Need to drain off curent.

[Jongig]

I built a attic elevator using DC gear motors. It's a great unit with a platform that is the attic floor and can lift 500 pounds. It has turned the attic into a great storage area for everything and it's easy to ride up/down with stuff.

Okay, now the problem. The two motors are 24-volt with breaks, wheelchair motors. They were perfect because any type of failure and they would just stop. They're connected to a regulated power supply.

With a heavy load the power supply while going down is not able to keep the voltage at 12-volts and the motors speed up too much. I need some way to manage the voltage so that if the voltage exceeds 14-volts the current is somehow managed. I thought of connecting resistors but I have no idea if this would work and then how much to use since the problem gets worse as the weight going down pulls on the motors making them go faster.

Thanks.

 

[BobK]

Limiting thr current can only make it worse. You need a power supply with a higher current rating.

Bob

 

[Bluejets]

If the load is driving the motors like you say, then from a failsafe point of view you will require mechanical speed regulation, possibly in the form of a centrifugal brake.

 

[dorke]

Agree with Bob,
What power supply are you using?
What is it's current rating?

 

[duke37]

It sounds as if the power supply can supply the current but cannot accept current if the motor is being driven by the load, this could damage the power supply.
The solution could be to limit the voltage to just above what is normally supplied. A powerful resistor or a super dooper Zener diode would be effective. You need to know the current which will have to be dumped.

It may be advisable to place a diode in series with the power supply to stop reverse current. The Zener would then come after this diode.

 

[73's de Edd]

Sir Jongig . . . . .

Considering the aspect that the elevator has to work to lift, but on the decline its a piece of cake due to gravity assistance.

Just for providing information and analytical evaluation . . . .

Can you take the elevator to top and disconnect the motor from the power supply completely and then put a jumper short across the motors disconnected terminals.
Then release the elevator . . . no load . . . to see just how much magnetic braking action then will be created.
This would answer if full or partial magnetic braking plus the added gear train drag / friction on the trip down could accomplish the downward travel task.
( Ignore all . . . . if this is being a single or dual jackscrew driven design, as I'm considering a gear train )

73's de Edd

 

[Jongig]

Limiting thr current can only make it worse. You need a power supply with a higher current rating.

Bob

The power supply is very adequate, going up with 300-400 pounds which is where the problem is the elevator goes up without any problem, going down it goes faster than I'd like.

 

[Jongig]

If the load is driving the motors like you say, then from a failsafe point of view you will require mechanical speed regulation, possibly in the form of a centrifugal brake.

The reason I think the motors are the key is because if the motors are shorted the elevator going down barely moves so if the electricity is regulated the motors would maintain a certain speed.

Is there anything as a electronic device that at the point the voltage hits 14 volts the current goes to a bank of resistors?

 

[Jongig]

Sir Jongig . . . . .

Considering the aspect that the elevator has to work to lift, but on the decline its a piece of cake due to gravity assistance.

Just for providing information and analytical evaluation . . . .

Can you take the elevator to top and disconnect the motor from the power supply completely and then put a jumper short across the motors disconnected terminals.
Then release the elevator . . . no load . . . to see just how much magnetic braking action then will be created.
This would answer if full or partial magnetic braking plus the added gear train drag / friction on the trip down could accomplish the downward travel task.
( Ignore all . . . . if this is being a single or dual jackscrew driven design, as I'm considering a gear train )

73's de Edd

There are two braking systems on the unit, first is what you're talking about. The unit has two switches up/down. When the down is pressed the mechanical brakes are released first and then 12 volts is applied to the motors, with 300 pounds on the platform you get a sense of freewheeling as it picks up too much speed. When the button is released the leads of both motors are shorted which provides a quick but not violent stop and then the mechanical brakes lock. I have tested leaving the mechanical brakes off and the elevator barely moves going down.
What I found with these motors is that you can't apply voltage to the motor and mechanical brakes simultaneously because the brakes need half a second to release, creates a huge current draw. The fix was a 1-second delay to the motors.

 

[Jongig]

It sounds as if the power supply can supply the current but cannot accept current if the motor is being driven by the load, this could damage the power supply.
The solution could be to limit the voltage to just above what is normally supplied. A powerful resistor or a super dooper Zener diode would be effective. You need to know the current which will have to be dumped.

It may be advisable to place a diode in series with the power supply to stop reverse current. The Zener would then come after this diode.

With 300 pounds on the unit as you go down the fan on the power supply spins faster and the light shines brighter. I should have measured the high voltage but the motors must be generating reverse voltage to the power supply. With me on the unit, 195 pounds, and going down the power supply seems able to regulate the voltage and keeps the platform from going too fast. When you hit the down button it starts to pick up speed but then stops going faster. With 300 pounds the PS can't regulate and the freewheeling starts. When the unit starts going fast I just let off the button and it stops so I press the down button on/off until I get down. During testing I thought about this effect but when I noticed that the power supply seemed to regulate the speed of the platform going down I thought I had gotten lucky, problem was that I didn't get over 240 pounds in my testing.

I don't completely understand how stopping the reverse voltage would help.

I thought about the two DC wires to each motor and had the idea if I could use a diode to pass current to a electronic device only when the motors are running in reverse/down. The device would pass current to a bank of resistors at the point of 14 volts. The voltage would have to be higher than the PS so as not to tax the power supply.

 

[Jongig]

Agree with Bob,
What power supply are you using?
What is it's current rating?

