Difference in Solenoid Rise/Fall time

[OCPumpGuy]

I am a mechanical engineer trying to understand how a solenoid works. I always thought the EE's had it easy. I mean a resistor is pretty linear in comparison to a hydraulic orifice. Apparently I have found an electrical non-linearity I don't understand.

I have a 12V DC solenoid. The resistance measures 6 ohms and the inductance is unknown (I'd guess around 30-40 mH). It is a wet pin solenoid for use in hydraulic valves. I wanted to do a "step response" test so that I could estimate the inductance based on the time constant from the step input (tau=L/R). My end goal is to include this solenoid in a dynamic model for an electro-hydraulic actuator.

To perform my test, I put a momentary switch in series with the solenoid (switching the high side). In order to dissipate the energy when the switch was opened, I put an IN5408 diode in parallel with the coil. I then activated and deactivated the switch about once every second for 10 seconds while recording the voltage across the coil and the current through the coil. The current was measured with a vektrex hall effect current sensor.

When I plotted the voltage and current, I was shocked. The response time on the rising side was significantly longer than the response time on the falling side (roughly a factor of 2). I'm not a EE, but I'm pretty good with dynamic systems. I'm having a hard time identifying the nonlinearity that is causing the rise time to be slower than the fall time.

I can post the actual test data if it would help (on Monday when I'm back in the office), but I've got to believe that an EE worth his salt knows exactly why this happens.

 

[itsmoked]

The 12V ON path has to overcome a spring, the resistance of the leads, and the supply's current and transient responses.

The OPEN case has spring assist and the current needs travel only thru the coil, the supply's response has no effect.

What is your supply?

Keith Cress
kcress -

http://www.flaminsystems.com

 

[OCPumpGuy]

Neither path is impacted by the spring. This is a proportional solenoid with a spring that moves the armature to the energized position. The valve spool and a spring attached to the spool push the armature away from the energized position. For these tests, the solenoid is removed from the valve body so the armature stays in the same spot for the entire test (i.e. the "zero air gap" position). At this point, the only transients I am evaluating are the electrical circuit.

I have suspected the supply dynamics. Its a variable voltage / current supply that goes to 36V (I think) and 3A. Its made by BK precision, not sure which model. I have the current knob cranked all the way up and the voltage knob is set for 12V. In the voltage trace, I see that the wave isn't perfectly square. It jumps up to nearly 12V in less than a ms, but while current is rising, the voltage is a little less than 12V. There is a very gradual slope as the voltage rises to the full 12V while current is also rising.

Do you have a recommendation for a power supply with faster dynamics? This valve is used on mobile equipment, so maybe I should use a large 12V battery as my supply, would it reduce eliminate the transient effects??

 

[Skogsgurra]

If you are looking for the non-linearity that causes this, you do not need to look far. The diode does that.

I assume (sorry Jeff) that your current measurement is taken close to the switch that you use to switch the solenoid. If that is so, you will record all current going to the solenoid at switch-on, but not the decaying current at switch-off. The latter is circulating between coil and diode and thus not measured by your vectrex transducer.

If you want to find the true behaviour of the solenoid when performing in an actual circuit, you need to measure in the actual circuit.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[waross]

Back to basics.
First, the magnetic strength of the solenoid is related to the AMP-TURNS. That means you should be looking at the current waveform, not the voltage waveform.
When a step change is made in the voltage across a solenoid coil, the response of the current is described by the time constant.
The time constant of a resistive/inductive circuit (solenoid coil) is the time it takes the current to reach 63% of the final value when a voltage is applied.
The time constant is fairly easy to calculate, Time Constant in seconds is Inductance divided by Resistance, but don't worry about calculating the time constant just yet.

Example:
100 volts is applied to a solenoid with 1 Ohm resistance.
The final current will be 100 Amps.
At the end of the first time constant the current will have risen to 63% of the final value or 63 Amps.
100 Amps minus 63 Amps equals 37 Amps to go.
At the end of the second time constant the current will have risen an additional 63% of the remaining 37 Amps or an increase of 23.3 Amps for a total of 86 Amps. 100 minus 86 equals 14 amps to go.
At the end of the third time constant the current will have increased 63% of 14 Amps or 9 Amps for a total current of 95 Amps.
The actual current will be a little higher due to rounding errors.
The current is often assumed to have reached 100% after five time constants.
In a perfect world, the charging current rises sharply and then tapers off.
The discharge current drops sharply at first and then tapers off.
In a perfect world the graph of the discharge current is the graph of the charging current inverted.
Now, lets get back to the real world.
The applied voltage from an infinite source will rise immediately to 100%. The current and the amp turns will rise on the curve described by the time constant.
The voltage drop from your source will extend the charging time a few percent.
The discharge voltage is generated by the energy stored in the magnetic field. As the current decays, the voltage will drop.
This is based on a circuit of zero impedance for the coil to discharge through. Usually a diode, which you have.
I would expect the voltage to rise almost immediately on energization and to drop to zero in 5 time constants on de-energization.
Try monitoring the current and see if the results are acceptable.
The bad news is that when you re-assemble the valve, this may all be meaningless.
The time constant is based on the induction of the coil.
The induction of the coil is influenced by the length of the magnetic circuit. The length of the magnetic circuit or path is based on the weighted sum of the the iron path and the air gap. Here comes the bad news, the weighting of the air gap may be in the order of 10,000.Yup Ten Thousand. (Based on the permeability of the iron core.)
The armature doesn't have to move very much to have a dramatic effect on the induction.
If the armature is free to move, The initial induction when the voltage is applied is fairly low. That gives a short time constant and fast action. Once the armature starts to move the induction increases dramatically and the time constant increases dramatically, but what the hey. The armature is moving and the magnetic inverse square law will slam it home and after that the slow increase to maximum current is moot.
When the current is interrupted, the induction is high and the time constant is much greater. Also when we are waiting for the armature to drop out, the magnetic inverse square law is working against us. It takes much less current to hold the armature in than it took to pull it in. The time delay on drop out may be more than you can live with.
There are a couple of tricks to speed up the drop out. One technique is to use a permanent air gap. Then your ratio of induction change as the armature moves is approximately the ratio of the air gaps and you avoid most of the 10,000 factor. Early contactors used a brass rivet in the pole face to prevent the pole faces from touching. After many cycles the rivets tended to be hammered flat and residual magnetism would cause the contactors to stick. (That old magnetic inverse square law again.)
Then someone realized that there could be two air gaps and many contactors now have a small slot in the back side of the core to introduce an air gap into the circuit.
Let us know how the current measurments go.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[OCPumpGuy]

