Chip with simple program for Toy

In article <69s4h7F34hffqU1@mid.individual.net>,
"Phil Allison" <philallison@tpg.com.au> wrote:

"krw" = just another autistic cunthead


Fuck off and die

- you festering pile of sub human excrement.




.... Phil
I guess we are blessed with another middle school misfit.
 
"krw" <krw@att.bizzzzzzzzzz> wrote in message
news:MPG.22a276c257ca4922989c6e@news.individual.net...
In article <483818ea$0$31724$4c368faf@roadrunner.com>,
daestrom@NO_SPAM_HEREtwcny.rr.com says...

How about a precision resistor placed inside a calorimeter :) The heat
dissipated is a direct measure of RMS current squared.

Calorimetry is notoriously difficult. Ask Pons and Fleishman. ;-)

Wasn't this done with some high-freq RF stuff at one time? Measured the
temperature or wattage of a resistor or something like that?

Yes, I used an HP true RMS voltmeter when I was in college. It was
a marvelously expensive widget and they didn't like mere students
using it. ;-) It's likely the best way to measure true RMS voltage
at high frequency. I suppose you could read current with the same
meter. ;-)
I worked on some true RMS panel meters many years ago, and they used a
heater and a thermocouple inside a glass bulb. There was a considerable
time delay to get an accurate measurement, and there was some ambient
temperature compensation required, but it was very accurate and worked at
DC to RF (at least to several MHz). But these sensors were rather expensive
and fragile.

A friend and I tried to design a true RMS meter using a lamp and a
photocell, but we found that there was a considerable aging effect on the
output of the lamp, and there was also some ambient temperature error that
needed compensation. I think we were finally able to solve the stability
and aging problems by using two lamp/photocell pairs in a sort of bridge
circuit, where one was driven by the measured signal and the other was
driven by a DC signal that also drove the meter. But the current draw for
both elements was more than the allowed specification, as it was a
self-contained meter with a range of about 4-8 VRMS.

For a bench instrument, such a method would be very practical. You just
need to make two well-matched lamp/photocell pairs, and some signal
conditioning, amplification, and limiting for the lamp, and then use an
op-amp or other means to drive the other sensor so that the photocell
outputs are identical. Then the DC current in the DC sensor matches the
true-RMS current in the other, for any waveform or frequency. But it does
have a rather narrow range of operation. You can't get very close to zero.
Probably 20% to 100% would be possible. The lamp must be driven hard enough
to become incandescent and be sensed by the photocell, which has a limited
spectral range. The heater/thermocouple could go lower, but becomes
relatively insensitive, because heat is proportional to I^2.

Paul
 
"Salmon Egghead = Fuckwit "


** Go drop dead - septic cunt.


...... Phil



"krw" = just another autistic cunthead


Fuck off and die

- you festering pile of sub human excrement.


.... Phil

I guess we are blessed with another middle school misfit.
 
"krw" <krw@att.bizzzzzzzzzz> wrote in message
news:MPG.22a2b088d0007291989c78@news.individual.net...
In article <69s4h7F34hffqU1@mid.individual.net>,
philallison@tpg.com.au says...


"krw" = just another autistic cunthead


Fuck off and die

- you festering pile of sub human excrement.

I bet you say that to all the guys.

--
Keith
I'm convinced that Phil has NTS (Newsgroup Tourette Syndrome). If it wasn't
so damn entertaining I'd get him some help.

Bob
--
== NOTE: I automatically delete all Google Group posts due to uncontrolled
SPAM ==
 
"cloudguitar"

I am trying to build a simple half-wave CW generator (aka Villard
cascade) for my Physics class.

At the moment, I have 4 stages composed of 1.5 nF capacitors (rated
for 2kV) and diodes with an internal resistance of 500ohm, connected
to a 12V 50 Hz AC supply.

** What is the impedance of a 1.5nF cap at 50Hz ????

The answer is in megohms.



....... Phil
 
"cloudguitar"

I am trying to build a simple half-wave CW generator (aka Villard
cascade) for my Physics class.

At the moment, I have 4 stages composed of 1.5 nF capacitors (rated
for 2kV) and diodes with an internal resistance of 500ohm, connected
to a 12V 50 Hz AC supply.

** What is the impedance of a 1.5nF cap at 50Hz ????

The answer is in megohms.



....... Phil
 
"cloudguitar"
"Phil Allison"
At the moment, I have 4 stages composed of 1.5 nF capacitors (rated
for 2kV) and diodes with an internal resistance of 500ohm, connected
to a 12V 50 Hz AC supply.

