I have device circuit that controls a valve. The circuit sends 9 volts dc to this valve and it returns a device to “home position”. As the device is returning it hits a limit switch that tells it is close to the “home position”. The voltage then changes to 7.5 volts dc. The valve changes and device slows down until it hits the “home position”.
I think the voltage change is to fast. I was considering adding a capacitor to the circuit so that the change from 9 volt to 7.5 would take around .5 of a second.
I tried muddling through the math it my limited understanding. The circuit designed to control this valve uses a 5 watt potentiometer. So using this I figure at 9 volts the most current the circuit is designed to handle is .55 amps.
Using Charge = Current * Time = .55*.5=.277.
Using Capacitance = Charge/Voltage =.277/1.5(Voltage Drop)=.184 Farad.
I am assuming I put the voltage at 1.5 because that is how much i want it to drop in a half of a second. Does this mean i need a .184 farad capacitor or am I doing this wrong.
I appreciate any help you can give me. Thanks.
To do any calculations regarding the valve timing requires that you know the actual current the valve is consuming. I don’t follow your reasoning about a 5 watt potentiometer.
Why do you think the voltage change is “too fast”? Did the circuit work before and now does not? What, exactly, is the problem?
You’ve got the right idea but you’re missing something (that you haven’t told us either). You’ve somehow come up with a 9/7.5 volt source supplying 0.55A. You’ve not said what the source output impedance (resistance) is and it’s current limit. Imagine your source w/zero output resistance and able to source or sink 1 billion amps. Put your cap in parallel w/the load and it’ll charge up to 9v. When the source goes to 7.5v, it’ll (not just the load) suck all the charge off that cap in near zero time.
You need a better source model/description.
so the valve closes at full speed (using 9V) until a first limit switch is hit
the valve actuation voltage drops to 7.5V unitl a second limit switch is hit
power is then removed
is this correct ?
do you have a part number for the valve ?
Are you sure that the drop from 9v to 7.5v isn’t because the actuator has hit the end stop, and the supply is drooping ?
-mark
markaren1:
Are you sure that the drop from 9v to 7.5v isn’t because the actuator has hit the end stop, and the supply is drooping ?
I think this is part of the design - he's changing it from 9V to 7.5V because he wants the valve to slow down. What he did not specify was the *how*.
Back to the OP. We need a more thorough explanation of the circuit. You told what the circuit is supposed to do, but not how you are doing it.
free-bee:
I think this is part of the design - he’s changing it from 9V to 7.5V because he wants the valve to slow down.
Perhaps but then the OP said the transition from 9v to 7.5v was happening too fast. He wants a cap to slow the transition down. Given closure is faster at higher voltages vs lower, it would seem the valve is slowing too much, too soon. His thinking re: dv/dt and charge pulled off the cap is correct ... in an isolated environment, which I'll guess this isn't in real life.
Alas this thread is like red haired stepchild … forlorn by it’s parents.