Chip with simple program for Toy

[Bob Eld]

"panfilero" <panfilero@gmail.com> wrote in message
news:f1496677-7937-43e6-9bbd-386d103b675d@p39g2000prm.googlegroups.com...

I was wondering if I take the RMS voltage of a messy looking square
wave, a noisy square wave, with some measurement device, and I want to
know the regular current of the thing can I take the RMS current and
divide it by the square root of 2 in order to get my regular current?

I know you can do this for sinusoidal ac signals but don't know if the
math still works out the same for other signals...

PS - I got my RMS current value by dividing the RMS voltage value by a
known resistance I have in my circuit

Much Thanks for any responses I get

Yes the RMS current is the RMS voltage devided by R.

No, the "regualr curent" What ever that is, is not the RMS current divided
by the square root of two. With a sign wave, the peak current divided by the
the square root of two is the RMS current, but ONLY for a sign wave.

With any other waveform, the the square root of two relation does NOT hold.
It must be calculated from first principles or measured with a true RMS
meter.

The equation for the RMS current is: Irms = ((int i(t)^2 dt )/ T)^1/2
Where i(t) is the current as a function of time, T is the time over which
the integral is taken, ( 0 to T ), usually one period of the wave.

That is, in words, the RMS current is the square root of the average time
integral of the current squared.

For a square wave of current Ip going plus and minus over a period of T the
RMS value is:

Irms =( ( (Ip^2 *T/2 + (-Ip)^2 * T/2)) / T)^1/2 = ( (Ip ^2)/2 +
(Ip^2)/2)^1/2 = (Ip ^2)^1/2

Irms = Ip, The RMS equals the peak value for a square wave. It very much
more complicated if noise is present.

 

[Bob Eld]

"BobG" <bobgardner@aol.com> wrote in message
news:d119b676-9277-4adc-af0c-39c002bbc8a8@w7g2000hsa.googlegroups.com...

The RMS of an on and off wave is the average. Try it with 1V and 0V.
Then you have to believe that it also works with 5v and 0v.

Thats right. In the example I gave the wave was plus and minus, not plus and
zero.

 

[Bob Eld]

"BobG" <bobgardner@aol.com> wrote in message
news:d119b676-9277-4adc-af0c-39c002bbc8a8@w7g2000hsa.googlegroups.com...

The RMS of an on and off wave is the average. Try it with 1V and 0V.
Then you have to believe that it also works with 5v and 0v.

Thats right. In the example I gave the wave was plus and minus, not plus and
zero.

 

[Phil Allison]

"BobG"

The RMS of an on and off wave is the average.

** Completely wrong !!

RMS current or voltage follow the square root of the duty cycle.

Eg: with a duty cycle of 0.5 ( square wave) the RMS value is 0.7071 of the
peak.



...... Phil

 

[Bob Eld]

"Phil Allison" <philallison@tpg.com.au> wrote in message
news:69pdnvF344snbU1@mid.individual.net...

"BobG"

The RMS of an on and off wave is the average.


** Completely wrong !!

RMS current or voltage follow the square root of the duty cycle.

Eg: with a duty cycle of 0.5 ( square wave) the RMS value is 0.7071 of
the
peak.



..... Phil

Vrms = ((vp^2) *T/2 + 0) *1/T) ^1/2 = ((Vp ^2) / 2 )^1/2 = Vp/ (2 ^1/2)...

Yep, for 50% duty cycle, Vrms = Vp/ (sqr 2) if the other half of the wave
is zero.

 

[Phil Allison]

"Bob Eld"
"Phil Allison"

"BobG"

The RMS of an on and off wave is the average.


** Completely wrong !!

RMS current or voltage follow the square root of the duty cycle.

Eg: with a duty cycle of 0.5 ( square wave) the RMS value is 0.7071 of
the peak.



Vrms = ((vp^2) *T/2 + 0) *1/T) ^1/2 = ((Vp ^2) / 2 )^1/2 = Vp/ (2
^1/2)...

Yep, for 50% duty cycle, Vrms = Vp/ (sqr 2) if the other half of the wave
is zero.

** Keep it simple:

For a rectangular wave,

V (or I) rms = sq.rt Duty Cycle ( expressed as a decimal fraction )

For a bi-polar, rectangular wave - first rectify the wave.



...... Phil

 

[Salmon Egg]

In article
<d119b676-9277-4adc-af0c-39c002bbc8a8@w7g2000hsa.googlegroups.com>,
BobG <bobgardner@aol.com> wrote:

The RMS of an on and off wave is the average. Try it with 1V and 0V.
Then you have to believe that it also works with 5v and 0v.

IIRC, although it is easy to check, the rms voltage of a series of
rectangular pulses will be the square root of the peak current times the
average current.

