Determining how many amps a capicitor is producing

[jwcornell]

I'm working on a project that will use a Reed switch to close a circuit on a charged capacitor that will open or close a solenoid. The problem I am having is in trying to determine the proper Reed switch to use.

Generally speaking, I've found Reed switches having the capability to handle .5, 1 and 1.5 switching amps up to 200 volts. Generally, the lower the amperage specification, the cheaper the switch.

I will be using a 2300 uF capacitor charged to about eight volts. I have no way to determine how long it will take to discharge this capacitor in the solenoid but tests so far would indicate that it's very fast (perhaps 1/100th of a second?) The specs on the solenoid says it needs five watts to operate at between 5 to 10 volts for ten milliseconds. Is there a way to determine how many amps this pulse is producing?

 

[MikeMl]

You would have to know the DC resistance of the coil, and its inductance. I'm guessing that it takes much longer to pull in a solenoid than it does to pull in a relay. Relays take on the order of 10 to 50msec, and they have much less mass than your solenoid. I'd be surprised if your solenoid will even twitch if the current pulse is shorter than about 50msec.

 

[jwcornell]

Thanks for the quick reply MikeMI.

The specifications for the solenoid state "Coil Resistance 5 +/- 0.03 W, Minimum Pulse 10 mS Duration at 6-9 Volts"

Does this help? Thanks in advance.

 

[ronsimpson]

"5 watts 5 to 10 volts"
If you had 5 volts on our capacitor (or any supply) and the solenoid used 5 watts thats 1 amp.
If you had 10 volts on our capacitor (or any supply) and the solenoid used 5 watts thats 0.5 amp.
The selenoid may have 20 ohms. Not much information.
I think the answer is 1/2 to 1 amp.
What voltage are you charging the capacitor to? This voltage will cause some current to flow. Test with a power supply set to that voltage.

 

[jwcornell]

Thanks ronsimpson. I will be using a solar cell to charge the capacitor so it will be a bit of trial and error to find out the voltage. I suspect that the final voltage will be around eight volts so using your equation, I get about .625 amps (8/5) Do you concur? Given the quick discharge of a capacitor, will a standard VOM be able to determine the voltage it's holding? Is there a way to determine the voltage on a capacitor before it discharges?

 

[crutschow]

I believe for the stated coil resistance of 5 +/- 0.03 W, the W stands for Ω (this symbol error occurs is some font conversions). Thus the solenoid would have 5Ω and the current would be 1.6A at 8V. The 2300µF cap would then give a time-constant of 11.5ms, which may be marginal if the solenoid requires a 10ms minimum pulse. But just charge the cap with a power supply and then connect it across the solenoid to see if it works.

A digital multimeter will be able to read the cap voltage without significantly draining the charge.

 

[MrAl]

Hi there,


With an 11.5ms time constant that means that the wave looks approximately like a straight line from some max to about one third of max. That means
that roughly for half that time (about 6ms) the wave is above two thirds of max and the other half of the time it is below two thirds of max. That puts the
average at about two thirds of max, or about 0.66 of whatever the peak current is. That basically means you do not get the full current over that time
period but a sort of average of two thirds of max approximately.

It sounds like you might have it right at 1.6 amps peak, if that's what it really is, but it's a good idea to go by the old saying,

"One test is worth a thousand expert opinions".

 

[MrAl]

Hi there,


With an 11.5ms time constant that means that the wave looks approximately like a straight line from some max to about one third of max. That means
that roughly for half that time (about 6ms) the wave is above two thirds of max and the other half of the time it is below two thirds of max. That puts the
average at about two thirds of max, or about 0.66 of whatever the peak current is. That basically means you do not get the full current over that time
period but a sort of average of two thirds of max approximately.

It sounds like you might have it right at 1.6 amps peak, if that's what it really is, but it's a good idea to go by the old saying,

"One test is worth a thousand expert opinions".

 

[jwcornell]

Thanks for your thoughts MrAl and crutschow. I have successfully fired the solonoid using a 1600uF capacitor after charging it with a 9 volt battery (probably around eight volts charge I suspect). The 2300 uF at 8 volts is closer to the calculated needs and will give me a bit of insurance to insure operation. It 1.6 amps would be the peak of the pulse, I would need to use a larger Reed switch than the one previously determined. Thanks for the thoughts on this.