I'm making a transporter platform that pushes human load. I'm using a 12V 40A battery with a PWM motor driver. When the platform pushes a human load at about 80% or so, it can take more than 10A, depending on the weight of the man standing on top. My question is, any way to lower the current so that i can integrate low rating components for other purposes, like battery indicator or a VU meter?

 

Can you change the drive ratio of the motor, like using a geared transmission or belt driven speed reduction unit? This will multiply the torque while reducing the speed.

 

You can do what the power companies do to transmit power: raise the voltage. If you double the voltage, the current will be cut in half. Of course, this changes your problem from finding parts with the requisite current rating to parts that are rated for the higher voltage.

 

If you take the course of action that Someonesdad suggested, you will need to use a motor that is rated for twice the voltage as well.

 

The motor and the gearbox comes in one, like a set. Changing mechanical parts will cost quite a fortune for us students as in Malaysia, parts like these don't come cheap. A raise in voltage will need a change in motor and a DC motor in Malaysia may cost about 1000 bucks.
What about cars? The fuse box seems to be able to step down well enough to power electronic components. Or does it mean that if my motor drains high current, my other components drain as high too if I parallel them?

 

The motor and the gearbox comes in one, like a set. Changing mechanical parts will cost quite a fortune for us students as in Malaysia, parts like these don't come cheap.

OK, so that is basically fixed in concrete.

A raise in voltage will need a change in motor and a DC motor in Malaysia may cost about 1000 bucks.

Ouch! forget that.

What about cars? The fuse box seems to be able to step down well enough to power electronic components. Or does it mean that if my motor drains high current, my other components drain as high too if I parallel them?

The fuse box does not "step down" current. The fuse box acts as the weakest link; if there is a problem somewhere, the cheap fuse protects the rest of the circuit by burning up.

OK young Jedi, your problem is to somehow limit the maximum current that can be drawn by the motor.

You can indirectly measure the power through the motor by a current sensing resistor. Ohm's Law says:
P= E^2/R

So, you need to figure out a way to decrease your PWM percentage when the power gets too high.

How do you propose to do that?

 

Hm... to lower the PWM percentage, probably I can use a fuse paralleling a resistor. When the fuse blow due to high current into the motor, the current will go through the resistor which lowers the voltage of the PWM output. But that will be a bit impractical as I will need to replace the fuse a lot of times. That's just my theory.

Just researched a bit on low-pass filters. I guess, I can set a limit for the PWM percentage with a low-pass filter so that there is a maximum voltage input to the motor.

 

Hm... to lower the PWM percentage, probably I can use a fuse paralleling a resistor.

No.
A fuse is to limit the maximum current before damage would occur.

When the fuse blow due to high current into the motor, the current will go through the resistor which lowers the voltage of the PWM output. But that will be a bit impractical as I will need to replace the fuse a lot of times. That's just my theory.

Why don't you determine a way to limit the maximum current to avoid blowing the fuse?

Just researched a bit on low-pass filters. I guess, I can set a limit for the PWM percentage with a low-pass filter so that there is a maximum voltage input to the motor.

Do you have any idea where you are going with this?

 

Um... I thought since the PWM output is a square wave and it has frequency, a low-pass filter may set a limit to it. Sorry if I sound kinda stupid in this. My lecturers teach us more on theory and answering questions. My college is exam oriented. So we hardly have practical experiences.

I just did a search in the internet. Perhaps http://www.physics.unlv.edu/~bill/PHYS483/current_lim.pdf is what I am seeking?

 

your VA (volt/ampere) consumption is a function of your total loads, which consist of moving a man at some speed, plus ancillary loads (meters, etc.), at some percent efficiency. You can manipulate the VA formula, but if V (volts) are implied via motor requirements, then your basically left with the efficiency component.

You've indicated that you are tied to your exsisting motor. That may be your limiting factor. What type of motor are you using, and are it's characteristics appropriate for the application?

If what you have to work with comes in a box, then your limited to that. If then a 100lb person draws 10A at some speed, you'll limit your wieght to 99lbs, leaving that little bit for ancillary.

 

Um... I thought since the PWM output is a square wave and it has frequency, a low-pass filter may set a limit to it. Sorry if I sound kinda stupid in this. My lecturers teach us more on theory and answering questions. My college is exam oriented. So we hardly have practical experiences.

