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.
Those amplifiers aren't perfect conductors either. They actually have something called "on-resistance" or "output resistance," RO(Fig. 2). This is actually a sum of resistances in the transistors and the internal resistors used for current sensing. This output resistance might be in the range of 0.2 to 0.5 Ω. So, take the design-maximum motor current, multiply it by the amplifier output resistance to get the "IR drop." Add that figure to the equation to get the voltage required at the input of the amplifier such that the output produces the design maximum terminal voltage to do the job at the motor shaft.
Working backward from the motor terminal voltage, through the amplifier to its input terminals, what you get is the minimum voltage required from the power supply. This is the voltage needed at the amplifier's input power terminals to run the motor in the way the ME intended.
If we allow that our 90-V motor might need 20 A for its peak acceleration, we can quickly see why a 90-V amplifier won't work--90 V/0.97 = 92.8 V (remember the maximum duty cycle?). And, 20 A * 0.3 Ω = 6 V. So, the input voltage to the amplifier has to be at least 92.8 V + 6 V= 98.8 V. Let's call that 99. From this you can see that your "90-V" motor will need something more than a 90-V power supply and a 90-V amplifier.
But, before you pick up the phone to order that 99-V power supply, check the datasheet for something called "regulation." This is the no-load voltage minus the full-load voltage, divided by the full-load voltage. Multiply this by 100 and you can express it in a percentage. Obviously, if the voltage didn't change as the load changed, the answer would be 0%.
You will not find this number on a linear power-supply datasheet. What you will find is a number more likely to be from 5 to 15%. This is caused by the internal resistance of the power supply. As long as you select your power supply based on the full-load output voltage, your system will have enough voltage to drive the motor. The effects of non-zero regulation will, however, affect the choice of servo amplifier.
What about your local electric power company? Read the fine print and you'll find that the 120 V (or 115 V, 100 V, 200 V, 230 V, or 240 V, depending on your country and locale) that is the nameplate voltage on your power delivery system isn't always accurate. It might vary as much as ±10%. Result: all of that stuff about minimum voltage needed at the amplifier terminals, and at the output of the power supply at full load had better be happening at low-line mains voltage, or your control system will be offline when the power dips. In fact, if you examine closely the operating range of equipment made for the U.S., you will commonly see a range of 105 to 132 V ac. This means that for a nominal 120-V supply, the output might dip (120 - 105)/120, or 12.5% from nominal during those brown-outs, and rise 10%, too.
In order to guarantee the minimum power-supply output voltage under the load calculated previously, you must now raise the voltage rating by 12.5% to be prepared for a low-line situation. This will handle the motor needs nicely.
All that's left now is to consider is the amplifier. From 120 V, the voltage can rise 10% to 132 V. We must be sure that the amplifier will be OK with this high-line, no-load voltage before we wrap and bag this design.
This is a key point of linear power-supply selection: when the power line sags, and the load is at a maximum, there must be enough voltage to drive the motor and amplifier. And, when the motor is idle, amplifier disabled, and the power line at its maximum possible voltage, the output will rise to a maximum that cannot damage or disable the amplifier. From a design point of view, pick the power-supply voltage based on expected line excursions and your minimum amplifier input voltage. Then, check the maximum voltage to see if it's OK for your amplifier's operating voltage range.
Back at the amplifier terminals, two things are happening. Most of the power entering those terminals is converted to watts at the motor terminals, but some of it just makes the amplifier hot. There we go again, those darn amplifiers just aren't perfect. But, they are pretty good, typically 90% or better. That is, 90% of the power going into the amplifier zips right out the other end, into the motor. Furthermore, the PWM action at the amplifier outputs acts like a dc transformer. The constant bus voltage is modulated to appear as a variable voltage from 0 to ±HV. So, the real wattage needed at the amplifier input becomes motor power plus amplifier losses. Divide this actual power by the needed bus voltage, and you'll get the dc current needed from the power supply.
