Back EMF on DC motor

[Al_The_Man]

You could also look at single ended hydraulic cylinder with pressure relief valve adjustment for the descent.
You would need a small resevoir, cyclinder and relief valve.
For this kind of load, it may be the better way to go.
Al.

[gar]

071127-9754 EST USA

rainman:

Before you do anything else short the motor and see if this is slow enough. Meaning is it too slow.

If your load is always 1 ton, then you have a constant down force, and a constant down torque at the motor. Torque is proportional to motor current. At a given generator speed you have a known output voltage. If it took 3 min to raise at 24 V input and you want to lower over the same time period, then you want to limit to something less than 24 V on descent because of the internal motor resistance.

If you know the current to raise at 24 V, R = 24/I is an estimate for the resistor for descent. Some of this is internal to the motor. Also efficiency has an influence on the resistance selected.

Power dissipation is from P = 24 * I. Also note that there is internal voltage drop in the internal resistance of the motor.

If your load is not a constant 1 ton, then you need a variable resistor. Al's method is equivalent to that.

Another method is a zener diode of approximately 24 V and adequate power dissipation.

Totally forget your 5 V supply method.

.

[vtxstar]

Whoa! I suspected something like this after seeing your drawing. I used to work with an x-ray table (for humans, not NDT) that used a relief cylinder for such a situation. It was basically a large shock absorber/damper connected between the motor mount and the load. It slowed the descent of the table due to gravity.

In this case, I think your actuator is misapplied. Either go with an upgraded actuator or add some gas shocks to help with the load. CLICK HERE for some information on this gadget. With that type of load, you need some mechanical assistance to balance the equation.

For instance, a linear actuator I have here in the shop has the following ratings:
Max. load: Push/Pull 10000 N/6000 N
Power Rating: 24 VDC Max. 7 Amp./5 Amp.
Duty Cycle: Max. 10% max. 6 min/hour

This means that this actuator can push 1.67 times more than it can pull (10/6 = 1.67). This actuator is one of two used to tilt a CT scanner gantry. It is attached to a lever arm at the gantry pivot. The lever arm provides the mechanical advantage for the actuator.

CLICK HERE for some beefy linear actuators and their ratings/specification links.

All this assumes you have an unlimited budget of course. If this is a garage project, hit the automotive supply houses for gas cylinders that reach the length you need and in the numbers you need to assist this overworked actuator.


-=Doug

[gar]

071127-0850 EST USA

rainman:

The reason your 5 V DC power supply won't work is that most regulated supplies are designed as an energy source and not to absorb energy. Typically these have a very low internal impedance when sourcing energy, but a high impedance in the reverse direction.

Whereas a DC servo motor drive is designed to work as both a source and sink for energy.

In the down direction you need an energy absorber. That is what a resistor or short across the motor does. When you have a short across the motor all of the absorbed energy is dissipated in the motor.

.

[pastera]

You need to put a diode in series with your power supply to protect it form absorbing energy and use a braking circuit set to activate 1-2 volts above the power supply voltage that will dump the regenerative energy.

You don't want to short the motor or you WILL burn it out. The energy needs to be dissipated somewhere - You don't want it all dissipated by the windings and that is what you get when you short the motor.

Can you get a current reading when raising?

Aaron

[gar]

071127-1307 EST USA

pastera:

I do not think the series diode is of great importance because the relay disconnects the power supply from the motor for descent.

On shorting the motor. If it was continuous, meaning long compared to the motor thermal time constant, then maybe a problem. If the descent does not cause a power dissipation in the motor greater than the motor's 100% duty cycle losses, then there is no problem. However, I think shorting will make the descent too slow so much of the energy to be dissipated will not be internal to the motor, but in an external resistor.

I did not find any indication of the height thru which this 1 ton is raised, or the HP rating of the motor. So I will make some assumptions to illustrate a method of calculation.

For reference: 1 HP = 746 W = 33000 FT-#/minute

Suppose the 1 ton is raised 4 ft in 3 minutes at 24 V DC to motor. This is 8000 ft-# of work in 3 minutes or 2667 ft-#/minute or at 100 % efficiency it is 0.081 HP or 60 W. The motor current is something greater than I = 60/24 = 2.51 A. Suppose the efficiency is 50%, then I = 5 A.

