, but that is not too difficult either. Good Luck,
Gary
g.shearer@v...
G. Shearer
Free Electron Laser Research Center
Vanderbilt University
Nashville, Tennessee
Original Message
From: "Craig Lefkowitz" <craig.lefkowitz@m...>
To: "Basic Listserv #2" <basicstamps@egroups.com>
Sent: Wednesday, July 26, 2000 6:43 PM
Subject: [noparse][[/noparse]basicstamps] AC Voltmeter
> I am attempting to build a Watt Meter. I have some basic questions
> regarding the principles of the hardware design.
>
> I was planning to measure the voltage across a resistor in series with a
> load in order to figure out the current flowing to the load. Knowing the
> current, I can figure out the power. The voltage across the resistor
would
> be fed to an A/D converter which would be hooked up to a BSII. Problem is
> the voltage across the resistor will be AC. I believe I need to get the
rms
> voltage across the resistor in order to get a true measure of the power.
If
> this is so, how will I be able to get an rms voltage?
>
> I need to be able to measure 0 - 25 A. I'd like to get as much resolution
> as possible. Also, the measurement doesn't have to be real precise; I'd
> like to keep the design as simple as possible. I can work with some fudge
> factors here - but not too much [noparse]:)[/noparse]
>
> Any ideas would be appreciated.
>
> --Craig
>
>
>
>
>> I am attempting to build a Watt Meter. I have some basic questions
>> regarding the principles of the hardware design.
Hi:
You're missing a very basic element here. It's quite possible to have 25
amps flowing and lots of voltage and have nearly ZERO watts!!
You must consider the phase angle between the E and I even if you choose to
ignore the true waveshape and assume a pure sine wave. The guru Don
Lancaster has addressed this item a number of times in his columns. I'd
suggest doing watt ;-) I did. Went to the power company and asked them for
a discarded
mechanical wattmeter like they use on your house.
Wayne
Wayne Roderick P.E. (EE) (ret) (NMRA life-1721)
CEO, Teton Short Line, Pocatello Idaho, USA
http://www.ida.net/users/tetonsl/railroad
Note new URL- /biz/ becomes /users/
e-mail tetonsl@i...
Wayne is right. I assumed a resistive load. If your load is inductive or
capacitive, more complexity is required to account for phase angle. Also,
if your load is nonlinear and creates a non-sine, the 0.707 doesn't apply.
What is your load?
Ray McArthur
> You're missing a very basic element here. It's quite possible to have 25
> amps flowing and lots of voltage and have nearly ZERO watts!!
>
> You must consider the phase angle between the E and I even if you choose
to ignore the true waveshape and assume a pure sine wave.
Also, I am measuring power from an outlet, so the waveform will always be
sinusoidal and 60Hz.
Original Message
From: rjmca [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=JC0OOUnAisb4FX3AiXtdCNjLG2kFnpRwWQc2TiSzSFPt01Anoq6Ry2skjyjkYgno4sVZV9Zu_v3FpA]rjmca@w...[/url
Sent: Thursday, July 27, 2000 3:26 PM
To: basicstamps@egroups.com
Subject: Re: [noparse][[/noparse]basicstamps] AC Voltmeter
Craig:
Wayne is right. I assumed a resistive load. If your load is inductive or
capacitive, more complexity is required to account for phase angle. Also,
if your load is nonlinear and creates a non-sine, the 0.707 doesn't apply.
What is your load?
Ray McArthur
> You're missing a very basic element here. It's quite possible to have 25
> amps flowing and lots of voltage and have nearly ZERO watts!!
>
> You must consider the phase angle between the E and I even if you choose
to ignore the true waveshape and assume a pure sine wave.
peak rectifier with the diode?