Power-Supply Definition Elbow room is great. A good design doesn't need to be more tightly defined than necessary, because tight tolerances cost money and narrow the range of operation. If motors had no internal resistance, amplifiers no losses, and power supplies perfectly regulated, none of this would be necessary. We choose not to wait for this to happen, but offer instead the step-by-step procedure for doing by the numbers what we have described above. We will not delve into the origins of the maximum revolutions per minute and accelerating torque at this time, but will assume that the mechanical types have dutifully left these data on your desk, along with a datasheet of the motor model of choice. Current RatingNow that you know the key voltages for your motion system, it's time to find the current-rating of the power supply. The easy way is to make this the same as the amplifier's peak current. But, in incremental motion systems, this will typically occur only a portion of the overall time, so using this figure will usually result in an oversized (and overpriced) power supply.Linear supplies are designed to tolerate short-term overloads of 200% to 300% of their continuous ratings. You can use this figure to find a rating closer to your actual needs. The real mechanical watts in your system are determined by the average motor speed, motor current (i.e.: torque), and duty cycle. Try to find a value for the average motor speed and current, multiply these, and get an average power. Take this number and divide it by the power-supply voltage, and you will have an estimate of the supply-current rating. This will generally be somewhat less than the peak amplifier's current value.
Now you have completed the first pass through the process. There were three windings for this motor, weren't there? Did this iteration result in a power-supply voltage range that fits with any servo amplifier you have seen lately? Yes? Jot down the model number, and proceed. If not, do it again! Fact: You are a techno-juggler with at least three pieces to keep in the air at a time. You must pick your parts so that the motor, amplifier, and power supply not only work, but work together simultaneously. But, if it was that easy, everyone would be doing it.
Suppose that you have done a fine job, and now have a motor winding, amplifier, and power-supply choice in hand. Are we there yet? For one motor, sure, but most real systems have multiple axes. Do we repeat this process for each axis, and then add up the watts and buy a power supply to handle all of these at once? Unless your loads are all pumps that run continuously at the design maximum load, the answer is "no." It's time for a major engineering fudge-factor.
Duty-Cycle Fudge FactorIt's called duty cycle (not the PWM type this time). In incremental-motion systems, those factory-automation engineers most frequently encounter, not all axes run all of the time. If your process is very deterministic, you might succeed in calculating the watts per axis, and adding these successfully. But, it's more likely that your machine runs different motion programs depending on a process that changes over time. The best way to sort this out is not on paper, but on the floor (of the R&D lab, in this case).Begin with a power supply sized for, say, one axis of a three-axis machine. Run your machine as it was intended. Come back in one hour, and place your hand on the power transformer. The copper in its windings is an excellent rms detector of current. We gave it an hour to run because of the long thermal time-constant of the transformer itself. If the transformer is hot, increase the power rating of the supply and try again. If it is cold, you oversized. Downsize the power rating of the supply and try again.
If the transformer doesn't get noticeably warm, you're probably not getting your money's worth. So, you can buy the confidence of over-specing, or play roulette with the winding temperature rating of the transformer. It's up to you. But, the final decision based on multi-axis, duty-cycle demands is tough to make on paper, so allow some lab or floor time to sort this out in the design cycle. The reward will be a power supply that saves you money, and reliably delivers the power you need.
Making Everything Work Are we done yet? If you know all about grounding, cabling, fusing, and handling regenerative energy, the answer is "yes." If not, continue.First, cabling. How do we connect all these parts together? Size the wire first. Use the charts in the wire catalogs, and any local electrical codes and wiring practices, if needed. Teflon-insulated wire is best for power wiring due to its very high-temperature and voltage ratings. Next best is cross-linked or irradiated PVC (polyvinyl chloride). Less common, and much cheaper than Teflon, it will withstand solder-iron temperatures without melting. And it's more flexible and easier to strip than the Teflon. The last choice is regular PVC. It's OK for lower voltage, in cooler-running or lower-powered equipment, but not a good choice for anything connected to the mains, or above, say 75 V dc, or so.
Next, PWM amplifiers put fast ripple currents into power supplies, in addition to the average currents they draw (the ones we have been discussing). To keep these currents from "talking" to adjacent cables, twist power-supply leads (one-to-three turns per inch) together. For newer CE-compliant applications, consider using shielded cables to reduce EMI emissions from these wires. Typically, these are going to be bigger wires (AWG 18 to 12), so it's best not to try soldering them, but use crimp-on connectors to mate with screw-lugs, or Euro-style compression connectors instead.
