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Current sensing using proportional valve 3

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markus506

Electrical
Apr 6, 2006
16
Hi everyone. First post here, and not even sure if I'm in the correct forum.

I am currrently working on a new product, whereas I am controlling a proportional valve using a PWM output. The PWM is controlled by a dsPIC. The actual PWM driver is an ST VND600SP. I am using the current sensing ability of the VND to determine output current. The problem I am having is how do I calculate the actual output current to the coil based on the PWM signal? Basically, without knowing an exact coil resistance, how do I figure out my output current in order to compare it to the sensed current to determine if any compensation is required?

Best regards,

Mark
 
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Hi Keith;
Hope I get this right.
The inductive reactance and the resistance act at 90 degrees to each other. If the impedance is 5 times the cold resistance the impedance will be 5.01 times cold resistance.
If the hot resistance is 125% of the cold resistance then the hot impedance will be 5.154 times the cold resistance of the coil. Our original 25% increase in resistance is now a 2.9% increase in impedance.
This may be acceptable.
respectfully

 
Yes, Keith, that VND device does vary greatly in its 'K' factor. However, there are other aspects of the chip that we like. Since the product is still in development, alot can change, hopefully this current sensing method will be one of them.

Note that most applications will use a PWM of aproximately 250 Hz...fairly low frequency.

And I'll make sure to check out that Maxim chip on Monday morning first thing.

Again guys, thanks for all the suggestions. Keep 'em coming!!

Mark
 
Back from vacation, so late to this thread, but puzzled by the direction it has taken. I want to bring up a very basic issue.

Mark: a device like this with PWM outputs provides a time-averaged voltage to the load. For example, with a 36V supply and a 50% duty cycle, it will apply a time-averaged 18V to the load, regardless of the impedance of the load. Now, the current that this voltage produces does depend on the impedance of the load, but the current-sense circuit (I presume some sort of "current mirror" in the chip) will detect the current reasonably accurately (I'm sure some devices better than others).

In control-loop terms, your control algorithm has a voltage output and current feedback, and you will need some sort of feedback law for compensation. Usually, PI (proportional plus integral) is sufficient. This is every-day stuff -- done all the time for both motors and hydraulic valves. If the resistance of the coil increases, as it will with temperature, the output voltage require to get a given current level will increase, but your feedback law should handle this automatically.

Because the resulting current waveform is a sawtooth, you do have to be careful about how and when you sample the current. Some here have suggested oversampling and averaging. You may also just be able to sample at the midpoint of your on-time and/or the midpoint of your off-time, which does a pretty good job of catching the average current level.

A valve that has built-in position feedback of the spool is a servo valve. These tend to be much more expensive and finicky (easily clogged) than proportional valves, although they tend to do better, particularly in the zero-crossing for bi-directional applications.

Curt Wilson
Delta Tau Data Systems
 
Well, Curt, better late than never!

I will admit (sheepishly) that I had not even considered using a PI controller in my firmware. I guess this is what I get for falling asleep during my controls classes. I will look in to this right away.

(start of ramble)
As for the feedback, the current sense circuit uses a 5.9K resistor and the voltage is measured across that and fed into an ADC. The resulting waveform is pretty flat, almost as flat as the PWM pulse. I am interrupting on the falling edge of the sense voltage so my sample voltage is approximately at the highest point of the waveform. But since I am using formulae to calculate the output current based on a number of factors, I am sensing current. (End of ramble).

Thanks again for joining in on the discussion, your post was most valuable. I'll let you all know how I progress.

Mark
 
Hmm...well, I tried to do some modelling of the coil I have here. However, the step response of the coil shows that it is behaving almost like a pure inductor (minimal resistance), thus a first order system. Is this something that I am able to compensate using a PI or PID contoller? I don't believe that this type of system can be over/under damped, etc. Excuse my ignorance, but this is the first time I have run into step responses since school, and, from memory, we only ever really dealt with 2nd order systems. Can someone shed some light on this for me?

Mark
 
If you are PWMing at a high enough frequency for the system to be responsive, the inductance will dominate the immediate response. So in the time frame of a PWM cycle, you can usually ignore the resistance and just get the slope of the current response from the inductance. Over a longer time frame -- multiple PWM cycles, where you can use the average value for each cycle, you will need to take into account the resistance of the coil.

Curt Wilson
Delta Tau Data Systems
 
Markus,

You might like to explore the servo valves produced by Moog or Bosch-Rexroth. Moog are hideously expensive, but are among the best in the business. Their website has a fair bit of technical information on how their valves and controllers work. It might trigger an idea.


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I don't suffer from insanity. I enjoy it...
 
Also sorry to be late in this thread.

In my world, proportional valves always have a feed-back signal called "actual position". It is usually a +/- 10 (or sometimes 5) V DC signal that accurately tells where the spindle is. 0 V meand centered and +10 V means 100 % open in the plus direction.

Have you looked for such a signal? It is really difficult to do any serious work without having position feedback from the valve. The hydraulics guys seem to rely heavily on actual valve position for many control tasks.

Gunnar Englund
 
Well Gunnar, it seems that the application we have, the guys are using just a cheap, PWM controlled flow control valve. This valve is supplying flow to the directional valve, which is on/off.

I agree that it would be much more convenient if the valve had a feedback signal for position, but since we are planning to interface to existing valves, it does not look like it is going to be an option.

Mark
 
skogs; I gather this is an application where a human is part of the immediate control loop. So accuracy is not going to matter 'all' that much. Mark maybe needs 3 or 3-1/2 bits of accuracy.

Keith Cress
Flamin Systems, Inc.-
 
Well Mr. Skogs,

3 bits -> 8
4 bits -> 16
delta = 16-8 = 8
1/2 delta = 4

3-1/2bits = 3 bits + 1/2 delta = 12

3bits -> 100%/8 = 12.5%

3-1/2bits -> 100%/12 = 8.3%

That's my theory and I'm sticking with it.[infinity]

Although actually now that I've revisited it, Mark really may only need 2-1/2bits to 3. Muwhahahaha.


Keith Cress
Flamin Systems, Inc.-
 
I think repeatability is important. With good repeatability, the absolute precision is not particularly important. Enough linearity to avoid excessively low P or high P at the low and high speeds will be nice. (P; proportional band, or ratio of output change to input change.)

markus506 said:
Basically, without knowing an exact coil resistance, how do I figure out my output current in order to compare it to the sensed current to determine if any compensation is required?
This may be the answer all along. If you can control on current, the amp turns will remain the same. When the coil heats up the output will rise to maintain the set current and thus automatically compensate for the change in resistance of the coil. The same current means the same amp turns and the same valve position.
I agree with skogsgurra on the position feedback. I do understand that it may not be possible to attach a feedback potentiometer.
respectfully
 
So, we have a discrepancy there. I reasoned like this: 2^3 = 8, 2^4 = 16. 2^3.5 = 11.3 which gives 100/11.3 = 8.8 percent.

Love this ;-)

Gunnar Englund
 
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