Hello, This is shri ganesh. we are involved in developing a robot which weighs around 25 kgs. Now i am confused in choosing a battery for the project. please give some technical aspects about it. The motor specifications for my project are 12 volts with each motor needing about 8 amperes. The torque specifications for my motor is 47 kgf cm at 8 amperes stalling current. Now , my doubt is if i give the 12 volts to my motor with current about 4 amperes, whether it can drive my load of 25kg with a motor of the above specifications ?? Or if i use a 24 volts battery with 7812 regulator, then can i get the driving of the load with 4 amperes? which battery should i choose for efficient driving of the load?

I think you misunderstand how a 7812 regulator works. For it to give 8A output, it requires 8A input. The voltage drop (12 volts in your example here) means that it has to get rid of 8*12=96Watts of heat, or equal to all of the power that the motors are using.

A switchmode power supply can give you 8A at 12v from 4A input at 24V. Probably more like 5A input, due to inevitable losses.

What's more important is how much weight and space have you allowed for the battery? This will drive a choice of battery chemistry:

- Lead-acid - heavy but very easy to work with. Doesn't burst into flame.
- LiFe - lighter for the same power (same amp-hours) but requires special chargers and battery protection
- LiPo - lightest commercially available battery technology. Requires special chargers and protection or it will burst into flame and burn down your house.

Once you have a battery chemistry picked out, then you need to stack up enough cells to give a voltage close to what you want. eg. a 4S LiPo will have about 16v when fully charged. A 6-cell lead-acid fully charged is about 13v.

Thanks for the reply ! what if i give current of 4 amperes to a motor ( this motor needs stalling current of 8 amperes at 12 volts for driving a load of 47kg) to drive a load of 25kg at 12volts ?? will the motor drive my load ?

and the battery size i need is obviously small. i need it around 15cm length.

It is not the size you need to worry about. Work out how much power is required and if the battery will last, IE, how many Ahrs are needed? Also does the battery have the capacity to provide the level of current required.

Weedpharma

weedpharma: It is not the size you need to worry about. Work out how much power is required and if the battery will last, IE, how many Ahrs are needed? Also does the battery have the capacity to provide the level of current required.

Weedpharma

No. i am not that worried about the size of the battery. I need 25 AH battery. and the above question of mine where i have asked whether i could control a motor(having a capacity to drive 47kg with stalling current of 8 amperes at 12 volts) to drive a load of 25 kg with current of 4 amperes??

i dont know much technicalities. please help me. And please give the difference between a 24volts battery with 20 amperes output current and 12volts battery with 20 amperes output current??

drive a load of 25 kg with current of 4 amperes??

This is a mechanical problem and the answer depends on wheel diameter, gear ratio, wheel and bearing friction, speed of robot motion, motor torque, whether the robot is climbing, etc. Only you can answer those questions.

You need to match the power and revs of you motor to the speed and drive method. You would not be able to use the motor to directly drive the wheels as the rotation speed of the motor is far too high.

As you gear the motor, you can reduce the wheel speed while maintaining the revs of the motor in its power band. It also gives more torque to the wheels.

You need to design the drive system to suit the motor.

A 24v battery at 20 A can supply 480 watts, a 12v battery at 20 A can only supply 240 watts.

If you get the gear ratio and motor power correct, the motor will drive ok.

Weedpharma

jremington: This is a mechanical problem and the answer depends on wheel diameter, gear ratio, wheel and bearing friction, speed of robot motion, motor torque, whether the robot is climbing, etc. Only you can answer those questions.

But a motor with the specifications -- stalling current of 8 amperes with stalling torque 47kgfcm . My doubt is that if i give 4 amperes to drive my load of 24kg , will it work with the above specifications of the motor??

As we have said before, it depends on the gearing. This is not known by us so we cannot give an absolute answer.

Weedpharma

A motor with 8A stall current will have a full-load current of something like 1 to 1.5A, otherwise the efficiency would be too poor.

So perhaps 10 watts mechanical power possible when operated continuously.

DC motors are always controlled with PWM, never with linear regulator (which wastes loads of power).

The 47kgf-cm (4.6N-m in sensible units) clearly shows that the motor already has reduction gearing, a 10W motor would have about 0.03Nm or so I reckon.