Fusing Most people think that the fuse is meant to protect their equipment. In fact, the prime function of the fuse is to protect the world from your equipment when it fails. The survival of your machine is secondary. At the minimum, you will want a fuse between the mains and your final choice of power supply. Given the inrush currents associated with capacitor-input supplies, this will be a time-delay fuse.Choose the current rating based on the power-supply nameplate rating, and add 25%. Remember, a fuse should only carry 80% of its rated current continuously. If you have a 500-W power supply operating from 120-V ac mains, the rated line current would be 500/120 or 4.17 A. Divide this by 0.8 to get 5.21 A. Make the choice between a 5- or 6-A fuse based on how much of that power-supply rating you're likely to use in practice.
The choice of a second tier of fuses to protect each individual amplifier is up to you. Depending on the cost or value you assign to an individual amplifier, you can let the pc-board etch be the fuse of final resort. And, you only need to put a single fuse in the mains wiring to the power supply, or fuse each amplifier individually. Given that the amplifiers usually shut down with internal (or external) short circuits, and that total, catastrophic failure is rare, most users decline the individual fuse per amplifier. It's your call.
Wiring And Grounding Don't daisy-chain wires from power supply, to amplifier 1, and then onto amplifier 2, etc. Instead, use a star system, with one twisted pair from each amplifier connecting to the terminals of the power-supply filter capacitor.Also, ground your system properly. Remember, there are two currents in those wires between the amplifiers and the power supply: slow currents, or the average currents that drive the motor, and fast currents, the ones you can't see, but exist due to the switching of the transistors in the amplifiers. The slow currents make measurable IR drops between the ends of the cables. The fast currents produce even bigger voltage spikes in the inductance of the wires.
Enter relativity. The voltage between the ends of the power wires are relative to those ends. Ground the negative terminal of the power-supply capacitor, and these voltages will appear between the amplifier ground terminals, and chassis (general electrical system) ground.
From the viewpoint of the amplifier circuits (remember relativity), the world just got very noisy. From where the control system sits, the amplifier looks really noisy. You can eliminate this noise by disconnecting the power-supply capacitor from ground, instead grounding each amplifier with its own short grounding conductor to a local star ground terminal. This way, the wiring noise will be seen between the capacitor negative terminal and ground, but there it won't matter to the amplifiers, or to the control system.
Regeneration Finally, because this is a motion-control system with stored mechanical energy, there's regeneration to consider. Your servo amplifier is a two-way street when it comes to power. Most of the time you deliver power to the load. But, once you accelerate a load to a given speed, or elevate it to some height against the pull of gravity, you have stored mechanical energy that must now dissipate to bring the load to rest. This energy flows from the load back into your system.The motor handles this two-way power delivery easily, as does your four-quadrant PWM servo amplifier. But, when you get to the power supply, you're looking into the business end of a rectifier that only passes power in one direction--from the power line into the filter capacitor. In the case of regeneration energy, it's a brick wall. The solution is a "dumper," or regenerative energy dissipater.
These devices connect across the filter capacitor terminals, and rapidly switch a high-power resistor across the terminals to dissipate the energy as heat. As more energy comes back from the load, the dumper switches with an ever-increasing duty cycle to mimic a resistor with a lower and lower resistance. The effect is to have a rheostat across the cap that is electrically variable to satisfy the Ohm's law solution R = E/I; where E is the bus voltage during regeneration, and I is the current coming from the motor during deceleration. As soon as the load is brought to rest, the "pump-up" of the power supply stops, the voltage drops, and the dumper turns off.
These devices are common accessories from amplifier makers, and found as part of the amplifier, or as external accessories. They are usually adjusted to a voltage that is below the amplifier's shut-down voltage, but above any steady-state, high-line, and no-load voltages. The peak power dissipated by these devices is very high, and they are intended to operate only during deceleration intervals when mechanical energy is to be dissipated.
