Stepper motor - back emf

[grevillea]

I am building a project with a stepper motor - NEMA23 76mm 18.6 kgf.cm holding torque. The motor may be loaded close to its holding torque limit (after allowing for reductions in torque limit due to half-stepping and power reduced driver operation during hold) and the load may generate back EMF. Motor rated current is 2.8A.

The motor is powered by 24V 15A switch mode power supply - a big overkill.
I am planning to also power logic circuits using a DC-DC converter to convert 24V to 5V (actually, 6.5-32V down to 5V).

I am concerned about the risk of voltage fluctuations on the power supply causing problems for my logic. So far, I have: 1N4004 as reverse voltage protection in the 24V line to the logic; 1.5KE27A transient voltage suppression diode (27V breakdown) across the input voltage (after the 1N4004) and in parallel with 1000uF electro cap and 10uF ceramic cap and then the DC-DC converter to generate 5V for the logic. Also a cap across the converter outputs as specified 100uF electro.

I have researched on the Internet and am aware that there are two ways the motor can generate back emf into my circuit - induction from the coils when a step is taken and generation of voltage from the rotation of the permanent magnet rotor.

Induction: The current chopping driver handles the current from the coils by using the 24V supply to drive the coils into the direction it wants the current flowing. When a step causes a direction change in the coil current, the driver has to reverse the current flow and it does this by attaching the 24V power across the coil, so you have 24V applied across the coil but the current (for a little while) flowing in the wrong direction. This may generate over voltage on the power rails if the power supply cannot absorb the reverse current. However, from what I have read, the voltage should not be very high unless the reverse current is blocked. The caps in the power supply would hopefully absorb it, or some may be absorbed by the caps in my logic input circuit. The inductance of the motor is 3.6mH and the coil resistance is 1.13ohm so based on a formula I founf on the Internet I calculate 380us time required for the current to reverse, and a charge transfer of 0.53mCoulomb. So, I think it should be OK. Am I safe?

Generation: This is harder to get a handle on. From what I have read, steppers are not good generators and need to be run at 1000's RPM to get mA out of them. So once again I believe that the generated current can be absorbed by the power supply and caps. However, I am unsure about the possibility that winding down a heavy torque load may generate current for a sustained period of time, thus possibly filling the caps and raising the voltage on the power rails. Does a switch mode power supply have any way to get rid of current that is forced back onto its inputs? Maybe my TVS diode would safely burn up excess power?

Any comments or encouragement appreciated.

 

[Diver300]

The energy in the 3.6 mH at 2.8 A is ½ * L * I² = 0.014 Joules

The supply voltage is 24 V, so each Joule is 1/24 coulombs, and 0.14 Joules is about 0.58 mCoulombs. That is pretty close to your calculation. It will increase the voltage on a 1000 µF capacitor by about 0.58 V, although you would see a bigger change due to the ESR of the capacitor.

I really don't know how much energy you would get back out of the stepper motor with a big inertial load. My guess is hardly any. Your TVS will absorb some, but they are not designed for continuous power dissipation.

I think that the normal way of dealing with the power, if there is any, is with a dynamic braking resistor. You could just connect the bottom end of the TVS to the base of an NPN transistor. Connect the emitter to ground, and the collector to a 6.8 Ω 100 W resistor. The other end of the resistor goes to the 24 V line. A base - emitter resistor of 1k or so would be good to stop leakage current turning on the transistor.

That will absorb as much power as you will ever generate with the stepper, and probably loads more, and only cost a tiny fraction of what the rest of the stuff costs.

 

[grevillea]

Diver300: Thanks for your reply. It has been very helpful. I am posting a detailed reply in case it is helpful to others in the future.

The key insight was to switch my thinking from voltage/current to energy. I repeated your calculations of the charge pushed back, and compared it with my previous calculations that are based on i(t) through the inductor. One of the differences is that the formula I used includes the phase resistance, so that explains the smaller value for the charge pushed back.

Taking the same idea of energy into the dynamic braking situation, I calculate that 18.9 kgf.cm corresponds to 1.85 N.m torque (gravity = 9.8). Being the holding torque, I believe that this is the maximum torque that the motor can exert on the load, whether driving or braking. (If the motor slips, there will be periods of time when the motor torque is in the direction of rotation, so the braking torque is expected to be further reduced.) Using 5 revolutions per second as the braking speed, and assuming that the braking torque is equal to the holding torque (clearly an overestimation), I calculate P = 2pi * torque = 58W. From curves i have seen elsewhere on the internet, the dynamic torque of a stepper degrades rapidly as the motor speed increases - since my power supply is 24V, the fall-off is quite rapid. So I have little reason to expect any more than 60W of dyanmic braking power. Do you agree with my logic and calculations?

