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Hello everyone,
I have downloaded this schematic from Circuit Cellar's website.
It is a wattmeter, and uses a 16C71 for A/D conversion.
Here goes the question:
Can anyone explain me how I and V are acquired ?
Correct me if I4m wrong:
I suppose the leds in the optocuplers, bright proportionally to V
aplied,
and I consumption, so the base of each transistor inside the optos will be
excited proportionally to I and V, etc, etc.
Is it something like this ??
Thanks a lot,
Gabriel.-
_________________________________________
Gabriel
ICQ#: 154107558
More ways to contact me: http://wwp.icq.com/154107558
_________________________________________
I'm not an EE, but I believe they are measuring the drop across shunt
(the .001 ohm resistor. Using that, and knowing what the voltage
supplied is (line voltage is a standard), you can calculate wattage. My
guess is that this circuit would need changing for use in Europe
(Circuit Cellar being an American publication I believe).
I wonder if the ADC could be replaced with a PIC with an internal ADC.
Flash PICs with ADC are plentiful now...That said, I don't know the
specs on that ADC, and perhaps it was chosen for a particular reason.
Josh
--
A common mistake that people make when trying to design something
completely foolproof is to underestimate the ingenuity of complete
fools.
-Douglas Adams
Gabriel Caffese wrote:
> Can anyone explain me how I and V are acquired ?
> Correct me if I4m wrong:
> I suppose the leds in the optocuplers, bright proportionally to V
> aplied,
> and I consumption, so the base of each transistor inside the optos will be
> excited proportionally to I and V, etc, etc.
> Is it something like this ??
Josh stated that 'line voltage is a standard'. The only thing that is
standard is line hertz: USA:60, Eroupe:50. The line voltage is different
at different location on the grid.If you turn on a heater in the house,
the line voltage will drop for another part of the house.
Also, power factor is important thing to measure. true power = E * I * PF.
So for 120v , 20 amp , PF .5 = 1,200 watts
Side Note: the power company needs to generate 20 amps, even though you
are using 10 amps.
> I'm not an EE, but I believe they are measuring the drop across shunt
> (the .001 ohm resistor. Using that, and knowing what the voltage
> supplied is (line voltage is a standard), you can calculate wattage. My
> guess is that this circuit would need changing for use in Europe
> (Circuit Cellar being an American publication I believe).
>
> I wonder if the ADC could be replaced with a PIC with an internal ADC.
> Flash PICs with ADC are plentiful now...That said, I don't know the
> specs on that ADC, and perhaps it was chosen for a particular reason.
>
> Josh
> --
> A common mistake that people make when trying to design something
> completely foolproof is to underestimate the ingenuity of complete
> fools.
> -Douglas Adams
>
> Gabriel Caffese wrote:
> > Can anyone explain me how I and V are acquired ?
> > Correct me if I4m wrong:
> > I suppose the leds in the optocuplers, bright proportionally to V
> > aplied,
> > and I consumption, so the base of each transistor inside the optos will be
> > excited proportionally to I and V, etc, etc.
> > Is it something like this ??
>
> --
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>
Well, I'd expect the best description to be in the accompanying
article, but here's a stab at it.
I believe they're using the two optos on the left to measure the
voltage. There anti-parallel each other so you can measure both phases
of AC. The optos on the right are measuring the voltage again after
the sense resistor. The sense value of 0.001 ohms gives one volt per
1000 amps (obviously the 14 ga wire used as a sense resistor would
light up like a light bulb well before we reach 1000 amps). The two
sense voltages are added through a high gain amp and buffered.
I haven't looked up the spec sheet, but since the PIC16C71 has onboard
A/D converters, my guess is that they choose the AD0831 for some other
reason. (precision, range, etc)
Considering they used a 25 watt resistor for MCLR and 9 volt voltage
dividers, I'm guessing there's a few errors on this schematic. In
light of the potentially lethal consequences of getting this circuit
wrong, I'd at least find/read the original article and possibly
contact the author for some clarifications.
Thank you very much for this description.
The complete pdf does not say anything about the principle of function, and
the author's email address is not valid anymore........
Well, I'd expect the best description to be in the accompanying
article, but here's a stab at it.
I believe they're using the two optos on the left to measure the
voltage. There anti-parallel each other so you can measure both phases
of AC. The optos on the right are measuring the voltage again after
the sense resistor. The sense value of 0.001 ohms gives one volt per
1000 amps (obviously the 14 ga wire used as a sense resistor would
light up like a light bulb well before we reach 1000 amps). The two
sense voltages are added through a high gain amp and buffered.
I haven't looked up the spec sheet, but since the PIC16C71 has onboard
A/D converters, my guess is that they choose the AD0831 for some other
reason. (precision, range, etc)
Considering they used a 25 watt resistor for MCLR and 9 volt voltage
dividers, I'm guessing there's a few errors on this schematic. In
light of the potentially lethal consequences of getting this circuit
wrong, I'd at least find/read the original article and possibly
contact the author for some clarifications.