I measured the amperage of the motors at it's full rating going up at about 30 amps, the PS is 40 amps if I recall correctly.

 

[BobK]

There is something I don't understand here. Are you saying that, to go down, you supply the motors with the same polarity as you do when going up, but lower voltage so that the motor slips backwards?

If so, the motors are not just stalled, they are going backwards. This will use more current than the stall current, which could easily be 3 times the running current, albeit at the lower voltage. I am guessing that the reason you cannot regulate down to 12V is because the motor is generating more voltage than that when rotated against its direction. When spun in the opposite direction, the back EMF becomes forward EMF, in other words the motor is acting like a generator

If a dead short stops the elevator completely, the way to slow it down is to use a resistor (a very high power resistor) instead of applying power to allow it to descend.

I also worry about your statement that all fail modes stop the elevator. If the motor comes completely disconnected (electrically) what stops it from going dropping at a speed limited only by how fast the motor shaft can rotate?

I can't see how this thing is safe without a mechanical brake.

Bob

 

[duke37]

I assume the motor uses one polarity for forward and the opposite polarity for reverse. Adding a diode of the correct polarity feeding a resistance will dump current in reverse. Set the resistor to drain 20A or so. it will need to be fat to dissipate 200W.

The zener diode trick will do the same thing and will need to be able to dissipate considerable power., it depends on the efficiency of the system when the motor is acting as a generator.

 

[(*steve*)]

Is there anything as a electronic device that at the point the voltage hits 14 volts the current goes to a bank of resistors?

Your power supply operates in a single quadrant, that is, it can regulate when it is supplying a positive voltage and a positive current. This is fine when the lift is going up.

When the lift is going down, the motors are being driven by the load, and the power supply sees a positive voltage and a negative current. There are some per supplies which will operate in this second quadrant, but most don't.

The simple solution is to load the motor (e.g. short it) when it has to go down, but this may slow it down to much or indeed prevent it from lowering at all in some conditions.

A relatively simple solution is to power the motors via your power source to raise the lift, and to use a shunt regulator to absorb the generated energy on the way down. Keeping a fixed voltage across the motor may give a relatively constant speed with varying load. You may want to have a way of varying there electrical load to cater for widely varying loads on your lift.

 

[Jongig]

Your per supply operates in a single quadrant, that is, it can regulate when it is supplying a positive voltage and a positive current. This is fine when the lift is going up.

When the lift is going down, the motors are being driven by the load, and the power supply sees a positive voltage and a negative current. There are some per supplies which will operate in this second quadrant, but most don't.

The simple solution is to load the motor (e.g. short it) when it has to go down, but this may slow it down to much or indeed prevent it from lowering at all in some conditions.

A relatively simple solution is to power the motors via your power source to raise the lift, and to use a shunt regulator to absorb the generated energy on the way down. Keeping a fixed voltage across the motor may give a relatively constant speed with varying load. You may want to have a way of varying there electrical load to cater for widely varying loads on your lift.

Your idea sounds good but in simple testing I'm not sure it'll work without applying voltage to the motors on the way down. With 200 pounds the unit needs voltage to go down. I'd like to build some form of shunt of voltage at 15 volts. If I somehow can drain off current when the voltage hits 15 volts I'd maintain a perfect speed. If the power supply puts out 14 volts then at 15 volts the circuit would include a shunt.

 

[davenn]

If I somehow can drain off current when the voltage hits 15 volts I'd maintain a perfect speed

as has been stated earlier and I will state it again as you don't seem to understand

you don't/cannot drain off current
The amount of current flowing in a circuit is a function of the voltage applied to the load resistance
Alter either of those two things and the current will change accordingly

 

[(*steve*)]

This thing you describe as "bleeding off current" is the two quadrant power supply I mentioned

If you require power to start the downward movement (which is, I think, what you're saying) then you can insert a diode between the power supply and the motor. This will prevent current flowing back into the power supply. Then, place a shunt regulator across the motor, with the voltage set just a little higher than the power supply. This will roughly simulate the effect of a two quadrant power supply and may be practically indistinguishable from it in practice.

A shunt regulator is a non-linear resistance. It exhibits a high resistance until the voltage across it reaches a particular value, then the resistance drops precipitously.

In essence, this is what I suggested previously with the added feature of being able to give the motor an initial kick to get it going.

 

[Alec_t]

The reason I think the motors are the key is because if the motors are shorted the elevator going down barely moves so if the electricity is regulated the motors would maintain a certain speed.

Agreed. Something based on the attached circuit should work. A FET is switched on/off rapidly with a PWM signal generator (details not shown). A DPDT switch controls motor direction.

During upwards movement of the elevator, the motor current passes alternately through the body diode of the FET (FET off) and the drain-source path (FET on).
During downwards movement of the elevator the body diode does not conduct, so motor current is intermittent via the drain-source path only. This is equivalent to intermittently 'shorting' the motor, thus providing dynamic braking. The duty cycle of the PWM determines the average motor current/braking-effect.

The motor voltage and current waveforms (time is not represented to scale) would be something like this :-

 

[Alec_t]

Hmmm, I'm not so sure that will work if the emf generated by the motor exceeds the voltage supply. Also, my bad, it's not equivalent to shorting to get regenerative braking. Back to the drawing board methinks.

 

[Alec_t]

Ok. This looks a better bet. The FET does now short the motor intermittently when the elevator moves down, so would give dynamic braking. The body diode is not used here :-