Original Message
From: G. Shearer [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=f2wQYLMTr_nvrzU3Hvy1_R3Nqoc42-NZyP9zjFwmg9ESAPFjRIfD-bbf2C8LNPl9snyoxhHyZ4JHa45QFIjoHQ]g.shearer@v...[/url
Sent: Thursday, July 27, 2000 8:36 AM
To: basicstamps@egroups.com
Subject: Re: [noparse][[/noparse]basicstamps] AC Voltmeter
Craig,
It is not too difficult to do what you want to do. The resistor idea is a
good start. Next rectify the AC voltage with a diode. Use a signal diode,
1N4148 or better. This will give the peak voltage at the diode output less
the junction drop of the diode. The diode junction drop will always be the
same, regardless of the voltage. You will have to add this value back in
after you obtain the result from the A/D converter. You can measure the
junction drop with a Fluke DMM or the like in the diode test mode. Once you
have the peak value, multiply it by 0.707 to obtain the RMS value. You will
have to write a fixed point math routine as the stamp can only do integer
math, but that is not too difficult either. Good Luck,
Gary
g.shearer@v...
G. Shearer
Free Electron Laser Research Center
Vanderbilt University
Nashville, Tennessee
Original Message
From: "Craig Lefkowitz" <craig.lefkowitz@m...>
To: "Basic Listserv #2" <basicstamps@egroups.com>
Sent: Wednesday, July 26, 2000 6:43 PM
Subject: [noparse][[/noparse]basicstamps] AC Voltmeter
> I am attempting to build a Watt Meter. I have some basic questions
> regarding the principles of the hardware design.
>
> I was planning to measure the voltage across a resistor in series with a
> load in order to figure out the current flowing to the load. Knowing the
> current, I can figure out the power. The voltage across the resistor
would
> be fed to an A/D converter which would be hooked up to a BSII. Problem is
> the voltage across the resistor will be AC. I believe I need to get the
rms
> voltage across the resistor in order to get a true measure of the power.
If
> this is so, how will I be able to get an rms voltage?
>
> I need to be able to measure 0 - 25 A. I'd like to get as much resolution
> as possible. Also, the measurement doesn't have to be real precise; I'd
> like to keep the design as simple as possible. I can work with some fudge
> factors here - but not too much [noparse]:)[/noparse]
>
> Any ideas would be appreciated.
>
> --Craig
>
>
>
>
The resistive load simplifies things. Since you should have line isolation,
consider a current transformer. Digikey has a 1000/1, 25 amp unit for $9.
You could rectify (full wave for winding balance) and peak detect the output
with a cap. You would use a full wave bridge rectifier in series with a 100
to 1000 ohm resistor on the secondary. The volts/amp output would need
calibration because of the diodes, but it would be a simple implementation.
Diode error would be greatly reduced by the 1000/1 turns ratio of the
transformer. Unfortunately, diode error is not a fixed voltage as can be
seen by looking at the V/I curve of any diode.
Regards,
Ray McArthur
> It's a purely resistive load (no phase difference between I and V).
>
> Also, I am measuring power from an outlet, so the waveform will always be
> sinusoidal and 60Hz.
Do not forget to add in the junction drop of the diode. Another point, this
will limit detection of voltages below the diode junction voltage.
Good Luck
Gary
Original Message
From: "Craig Lefkowitz" <craig.lefkowitz@m...>
To: <basicstamps@egroups.com>
Sent: Thursday, July 27, 2000 6:31 PM
Subject: RE: [noparse][[/noparse]basicstamps] AC Voltmeter
> When you say rectify with the diode, do you mean I need to build a full
wave
> peak rectifier with the diode?
>
>
Original Message
> From: G. Shearer [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=KA1GMkm8UxUl6b7f9qoOhIrRMx1q2u1w6tULrf3m0_ARCbxqILMKx3nHq4y8Z9PdQ2fBnsGHCa-veVhw4CSz2XGP]g.shearer@v...[/url
> Sent: Thursday, July 27, 2000 8:36 AM
> To: basicstamps@egroups.com
> Subject: Re: [noparse][[/noparse]basicstamps] AC Voltmeter
>
>
> Craig,
> It is not too difficult to do what you want to do. The resistor idea is a
> good start. Next rectify the AC voltage with a diode. Use a signal diode,
> 1N4148 or better. This will give the peak voltage at the diode output less
> the junction drop of the diode. The diode junction drop will always be the
> same, regardless of the voltage. You will have to add this value back in
> after you obtain the result from the A/D converter. You can measure the
> junction drop with a Fluke DMM or the like in the diode test mode. Once
you
> have the peak value, multiply it by 0.707 to obtain the RMS value. You
will
> have to write a fixed point math routine as the stamp can only do integer
> math, but that is not too difficult either. Good Luck,
> Gary
>
> g.shearer@v...
