energising solenoid using a Capacitor

[cubanflyer]

Hi

I have a small circuit that is using a logic level gate Mosfet to fire a solenoid. The solenoid is powered from a 9v battery and will only be energised for about 5-10ms.

I am planning on using a capacitor to supply the initial power booster to energise the soleniod. The problem I have is calaculating the capacitor size, I could use trial and error but would rather get it right from the start.

I am after a formula to calculate this value if any one can point me in the right direction.

 

[klystron]

The charge in a capacitor equals the voltage across x capacitance
The supplied current is the charge / time duration.
What is the expected current required from the capacitor?
Take note that the tolerances on electrolytic capacitors can be + 50 - 20 %

 

[cubanflyer]

The charge in a capacitor equals the voltage across x capacitance
The supplied current is the charge / time duration.
What is the expected current required from the capacitor?
Take note that the tolerances on electrolytic capacitors can be + 50 - 20 %

So using this is for example

if I have a 4700uf capacitor @ 9v and I supply a 5ms pulse to the solenoid, I should be able to deliver
8.46amps to the solenoid, in an ideal world.

The solenoid draws 2.8amps at 9v when energised, I have not been able to measure the current as it starts to energise.

 

[awright]

If the solenoid draws 2.8 amps steady state from your 9v supply, you can only get 8.46 amps through the solenoid by feeding it more than 9v. Assuming that you were using a power supply with adequate current capability so that it's voltage did not droop during your test, you will not be able to increase the solenoid current by adding a capacitor of any size. You will have to boost the voltage on the capacitor with a DC-DC converter if your supply is limited to 9v and you want to get 8.46 amps using your present solenoid.

Analysis of the voltage/current transient requires knowledge of the resistance and the inductance vs. time of the solenoid since the inductance will vary strongly as the core moves into the solenoid. Such analysis is way beyond me. I think experimentation is in order here. Even I could handle that.

You may want to consider using a solenoid designed for a much lower voltage and hit it with 9v on a capacitor of size determined experimentally. Charge the capacitor from the 9v supply through a resistor sized to limit the steady-state current and the drain on the power supply to a safe value but which can charge the capacitor quickly enough to be ready for the next pulse with some safety margin. There are also fairly simple transistor circuits that turn on to charge the capacitor but turn off during discharge of the capacitor through the solenoid. This allows rapid charge of the capacitor but prevents excessive current demand on the power supply during discharge.

A solenoid with a lower voltage winding than your present one will have lower inductance and resistance and will require more than 8.46 amps to do the same work, but that reduces to a capacitor selection. You will probably have to use a higher current MOSFET also.

awright

 

[FvM]

Awright has analyzed the situation exactly. If the solenoid inductance would be constant, the current could be described by a simple 2nd order differential equation. But it isn't constant.

As a first condition, the boost capcitor voltage needs to be > 27V. Practically, it must be larger (e.g. 40V) to maintain the current over some amount of time without making the capacitance unsuitable high. According to the coil resistance of about 3 ohm, you'll need at least 2200 uF to achive a time constant in the intended order of magnitude.

 

[chuckey]

If the DC current is 2.8A then when the armature is out off the solenoid it has its lowest inductance, so as the current is switched on the max current will be less then 2.8A, the rising edge of driving pulse being opposed by the back emf of the inductance of the coil. As the armature moves in the inductance increases, further increasing the back emf and decreasing the current flow. Eventually the armature goes all the way in, the inductance stabilises, the back emf falls off to nothing and the current settles at 2.8A. So depending on the various time constants there will be a finite delay if you drive the solenoid from a constant voltage source. I doubt if you will get any where near your 10mS actuation time. To make the solenoid acts swiftly, you need to drive it with a constant current source with a lot of voltage behind it, i.e. 24V.
Frank

 

[elactis]

Hi, If you are using a 9V solenoid, you need to charge the capacitor to ~3-4 times higher voltage. During operation, the current will rise in the coil and the voltage will drop in the capacitor. The size of the capacitor should be such that the voltage on the coil should be still more than 9 V when switching. The inductance change is not an issue. Once a plunger starts to move the magnetic force increases very fast and the value of the current is not important. The easiest estimation you can get from the magnetic energy stored in the inductance. Emag=L*I*I/2. You can easily calculate the inductance from the current rise and the resistance. You need to provide ~2 times this energy with the capacitor that will have a voltage drop from the initial value to ~9V. If I well understood, R=3 Ohm, the current rises ~5ms which gives 15mH inductance. The energy stored in the capacitor has to be E~120mJ with still 9V left. If you pump up the voltage to 19V (easier to calculate) then the capacitor should be much larger than: C=2*E/(deltaV^2)~=50mF.
To give you an idea, I used a similar solution for a valve with the following conditions: 24V, small solenoid with 2ms response time and 36 Ohm resistance, 1W switching power. 22uF was just enough to switch at 6V. From the above formula we would need ~10uF.