** What is the impedance of a 1.5nF cap at 50Hz ????

The answer is in megohms.
I would use the CW to charge some other capacitors. Therefore, a low
current output shouldn't be great a problem.

** Very stupid idea.

Each cap in the multiplier sees only a fraction of the output voltage - so
you can use readily available caps.




...... Phil
 
"cloudguitar"
"Phil Allison"
At the moment, I have 4 stages composed of 1.5 nF capacitors (rated
for 2kV) and diodes with an internal resistance of 500ohm, connected
to a 12V 50 Hz AC supply.

** What is the impedance of a 1.5nF cap at 50Hz ????

The answer is in megohms.
I would use the CW to charge some other capacitors. Therefore, a low
current output shouldn't be great a problem.

** Very stupid idea.

Each cap in the multiplier sees only a fraction of the output voltage - so
you can use readily available caps.




...... Phil
 
In article <MPG.22a34f726b9532dd989c79@news.individual.net>, krw <krw@att.bizzzzzzzzzz> wrote:
In article <69sdsuF34dl0sU1@mid.individual.net>,
philallison@tpg.com.au says...

"Salmon Egghead = Fuckwit "


** Go drop dead - septic cunt.


I think that means he loves you.

Mayb -- just maybe -- if everybody would stop responding to Phyllis, he'd go
find some other sandbox to poop in.
 
In article <MPG.22a34f726b9532dd989c79@news.individual.net>, krw <krw@att.bizzzzzzzzzz> wrote:
In article <69sdsuF34dl0sU1@mid.individual.net>,
philallison@tpg.com.au says...

"Salmon Egghead = Fuckwit "


** Go drop dead - septic cunt.


I think that means he loves you.

Mayb -- just maybe -- if everybody would stop responding to Phyllis, he'd go
find some other sandbox to poop in.
 
Don Kelly wrote in message <595_j.164097$Cj7.146390@pd7urf2no>...
----------------------------
"krw" <krw@att.bizzzzzzzzzz> wrote in message
news:MPG.22a276c257ca4922989c6e@news.individual.net...
In article <483818ea$0$31724$4c368faf@roadrunner.com>,
daestrom@NO_SPAM_HEREtwcny.rr.com says...

"Phil Allison" <philallison@tpg.com.au> wrote in message
news:69pnc3F33qck8U1@mid.individual.net...

"Don Kelly"

Bob Eld and Phil Allison have dealt with this topic except for the
question "how did you measure the rms voltage?".
If you are using a true rms meter which samples at a high enough
rate-
then you are OK.

** ?????????

The " true rms" function on a DMM is *not* obtained by sampling. It
is
normally obtained by the use of a " true rms to DC" converter IC -
generally one made by Analog Devices like the AD736.
-----------
If it is not sampling- then what is it? Thermal? There is a little black
box in the device which is just that- a little black box.
I presently have problems getting the applicable pdf from the Analog device
site.
--

Don Kelly dhky@shawcross.ca
remove the X to answer



I've got an old AD catalog with "little black boxes" in it. From what I
remember the cat has precision instumentation amps, log converters and ???.
If you have a part number, I'll see if it's in there.
bg
 
Don Kelly wrote in message <595_j.164097$Cj7.146390@pd7urf2no>...
----------------------------
"krw" <krw@att.bizzzzzzzzzz> wrote in message
news:MPG.22a276c257ca4922989c6e@news.individual.net...
In article <483818ea$0$31724$4c368faf@roadrunner.com>,
daestrom@NO_SPAM_HEREtwcny.rr.com says...

"Phil Allison" <philallison@tpg.com.au> wrote in message
news:69pnc3F33qck8U1@mid.individual.net...

"Don Kelly"

Bob Eld and Phil Allison have dealt with this topic except for the
question "how did you measure the rms voltage?".
If you are using a true rms meter which samples at a high enough
rate-
then you are OK.

** ?????????

The " true rms" function on a DMM is *not* obtained by sampling. It
is
normally obtained by the use of a " true rms to DC" converter IC -
generally one made by Analog Devices like the AD736.
-----------
If it is not sampling- then what is it? Thermal? There is a little black
box in the device which is just that- a little black box.
I presently have problems getting the applicable pdf from the Analog device
site.
--

Don Kelly dhky@shawcross.ca
remove the X to answer



I've got an old AD catalog with "little black boxes" in it. From what I
remember the cat has precision instumentation amps, log converters and ???.
If you have a part number, I'll see if it's in there.
bg
 
In article <MPG.22a3c36b27c95ab2989c7a@news.individual.net>,
krw@att.bizzzzzzzzzz says...
In article <69uaidF34vba9U1@mid.individual.net>,
philallison@tpg.com.au says...