> G. Shearer
> Free Electron Laser Research Center
> Vanderbilt University
> Nashville, Tennessee
>
>
Original Message
> From: "Craig Lefkowitz" <craig.lefkowitz@m...>
> To: "Basic Listserv #2" <basicstamps@egroups.com>
> Sent: Wednesday, July 26, 2000 6:43 PM
> Subject: [noparse][[/noparse]basicstamps] AC Voltmeter
>
>
> > I am attempting to build a Watt Meter. I have some basic questions
> > regarding the principles of the hardware design.
> >
> > I was planning to measure the voltage across a resistor in series with a
> > load in order to figure out the current flowing to the load. Knowing
the
> > current, I can figure out the power. The voltage across the resistor
> would
> > be fed to an A/D converter which would be hooked up to a BSII. Problem
is
> > the voltage across the resistor will be AC. I believe I need to get the
> rms
> > voltage across the resistor in order to get a true measure of the power.
> If
> > this is so, how will I be able to get an rms voltage?
> >
> > I need to be able to measure 0 - 25 A. I'd like to get as much
resolution
> > as possible. Also, the measurement doesn't have to be real precise; I'd
> > like to keep the design as simple as possible. I can work with some
fudge
> > factors here - but not too much [noparse]:)[/noparse]
> >
> > Any ideas would be appreciated.
> >
> > --Craig
> >
> >
> >
> >
>
>
>
>
>
>
>
I agree that there is a V/I curve for any diode. Looking at this curve, what
do you think the error would be for say a voltage swing of 5 volts and a
load impedance of >1M? Using an ADC0831 or LTC1298 as the A/D converter
would present the aforementioned scenario. Looking at the full wave bridge
method, on any half cycle, there would be 2 diodes conducting, dropping
voltage and introducing the errors you mentioned. Are the errors produced by
the V/I curve significant or negligible. I would be more concerned about the
voltage drop as it relates to the minimum detectable voltage. A peak
detector might be in order to be able to take care of the proper sampling
time of the ADC and at the same time, be able to scale the detected voltage
to the limitations of the ADC. If used, the circuit would require
calibration by taking readings at various input voltages and performing a
linear curve fit to obtain the calibration factor. I guess it all hinges on
what accuracy Craig requires for his project. Craig might comment on this
point.
Best regards,
Gary
G. Shearer
Free Electron Laser Research Center
Vanderbilt University
Nashville, Tennessee
Original Message
From: "rjmca" <rjmca@w...>
To: <basicstamps@egroups.com>
Sent: Thursday, July 27, 2000 8:54 PM
Subject: Re: [noparse][[/noparse]basicstamps] AC Voltmeter
> Craig:
> The resistive load simplifies things. Since you should have line
isolation,
> consider a current transformer. Digikey has a 1000/1, 25 amp unit for $9.
> You could rectify (full wave for winding balance) and peak detect the
output
> with a cap. You would use a full wave bridge rectifier in series with a
100
> to 1000 ohm resistor on the secondary. The volts/amp output would need
> calibration because of the diodes, but it would be a simple
implementation.
> Diode error would be greatly reduced by the 1000/1 turns ratio of the
> transformer. Unfortunately, diode error is not a fixed voltage as can be
> seen by looking at the V/I curve of any diode.
>
> Regards,
> Ray McArthur
>
> > It's a purely resistive load (no phase difference between I and V).
> >
> > Also, I am measuring power from an outlet, so the waveform will always
be
> > sinusoidal and 60Hz.