Why did the author choose a external ADC.... ---->
"
I initially chose the PIC16C71
because of its low cost and onboard
four-channel eight-bit ADC. However,
you can save even more by replacing the
’16C71 ($12.30) with a separate ADC,
National’s ADC0831 ($3.29), in combination
with the ADC-less PIC16C61
($6.15).You save $2.86, but you need
space for one more 8-pin DIP.
"
This circuit is MUCH cleverer than the descriptions so far have given it
credit for.
It implements a true four quadrant multipler.
Note that the 25w resistors are not major errors - they are meant to be
0.25w and the decimal point has become hard to see. A good example of why
leading zeroes should be used.
Sadly there doesn't seem to be any trace on the web of the original article.
its very sad that the cct Cellar article didn't provide a proper or even
superficial description as this is the real genius of the design.
Consider what happens when mains is at full positive half cycle into "hot"
terminal.
Top left and bottom right optos will be on driven by same 100k resistor R7.
If there is no current flow then both optos are driven equally and the R5/R6
midpoint is maintained at half supply. The U5A integrator is not driven off
midpoint.
The integrator is "leaky" and is effectively an amplifier with a gain of
between about 15 and 200 but with its response smoothed by the 4.7uF C5.
Now, if current is flowing the rh end of R8 is more positive than left hand
end and there will be less voltage across the bottom left opto than across
the top left opto. The midpoint of R5/R6 is driven towrds ground and the
integrator ramaps positively. The amount of midpoint offset is controlled by
the product of the drive curret - provided by the input mains voltage, and
the amount by which this voltage is offset which is caused by the current
flow. Increasing either voltage or current will linearly increase the offset
and thus the integrator drive.
When the voltage reverses the opposite diagonl of optos conduct. Both
voltage and current sense are swapped in the two optos so a positive product
still occurs.
If voltage and current are out of phase there will be a period when the
product is neagtive and the integrator input will be redcued. Overall a true
VI summing is provided across the cycle. (What it doesn't (seem to) do is
provide an RMS sum as the amplifer/integrator does not track the cycle. I'd
have to think about this further).
The multiplier gives a true (more or less) VI product across the entire
waveform.
I consider this circuit to be utterly brilliant. It may not produce an
overly accurate result but for very very simple implementation it is very
hard to beat for cost. Where are my quad optos ... ? :-)
> I'm not an EE, but I believe they are measuring the drop across shunt
> (the .001 ohm resistor. Using that, and knowing what the voltage
> supplied is (line voltage is a standard), you can calculate wattage. My
> guess is that this circuit would need changing for use in Europe
> (Circuit Cellar being an American publication I believe).
Unfortunately that is not, by itself enough, you also must measure the
phase difference between the V and I waveforms. The reason is power factor.
If you just multiply V and I of an AC power system you get VA, power is
VA*pf (pf = power factor). TTYL
On Thu, 2003-10-23 at 07:34, Russell McMahon wrote:
> Sadly there doesn't seem to be any trace on the web of the original article.
> its very sad that the cct Cellar article didn't provide a proper or even
> superficial description as this is the real genius of the design.
I looked for that too when I originally found this article... Many
people have cited that paper -- it seems to be a popular project in
Engineering faculties. However the online back issues of "Electronic
Design" only go back a few years.
If anyone can find this paper, either online or scanned in, I'd
certainly appreciate a copy.
Apologies to all. I thought my reply was inferior before I sent it,
but it was late, I was tired, blah blah blah, [insert excuse here].
With the 25w resistors, I have to disagree. Considering the decimal
point for the 5.1k resistors is clearly visible, I have to assume this
was a typo in the original schematic. BUT my calculations for R7 show
almost 6 watts, so a 25 for that one isn't really out of line. For the
rest of them I agree that 0.25w should be fine.
> This circuit is MUCH cleverer than the descriptions so far have given it
> credit for.
> It implements a true four quadrant multipler.
>
> Note that the 25w resistors are not major errors - they are meant to be
> 0.25w and the decimal point has become hard to see. A good example of why
> leading zeroes should be used.
>
> Sadly there doesn't seem to be any trace on the web of the original
article.
> its very sad that the cct Cellar article didn't provide a proper or even
> superficial description as this is the real genius of the design.
>
> Consider what happens when mains is at full positive half cycle into "hot"
> terminal.
> Top left and bottom right optos will be on driven by same 100k resistor
R7.
> If there is no current flow then both optos are driven equally and the
R5/R6
> midpoint is maintained at half supply. The U5A integrator is not driven
off
> midpoint.
>
> The integrator is "leaky" and is effectively an amplifier with a gain of
> between about 15 and 200 but with its response smoothed by the 4.7uF C5.
>
> Now, if current is flowing the rh end of R8 is more positive than left
hand
> end and there will be less voltage across the bottom left opto than across
> the top left opto. The midpoint of R5/R6 is driven towrds ground and the
> integrator ramaps positively. The amount of midpoint offset is controlled
by
> the product of the drive curret - provided by the input mains voltage, and
> the amount by which this voltage is offset which is caused by the current
> flow. Increasing either voltage or current will linearly increase the
offset
> and thus the integrator drive.
