[SJ781]

Is the torque constant on a dc sevo motor? Say it makes 1000rpm at 50v and 500oz/in torque at 10A. If you take and lower or raise the voltage to change the rpms but keep the current the same does the torque change?

[balsaman]

well,

Yes. A servo drive will adjust the voltage to the motor to maintain the rpm it is comanded to go. If you increase the load the voltage will increase to maintain the speed. So lets say you want to go 30 ipm, and your axis hits some heavy cutting, the voltage will increase to maintain 30 ipm. If we keep increasing the load, the voltage and (current) will increase until you reach the point where the motor can no longer maintain the speed, at which time the motor will be running at it's rated voltage and current (or more) and the drive will fault.

So, just because the motor is running at less than it's rated speed, it does not mean it's running at less than it's rated voltage and current. Those two are load dependant, not speed dependant.

Eric

[dpaulson]

I thought that most dc servo drivers are PWM type and to raise and lower the speed, the pulse width is adjusted. It would still output 50v regardless of what speed it is turning.

[vacpress]

i think that the drives are both PWM controlled, and "current compensated".


i believe pwm signals can have various frequencies, but still have the amperage altered to try and force a closed-loop feedback device to reach its goal.

someone who knows something comprehensive should comment on this topic.

[balsaman]

If the pulse width is adjusted, that is effectively adjusting the voltage

Eric

[vacpress]

balsaman - is the amperage controlled seperately from the voltage? or is it on some linear scale or similar preset curve?

[balsaman]

The amperage is not controlled. The motor just takes what it needs up to the limit of the drive. If you take a motor and load it, the motor draws more current. The only way to force a motor to draw more current is to load it.

Eric

[balsaman]

horse power is volts times amps. Lets say you have a load that required .1 hp to move. Lets say the motor is running at 100 volts. Now change the load so it needs .5 hp to move. If you are still supplying 100 volts then the motor will draw 5 times the current to move it.

Eric

[arvidb]

balsaman, I'm not sure I agree totally with what you are saying.

First, power (watts) is voltage (volts) times current (amps). To get horsepower (which is also a measure of power), divide by about 735 (or about 745 depending on if you are using SI horsepower or imperial horsepower - or better yet, stay with watts ).

Second, I believe that most servo drivers control current, and not voltage to the motor. It works something like this:

1) You tell the servo to go to a certain position (by sending it a number of step and dir pulses, for example). If the servo sees that it's at some different position than the commanded, it calculates what speed it should use to get to the right position "in reasonable time". This is called the "position loop". So, input to this loop is the commanded position and the actual position, and from this a speed is calculated.

2) Now the servo knows how fast it should turn the motor. It compares this speed reference with the actual speed, and if it sees that it is off, it calculates how much torque will be needed to reach the commanded speed. This is called the "speed loop". Here, input is commanded and actual speed, and output is a torque command.

3) Since motor torque is proportional to motor current, the servo drive applies the voltage needed to get the commanded current. It is the current that is controlled, but the only way to control current is by adjusting the voltage. This is usually done with PWM.

Since it is the current that is controlled, it's usually ok to use a somewhat higher input voltage to the driver than the motor's rated voltage. The drive should have a current limit, and if this is set correctly (by you) the current will not have any chance to become high enough to damage the motor, even if the PWM peak voltage is higher than the motor's rated voltage.

Hope this makes things clearer, instead of the opposite

Arvid

[ESjaavik]

Arvid: It's very well explained. Quite unexpected coming from a Swede. :-)

The question is a bit fuzzy, but assuming SJ781 wants to know if the *maximum available* torque is constant, then the answer is:
Yes, up to the design RPM, usually found on the label. From that point on to max RPM the HP is constant and as a result the torque falls off. It may be a point to note that if the motor is at standstill the HP is zero, no matter how high the torque is. And this is theory. Usually a motor doesn't follow the ideal curve.

[balsaman]

Horsepower is the same as watts which is power. You are correct that you need to multiply watts x 746 to get HP.

Current is controlled by controlling voltage, as you say. The drive only can control voltage tho and the current is the result, which is what I was trying to explain...perhaps badly.

Eric

[ESjaavik]

The instant voltage from a chopping servo drive is zero or full voltage. The integrated voltage is varying. What most chopper drives try to do is to keep a target current flowing into the motor. That's also one reason it does not need a regulated voltage power supply: it doesn't care about the voltage.
If the voltage drops, it will compensate by holding it longer in the on-state to get the target current.

[SJ781]

Ok now for the next question.......What are the formulas for figuring motor data? How do you figure watts if you know nominal volt, rpms, and amps. How do you figure torque. and so on and so forth provided there are formulas for figuring such.

