Torque of the motor

[sarang1_in]

This is the circuit I built for my application, but at low RPM the motor is stopping and unable to deliver the torque.
kitsrus.com/pdf/k166.pdf
Please suggest something to improve the torque.

 

[Nigel Goodwin]

This is the circuit I built for my application, but at low RPM the motor is stopping and unable to deliver the torque.
kitsrus.com/pdf/k166.pdf
Please suggest something to improve the torque.

Add a gearbox!.

If you reduce the power to the motor to slow it down, the motor has GOT to be less powerful.

By implementing a speed servo loop (not to be confused with R/C servos!) you can make it perform better (as it automatically increases and decreases the power to keep the speed at the value you set) - but a gearbox will perform many times better.

 

[stevez]

There are several things to keep in mind.

There is likely to be a range of shaft speeds or motor speeds that meet your requirements. If the addition of a gearbox will still result in the full speed performance that you want then a simple reducing gearbox might be the solution. Keep in mind that while gears can be very efficient that there is some loss - often relatively small.

It would appear (google on dc motor torque) that with full power applied, the torque delivery capability of a DC motor is greatest at zero or close to zero RPM. Nigel's speed servo loop suggestion would result in more power being applied, probably up to full power, if the shaft speed dropped below the setpoint- so you would have maximum motor torque, given the rpm, at all times.

Back to the low speed torque of a DC motor. The instantaneous torque is dependent on rotor position. At higher speeds the intertia of the rotor smooths out any variation. At low speed there is likely to be a maximum torque position and a minimum torque position. It would seem then that at very low or zero rpm, the minimum torque might be a limiting factor - such as when starting up from zero rpm. Those who apply DC motors probably understand this quite well.

Does the load on the motor change with RPM? Keep the low speed or startup torque demands of the load in mind. Sometimes the starting or breakaway torque can be higher.

 

[Styx]

torque is basically limited to current.
sounds like whatever you are using to contol the machine - soz not looked yet -= does not have a current-loop. that will solve the problem, although you will prolly need a speed-loop surrounding the current loop as well

 

[chiba]

I would have figured the motors winding resistance and maybe the rated peak power of the motor would determine what current you could run. Just run the highest allowable voltage that you can without exeeding the IRFZ44's current rating or motors peak power rating. I believe the quad op-amp is limiting this circuit to 32v, are you anywhere near that?

 

[sarang1_in]

Thanks for the reply
I have modified the circuit. Following are the details. I have replaced the MOSFets. Now using IRF9540 and IRF540,
using seperate supply for LM324. Regulated 12V. and taking that output with simple pullups of BC547. (Pin no 8 and 14 are given to base of BC547 via a 1.2K resistor)
I have implemented a current control with the help of NAND gate and comparator. The collector of BC547 is given to a NAND gate CD4011 and the second input is from a comparator IC LM393 pin no 1.
Inputs of LM 393 are fed with a divider on pin 3, and current feedback voltage dropped across 0.1 ohm 5W resistor on pin 2.
I have powered up the bridge with 36V, and above logic is working well.
Now I want to maintain the torque automatically. So the pot will continuously oscillate from dead position to + - 15 deg. (In a fixed band)
But still the torque is not coming as required. I tried for a gearbox, but I don't have a space. Also I want to solve the problem electronically.
The rating of motor is 36 V, 180 W, I am limiting the current to 6A via the comparator.
Now in addition to this, what I will have to do?

 

[Nigel Goodwin]

Also I want to solve the problem electronically.
The rating of motor is 36 V, 180 W, I am limiting the current to 6A via the comparator.

If you're limiting the current to 6A you're severely restricting your low speed torque, startup current is going to be many times that.

 

[audioguru]

Just measure the resistance of the motor and use Ohm's Law to calculate the massive current that it needs to start running, or for very low speed.