>
> When the voltage reverses the opposite diagonl of optos conduct. Both
> voltage and current sense are swapped in the two optos so a positive
product
> still occurs.
> If voltage and current are out of phase there will be a period when the
> product is neagtive and the integrator input will be redcued. Overall a
true
> VI summing is provided across the cycle. (What it doesn't (seem to) do is
> provide an RMS sum as the amplifer/integrator does not track the cycle.
I'd {Quote hidden}
> have to think about this further).
>
> The multiplier gives a true (more or less) VI product across the entire
> waveform.
> I consider this circuit to be utterly brilliant. It may not produce an
> overly accurate result but for very very simple implementation it is very
> hard to beat for cost. Where are my quad optos ... ? :-)
>
>
> Russell McMahon
>
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>
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> Gee Russel, quite brillant on your part for working that out!
Thanks.
It might even be right :-)
The man who designed it was very clever.
It's an opto implemenmtation of a "Gilbert cell".
Its main problenm is probably the matching of the optos.
But it should do a reasonable job as a comparison instrument.
RM
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> Gee Russel, quite brillant on your part for working that out!
Thanks.
It might even be right :-)
The man who designed it was very clever.
It's an opto implemenmtation of a "Gilbert cell".
Its main problenm is probably the matching of the optos.
But it should do a reasonable job as a comparison instrument.
RM
Hello Russel,
I think you get the point. There is a basic application note from
National Semiconductor AN265 from february 1984 "Electronic Watt-Hour-Meter"
which uses matched transistor pairs (LM394)for the multiplier (no optos).
Googling for the original article from oct. 1994 gives no usefull result
for me,
but it seems that the author, Steve Woodward is at UNC, Chapel Hill.
Now looking for a hardcopy...
Thomas
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>There is a basic application note from
> National Semiconductor AN265 from february 1984 "Electronic
Watt-Hour-Meter"
> which uses matched transistor pairs (LM394)for the multiplier (no optos).
Thanks.
Google says that here's the original for anyone interested.
Realy nice doc ! I'v learnt a lot, such as this, and I quote
directly from the app note :
"A complete shave, including the 4 areas listed, costs 0.00692 cents/day
or 8.68 cents per year. If this is too high, you can economize
by growing a beard."
:-)
Jan-Erik.
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Great, I have managed to track down the original of the magazine in our
library.
One & a half page article, including circuit. The circuit as given in the
article states it converts the power reading to 1-pulse/joule, (1 Hz/Watt)
frequency output.
...
The principle of operation of the opto-bridge multiplier is explained in an
earlier Idea for Design ("Optical Isolator computes Watts", Electronic
Design, Oct 14, 1994, p102).
...
In the prototype, the 0.002 ohm copper shunt consists of 5.8 inches of 16
gauge copper wire. Coppers 3900 ppm/C co-efficient of resistance serves to
temperature compensate most of the scale factor variation due to LED and
photo-transistor temperature co-efficients. Zero stability (voltage feed
through) of the wattmeter is very sensitive to temperature gradients in the
opto-isolators. They should be carefully shielded from drafts.
He includes a bit of circuitry to send the pulse onto an RS232 connection,
so that it will act as a start bit to a character (the pulse is 400uS wide).
He then includes a QBasic program to detect the rate of characters sent, and
display this. The characters, of course, have no useful data in them, apart
from the rate at which they get sent.
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>The principle of operation of the opto-bridge multiplier is
>explained in an earlier Idea for Design ("Optical Isolator
>computes Watts", Electronic Design, Oct 14, 1994, p102).
Found a copy of this one as well. It is by the same guy. Must take these
away and digest them now.
The only concern I see for these is the non-linearity of the LED versus the
supplied voltage. Not sure how the circuit gets around this.
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"Alan B. Pearce" <EraseMEA.B.PearceRL.AC.UK> wrote:
> The only concern I see for these is the non-linearity of the LED versus
> the supplied voltage. Not sure how the circuit gets around this.
The LEDs are driven by current, not voltage. The 100K resistor makes sure
of that. Optoisolators are fairly linear with respect to current.
Well, it does show a single-package quad optoisolator, but I don't know
whether those are usually monolithic internally. Even if they're not, the
individual pieces can probably be expected to come from the same batch,
maybe even the same wafer.
Does it not use the non-linearity for multiplication. I am busy with a
Electricity meter design at the moment, but I don't think this will be
accurate enough for a product that must be certified.
> Does it not use the non-linearity for multiplication. I am busy with a
> Electricity meter design at the moment, but I don't think this will be
> accurate enough for a product that must be certified.
No - it implements a traditional Gilbert cell (which I stupidly failed to
recognise when I waded through my essentially correct operational
decsription the other day).
The Gilbert Cell in its purest is a pure current mode device but can be
operated with voltage inputs in a voltage input transconductance mode as
long as voltage swings are small. I haven't gone back and thought through
the implications of using optos. OTC it seems that there should be a
reasonable degree of linear current gain between input LED and output
transistor current (CTR) and it may be closer to "real" than using voltage
driven transistors.
*** It's well worth reading Gilbert's description of his invention here if
you haven't seen it already: ***
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