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Power consumption of vibratory pump when pumping against closed valve

Hey all,

I'm currently in the process of modifying my coffee maker (a Krups Oblo) to be controlled with an integrated Arduino. The goal is to be able to have the machine heat up, start pouring coffee, then stop after a certain time, all with a single press of a button. To achieve this, I want to emulate the inputs the existing control circuitry expects.

To start the machine, the user flips a lever, which opens a valve to allow for water flow and sends a 5V pulse. So far, so good. However, to stop the machine, the user returns the lever to the initial position, closing the valve. No pulse of any kind is generated when doing this, leading me to believe that the amperage going to the pump, now pumping against a closed valve, is measured to detect this user input.

The pump is powered by 230 VAC. I've tried recreating these conditions by opening the valve, have the Arduino send a pulse (starting the pump), then using a changeover relay to redirect the power going to the pump to a few power resistors. I've used a few combinations of resistance values, starting by dissipating about half the pumps rated power, to going just over that value. None of the resistance values seemed to trick the control circuitry into thinking the pump was pumping against a closed valve. Other than the phase line and a common ground line, the pump is disconnected from the rest of the electrical circuitry.

The pump is an ULKA vibratory pump, type EP5GW (closed thing to a datasheet I could find). The AC current passes through a rectifying diode, so the pump uses half a sine wave to perform the pump stroke. The plunger makes the return stroke with a spring.

Any help or insight would be greatly appreciated. If some things need more clarification, I'd be more than happy to provide that ;).

Cheers,

TheCoffeeGuy
 
It appears to be a simple pulsating piston pump, if the coil is DC then there should be no difference in current, in AC the current would vary if the piston had not moved over due to reduced inductance.
If it were an impeller pump the current would reduce regardless when the outlet were blocked.
M.
 
I'm not sure I get it either Martaine ;)

@Minder
The pump is a piston type. As far as I can tell, the only way to detect a closed valve is by measuring the current. Is there any way to predict the peak current draw when the piston is unable to move?
 
Oh dear Uncle Chris,
That's hilarious:eek:
I didn't need my medical history posted!
Perhaps my annual round robins to you will omit my medical conditions and just concentrate on yours!:p:D

Marty
 
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