If the motor does slip, and the back EMF direction agrees with the coil current direction, then the motor will be driving the power rails from positive to negative, potentially drawing more current from the power supply than the 2.8A required for the nominal coil current. This introduces the risk of putting negative voltage onto the power rails, but with a 15A power supply and only one motor I don't think i have to worry about negative voltage actually being generated. However, in this situation, the energy will be burned up in the motor coil, power supply, and wiring. Fortunately, it is a transient event, and a slipping motor means loss of control, so something that the application will correct.

I have limited space for the braking resistor (and heat sink), so I need to carefully consider the design parameters.

Analysing the circuit that you suggested, the NPN transistor bears the load at low current drains (right?) and the resistor's load increases as the current drain increases. With 6.8 ohm resistor, full power is absorbed by the resistor when current is 3.67A (assuming 25V clamping voltage which is close enough to the specs of the TVS diode). The power dissipation is then 91W. (However, with maximum coil current of 2.8A, surely there can never be more than 2.8A of braking current? The motor would surely have to be slipping?)

Maximum power load on the transistor occurs at 1/2 current, i.e. 1.84A and is 23W.

Redesigning for 60W braking power, full load is 2.4A and a resistance of 10ohm is appropriate. Maximum power on the transistor is 15W when the current is 1.2A.

I am inclined to go somewhere in between these two designs. Key considerations are the physical size of the resistor and the power load on the transistor which must be heat sinked to my aluminium enclosure (don't know the thermal resistance of the enclosure - 1.6mm thick aluminium) because of space limitations. At 10 ohm there are resistor options in a TO220 style package as well as much larger options that would not require heat sinking for the resistor. It also helps that in my application braking at full power could not continue for more than a few seconds (and is very unlikely at all), and the (big) power resistors that I have looked at can handle overload for several seconds without problems. I do not have the physical space for 6.8ohm 100W power resistor, but I dont think I will need it.

Again, thanks for your help.

Regards
Grevillea

 

[Mr RB]

What is the load, and what top speed and deceleration rate? If the decel is not too severe the current returned to the PSU will be less than the wasted current used by the (always inefficient) stepper motor so the PSU voltage will not rise.

If you have a hard deceleration the current returned to the PSU is generally absorbed by the other stepper motors in the system.

 

[Diver300]

I think that you have a full bridge arrangement, with diodes, so then back emf or power from dynamic braking will generate current that will tend to increase the supply voltage, not drive it from positive to negative as you said.

I would agree that 60 W is generous. My suggestion of 6.8 Ω was a quick calculation for worst case based on the 2.8 A, so you could almost certainly go for more resistance and less power.

You do not need a big heat sink on the transistor, as it won't be held at the maximum power point for any length of time. You could even add hysteresis so that the braking resistor is either on or off, say turning on at 25 V and off at 24.5 V, and then the transistor dissipation will be minimal.

I suggest that you try it out, as you might not need any thing more than the TVS, and if you do need a resistor and transistor, it might well be as large as 100 Ω.

As RB implies, if you have several stepper motors, you only need one braking resistor on the power supply.

 

[grevillea]

Thanks for further comment.

The application is driving a single stepper motor that will operate a winch (a small one for small loads of only a few kg). Worst case I can imagine is lowering the load at limiting torque for a period of 5s - limited by the operating distance. Sadly, there are no other stepper motors in the system to consume the back EMF.

I am wary to 'try it out' because if I do get overvoltage on my logic and blow up $300 control board, it will take a long time (and $) to replace. I am in favour of conservative design. Also, the operating conditions are not fixed, so it would be difficult to test all the different possible load situations that might be encountered in the future. TO add complexity, the winch is not yet built, so I cant test it yet.