DC motors have current proportional to torque and voltage proportional to angular velocity, roughly speaking, so you use PWM to control the effective voltage to set the speed and the load will cause the motor to pull enough current (at the expense of a small slow-down).

Perhaps you could actually let us know exactly what motor you have?

MarkT: A motor with 8A stall current will have a full-load current of something like 1 to 1.5A, otherwise the efficiency would be too poor.

So perhaps 10 watts mechanical power possible when operated continuously.

DC motors are always controlled with PWM, never with linear regulator (which wastes loads of power).

The 47kgf-cm (4.6N-m in sensible units) clearly shows that the motor already has reduction gearing, a 10W motor would have about 0.03Nm or so I reckon.

DC motors have current proportional to torque and voltage proportional to angular velocity, roughly speaking, so you use PWM to control the effective voltage to set the speed and the load will cause the motor to pull enough current (at the expense of a small slow-down).

Perhaps you could actually let us know exactly what motor you have?

Mark T , thanks for the reply. the motor which is used for our project is present in this link" http://www.nex-robotics.com/products/motors-and-accessories/300-rpm-side-shaft-heavy-duty-dc-gear-motor.html " .

And if thats the case where in i can use less voltage for less speed, what it the guarantee that it wont get affected with the torque capacity of the motor ?(i need nearly 25 kg of load to be pulled by the motor) because higher the voltage , higher the pushing force , so that higher rate of flow of charge will take place, so that higher currents will flow to meet the torque capacity. Am i right ?? please explain this concept clearly.

We had this same confusion here a few days ago, from someone else. They were thinking of the physics professor's diagram of the torque-vs-speed diagram of a motor. That shows more torque at lower speed for common DC motors. While that is true, it doesn't help solve this problem.

Imagine you have your motor in your robot turning the wheels and driving it across the carpet. What torque is it providing? We don't really care and it's difficult to measure. How fast is it going? Well that's easy to measure - you can see it.

Let's say it's going too fast on the carpet and you want to slow it down. To slow down you need LESS TORQUE. If it had more torque, then it would speed up until the drag at high speed matched the torque available. This is exactly the opposite of the professor's diagram.

Getting less torque is easy - just reduce the voltage driving the motor and it will slow down until the drag equals the torque available at that lower voltage.

Speed control is different from torque control. If it is able to go "too fast" at full voltage then it's very easy to go slower.

A DC motor with permanent magnets (not a series wound motor) has torque proportional to current

(less friction losses due to the commutator).

The speed is basically proportional to voltage, except that at higher currents some of the voltage

is dropped across the internal resistance of the winding and unavailable for driving the motor.

So for a fixed voltage you get a steep curve (line) of torque v. speed - the motor tries to hold

the speed constant, but fails a bit (worse for smaller motors usually).

For torque control you need to control the current, not the voltage.

Thanks for the reply !

weedpharma: The advantage of using a 24v battery is that you can use a more powerful motor without increasing the current. It is easier to control a smaller current than larger current.

Weedpharma

i still couldn't understand the advantage of 24 volts battery over 12 volts battery with the same output discharge current. Also please explain in terms of power being used up.

Power = voltage x current

If you have a 20A current for 12v you have 240W. If you have a 20A current for 24v you have 480W. To have the same power for the 12v you have to double the current to 40A.

Simple ohms law.

Look at specifications for switches. It is easier to get switches that are able to control 20A than to control 40A. There are also lower losses in the cables etc as

Power = current squared x resistance.

Weedpharma

weedpharma: Power = voltage x current

If you have a 20A current for 12v you have 240W. If you have a 20A current for 24v you have 480W. To have the same power for the 12v you have to double the current to 40A.

Simple ohms law.

Weedpharma

ok . I could understand that lesser currents can be controlled easily.

what does power tell us ? please explain . Because 24 volts explain that it has the force equivalent to energy of 24 volts to push a charge that the rate of flow of charge can be maximum. But here, the rate of flow of charge is also same as that of the 12 volts battery . same 20 amperes. How does power makes a difference ? please explain.

Power = current squared x resistance. but power = voltage squared divided by the resistance also nah ??

Put a push bike with a low power rider in a race against a high powered racing car.

This is what power does.

Weedpharma

okay thanks.