By now, you should know enough to pick your motor winding, select and size a power supply, roughly spec a servo amplifier, and wire the whole thing together. You're now ready for your next assignment--making it all work! But that's another story.
Linear Power-Supply Regulation And Internal ResistanceThe full load current, in terms of the voltage across the load resistance, is found through the equation:
IL= VFL/RL
The load current, in terms of the load current and internal, or no-load voltage is given as:
IL= VNL/(RI + RL)
Combine the above two equations to get:
VFL/RL = VNL/(RI + RL)
Solve for RI when load resistance, no-load, and full-load voltages are known to get:
RI= RL x \[(VNL - VFL)/VFL\]
However, these data are not usually shown on a datasheet. It's more likely you'll find the output (full-load) voltage, current, and percent regulation. So, look again at the diagram above and see that VFL is actually the internal no-load voltage minus the drop across the internal resistance:
VFL = VNL - (IL x RI)
Rearrange to solve for RI:
RI= (VNL - VFL)/IL
Now, remember that percent regulation was defined as:
Percent regulation = ((VNL - VFL)/VFL) x 100
Insert the above term into the equation for RI and reduce it to give:
RI = (percent regulation/100) x (VFL/IL)
The result is a handy equation to model the effective internal resistance of a power supply based on specification-sheet data--namely, full-load voltage and current, and percent regulation. (Fig. 3)
Worst-Case, Power-Supply Output Voltage
Lowest Output Voltage (VLO(DC)) =
(VLOWLINE/VNOM) x VFULL_LOAD
Highest Output Voltage (VHI(DC)) =
(VHILINE/VNOM) x VFULL_LOAD x (1 + %R/100)
where:
VLOWLINE = low-line mains voltage
VNOM = nominal mains voltage
VHILINE = high-line mains voltage
VFULL_LOAD = full-load supply output voltage
%R = power-supply regulation in percent
Example:
A power supply has these specs: 65 V dc, 8 A, and 5% regulation.
VLOWLINE = 105 V ac
VNOM = 120 V ac
VHILINE = 132 V ac
Lowest Output Voltage = 56.9 V dc
Highest Output Voltage = 75.1 V dc
Compare the minimum required voltage at the amplifier input terminals with the Lowest Output Voltage to check adequacy.
Compare the Highest Output Voltage with the amplifier's operating voltage range to ensure that the amplifier will not shut down due to an overvoltage condition.
Power-Supply Definition Procedure1. Find the maximum motor armature voltage:
VAM = VDM x Ke
where:
VAM = maximum armature voltage
VDM = design-maximum rpm (in krpm)
Ke = V/krpm (volts per rpm/1000)
Add at least 10% to compensate for manufacturing tolerances, and to include a little operating headroom for the control system.
2. Find maximum IR drop (VRM) in motor resistance:
IDM = TDM/Kt
where:
IDM = design-maximum current
TDM = design-maximum torque
Kt = torque constant
TDM and Kt should use same units for torque
VRM x 1.5 x IDM x RA
where:
VRM = maximum IR drop
RA= armature resistance
1.5 = addition of 50% to VRM to cover hot-motor scenario.
3. Find design-maximum motor terminal voltage:
VTM = VAM + VRM
where:
VTM = design-maximum motor terminal voltage
Multiply this by maximum current to get armature wattage. Now you know the maximum power input to the motor terminals in watts.
4. Find minimum power-supply voltage at amplifier input terminals:
HVMIN = (VTM/DM) + (IDM x RO)
where:
HVMIN = minimum power-supply voltage at amplifier input terminals
DM= maximum duty cycle (e.g. 0.97 for 97%)
RO= amplifier "on" resistance
5. Find full-load, power-supply output voltage:
VFL = HVMIN x (VNOM/VLOWLINE)
where:
VFL = full-load power-supply output voltage
VNOM = nominal mains voltage
VLOWLINE = low-limit mains voltage
In the power-supply catalog find a device with a full-load output voltage near to this. Take this number, and calculate the high-line, no-load voltage(VHI(DC)) according to the procedure already outlined (see "Worst-Case Power-Supply Output Voltage,"). This is the high-line voltage that the amplifier must tolerate without going into shutdown.