>
>
>
>
>
Original Message
From: G. Shearer [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=hA0vSXPsIfLxZj4OIvadyY2REFdYjl0mO1wCdd5Ltmyjujt7fANrmxZk6XkL1QsaADxXKb-PZlfmDs9YPrwlLg]g.shearer@v...[/url
Sent: Friday, July 28, 2000 9:15 AM
To: basicstamps@egroups.com
Subject: Re: [noparse][[/noparse]basicstamps] AC Voltmeter
Ray,
I agree that there is a V/I curve for any diode. Looking at this curve, what
do you think the error would be for say a voltage swing of 5 volts and a
load impedance of >1M? Using an ADC0831 or LTC1298 as the A/D converter
would present the aforementioned scenario. Looking at the full wave bridge
method, on any half cycle, there would be 2 diodes conducting, dropping
voltage and introducing the errors you mentioned. Are the errors produced by
the V/I curve significant or negligible. I would be more concerned about the
voltage drop as it relates to the minimum detectable voltage. A peak
detector might be in order to be able to take care of the proper sampling
time of the ADC and at the same time, be able to scale the detected voltage
to the limitations of the ADC. If used, the circuit would require
calibration by taking readings at various input voltages and performing a
linear curve fit to obtain the calibration factor. I guess it all hinges on
what accuracy Craig requires for his project. Craig might comment on this
point.
Best regards,
Gary
G. Shearer
Free Electron Laser Research Center
Vanderbilt University
Nashville, Tennessee
Original Message
From: "rjmca" <rjmca@w...>
To: <basicstamps@egroups.com>
Sent: Thursday, July 27, 2000 8:54 PM
Subject: Re: [noparse][[/noparse]basicstamps] AC Voltmeter
> Craig:
> The resistive load simplifies things. Since you should have line
isolation,
> consider a current transformer. Digikey has a 1000/1, 25 amp unit for $9.
> You could rectify (full wave for winding balance) and peak detect the
output
> with a cap. You would use a full wave bridge rectifier in series with a
100
> to 1000 ohm resistor on the secondary. The volts/amp output would need
> calibration because of the diodes, but it would be a simple
implementation.
> Diode error would be greatly reduced by the 1000/1 turns ratio of the
> transformer. Unfortunately, diode error is not a fixed voltage as can be
> seen by looking at the V/I curve of any diode.
>
> Regards,
> Ray McArthur
>
> > It's a purely resistive load (no phase difference between I and V).
> >
> > Also, I am measuring power from an outlet, so the waveform will always
be
> > sinusoidal and 60Hz.
>
>
>
>
>
A current transformer is a weird beast; it behaves like a current source on
the secondary when reasonably loaded. If the volt/amp numbers in the
Digikey catalog are correct, it appears that the major losses in their
TE1025-ND is ~2K secondary shunt, (it's 1000/1). This implies that 1.4
volts to start conduction of 2 diodes would be developed at 0.7 amps
primary. A diode bridge loaded with 200 ohms should provide ~0.2 volts /amp,
which is 5 volts at 25 amps. This could be peak detected with a cap on the
bridge output. Better performance could be achieved with a higher quality
(bigger), transformer, but Digikey didn't list any. BTW, a full bridge is
needed to keep DC out of the transformer to prevent saturation.
The IC that Steve Premena suggested looks like an excellent, more complex,
approach, but a current transformer should still be used for current
sampling and line isolation.
Just some thoughts,
Regards,
Ray McArthur
> I agree that there is a V/I curve for any diode. Looking at this curve,
what
> do you think the error would be for say a voltage swing of 5 volts and a
> load impedance of >1M? Using an ADC0831 or LTC1298 as the A/D converter
<craig.lefkowitz@m...> writes:
> I am attempting to build a Watt Meter. I have some basic questions
> regarding the principles of the hardware design.
>
> I was planning to measure the voltage across a resistor in series
> with a
> load in order to figure out the current flowing to the load.
> Knowing the
> current, I can figure out the power. The voltage across the
> resistor would
> be fed to an A/D converter which would be hooked up to a BSII.
> Problem is
> the voltage across the resistor will be AC. I believe I need to get
> the rms
> voltage across the resistor in order to get a true measure of the
> power. If
> this is so, how will I be able to get an rms voltage?
Ckeck out the AD637 from Analog Devices. This is a true RMS to DC
convertor. Kinda pricey. The last bunch we bought at work set us back
$44 each. Jameco has the older AD536 for $16 -- it may work just as well
for your application as the AD637. These chips are essentially the guts
behind Ohio Semitronics voltage and current xducers.
