[Design Application]
Power-Supply Considerations For Servo Amplifiers
Specifying To Real-World, Power-Supply Requirements Can Greatly Reduce Initial System Costs, While Ensuring A Successful Design.
The "linear" power supply is simple: just a transformer, rectifier, and capacitor. But, selecting one to power a servo amplifier and motor can be anything but simple. What follows is a short tutorial outlining some of the problems motion-control engineers encounter and the solutions available for them.
While a control-system engineer thinks in terms of control/amplifier/motor/load, it's common for the motor to be chosen by a mechanical engineer (ME) who is more in touch with the actual pieces of an automatic machine. The ME makes choices based on mechanical units such as torque needed for acceleration, maximum rpm, and continuous power needed at a shaft, and so forth. These considerations yield a motor, rating, case size, and rotor inertia.
What's missing from this picture is the copper. Magnetic fields produce torque in the motor, and these are produced by ampere-turns of copper in the motor. It is this magnetic-field strength in the motor that gives it the power rating at the shaft.
The field strength is a function of the number of turns of wire multiplied by the current in the wire. This produces the motor constant (Km), which can be expressed in units of torque per ampere of winding current (Kt), or as volts per rotational velocity units (Ke). In the English system, one sees Kt as lb-in./A, or oz-in./A, and in the SI system, these become Nm/A. The rotational units (Ke) are V/krpm (volts per thousand rpm) in English, or V/rad/second in SI.
Now, ampere-turns as a product, is governed by two factors: the current in the wire, and the number of turns. Look in a motor catalog, and you'll see the real-world embodiment of this fact as a choice of windings for a particular case size and power rating. You won't see the number of turns listed, it's hidden in the motor constant. Ordinarily, there will be from two to four windings for a given motor, with different motor constants for each. Since torque = torque constant * current (Kt * I), for a constant shaft torque rating, it will take more or less current to do the job. Or, high current multiplied by few turns produces the same strength field in the motor as low current in a lot of turns. So which winding is the right one?
Enter the electrical engineer (EE), who is informed by the ME which motor must be used for the latest and greatest machine, and who must produce a control and drive system for it. How do you choose between the different windings available?
Choosing The Windings Begin with the maximum expected speed of the motor, then add a fudge factor for possible variations, to produce your design-maximum revolutions per minute. Divide this number by 1000 to get "krpm," run your finger across the columns in the motor chart where you find Ke (back-EMF constant in V/krpm), and do some quick multiplications. The result is a range of voltages that are the motor back EMF (BEMF) at your design-maximum rpm.
Next, have your ME give you the maximum torque expected during acceleration to the maximum rpm. Divide this (don't lose track of units) by the torque constant, and the result should be the peak current in amperes. Multiply this by the motor resistance to get the IR drop across the motor windings.
If you're going to be thorough, don't forget that the motor will heat up if you're driving it hard. Without extending this tutorial into all of the calculations needed to compute the motor rms (root-mean-square) current, let's just assume that you heat it up to the motor's design TMAX, frequently 150°C. If you start at room temperature, 25°C, this is a change of 125°C, with the result that the armature resistance can increase by as much as 49%! So now you have the "worst-case" terminal voltage for the system, which is the voltage required to accelerate the motor at the maximum rate up to the maximum velocity expected.
Mechanical considerations should give you maximum accelerating torque, TX, and the top speed, UX. From these, calculate the maximum motor terminal voltage that must be delivered by the power supply and amplifier combination (VTX) (Fig. 1):
VTX =
VAX + IX(RA + RTH) + LA(dIX/dt)
where VAX = maximum BEMF, or UX * Ke, and IX = TX/Kt.
For the power-supply selection, ignore the armature inductance, because the voltage across this will depend on the modulation type. It can be assumed to have an average value of zero under steady-state conditions.
Choosing a power supply is like most other decisions made in pursuit of a goal. All of this motor talk is about defining that goal. Congratulations, you just did that. Now you know the terminal voltage that must be delivered to the motor by the amplifier/power-supply combination if your machine is going to work. Next stop--the servo amplifier.
The Servo Amplifier Commonly a pulse-width-modulated (PWM) type servo amplifier controls the voltage across the motor to produce a current in the winding. This current, produces torque, which divided by inertia, produces acceleration. When the acceleration is integrated with respect to time, it produces velocity, and then position, and so on.
Suppose that we calculated our design-maximum motor terminal voltage to be 90 V dc. Do you choose an amplifier with a 24- to 90-V dc operating voltage range? You could, but it would be cutting it close to use an amplifier that is about to go into overvoltage shutdown just as it outputs the voltage that the motor needs. The problem is, the amplifier is not just a piece of wire connecting the power supply and the motor.
In a PWM amplifier, MOSFET or IGBT devices are switching away, but rarely get to turn on all of the time (achieve a 100% duty cycle). So, our maximum design voltage might occur at, say, 97% duty cycle, therefore add 3% to the input voltage to the amplifier.
I AM A STUDNT OF B.TECH 4TH YEAR.MY PROJECT IS PC CONTROLLED AUTOPILOT HELICOPTE.MAX WEIGHT OF MY HELICOPTER IS 5.KG. PLEASE PROVIDE ME DETAILS AND PARTS OF THIS PROJECT.HOW IT MAKE IN EASY WAY OF MIN AMOUNT?
Anonymous -September 28, 2009
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