Maker Pro
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60Hz noise making it's way into my test

S

SklettTheNewb

Jan 1, 1970
0
I am NOT an engineer, I'm helping my Dad out with a PC based data
acquisition program to test a product his company makes.

With that little disclaimer out of the way, here is the problem:

I'm reading voltage values of his device, it outputs an AC square wave
signal at 4000Hz. The voltages vary from 0 - 25 Vcc Due to this
variance in voltage, I wasn't able to use the counter on the
acquisition board, so I need to resort to counting rising edges in the
collected voltage data, then determine from that the frequency. This
is all working pretty well.... as long as I power the device being
measured with batteries. If I use the wall adapter, I'm getting a
pattern in the results that repeats at 60Hz. This made me suspect the
power supply so that's when I tried the batteries and it worked
correctly.

So, my question is: What do I need to add to my DC power supply to
remove that 60Hz that is making it's way into the system?

I really have no idea... when I put the power supply on the scope, I
don't see the signal, but sure enough.. it's there somehow as it shows
up in the data I'm capturing.

You can see a screen shot of it here:
http://steventara.dnsdojo.com/images/pattern.gif

Our power supply puts out 3.22 Vdc @ 1.5 Amps

Hopefully this is something that might not be too hard to fix.

best Regards,
Steve Klett
 
J

John Popelish

Jan 1, 1970
0
SklettTheNewb said:
I'm reading voltage values of his device, it outputs an AC square wave
signal at 4000Hz. The voltages vary from 0 - 25 Vcc

Is that 0 to -25 or 0 to +25 volts?
Due to this
variance in voltage, I wasn't able to use the counter on the
acquisition board, so I need to resort to counting rising edges in the
collected voltage data, then determine from that the frequency.

Does this mean that you are reading the voltage with an analog input,
and finding the edges in that data?
This
is all working pretty well.... as long as I power the device being
measured with batteries. If I use the wall adapter, I'm getting a
pattern in the results that repeats at 60Hz. This made me suspect the
power supply so that's when I tried the batteries and it worked
correctly.

So, my question is: What do I need to add to my DC power supply to
remove that 60Hz that is making it's way into the system?

I really have no idea... when I put the power supply on the scope, I
don't see the signal, but sure enough.. it's there somehow as it shows
up in the data I'm capturing.

You can see a screen shot of it here:
http://steventara.dnsdojo.com/images/pattern.gif

Our power supply puts out 3.22 Vdc @ 1.5 Amps

Is one side of the supply connected to signal ground of your data
acquisition board?
 
S

SklettTheNewb

Jan 1, 1970
0
Hi John,

I would like to reply inline, but I'm using the google website for this
and it won't show me the orginal message. Apologies for my bad
formatting.

To answer your questions:

"Is that 0 to -25 or 0 to +25 volts?"
0 to +25 volts

"Does this mean that you are reading the voltage with an analog input,
and finding the edges in that data?"
Yes, this is exactly what I'm doing

"Is one side of the supply connected to signal ground of your data
acquisition board?" This I'm not sure about. I do run the supply
through the board to check current draw (cheap short check), but I'm
not sure if I'm using signal ground. I don't have it in front of me at
this time, but I will look into that. Sounds promising.

Thank you for your post!
-Steve
 
SklettTheNewb said:
Hi John,

I would like to reply inline, but I'm using the google website for this
and it won't show me the orginal message. Apologies for my bad
formatting.

THis is also posted from Google Grpups. You have to select "Show
options" at the top, instead "Reply" at the bottom, then "Reply" at the
top, to have the previous post quoted. It is not well explained on the
site.
To answer your questions:

"Is that 0 to -25 or 0 to +25 volts?"
0 to +25 volts

Okay. What voltage swing does the counter input need?
 
S

SklettTheNewb

Jan 1, 1970
0
THis is also posted from Google Grpups. You have to select "Show
options" at the top, instead "Reply" at the bottom, then "Reply" at the
top, to have the previous post quoted. It is not well explained on the
site. :0)


Okay. What voltage swing does the counter input need?

From what I see on the spec sheet:
"input low voltage: 0.8v max"
"input high voltage: 2.0v min"

I'm figuring that this counter wants a DC signal, so I would need to
rectify the AC square wave (from what I have read and been told).

 
S

SklettTheNewb

Jan 1, 1970
0
SklettTheNewb said:
"input low voltage: 0.8v max"
"input high voltage: 2.0v min"

I'm figuring that this counter wants a DC signal, so I would need to
rectify the AC square wave (from what I have read and been told).
No, I am wrong here. please ignore this.
 
J

John Popelish

Jan 1, 1970
0
SklettTheNewb said:
"input low voltage: 0.8v max"
"input high voltage: 2.0v min"

I think this means that any voltage below .8 volts (but above zero)
will be interpreted as a logic low, and any voltage above 2 volts (and
probably below 5) will be interpreted as a logic high. Any voltage
between .8 and 2 is not guaranteed to be interpreted as one state or
the other.
I'm figuring that this counter wants a DC signal, so I would need to
rectify the AC square wave (from what I have read and been told).