"Doug Miller"


** Go drop dead - septic cunt.


NO, Doug. He'd just start repeating himself.
Yeah, he's not real imaginative, is he?
 
The RMS voltage of a true and symtrical square wave is just the voltage from
ground of the "top" of the wave.

A "true RMS" meter will measure this accurately.

However, most AC voltmeters don't respond to true RMS but something else
like "average of absolute voltage" or "peak". They are calibrated in RMS
using the assumption that the wave form is a sine wave.


** Posted from http://www.teranews.com **
 
"cloudguitar" <cloudguitar@gmail.com> wrote in message
news:1c4ca591-e62c-42b5-a2d2-f47fd624ea4a@2g2000hsn.googlegroups.com...
Hi all,

I am trying to build a simple half-wave CW generator (aka Villard
cascade) for my Physics class.

At the moment, I have 4 stages composed of 1.5 nF capacitors (rated
for 2kV) and diodes with an internal resistance of 500ohm, connected
to a 12V 50 Hz AC supply.

I simulated the circuit with Qucs and, as I was expecting from
calculations, the no-load output would have been around 100V DC. What
happens is that my multimeter reads 2.5V DC, instead.

I suspect I am doing something wrong with voltage probing. Referring
to this circuit schema:

http://upload.wikimedia.org/wikipedia/en/4/4c/Cockcroft_Walton_Voltage_multi
plier.png
I connected the red probe to the output of the last stage (the circle
in the image) and the black probe to where the ground should be in the
schema. The circuit is open, and the voltmeter should close it by
connecting the last stage to the generator - also, there is no
connection with the AC supply ground.

Unfortunately, I am a newbie and I cannot figure out what is
happening.
Any idea?

Thank you very much for your time,

Claudio A.
The reactance of a 1.5nF cap a 50 Hz is 2.1 MOhms. Of course, the caps all
charge in series so the impedance out at the end is very high. Reverse
current in the diodes plus you meter's impedance loads the circuit down
giving the output you observe. Also the forward voltage drop of the diodes
adds and reduces the output by that amount.

You simply must use larger capacitors at 50 Hz or use a higher frequency. If
you are starting with 12 Volts, you can use electrolytic caps of several
microfarads. The thing you need to determine is what voltage are you trying
to multiply and how much current do you want to deliver?

The reactance of the capacitors must be consistent with the max. current
because of the voltage drop it creates.

Keep in mind that this type of voltage multiplier is usually used for very
high voltages, I've used them at 200kV. With high voltage, a few volts
across the caps is totally insignificant. At 12 volts it's a different
story.
 
"cloudguitar" <cloudguitar@gmail.com> wrote in message
news:1c4ca591-e62c-42b5-a2d2-f47fd624ea4a@2g2000hsn.googlegroups.com...
Hi all,

I am trying to build a simple half-wave CW generator (aka Villard
cascade) for my Physics class.

At the moment, I have 4 stages composed of 1.5 nF capacitors (rated
for 2kV) and diodes with an internal resistance of 500ohm, connected
to a 12V 50 Hz AC supply.

I simulated the circuit with Qucs and, as I was expecting from
calculations, the no-load output would have been around 100V DC. What
happens is that my multimeter reads 2.5V DC, instead.

I suspect I am doing something wrong with voltage probing. Referring
to this circuit schema:

http://upload.wikimedia.org/wikipedia/en/4/4c/Cockcroft_Walton_Voltage_multi
plier.png
I connected the red probe to the output of the last stage (the circle
in the image) and the black probe to where the ground should be in the
schema. The circuit is open, and the voltmeter should close it by
connecting the last stage to the generator - also, there is no
connection with the AC supply ground.

Unfortunately, I am a newbie and I cannot figure out what is
happening.
Any idea?

Thank you very much for your time,

Claudio A.
The reactance of a 1.5nF cap a 50 Hz is 2.1 MOhms. Of course, the caps all
charge in series so the impedance out at the end is very high. Reverse
current in the diodes plus you meter's impedance loads the circuit down
giving the output you observe. Also the forward voltage drop of the diodes
adds and reduces the output by that amount.

You simply must use larger capacitors at 50 Hz or use a higher frequency. If
you are starting with 12 Volts, you can use electrolytic caps of several
microfarads. The thing you need to determine is what voltage are you trying
to multiply and how much current do you want to deliver?

The reactance of the capacitors must be consistent with the max. current
because of the voltage drop it creates.

Keep in mind that this type of voltage multiplier is usually used for very
high voltages, I've used them at 200kV. With high voltage, a few volts
across the caps is totally insignificant. At 12 volts it's a different
story.
 

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