I don't understand the concept of the transistor as switching on and off. The transistor I have selected (but can easily change) is NPN D44H11G. It has current gain of 60.
I would have assumed that as the voltage on the TVS diode rises, the current flowing through it increases (very non-linearly) and that drives the base of the transistor, so that the tranny dumps additional current at 60x the current that is flowing through the TVS diode. I have been expecting the transistor to behave as a linear amplifier, but perhaps the TVS diode itself acts as a switch? The voltage drop across the TVS diode is very stable as the current increases.

I like the idea of hystersis but have no idea how to implement it. A Schmitt trigger circuit needs two transistors from what I found so far on the internet, but that seems unnecessarily complex for this application. Perhaps a capacitor could capture some current flowing through the TVS diode and hold it to drive the transistor even when the voltage drops again? But too much and the tranny would try to dump the power supply, I think. I would very much appreciate a pointer in the right direction here.

Thanks,
Grevillea

 

[shortbus=]

For use as a winch, wouldn't it be better to use a small worm gear speed reducer between the stepper motor and the lift cable?This is the basics of a winch. The worm gear mechanism is self locking and will not require the motor to be kept on to hold the load. The use of 'holding torque' in a stepper will cause a lot of heat in the motor.

Here is a link to a worm gear reducer made for stepper motors; https://www.directindustry.com/prod/jvl/right-angle-worm-gear-reducers-22516-473216.html They are also on Ebay from time to time.

 

[Mr RB]

If the PSU does pump up under decel it will only be a few volts. Your "logic" should be well and truly protected by it's own 5v regulator etc.

If you get PSU overvoltage it will only be the main PSU and the stepper driver that is affected. Like Shortbus said a worm drive will return much less current under deceleration as it will resist being backdriven. Steppers are quite inefficient so given the amount of gearing you will need for a size23 stepper on a winch you may not get the PSU voltage pumping up at all (ie returned current to PSU is less than the current used by stepper and driver).

 

[grevillea]

Thanks for suggestions re gearing the stepper motor. I may do that in the future if it turns out to be necessary, but for now I am reasonably confident that the stepper alone will do what I want.

re Holding current. 2.8A at 1.13ohm phases resistance is 9W power dissipation in the coils at maximum holding torque. I have no idea how much extra power may be dissipated through the variations in the magnetic field (chopping current driver) which presumably heat the iron. The 2M542 has the option to reduce holding current, and I will probably use that option. A 50% reduction in current halves the torque (it is still enough for my application) and cuts the power to 25%.

re back EMF voltage limits. I just did a simple test turning a NEMA23 by hand into my oscilloscope. Nice sine wave traces, with up to +- 15 V (i.e. 30V p-p). I don't think that I had reached the speed limit for my motor since I was just spinning the rotor as fast as I could be hand. Added to 24V, 15V would be enough to exceed the input voltage limit on my DC-DC converter, so I'm definitely going to have some overvoltage protection. This is sort of what I expected - pumped into an open circuit, changing magnetic fields can generate large voltages. Of course, the 'logic' circuits (and cooling fan) draw some current at all times, but they would not consume anything like 60W.

Space is limited for big fat 100 W resistors, so I have selected Arcol HS50 10 ohm which fits tidily in front of my fan and heat sinks to the case. It is rated to 50W but can tolerate overload by a factor of 2X for 3 minutes, so that's as good as 100W for me. I'll stick with the NPN I picked out - it is rated to 50W so should be happy heat-sinked to my case, especailly if it never sees more than 15W.

Thanks for all your help.

Regards
Grevillea

 

[Mr RB]

...

so I'm definitely going to have some overvoltage protection. This is sort of what I expected - pumped into an open circuit, changing magnetic fields can generate large voltages. Of course, the 'logic' circuits (and cooling fan) draw some current at all times, but they would not consume anything like 60W.
...

It's not definite at all! It will only pump up the PSU voltage if the average return current to the PSU is greater than the average current consumed by the stepper.

As steppers are quite inefficient that will only happen with very hard deceleration, with any gentle deceleration the amount of current "generated" will be less than the current consumed by the stepper and so it will just draw less current from the PSU under deceleration, but not supply current BACK to the PSU.

If you have a known load mass and known max speed and you can set the desired deceleration under electronic control you won't get PSU voltage pumping up unless your deceleration is above a certain level.

 

[grevillea]

Mr RB: Indeed you may be right that overvoltage protection will not be used, but I intend to implement it anyhow. If there is time, I hope to do some tests and report back whether in fact the overvoltage circuitry does draw any significant amount of current, and under what circumstances.