soft start for powerhead

[clintyiu]

I read in the other forum that you cannot soft start a powerhead with cheapo capacitor/voltage reduction etc.

So I searched and searched for a soft start module that will work with a single phase shaded pole motor. The only circuit I can find will work with single phase motors but they didn't say whether they can handle shaded pole systems....and they cost a fortune.

If what I read is true then how do those commercial wavemakers provide a soft start feature. Has anyone tinkered with a simple capacitor soft start for their powerheads?

[titus]

Hello,

I wonder how does ramping the supplied voltage from zero to the rated level would work.

[DJ88]

I have been looking into this myself and the voltage is not what is variable. That is why wavemakers are so frigging expensive. They ramp the current up and down for a soft start. Not voltage. I'll find a thread on RC where I was told this.

[titus]

Hello,

<ul type="square">[*]I wonder what happens when we feed it V(t) = 120 * sin(60 * t)[*]I can only think of changing current by changing the inline resistance, or supplied voltage.[*]Man this is like back to the area where we think about current sourcing amplifiers, tubes or transistors?[/list:u:31a2577a24]

[reefburnaby]

Hi,

I am being a bit picky, but isn't it V(t) = sqrt(2) * 120 * sin(2 * pi * 60 * t) ?

As for soft starting, motors look like inductors and during the initial power up...it looks like a dead short. So, it needs to time energize and create a back emf. Usually, the electromagnetic strength is proportional to the AC current. So, during the inital power surge, the electromagnets are pulling and pushing the stator very hard -- hence the rough start. Once the back emf kicks in, less current flows through the electromagnets and the motor returns to its normal operating point.

So, if you can control the current (i.e. slow down the in rush current), then the motor will start up with less stress. You can use voltage to control the in rush, but the circuit is even more complex. The voltage envelope needs a x^2 like shape to work...rather than x like shape for current control. Another voltage method is by varying the frequency of the AC power supply (VFD), but that's even more complex.

Hope that helps.

- Victor.

[titus]

Hello,

Sorry Victor,you are right I forgot about the peak to rms ratio and I ignored the 2*pi part.

Hey why does it have to be V(t) = t^2 shape instead of V(t) = t? If it is V(t) = t^2, we can do use an op-amp integrator to drive a transistor.

[DJ88]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote

Hey why does it have to be V(t) = t^2 shape instead of V(t) = t? If it is V(t) = t^2, we can do use an op-amp integrator to drive a transistor.

I Kind of see where you are going Titus. Victor.. why? hmm? Using some form of integrator may work quite nicely. Nothing fancy. I hope I am reading this right.

hmm time to go do some more reading.

If this were possible for a reasonable price you could set up a wave making system where you can vary the output of the PH's could you not? nothing massive but a bit. Great option for a wave maker. Don't shut the PH's off but just slow em down so to speak. As long as you are over the EMF threshold they will spin.

Another application could be using MAG's for closed loop systems. Like what I want to do. Instead of getting the chatter on start up with using a plain old timer, hook them into a "wave maker" of sorts, with an integrator for start up, and then let the switching unit start and stop the pumps for random water flow on a larger scale in the tank.

Anyone agree? Or am I in left field??????

[titus]

Hello,

Well I don't know why it has to be a square function but nevertheless integration would give you a rising and decaying function with time constants that you can easily adjust.

[reefburnaby]

Hi,

Rather than digging in to the equations, lets get some trends.

1) Using a Step response to turn on the power supply. In this case, the current will overshoot and have exponential decay function from its peak current.

2) Using a Ramp response. In this case, the current will be a sort of a step response (not quite...but close). This is what a capacitor or a RC response will do -- which we know that it doesn't work very well.

3) Using a x^2 like/sin function. In this case, the current will change more linearily with time. I am just describing the general shape, but this is what I was referring to as a x^2 function.

Either that...or my wires are really crossed from writing too much.

- Victor.

[ 22 February 2002: Message edited by: reefburnaby ]

[ 22 February 2002: Message edited by: reefburnaby ]</p>

[titus]

Hello,

How about this:
V(t) = 1.414 * 120 * (1 - exp(-t/TC)) * sin(2*pi*60*t)?

This is something we all can do with an integrator-and-dump circuit preceding a transistor current driver. We'll let Darren to actually put it together. [img]images/smiles/icon_biggrin.gif[/img]