I think you might be able to just divide the 0 to +25 volt wave to a
something like a 0 to +5 volt amplitude with to resistors and
connected that smaller signal directly to the counter. A pair that
might work would be 10k to the 25 volt signal, and 2.4k to ground.
The junction of those two resistors should provide a 0 to +4.8 volt
signal.
 
S

SklettTheNewb

Jan 1, 1970
0
John said:
I think this means that any voltage below .8 volts (but above zero)
will be interpreted as a logic low, and any voltage above 2 volts (and
probably below 5) will be interpreted as a logic high. Any voltage
between .8 and 2 is not guaranteed to be interpreted as one state or
the other.


I think you might be able to just divide the 0 to +25 volt wave to a
something like a 0 to +5 volt amplitude with to resistors and
connected that smaller signal directly to the counter. A pair that
might work would be 10k to the 25 volt signal, and 2.4k to ground.
The junction of those two resistors should provide a 0 to +4.8 volt
signal.


I really want to do that :0)
We had hired a consultant to make a basic application to show how to
interface with the acquisition hardware. I could have figured it out,
but decided that it would be more efficient to pay someone to "jump
start" things. Anyway, long story short, for a reason that I can't
remember at this time, the consultant said he couldn't use the counter.
He told us we would need to calculate it based on the analog voltage
values read in.

We are squeezed for time so accepted this. So as of this writing, I'm
just trying to get the system that he delivered working and in use.
Then, when I have a little time, I want to do exactly what you said,
divide things way down and try and use the counter, which I would hope
is much more accurate that counting analog edges like I am now.

I will look into the ground issue you had mentioned. Thanks for
reassuring my suspicions that the counter is a good way to go or
atleast worth another look.

-Steve
 
C

Chris

Jan 1, 1970
0
SklettTheNewb said:
I am NOT an engineer, I'm helping my Dad out with a PC based data
acquisition program to test a product his company makes.

With that little disclaimer out of the way, here is the problem:

I'm reading voltage values of his device, it outputs an AC square wave
signal at 4000Hz. The voltages vary from 0 - 25 Vcc Due to this
variance in voltage, I wasn't able to use the counter on the
acquisition board, so I need to resort to counting rising edges in the
collected voltage data, then determine from that the frequency. This
is all working pretty well.... as long as I power the device being
measured with batteries. If I use the wall adapter, I'm getting a
pattern in the results that repeats at 60Hz. This made me suspect the
power supply so that's when I tried the batteries and it worked
correctly.

So, my question is: What do I need to add to my DC power supply to
remove that 60Hz that is making it's way into the system?

I really have no idea... when I put the power supply on the scope, I
don't see the signal, but sure enough.. it's there somehow as it shows
up in the data I'm capturing.

You can see a screen shot of it here:
http://steventara.dnsdojo.com/images/pattern.gif

Our power supply puts out 3.22 Vdc @ 1.5 Amps

Hopefully this is something that might not be too hard to fix.

best Regards,
Steve Klett

Hi, Steve. The trace and your description indicate that you're
probably getting 120Hz ripple on your wall wart power supply -- that
makes it a good candidate for replacement. If you can find the spec on
the wall wart (usually on the back of the supply, on the side where the
plug is), that will give you a good idea where to start. Possibly your
wall wart is/was a 5VDC regulated power supply? It kind of looks like
the input cap in the wall wart went south and opened up. Replacements
are commercially available at less than $20 for 5VDC regulated (make
sure it's a linear regulated s3upply) at 1 amp.

Jameco sells their P/N 168605CB 5V regulated supply for $12.55 -- buy
2, and you've got your $25 minimum.

www.jameco.com

If it worked well before, you should be able to get it working again.
Unfortunately, you didn't include any more information, so I guess it
would be tough to be more specific. Sorry about that. Feel free to
post again with as much information as you want.

Good luck
Chris
 
P

Pig Bladder

Jan 1, 1970
0
Hi John,

I would like to reply inline, but I'm using the google website for this
and it won't show me the orginal message.

Yes, it will.

Don't click the "Reply" link at the bottom of the post. Look up at the
top of the post, and click the "Show Options" link. Then, click _that_
"Reply" link (toward the left of the list of options). This will quote
context.

Then, snip irrelevant material, scroll to the bottom of the post, and
put your reply at the bottom. This maintains the natural order of reading
a thread.

Good Luck!
Rich
 
S

Sandbox Moderator

Jan 1, 1970
0
Then, snip irrelevant material, scroll to the bottom of the post, and
put your reply at the bottom. This maintains the natural order of reading
a thread.

This was supposed to be attributed to "Sandbox Moderator", not "Pig
Bladder".
 
P

Pig Bladder

Jan 1, 1970
0
.
but decided that it would be more efficient to pay someone to "jump
start" things. Anyway, long story short, for a reason that I can't
remember at this time, the consultant said he couldn't use the counter.
He told us we would need to calculate it based on the analog voltage
values read in.

We are squeezed for time so accepted this. So as of this writing, I'm
just trying to get the system that he delivered working and in use.

If it isn't working and in use, his job isn't done.

If you've already paid the consultant, and he didn't get the job done,
then you're a fool and he's a swindler.
 
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