You too Craig. Ray makes some great and very valid points. As far as the
peak detector is concerned, a couple of op amps would improve the detection
and provide another possible method of scaling. Horowitz and Hill go into
great detail about the use of the op amps and the single diode peak
detector. The diode is after the first and before the second op amp so as to
provide a high impedance input as well as to increase the minimum operating
voltage of the entire peak detector circuit (formerly limited by the diode.
I guess a great question would be to have Craig enlighten us helpers on
approximately what is the total design objective. While loving to exchange
ideas with others, I would like to be able to properly suggest a circuit
that would fill the bill for Craig. I guess that I am as guilty, if not
more, than others for making asumptions as to what all of the design
objectives really are. I have to check out the curent transformers from
digi-key, I have to use something to monitor curent in the bias and filament
voltages to both a klystron and thyratron operating at about the 25 MW (not
mW) level in the newest pulse power unit that we are constructing to provide
drive RF to a liniar accelerator. We have specified current transformers for
the readbacks however not which ones we will be using. I will talk more
about it if you are interested.
Best regards,
Gary
g.shearer@v...
Free Electron Laser Research Center
Vanderbilt University
Nashville, Tennessee
Original Message
From: "rjmca" <rjmca@w...>
To: <basicstamps@egroups.com>
Sent: Friday, July 28, 2000 5:33 PM
Subject: Re: [noparse][[/noparse]basicstamps] AC Voltmeter
> Gary:
> A current transformer is a weird beast; it behaves like a current source
on
> the secondary when reasonably loaded. If the volt/amp numbers in the
> Digikey catalog are correct, it appears that the major losses in their
> TE1025-ND is ~2K secondary shunt, (it's 1000/1). This implies that 1.4
> volts to start conduction of 2 diodes would be developed at 0.7 amps
> primary. A diode bridge loaded with 200 ohms should provide ~0.2 volts
/amp,
> which is 5 volts at 25 amps. This could be peak detected with a cap on
the
> bridge output. Better performance could be achieved with a higher quality
> (bigger), transformer, but Digikey didn't list any. BTW, a full bridge is
> needed to keep DC out of the transformer to prevent saturation.
>
> The IC that Steve Premena suggested looks like an excellent, more complex,
> approach, but a current transformer should still be used for current
> sampling and line isolation.
>
> Just some thoughts,
> Regards,
> Ray McArthur
>
> > I agree that there is a V/I curve for any diode. Looking at this curve,
> what
> > do you think the error would be for say a voltage swing of 5 volts and a
> > load impedance of >1M? Using an ADC0831 or LTC1298 as the A/D converter
>
>
>
>
>
>clipped from Craigs query
> >> I am attempting to build a Watt Meter. I have some basic questions
> >> regarding the principles of the hardware design.
>
>Hi:
>
>You're missing a very basic element here. It's quite possible to have 25
>amps flowing and lots of voltage and have nearly ZERO watts!!
>
>You must consider the phase angle between the E and I even if you choose to
>ignore the true waveshape and assume a pure sine wave. The guru Don
>Lancaster has addressed this item a number of times in his columns. I'd
>suggest doing watt ;-) I did. Went to the power company and asked them for
>a discarded
>mechanical wattmeter like they use on your house.
>
>Wayne
Just my 2 cents -
Or alternatively there are pre-packaged units which might be eplored. E-Mon
is one such solid state unit, ordinarily used for sub-metering:
[noparse][[/noparse] http://www.emon.com/ ]
Hope that gives you some idea as to what commerical unit offerings look
like. No comparison presumed or inplied; just by way of example.
Regards,
Bruce Bates
>Wayne Roderick P.E. (EE) (ret) (NMRA life-1721)
>CEO, Teton Short Line, Pocatello Idaho, USA
>http://www.ida.net/users/tetonsl/railroad
>Note new URL- /biz/ becomes /users/
>e-mail tetonsl@i...
measure 110 AC RMS. I used an RMS DC converter with an output of 1 volt being
max. This limits the input resolution to 1 volt to the stamp. Using 1 volt
brings the resolution to less than .001 volt DC. Would be alot better if it used
5 volts. Anybody know a different part or another way to get the results?
Jeff
[noparse][[/noparse]Non-text portions of this message have been removed]