# Calculate capacitance needed to actuate a solenoid one time

A 1.2W solenoid has to be actuated for 500 msec, using only a (pre-charged) capacitor. The solenoid voltage range is 4.5-6 V. I have to calculate the needed capitance.

https://www.engineeringtoolbox.com/capacitors-energy-power-d_1389.html states that the time to discharge a capacitor at constant power load can be expressed as: $$\Delta t=\frac{\frac{1}{2}C(U_s^2-U_f^2)}{P}$$ Δt = discharge time (s)

Us = start voltage (V)

Uf = final voltage (V)

P = power (watts, W)

Which results in: $$C = \frac{2P\cdot\Delta t}{U_s^2-U_f^2} =\frac{2\cdot1.2_{watt}\cdot0.5_{sec}}{6_{volt}^2-4.5_{volt}^2}=76mF$$

Is this correct?

## Edit

Thank you all for the feedback! Plenty food for thought.

I understand much depends on the characteristics of the solenoid (including the mass that it must move etc.). This variable should of course not be ignored, but it does add complexity which makes it hard (for me at least) to grasp the basics. Therefore I'd like to simplify the problem by assuming the solenoid behaves like a simple resistor (as suggested by @Transistor).

Based on 1.2W at 4.5V, this resistor will be $$\R=\frac{U^2}{P}=\frac{4.5_{volt}^2}{1.2_{watt}}=16.875\Omega\$$

1. What is the right way to calculate the capacitance based on this? Should I go with $$\V_C=V_S\cdot e^{-t/RC}\$$, where

• Vc=voltage across the capacitor
• Vs=supply voltage
• t=time since the removal of the supply voltage
• RC=time constant of the RC discharging circuit

Which results in $$\C=-\frac{t}{R\ln(\frac{V_C}{V_S})}=-\frac{0.5}{16.685\ln(\frac{4.5}{6})}=103mF\$$

2. What is the effect of the capacitor' ESR on the calculation?

• One problem is that with varying voltage your load voltage and current will change as the capacitor discharges so the power in the solenoid won't be constant. You might be better off calculating the solenoid's resistance - presumably it dissipates 1.2 W at 4.5 V (the drop-out voltage) and then just treat the setup as an RC discharge problem discharging C through R from 6 V to 4.5 V. Oct 13 at 18:50
• My intuition says that 76mF will definitely work, but you may be able to get away with significantly less. Note that the exact electrical actuation requirements may be affected by the operating pressure/force. If it were mine, I'd buy 8pcs of 10,000µF/10V, solder in parallel, and try it. Then try 7pcs, then 6. Oct 13 at 19:43
• My guess is conservation of energy with matched LC energies and maximum energy transfer theorem dictates transfer efficiency will be 50% so double the size of C. Oct 13 at 19:48
• Electromagnet may hold even with voltage lower then 4.5V. capacitor has tolerance -25%-+80%. So you calculation approximate. Oct 13 at 20:11
• @TonyStewartEE75 you suggest to double the size of because of the 50% transfer efficiency, however my understanding is that this only plays a role when charging the capacitor, not when discharging? Oct 14 at 19:12

The conservation of energy with matched C energy is what you stated that we recognize. But maximum power theorem dictates transfer efficiency will be 50% if (real part of ) impedances are matched. But here there is reactive energy transferred. You don't always want to transfer the most power in the fastest time,T , but here the time is given as 1.2W for 500 ms. This could be 12V into 10 Ohms with 1.2A that decays from back EMF at max velocity between start and stop end points. Normally a solenoid is spring loaded. So making contact at the end point is not enough, usually it must hit the endstop with a certain velocity to force an action with inertia. This detail was omitted.

It might look like an LC resonator with losses but the solenoid would look like a linear motor with a parallel resistive load being the opposing forces and a series resistance for conduction losses that also ramps then limits current then opposes current from velocity generated back EMF then quickly ramps current to limit at endstop when held by constant voltage. and causes the current to ramp up then limit then back EMF opposes time to .

The losses depend on the current limiting resistance. But the energy can be increased with voltage with a smaller Cap since Ec=CU^2/2 or 1/2 CV^2 and I^2R is the power loss.

Using your Us, Uf values may be reasonable but this leaves about half of the energy in the final state. $$\(6^2-4.5^2)/6^2 = 44\$$% Where as starting from twice the voltage stores 4 times energy but also 4 times the resistive loss. Is what you assumed, however a cap will keep discharging even if the solenoid position retracts from the opposing force current exceeding the discharging cap current the resulting in a full retraction and cap, discharge.

So without detailed specs on Solenoid and forces involved , I would estimate up to twice the energy is needed and then use the lowest ESR cap possible at the highest voltage practical. E.g. 2x

In the end, a small rechargeable battery may be better or at least trial and error Q=CV = 76mF * 12V to determine what impact velocity you need .

Why you need this is a bigger question, that may reveal assumptions that need examination.

• Maximum energy transfer doesn't really apply here (there isn't a "maximum energy transfer" theorem -- there is a maximum power transfer theorem). The source resistive impedance is zero (it's all capacitive); It is perfectly possible to transfer ALL the capacitor energy to the inductor (connect them and see the resonance). Oct 14 at 0:12
• Ok @jp314 yes matched resistance for max power principle applies to power, but here energy is transferred here. Neither the cap storage or solenoid are lossless and is reactive energy useful? The stored reactance must be converted to real work.. Ideally it should be critically damped by the work load to a stop and not oscillate. There is also a thermodynamic principle that energy conversions cannot be both fast and efficient. Oct 14 at 0:55
• There is no 'reactive energy'. It is perfectly possible to transfer ALL the capacitor's energy to the solenoid (even if is is lossy) -- just connect them, & if underdamped, apply a short across the cap when the voltage rings down to 0 (and inductor current is at a peak). The inductor's current will then L/R decay. If underdamped, you have to wait an infinite time... Note that the solenoid losses include both the real R losses and the losses from the change in inductance as the solenoid actuates (converts electrical energy to mechanical energy; sometimes mechanical POTENTIAL energy)) Oct 14 at 5:32
• @jp34 There are 2 reactive elements, L, C with velocity changes and BEMF. How can there not be reactive energy in L? I agree all the "Ec" is used. Note that a mechanically loaded solenoid with kinetic energy is real R in shunt as explained. Oct 14 at 14:30

The previous calculations of 76 mF are reasonable -- if the capacitor is charged to only ~ 6 V, but basically you need to store a certain amount of energy -- 1.2 W for 500 ms == 0.6 mJ. You could store this in a high V capacitor (say 16 V) and use a DCDC converter to convert that to a voltage that can drive the solenoid (say 4.5 V).

Now, the 1.2 W doesn't fully make sense -- because it is current and time that energize the solenoid. Generally you'll be better off applying a slightly higher voltage (for a shorter time) than a low voltage for a much longer time (e.g. if you applied 4.4 V it could take an infinite time to energize).

Anyway, a 16 V capacitor supplying 0.6 J @ 5 V via a 90 % efficient DCDC converter would need:

½C(16^2-5^2)*0.9 = 0.6; ==> C = 6000 uF.

This is more reasonable, but still large. You would need additional margin for capacitor degradation etc.

• Nice suggestion! there is limited space available on the PCB, therefore I'm considering a supercap + boost converter Oct 14 at 19:15