[Stevie]

Great thread

Now for my question and not to steal SJ781 thread

I have a Copley 503 drive amplifier and a QMC servo (brushless)
The drive needs 5Amps cont; with a peak of 10Amp
If I had 3 of these drives would i need a 30Amp unregulated power supply; or as the peak would not be used 99% 0f the time could I get away with say a 20Amp supply

[arvidb]

I'd say you need a 15 A continous supply with a peak capacity of 30 A for as long as the peak is needed

I guess it depends on how long the peak will last, and how important it is for you to really be able to get 100% peak torque simultaneously on all motors.

I think a good compromise is a 20 A supply with lots of capacitance. Perhaps you can use a 20*1.4 A transformer, put 20*1.4 A slow blow fuses between transformer and rectifier, and then put 30 A fuses after the capacitors? This would overload the transformer during the rare peaks, for as long as the slow fuses holds up.

Hmm, it would be nice to get other people's comments on this though. I'm not really sure what I'm talking about Maybe it'll be cheaper and more reliable to just use a 30*1.4 A transformer and less capacitance...

Also, since this is a PWM amp, remember that power is conserved; that is, if you use a 50 V supply @ 10 A then you can draw some 20 A @ 25 V at the amp outputs (minus losses).

Arvid

[Stevie]

I think a good compromise is a 20 A supply with lots of capacitance. Perhaps you can use a 20*1.4 A transformer, put 20*1.4 A slow blow fuses between transformer and rectifier, and then put 30 A fuses after the capacitors? This would overload the transformer during the rare peaks, for as long as the slow fuses holds up.

Wish i understood that lot; it sure sounds good

just what are the * for

[arvidb]

When you rectify the AC from the transformer, you get a DC voltage that is 1.4 times the AC voltage. (Really the square root of 2 times the AC voltage, minus about a volt or so that's eaten by the rectifier diodes).

So let's say you have a 30 VAC transformer rated at 10 A. This is 300 VA = 300 W into a resisitive load.

Now let's say you rectify your 30 VAC to get about 42 VDC. But the transformer is still only 300 VA, which means you no longer have 10 A of current. You can only draw 10/1.4 or about 7 A after the rectifier (7A*42V = about 300W). You can't get something for nothing!

So the *'s in my post was to adjust the AC current so that you would have 20 ADC after the rectifier.

Arvid

[ESjaavik]

Arvid: Usually what you write is well explained and correct, but here I would dig deeper.

You can draw more than 7A (in your example). The voltage will drop as you draw more current, so you can draw some more current again dropping the voltage. But the sum of voltage and current should not exceed 300VA continously.

So it is still true that you don't get something for nothing.

So let's bring time into the picture.

Actually you can probably draw 20A maybe even 40A for short bursts if your average current draw is lower than 10A. But when you do so, the voltage may drop enough to fault out the drive. If you know you have this load pattern you may use 40A fuses and a thermo switch on the transformer to protect it if the high current bursts lasts too long and thus put it in danger of destruction.

Let's say you make a PCB drilling machine using servo drives. It will typically run with Z-axis idle when moving from one hole coordinate to another using max acceleration, then highest speed, then max retardation. This will be done in a short time, and during the retardation phase it will feed power back into the reservoir capacitors. Then while drilling, X and Y will consume minimal power, and Z moves (fairly) slowly down. Then quickly up. This is a pattern with short bursts of power demand, recouperating some of it (to the cap's), and rests in between. So the peak current demand after the capacitors can be high, while the average is not. Weight and cost can be saved by not designing only for peak current, but also for average.

This I believe is relevant for many hobby mills/lathes where the cuts proceeds at slow feeds, but where high speed movements are desired between cuts. These movements will be of short duration and with rests in between. So it will be overkill to size the transformer for continous high speed movements.

SJ781: If you know torque at a certain RPM, you can find output power. Multiply torque in Nm with RPM, divide by 10 and you have output (shaft) power in Watt.

If you find the voltage and current rating you can multiply them to find the (input) wattage. The output will be less than this.

If there is a max Amp given, don't ever exceed this, even for fractions of a second! It can demagnetize the motor. One way of exceeding it is to connect the motor to a high capacity battery, even if there is no load on the motor. So don't. You can however run it close to the max current for short periods. But it will heat up the motor fast.

There are formulas for figuring out whatever you like to know. The problem usually is you lack the data to plug into them. :-)

[Stevie]

Great explinations guys

I'll have to look for the right transformer then

[arvidb]

Einar: thanks, and you are correct of course. Good idea to use a thermo switch!

Arvid