No, it's not a problem.
Basically, your PWM output IS a square wave (or mostly rectangular). If you run that output through a low-pass filter (R in series, C to groun), you will wind up with a roughly triangularly-shaped waveform.

The larger the values of R and C, the flatter the triangle becomes. However, it's better to use several stages of filtering; each successive filter stage takes out more of the triangular shape until it's basically a DC level with a little ripple on it.

However, you need to watch the phase shift (time lag) caused by the filtering. You'll want fairly rapid response at the filter output, but you want to get rid of most of the noise.

If you compare the PWM ratio with a voltage across a sense resistor (which can simply be a length of wire; wire does have resistance) you will get an idea of how hard your motor is working.

I just did a search in the internet. Perhaps http://www.physics.unlv.edu/~bill/PHYS483/current_lim.pdf is what I am seeking?

No, that is a linear-type current regulator.

Basically, if the voltage across a sense resistor becomes too high, you know that the PWM ratio needs to be decreased.

 

I see what you mean. Unfortunately, I only have the only DC motor with a gearbox attached to it. As I mentioned earlier, getting a DC motor for such an application will be too expensive. I can't say whether the characteristics are appropriate or not as I am using the motor and the gearbox of a golf trolley, those that pushes the golf bag through the golf course while you walk along it. It's a 12V DC motor, as far as I know. Tried with my old PWM motor driver and it works pretty well until something is wrong with the circuit -_-. The motor was not hot. So I guess it suits this application quite well.

It's quite hard to measure the current consumption, I think? That's where I'm stuck. I only know how to reduce current with resistors but the loads must be constant. There may be other ways but I think I may have forgotten about them xD.

 

your (VA) is translated by the motor/gearbox into torque/speed, again, at some effeciency. If you want to reduce average current, then your going to reduce torque. Conversely, if you reduce voltage, your going to reduce speed (and torque). Remember, you don't get something for nothing. You can increase the voltage to the motor, but doesn't serve your desire to reduce current.

If you have a 10 amp draw from a 40 amp battery, why do you need to reduce the current to the motor?

To get a load from point A to point B requires a certain amount of work. Your battery will have a certain capacity to perfrom this task, and may or may not have something left. In a comparision between the task of moving the human load, and that of indicating, I'm guessing there is a conflict of intent.

Kind of like, I'm sorry teacher, it used to carry you over there, but now that we have a power meter, you need to walk the last bit. Do I still get an A?

Or, are you asking how you can get a DC voltage relative to the motor speed?

 

Lol perhaps I did not explain well. I don't want to reduce the current to the motor. I just want to reduce the current for other devices other than the motor. I am thinking of using LM3916 to "show" the speed but the shop gave me 3915 because they don't have 3916. Actually it's just showing how many volts are supplying into the motor. 10 LEDs (1 red, 3 yellow, 6 green) show 0% to 100% of the speed. Just for indication, no need to show actual speed. So I guess I am asking how to get a DC voltage relative to the motor speed.

What I am trying to do is when the user turns the pot, the "speed" bar rise accordingly.

 

Well, you can always use a bigger battery, or one with a higher Amp-Hour / size ratio (Lithium). Battery technology is constantly being improved these days, so watch for new batteries coming onto the marketplace. Because the motor drive system is a given, you are limited somewhat in your options for improving battery life. More slippery synthetic lubricants for your gearbox will help too. As an afterthought, has the PWM programming been optimized for maximum battery life?

Regards, DPW

 

No, it's not a problem.
Basically, your PWM output IS a square wave (or mostly rectangular). If you run that output through a low-pass filter (R in series, C to groun), you will wind up with a roughly triangularly-shaped waveform.

The larger the values of R and C, the flatter the triangle becomes. However, it's better to use several stages of filtering; each successive filter stage takes out more of the triangular shape until it's basically a DC level with a little ripple on it.

However, you need to watch the phase shift (time lag) caused by the filtering. You'll want fairly rapid response at the filter output, but you want to get rid of most of the noise.

If you compare the PWM ratio with a voltage across a sense resistor (which can simply be a length of wire; wire does have resistance) you will get an idea of how hard your motor is working.

Thanks for your help. Perhaps I should try with filters. Seems the best option I got for a low cost.