Hmmm... I know about all the time constant stuff and the basic physics and as stated in the first post, I am measuring current. Also, the design of this coil is such that it is unaffected by installation in the valve body. There is a steel frame around the coil that provides the magnetic circuit - I've already tested and verified this. Also, I have verified that moving the armature within its stroke has very little impact on inductance. This is not a simple on-off solenoid where there are much greater changes within the stroke. It is designed for CONSTANT force over the entire stroke range.

My current measurement does detect ALL the current on both rise and fall since it is placed between the diode and coil.

I already knew that the diode would cause some difference between rise and fall, but not this much. I used a rectifier diode that only has ~0.5 V breakdown voltage (I think that's the right term...). The difference shouldn't be that big, at least that's my guess, I'll include the diode in my dynamic model and verify it. Since I am measuring current and voltage I can see that on the fall, voltage goes negative equivalent to the diode breakdown voltage. I assume that if I added a zenner diode biased opposite to the primary diode, the voltage would be more negative on the fall and drop-out time would be reduced. Right? I'm not all that concerned about reducing the drop out time right now since the rise time is so much longer than the fall (ideally for this application, they would be the same).

I'm thinking the first reply about the power supply is dead on. I've read a little about the power supplies on the BK website, and there is a "recovery time" published. I'd guess that I have one of the slow ones and that is my problem. What I couldn't find is how recovery time is defined. It looked like the times ranged from 100 us to 500 ms - what exactly does that mean? Is there an industry standard test method? What inside the power supply actually contributes to the recovery time?

I'm going to try and test the power supply theory when I get back in the office. I assume I could replace the coil with a 6 ohm power resistor and repeat the test. If the power supply is limiting the rise time, it should have that same limitation with a purely resistive load, right?

 

[Skogsgurra]

I think you know where your towel is.

One thing, it is not 'breakdown voltage'. It is 'diode forward voltage drop'.

The rest is so "circumstance sensitive" and "correct term" dependent that it will be very difficult to discuss it without getting into a lengthy thread.

Search the IEEE library. They usually have lots of quite detailed papers on any subject.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[itsmoked]

BTW I would use a large car battery to rule out the supply issues or to get you "automotive results" anyway.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[BobM3]

What solenoid are you using? Maybe the magnetics are doing something wierd.

 

[2dye4]

OC

The relationship is

V=L(Di/Dt)
Where V is the voltage across the coil.
L is the inductance
i is the current.

Your issue is that for the ON transition the coil sees the supply voltage and this is the V in the equation.

For the OFF transition the V is the diode drop, that is why the current decays slowly.

Either case you should measure the voltage across the coil and the current through the coil at the same time.

It helps to have a scope that you can export data from, then you can estimate the current derivatives in excel.

To estimate the derivative at a particular voltage use the central difference formula.

Take a point prior to the voltage measurement and one after the voltage measurement so that the center of the points corresponds to a point of the voltage curve.
dt= time of interval between the current measurements.
di= current difference between the two points.

L= V*dt/di

 

[OCPumpGuy]

Well guys, thanks for all the help - I almost caught that wild goose ;-) Three power supplies and a battery later, I've determined that the problem is not the power supply. I even threw in a giant cap in parallel with the power supply!!

I had to go back to the good old transactions of the ASME to find the answer.

The Modeling and Simulation of a Proportional Solenoid Valve
N. D. Vaughan and J. B. Gamble
J. Dyn. Sys., Meas., Control -- March 1996 -- Volume 118, Issue 1, 120 (6 pages)

The issue is that the iron core of the solenoid and fringing effects of the air gap cause a nonlinear function of inductance (L=f(i, x, di/dt)). Of all the factors (current, armature position, and current change rate), current has the largest impact. There is a second article that is an easier read, but I think there are some errors in the physics. It does, however, show plots of the response that are very similar to what I was seeing. See page 6 of the illustrations link.

http://homepage.ntlworld.com/ken.elmer/elmerf/papers/paper6.pdf

http://homepage.ntlworld.com/ken.elmer/elmerf/papers/illustr6.pdf