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Old 15th January 2004, 12:01 AM   (permalink)
Default Current Draw Counter (don't really know how to describe)

What I am trying to acomplish is this:

I have a motor and gear assembley, as the motor is running a load is place and removed creating one high then low current draw. This high then low is considered one cycle. What I want is to create a circuit that cuts power after 3 of these cycles then willrepeat it after the power to the motor is turned or and re-applied.

is this possible? if so how
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Old 15th January 2004, 02:01 AM   (permalink)
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It depends on many factors. Need more info.
Measure current - possible. Detect current difference with or
without load - might be tricky for some motors and small loads.
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Old 15th January 2004, 02:42 AM   (permalink)
Default

Yes, it is possible. A hall-effect device can measure the current and a counter can tell when 3 cycles have happened. You need to post more details for a better answer.
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Old 15th January 2004, 03:06 AM   (permalink)
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not sure what else you wanted, the power source is a 9.6v 1700mah battery the draw when the load is amplied peek my multi-meter at 10 amps the draw without the load was aprx 5 amp and entire 3 cycles take maybe .25 seconds.
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Old 15th January 2004, 03:45 AM   (permalink)
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how do you measure 3 cycles in 1/4 second? Or did you mean 25 seconds? With 10 and 5 amps, the hall effect would only need to be close to the wire. The system consists of hall-effect, amplifier or comparator, counter, fet driver, MOSFET power switch. I could design it, but I am too busy right now.
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Old 15th January 2004, 04:12 AM   (permalink)
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The current highs and lows are est. from previous data I have the gap between the draw with the load on it is rather large maybe 10 amps. Unfortunetly this all makes a small amoutn of sense to me with me limitted knowledge of this topic. it's not 25 seconds the device makes aprx. 900 cycles in 1 minute.
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Old 15th January 2004, 07:03 PM   (permalink)
Default

Ok, the numbers you have are what I was hoping for.

I would use small resistor in series with load (motor)
to create voltage drop. This is proportional to the current
and it can be amplified (high gain opamp or comparator
will do). This can be feeding 4017 cunter that controls
switching transistor (mosfet or cascaded darlington or just
a transistor with a relay).
The low value resistors are not hard to find just lk for
5-10miliOhm and make sure they are fine for 10A current.
There are precision shunts for industrial use but they are
probably overkill for this application (too large and too expencive).
With comparator you should be able to get the trip point at
about 7.5-8 Amp. I guess everything else is obvious.
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Old 15th January 2004, 08:47 PM   (permalink)
Default

I agree with the above poster. You can get medium power, low resistance resistors. I have a 1.2 ohm one, at 4 Watt for doing just that on a home made power supply. It's a big enough resistor.

The way to measure the current increase is to watch for a voltage increase across the resistor. Even a BC108 transistor will achieve this, and with a variable resistor you could trim it...
One thing I believe will be a difficulty is that this method of measurment is not going to be 100% accurate, due to spikes on the motor, and a lack of defined linearity.

Best of luck anyway.
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Old 15th January 2004, 10:23 PM   (permalink)
Default THANX!

Thank You for all your help/
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Old 20th January 2004, 10:39 PM   (permalink)
Default

I am afraid 1.2Ohm will not cut it. :?

At 9.6V (let's call it 10V just for simpler math) and 5Amp current
load is 10/5=2Ohm. With stall current of 10A, load resistance is
down to 1Ohm.

Power used by the tool (DC motor) is 10*10/2=50Watt continuous
and 100W stall. If you put 1.2Ohm in series current will be limited
to about 5Amp all the time. At 5A current through 1.2 Ohm resistor,
you get 6V dropon resistor and only 4 (or 3.6V) on motor.
The resistor would burn some 6*6/1.2=30Watt (and that's 650% over
it's rating) while motor would get only ca. 20 Watt.

Compared to 50Watt he has now, that's dramatic power loss and
huge waste of batteries. Besides 30Watt is lot's of heat, I have
soldering irons as low as 8 and 15Watt ;-)

Resistor that would be used for this application must be VERY low.
I would use 0.005 Ohm / 2Watt. They are available through regular supply
channels. At only 5mOhm and at full 10Amp stall current,
voltage drop is only 0.05Volt and heat disipation is just 0.5Watt.


Regards,
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Old 20th January 2004, 11:58 PM   (permalink)
Default

I kinda solved the problem, I aquired some Allergo Hall_effect current sensors from a friend... should make the process much eaier
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