Bill

 

[Don Kelly]

----------------------------
"BobG" <bobgardner@aol.com> wrote in message
news:a0bfd275-76bf-4a30-8aee-fb158186b6bd@34g2000hsh.googlegroups.com...
On May 23, 7:11?pm, krw <k...@att.bizzzzzzzzzz> wrote:

The average
of the voltages doesn't tell you much (the average of your wall
outlet is zero).
============================================

Not after sampling and squaring. The power is all positive, like the
squared samples.

-----
But the average of the squared values is not the average value of the
voltage and taking the root of the average of V*2 will not, in general, give
you the average of V.

It will give the rms value which was devised originally as a way to get an
"equal Power" to DC with the same load- as krw indicates.

--

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

 

[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.
If you are using a cheaper multimeter on the normal "rms" scale, then it
simply assumes that the AC measurement is sinusoidal and scales from the DC
average accordingly and will not give the correct value except for a
sinusoid or for an On/Off cycle with equal on and off periods.

--

Don Kelly dhky@shawcross.ca
remove the X to answer
----------------------------
"panfilero" <panfilero@gmail.com> wrote in message
news:f1496677-7937-43e6-9bbd-386d103b675d@p39g2000prm.googlegroups.com...

I was wondering if I take the RMS voltage of a messy looking square
wave, a noisy square wave, with some measurement device, and I want to
know the regular current of the thing can I take the RMS current and
divide it by the square root of 2 in order to get my regular current?

I know you can do this for sinusoidal ac signals but don't know if the
math still works out the same for other signals...

PS - I got my RMS current value by dividing the RMS voltage value by a
known resistance I have in my circuit

Much Thanks for any responses I get

 

[Phil Allison]

"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 you are using a cheaper multimeter on the normal "rms" scale, then it
simply assumes that the AC measurement is sinusoidal and scales from the
DC average accordingly and will not give the correct value except for a
sinusoid or for an On/Off cycle with equal on and off periods.

** Most DMMs use a precision rectifier circuit followed by RC averaging of
the output and then DC scaling so that the final reading equals the rms
value of a sine wave input ** within** the particular meter's specified
frequency range.

Mostly, this range is very narrow - ie from about 30Hz to a mere 1kHz (with
+/- 1% accuracy).

So, unless you KNOW the energy spectrum of your wave falls inside this
narrow band, the reading can be way wrong -ie well below the actual value.

Unfortunately, most " true rms " DMMs have the same frequency range issue -
some of the more expensive ones work out to 20kHz or 100 kHz with good
accuracy.

The only * low cost * solution is to examine the wave on a scope and then
compute the rectified average and rms values - not possible with any
accuracy if the wave is noise like.



...... Phil

 

[ZACK.]

2N3053- npn,s
ive used these as power outputs
dont know the pnp match.
some amps use the same on pushpull.


"Ron Hubbard" <ryon@quik.com> wrote in message
news:4bbeac40-9d4d-4b84-9700-e2b590f0b215@c58g2000hsc.googlegroups.com...

A long while ago Radio Shack once had some power PNP ad NPN
transistors that both came in a TO-5 metal casing. Those were nifty
little transistors. Does anyone remember what those transistors were?

Ron

 

[ZACK.]

2N3053- npn,s
ive used these as power outputs
dont know the pnp match.
some amps use the same on pushpull.


"Ron Hubbard" <ryon@quik.com> wrote in message
news:4bbeac40-9d4d-4b84-9700-e2b590f0b215@c58g2000hsc.googlegroups.com...

A long while ago Radio Shack once had some power PNP ad NPN
transistors that both came in a TO-5 metal casing. Those were nifty
little transistors. Does anyone remember what those transistors were?

Ron

 

[ian field]

"Dave.H" <the1930s@googlemail.com> wrote in message
news:059b3129-7594-4053-9dcb-77abc768b5bf@q24g2000prf.googlegroups.com...

On May 25, 2:03 am, Don Bowey <dbo...@comcast.net> wrote:
On 5/24/08 2:15 AM, in article
f0c62a65-5604-4e8e-b0b3-2c54f0b69...@x1g2000prh.googlegroups.com,
"Dave.H"

the19...@googlemail.com> wrote:
I found a website that has construction details for a radio using the
MK 484 IC. The site doesn't seem to say what variable capacitor value
to use, I happen to have a fairly new 365 pF unit lying around, that I
could use. I would also want to know if any type of ferrite coil
would work, or would it be better to wind my own?

Thanks, Dave
Australia

The component values depend on what frequency band you wish to tune.

Here's a link that may help you.

http://www.rapidonline.com/resources/docs/82-1026.pdf

Note that the MK484 IC is rated only to 3 MHz.

I want to tune the AM band 530 kHz too at least 1500 kHz. There's not
too many AM stations in the high end of the band, (highest one is at
1431 kHz, the one I usually listen to) so I'm not worried about the
radio tuning any higher than 1500 kHz

Pinch the tuning cap and ferrite rod out of an old transistor radio.

