Solenoid constant time

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Hello,

I have a small problem.
Solenoid has a resistance of 14.2 ohms and an inductance 0.056H is connected in series with a resistance 2.2Ohm
Thus calculated is the Tau = L / R = 0,056 / (14.2 + 2.2) = 0.0034s
But the measured value is 0.02s
The measurement is done on an oscilloscope for resistance.
Source is a classic laboratory.

If I measure only air-core coil corresponds to the time constant.
How is it possible that tau is based on much more?

Solenoid Type:
Code: 
 http://docs-europe.electrocomponents.com/webdocs/001b/0900766b8001b732.pdf [/ code]

Thank you for your response.
 

Have you measured the internal resistance of your power supply? It needs to be insignificant compared to your test circuit, or else it can throw off your results.

Of course it depends on which component you take a measurement across. Also where you place the standard 63% mark.

Your calculated time constant is correct:
.056 / 16.4 = .0034 sec.

Supposedly, a substantial supply resistance will shorten your time constant. However there is a simulation setup I'm testing, which can be interpreted as lengthening the time constant instead. This may be similar to the questionable result which you report.
 

if the solenoid has moving metal parts associated with it, then the L will change if these move during your measurement...
 
if the solenoid has moving metal parts associated with it, then the L will change if these move during your measurement...
Click to expand...
Specifically, the inductance increases when the solenoid engages because the magnetical path closes. I presume you are measuring the low initial inductance. You can visualize the inductance change by watching the current waveform, which is considerably different from a simple exponential curve.
 
Hi guys,

Do you have any idea how to design a driver for solenoid (30mH, R=1.5ohm) so that the current waveform through the solenoid can be square wave?
Solenoid lift frequency: 150Hz and I pp= 1.2 A

Best regards,
 

The current through an inductor can be at best trapezoidal, not a squarewave. You need to specify the required current rise time. You can easily calculate the voltage to drive the current slope, V = Ldi/dt + RI. E.g. nearly 40 V for 1 ms rise time.

Then implement a constant current driver. Falling edge can be adjusted with a zener clamp across the coil, of course the driver transistor must have sufficient voltage rating.
 
Also there is back EMF once motion starts which raises effective impedance and L is non-linear, especially near saturation current and stroke position. Then there is the net force and mass which affects acceleration, then distance to endstop that affects time duration.

Then with AC inrush current gives more force in the closed position with higher core coupling than the open position.



Then there are remenance , plunger type and thermal effects.


I can't think of at least 3 equations needed for these unknowns, some which include V=L dI/dt + I dL/dt + f(velocity, momentum) + ESR* I +

so when buying, follow the datasheets.
 
The current waveform can never be square, only the drive voltage, with a free wheel diode (45V schottky) the drive pwm can go to 70% say, the current will be DC + a small triangle ramp, the size of the triangle determined by driving frequency.

Easy enough to perfect on the bench....
 
Thank you guys for helpful replies.

Do you have any recommendation for constant current driver?

BRs
 

So you want it to switch faster 5x ?

Charge up a Storage Cap to a higher voltage and discharge it.
4ms? /16R =0.25F at 60V ... too big
energy = 1/2 CV² = 450 Watt-seconds.. way too big, but that's what it takes.

Why 5x as long as L/'R?
Depends how you measured the Solenoid time.
If you measured displacement end to end....

Like charging a Cap, steady state is arrived at ~5x Time constant !
 

The problem is that the voltage across the solenoid is only 12 V.
Click to expand...
Sounds like a helpless case. You absolutely need a high voltage source to drive the rising inductor current.
The energy stored in the inductor is 0.5 LI², some 10 mWs in this case. A storage capacitor in a 100 µF range should be sufficient, but it must be supplied by boost converter if you have only 12V power supply available.

You'll switch between constant voltage for the rising current and constant current driven by 12 V for economic operation.
 

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