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average value of current pulses?

i am trying to figure out the average current consumption of a pump. it
draws 13ms pulses of about 500ma about 15 times per second at randomly
spaced intervals.

i put a 0.1ohm sense resistor in the positive lead of the battery which
powers the pump, and i put an rc integrator across this resistor (10k
and 1000uF capacitor).

does this make sense as a way to average out these pulses and get a
voltage which represents the long term (relative) average of the
voltage across the sense resistor?
 
A

Andrew Holme

Jan 1, 1970
0
i am trying to figure out the average current consumption of a pump. it
draws 13ms pulses of about 500ma about 15 times per second at randomly
spaced intervals.

i put a 0.1ohm sense resistor in the positive lead of the battery which
powers the pump, and i put an rc integrator across this resistor (10k
and 1000uF capacitor).

does this make sense as a way to average out these pulses and get a
voltage which represents the long term (relative) average of the
voltage across the sense resistor?

Yes, but the time constant doesn't need to be so long. With those values,
it's going to take almost one minute to fully charge the capacitor. I would
try 10k and 100uF.
 
B

Bob Eld

Jan 1, 1970
0
i am trying to figure out the average current consumption of a pump. it
draws 13ms pulses of about 500ma about 15 times per second at randomly
spaced intervals.

i put a 0.1ohm sense resistor in the positive lead of the battery which
powers the pump, and i put an rc integrator across this resistor (10k
and 1000uF capacitor).

does this make sense as a way to average out these pulses and get a
voltage which represents the long term (relative) average of the
voltage across the sense resistor?

This will work but it may give difficulty. The peak voltage is only 50mV
across the .1 ohm resistor and the average voltage will be less than 10mV.
That's a small number prone to errors with an electrolytic capacitor. I
would use a film capacitor of several microfarads with hundreds of "k" in
the resistor to get a time constant about one second. It is now 10 seconds,
longer than necessary. You may want a little more sense voltage to work with
by making the .1 ohm resistor .2 or even .5 ohms.

What are you going to do with the average value for current? If it is going
to feed any type of processor or controller for indication or control, you
might consider doing A to D conversion and numerical averaging in the
digital domain in that processor. It could be a single chip solution without
any analog integration, large caps or other parts.
 
P

Phil Allison

Jan 1, 1970
0
** Groper alert !

i am trying to figure out the average current consumption of a pump. it
draws 13ms pulses of about 500ma about 15 times per second at randomly
spaced intervals.


** How ambiguous.

What exactly are these " randomly spaced intervals " ??

Every other Tuesday afternoon if it rains ??




....... Phil
 
J

John Larkin

Jan 1, 1970
0
** Groper alert !




** How ambiguous.

What exactly are these " randomly spaced intervals " ??

Random intervals can't be described exactly. That's because, I think
the theory goes, they're random.

John
 
R

redbelly

Jan 1, 1970
0
i am trying to figure out the average current consumption of a pump. it
draws 13ms pulses of about 500ma about 15 times per second at randomly
spaced intervals.

So, each pulse has a charge of 13x0.5 mC = 6.5 mC.
There are about 15 of these pulses per second, so:
15 x 6.5 mC/s = 98 mA.

The average current is about 100 mA.

Mark
 
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