At 100% efficiency coming down the current would be 2.5 A at 24 V or a load resistance of 9.6 ohms. But there are mechanical losses so the breaking direction current would be less than 2.5 A. Thus a 10 ohm resistor would cause a slower descent than rise. Rating of this resistor could be 50 W for two reasons. First, its thermal time constant and overload tolerance, and second the current would be less than the assumed 2.5 A.

Al's circuit with the 5 ohm resistor would be more than adequate, because we are probably looking at something like 20 ohms to get the same descent rate. All this is based on my above height assumption and an estimated 50% efficiency.

More facts would help clarify the optimum circuit. If the load changes substantially one needs an automatically adjusting load, such as Al's circuit.

.

[pastera]

The motor only dissipates the heat from it's losses while raising the load. By shorting it on descent, it must dissipate the entire load energy minus screw losses most probably a much higher number.

My suggestion of a series diode and brake circuit was not in reference to the posted circuit because as shown it will not work.

D5,D8 are 22 volt zeners. While it is fine to run the lm311 on a 22 volt supply it places the junction of R7,R8 at 13.2 volts. D5 reduces the motor generated voltage by 22 volts so the minimum voltage that the circuit could activate at would be ~35 volts
You would also need a pullup on the junction of R3/U1-7 to turn on Q1

There is no hysteresis, so U1's output will rapidly switch at high values of R1 (R1 and C1 form a low pass filter). This may switch Q1 at that rate or it could cause Q1 to operate in its linear region (R3 and gate capacitance form a low pass filter).

Also, two diodes should be added one connected to the 24 supply and R9 the other to the motor buss and R9. This would allow the motor to power the braking circuit in the event of a power failure during operation.

Aaron

[gar]

071128-1132 EST USA

pastera:

I do expect more dissipation in the motor with it shorted for descent, than compared to raising. But I do not expect his operation to be a non-stop up and down motion and thus my guess is that even if run shorted for descent it is likely that the motor would not be overheated.

But also I think shorted would produce too slow a descent and an external resistor would be used to control the rate.

The big problem here is that rainman needs a quick and simple experiment to use to point him in the correct direction. Shorting the motor for one cycle should cause no harm, or he does not need to fully raise the load to test the effect of shorting or any suitable resistor. So after my above calculations in my previous post I might now suggest trying a 10 ohm resistor in place of the short if the raised height is anywhere close to my guess of 4 ft.

rainman:

We need a little more information. How high do you raise the 1 ton in 3 minutes? What is the motor HP rating? What is the current into the motor while raising? What access do you have to resistors? A 120 V 1400 W heater is about 10 ohms.

.

[Al_The_Man]

Originally Posted by pastera D5,D8 are 22 volt zeners. While it is fine to run the lm311 on a 22 volt supply it places the junction of R7,R8 at 13.2 volts. D5 reduces the motor generated voltage by 22 volts so the minimum voltage that the circuit could activate at would be ~35 volts You would also need a pullup on the junction of R3/U1-7 to turn on Q1

I am not sure how those values got on the schematic, D8 should be 1N4742 e 12v zener.
I have been producing the 240 v generator regulator for 15yrs now using this basic circuit, I do have a driver pair operating the original Hexfet as it is 400v 35amp type.
I just modified a couple of components to indicate a direction to go in for a possible solution, I have not tested it using these components.
Al.

[pastera]

Al - I thought that the errors were typos but I didn't want someone to build the circuit and find out the hard way.

I would still add a pair of diodes to allow the motor to power the brake in the event of a power failure - this would work with or without the relay.

Aaron

[coldarc]

Originally Posted by rainman Thanks for the replies. Yes it is linear actuator, the position is not that accurate, I have 2 reed switches that is mounted on the side of the actuator for positioning control (2 positions only) It took me 3+ mins to extend to the max and 40 secs to retract fully (with 1 tonne load) using 24VDC PS (back EMF til 60VDC) With that kind of load, it was pretty scary to see the descent, not to mention about the sound the actuator created. I'm caught in the situation whereby I can only solve the problems in terms of pure electronics. I have yet to try out AI_the_Man method and the short method. Desperately I will have to temporary stick to 5V PS and upgrade the capacitor and finger-crossed it may last. Rgds, Rainman

use a neon bulb in paralell to the power supply, the neon will eat up the back emf. if that doesn´t work you can use a resistor connected to a large capacitor or a bank of capacitors to buffer the output of the supply and add the load to the cap.
this way the resistor will protect against the spikes and you save energy at the same time.