 

[sarang1_in]

I measured the resistance of motor, and it is 20 ohms. So if I am giving 36V to the motor then the startup current will be 36/20, and should not be more than 2A. Is it right?
Also when the motor will be rotating, the resistance will vary depending on the Inductive reactance...(Please correct me if I am wrong)
Now though I am controlling the voltage at lower side, I am allowing good enough current. Or is it that I will have to change the ref. vtg as per the torque requirement?
I am now slight ly confused with what I am following and what I want to achieve.
What do you sugest?

 

[audioguru]

Hi Sarang,
Please post the schematic with your modifications for current control.

I think your motor barely has enough torque to start driving the load at a slow speed. I think it will work like this:
1) With the motor stopped, you turn-up the speed control but the motor doesn't start turning.
2) You turn-up the speed control more, but it still doesn't start turning.
3) You turn-up the speed contol quite high and the motor starts turning but goes faster and faster.

Your current control should automatically give more power to the motor to start it turning because the current will be very high. And your current control should limit the motor's speed after it starts turning because the current will drop. It looks like it will work. :lol:

 

[Nigel Goodwin]

I measured the resistance of motor, and it is 20 ohms. So if I am giving 36V to the motor then the startup current will be 36/20, and should not be more than 2A. Is it right?

No, it's not - a motor probably isn't something that you can measure easily on a meter? - but as the motor is rated at 180W at 36V that's (very roughly) 6A, under normal load. Under stall conditions, which starting from cold is only a short version of, I would expect at LEAST ten times that, if not more?.

 

[Styx]

I measured the resistance of motor, and it is 20 ohms. So if I am giving 36V to the motor then the startup current will be 36/20, and should not be more than 2A. Is it right?

No, it's not - a motor probably isn't something that you can measure easily on a meter? - but as the motor is rated at 180W at 36V that's (very roughly) 6A, under normal load. Under stall conditions, which starting from cold is only a short version of, I would expect at LEAST ten times that, if not more?.

I 2nd this!!!!
Stall torque consists of two things

1) what torque the machine/MDE must be able to support without rotating - this is more design specific

2) breakout torque (more what Nigel is talking abt). Basically starting out how much current (ie torque) is required to start rotor rotation.
Two things influence this

cogging & torque ripple - more of a problem for BLDC and SR machines but induction machine, Sycrounous machines and brushed DC also suffer from this to an extrent - bascially there are points on a machine rotation where the developed torque is lower then other points (equally higher then others). you must be able to source enough current to cope just incase yr machine starts in one of the low-torque points

Inertia - THIS!!! is the major thing. It takes alot to get the rotor moving, even more so if it is attached to a load - once it is rotating, such inertia helps.

If you want to make sure it will work - hook up the machine to the load and use a DC source and a variable resistor (with suitable power rating) and find out what is the minimum current needed to start the machine - then give a bit of overhead ontop of that

 

[chiba]

20 Ohms seems high for a 36V 180W motor? I'm guessing that rating may have come from a marketing department?

I just had a brain fart, in the chart below how would one set a current limit based on a known peak power, a set voltage and know resistance?

P=I*V
or
P=I^2*R

For axample, the first motor 680W, 12V, 0.05Ohm

p/v= 680/12 = 56.67A
or
sqrt(P/R) = sqrt(680/0.05) = 116.62A


I'd probably err on the side of common sence but remain wiereded out

 

[sarang1_in]

Hi all.
Sorry for little(more) late.
Here I am attaching the circuit I have used. The only part not shown is that the output of nand gate is fed to the transistor and transistor's collector is pulled up to power VCC and given to MOSFETs.
The current set pt. is set to 1.00V ie 10A.
Everything is ok and working under no load conditions. But I am not getting torque at lower speed. That is just becase I am not applying full voltage, but my application is more related to holding the shaft rather than rotating it with high speed. I am using two pots for this, one is same as given in kit 166 and one added by me to increase the dead band in reversal. So that I will get the shaft movement from 15-0-15 to 100-0-100 degrees.
Please suggest me the necessary changes in the circuit.
Also Nigel, I want to know about speed servo loop, will it help my application?.
One more observation is that after sudden reversal motor takes a heavy jurk, I want to smooth out that effect.
How can I get more or say settable torque at low or zero speed?
Please Suggest.
Thanks