
#1
Apr912, 03:09 PM

P: 7

Hello!
I'm trying to activate a solenoid with a linear ramp in voltage of 0  24VDC in 2 seconds. An RC series circuit helps achieve soft start but in an exponential way (with 63% of supply voltage in 1 time constant). Can I use a single circuit for charging and discharging? (linear ramp down also preferred). Ideal ramp down time should be less than 1 second. Trying to figure out if this circuit is possible with resistor/ capacitor elements and without the use of an expensive proportional amplifier. Any help is much appreciated! 



#2
Apr912, 03:41 PM

P: 834





#3
Apr912, 03:54 PM

P: 834

Here is a derivation I did for it a while ago.




#4
Apr912, 04:24 PM

P: 7

RC circuit for linear ramp
Thanks DragonPetter for the quick response! This equation looks more like the one for RC discharging circuit. And to achieve constant current source, I connect a resistor between voltage source and capacitor?




#5
Apr912, 04:26 PM

P: 1,784

How much is your maximum current?




#6
Apr912, 04:27 PM

P: 7

The coil can take upto about 1.6A..




#7
Apr912, 04:31 PM

P: 834

One possible way to do what you want is to charge a capacitor up to the desired start voltage, and then to put a current source in series between the capacitor and the solenoid, where the current source's supply is from the capacitor. I don't know what problems you may have doing this in a practical circuit tho. I see a problem in that the voltage at the inductor would be 0 if it is given a constant current, which defeats your goal to have a ramp voltage across the solenoid. You could still use the linear V discharging cap as a reference voltage. 



#8
Apr912, 04:57 PM

P: 7

@DragonPetter: You're right. I actually have to use a 8Ω resistor anyway to have a current of 1.6A at the solenoid coil (7Ω) from a 24V DC source. So my circuit currently has a 24V source in series with an 8 ohm resistor, a .056 F capacitor and the coil. Of course linear ramp is difficult here. Don't have much experience with opamps, will they help?




#9
Apr1012, 01:35 AM

HW Helper
P: 4,706

What are you using the solenoid for, eegeek, and why do you want a softstart?




#10
Apr1012, 08:13 AM

P: 7

@NascentOxygen: The solenoid is used for steering a large equipment and a sudden step input might result in jerk resulting from force needed to overcome the moment of inertia (infact an exponential voltage coming from series RC circuit might do the same). Hence the soft start..Do you have a solution?




#11
Apr1012, 12:32 PM

P: 4,664





#12
Apr1012, 02:07 PM

P: 652

I concur. Cool little display animation with the integrator amp there Bob S.
To add to what Bob s is saying: The TRANSFER FUNCTION you are looking for is 1/JωRC. The gain of the amp shown is RF/RA (1/JωC)/R = 1/JωRC. Test your limits at max frequencies. When ω=0..........gain will be infinite or saturated in steady state.........The time it takes to get there is the RC constant....... When ω=infinity.............gain will equal zero. Reminder: db is 20 log Gain........that's how that peske minus sign goes away as far as gain goes.....as far as output voltage, it absolutely plays a part. Look at this (1/JωRC) in regards to sin and cosine. We know that sin is the intregal of cosine. Take your cosine and flip it across the horizontal acces according to the minus sign in the formula. Now phase shift it 90 according to the J in the denominator. You now have the Sin. To someone smarter than me......Does what I just wrote make sense? I kinda made it up as I went along.....lol. 



#13
Apr1012, 02:35 PM

P: 7

Thanks all for the helpful replies...




#14
Apr1012, 03:07 PM


#15
Apr1012, 04:59 PM

P: 834

I believe a constant current charge or discharge from a capacitor also gives a truely linear voltage ramp as well, but in this case you don't use a control voltage, but rather a control current. http://www.maximic.com/appnotes/index.mvp/id/3655 http://www.ti.com/lit/ds/symlink/lm555.pdf (page 10) http://www.vk2zay.net/article/196 (about a third down) Those circuits are using a constant current charge/discharge on a capcaitor to generate their ramps. What would make them not truely linear (within their operating ranges)? 



#16
Apr1012, 08:47 PM

HW Helper
P: 4,706

@yungman You were quick off the mark with that circuit. A couple of things as I understand it: his supply is 24V, and the solenoid requires a max drive of 11.2V (1.6A into 7Ω). So possibly include a resistor in the drain of the MOSFET to allow a lower power device.




#17
Apr1012, 11:01 PM

P: 3,842

Oh yeh, anything to lower the power. In fact I am kind of concern with the power dissipation also. I just throw the circuit out as an idea, definitely need some messaging.




#18
Apr1112, 01:17 PM

P: 7

@yungman: Thanks! Would you mind a quick sketch for this novice?
"I still don't quite know what you are trying to do. Do you mean you want to achieve a soft start by using a linear Ramp to ramp the voltage from 0 to 24V across the solenoid to create a soft start. Then when you want to turn off the solenoid, you want to ramp down the voltage to achieve soft turning off?" Yes, this is what I want to do. 


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