 

[Paul E. Schoen]

"Jamie" <jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote in message
news:lkZZj.1748$IF3.151@newsfe07.lga...

cloudguitar wrote:

On May 24, 8:32 pm, Jamie
jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:

cloudguitar wrote:

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_Voltag...
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.

Is your meter a 10 meg ohm input type ?
also, 1.5 nf is small for 50 hz. you may want to
try larger caps..
Also, you don't need 2kv caps at the voltage you're
starting at.

http://webpages.charter.net/jamie_5"


Thanks for your reply.

yes, my meter is a 10MOhm multimeter. I was planning to add 4 stages
(reaching 8 stages) and to switch to a 230V 50Hz supply, so 2kV should
be barely enough.
Do you think that I should try with 230V right now?

I also have 15nF @ 3kV caps, do you think the would be better suited?
I would try the larger caps first.

Also, check the forward voltage drop of your diodes. The 500 ohm resistance
figure you give does not make sense. Maybe you mean 500 mV at 1 mA, but if
you have high voltage diodes they may be several junctions in series, which
would explain a low output with 12 VAC.

Paul

 

[Paul E. Schoen]

"Jamie" <jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote in message
news:lkZZj.1748$IF3.151@newsfe07.lga...

cloudguitar wrote:

On May 24, 8:32 pm, Jamie
jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:

cloudguitar wrote:

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_Voltag...
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.

Is your meter a 10 meg ohm input type ?
also, 1.5 nf is small for 50 hz. you may want to
try larger caps..
Also, you don't need 2kv caps at the voltage you're
starting at.

http://webpages.charter.net/jamie_5"


Thanks for your reply.

yes, my meter is a 10MOhm multimeter. I was planning to add 4 stages
(reaching 8 stages) and to switch to a 230V 50Hz supply, so 2kV should
be barely enough.
Do you think that I should try with 230V right now?

I also have 15nF @ 3kV caps, do you think the would be better suited?
I would try the larger caps first.

Also, check the forward voltage drop of your diodes. The 500 ohm resistance
figure you give does not make sense. Maybe you mean 500 mV at 1 mA, but if
you have high voltage diodes they may be several junctions in series, which
would explain a low output with 12 VAC.

Paul

 

[Phil Allison]

"krw"
Phil Allison
"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.

Slick, though not perfect.

** Err - so just like you ?

If you are using a cheaper multimeter on the normal "rms" scale, then it
simply assumes that the AC measurement is sinusoidal and scales from
the
DC average accordingly and will not give the correct value except for a
sinusoid or for an On/Off cycle with equal on and off periods.


** Most DMMs use a precision rectifier circuit followed by RC averaging
of
the output and then DC scaling so that the final reading equals the rms
value of a sine wave input ** within** the particular meter's specified
frequency range.

Normally <> is not

** ?????

It *is* used, though perhaps not in handheld DVMs.

** Bollocks .

Mostly, this range is very narrow - ie from about 30Hz to a mere 1kHz
(with
+/- 1% accuracy).

So, unless you KNOW the energy spectrum of your wave falls inside this
narrow band, the reading can be way wrong -ie well below the actual
value.

Unfortunately, most " true rms " DMMs have the same frequency range
ssue -
some of the more expensive ones work out to 20kHz or 100 kHz with good
accuracy.

The only * low cost * solution is to examine the wave on a scope and then
compute the rectified average and rms values - not possible with any
accuracy if the wave is noise like.

How do you get better than 1% accuracy on a scope? Sampling works
to any accuracy one desires and can go fairly high in frequency.
Even higher if the waveform is repetitive.

** You are not paying attention to the point at issue - fuckwit.

BTW:

Fix your stupid settings so the name of the poster you arr replying to is
left visible.



....... Phil

 

[Phil Allison]

"daestrom"

*PROVIDED* you measured that voltage with a true RMS voltmeter, then yes,
you have the true RMS current.

** It ain't that simple.

The vast majority of so called " true rms " AC ranges fitted to DMMs are
of very limited bandwidth - typically 30Hz to 1kHz within 1% accuracy, a few
more expensive examples go to 20 kHz or even 100kHz.

So, unless the voltage wave under test falls within the particular meter's
bandwidth, the reading will be in error and likely very seriously so.



...... Phil

 

[Phil Allison]

"krw" = just another autistic cunthead


Fuck off and die

- you festering pile of sub human excrement.




..... Phil

 

[Don Kelly]

----------------------------
"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

 

[Phil Allison]

"Don Kelly"

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.

** Analog Devices have been supplying " true rms to DC converter " ICs for
about 30 years.

The data sheet for the AD636 will give you some idea how it computes the DC
value of an AC input.

http://www.farnell.com/datasheets/57885.pdf

Use Google for more info.

Don't be so damn lazy.



...... Phil