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Topic: DC motor speed (Read 3839 times) previous topic - next topic

delodephius

I have built a small two-wheeled robot car. The two small DC motors are powered by four 1.5V AA batteries (so 6V). The motors are controlled through a L293D chip. Arduino is not connected to the same power circuit, it uses it's own 9V from a PP3 battery.

The motors seem to be turning rather slow, so how can I increase their speed? Do I need more voltage, let's say I replace the 4x1.5V batteries with one 9V PP3 battery, or should I add more AA batteries in series to increase both the voltage and current? If that is how it works? I'm still a beginner so I'm not quite sure how it all works.

Thank you in advance,
Tomas

jackrae

If you must use AA then rig up 6 in series to create a 9volt supply.  The PP3 will be useless at powering the motors.

delodephius

Actually, my mistake, it's a 6F22 9V battery. Does that make any difference?

dc42

The L293D is an old bipolar motor driver chip with about 3V voltage drop. So only about 3V of your original 6V will be getting through to the motors. Either use a higher voltage battery (e.g. 6xAA in series), or get a better motor driver chip (e.g. http://www.vishay.com/docs/70007/si9986.pdf).
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jackrae

6f22 is still useless at motor driving.  These are designed for very low current demand, which a motor is not.

delodephius


6f22 is still useless at motor driving.  These are designed for very low current demand, which a motor is not.

I connected two 9v batteries, in paralle,l to the pin8 on the L293D and now it is moving about twice faster. But I'm waiting for a casing for 10x1.5V AA batteries, the pin can take up to 36V.

I also wanted to ask, what does the motor go faster? More volts or more amps?

dc42


I also wanted to ask, what does the motor go faster? More volts or more amps?


The no-load speed is roughly proportional to the voltage. The torque is proportional to the current.
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delodephius



I also wanted to ask, what does the motor go faster? More volts or more amps?


The no-load speed is roughly proportional to the voltage. The torque is proportional to the current.

Oh, thank you.

MarkT

And the torque you see at the shaft is reduced by that torque lost to friction in the commutator and bearings...

A rough rule of thumb is usable torque is proportional to current minus no-load current in a simple
DC motor (permanent magnet stator, to be precise).

In theory speed is proportional to voltage at the terminals less IR losses in the windings.

Large well engineered PM DC motors behave close to the theory IMO.
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delodephius

Another silly question: if the motors spend 4.5V each, do I have to supply the circuit with 9V?

Also, now I have the 10xAA case for 1.5V batteries, but that would give 15V which I'm afraid is too much, even if the IC drops 3V from the motors, that'll be 12V. Is that too much then? Because when I supplied them with a 9V battery, so -3V means 6V, they worked fine.

MarkT


Another silly question: if the motors spend 4.5V each, do I have to supply the circuit with 9V?


Each motor and its controller are in parallel (I hope), sharing ground and supply, so 4.5V. 

Things in parallel add currents (but share voltage), things in series add voltage (but share the same
current).  Clearly you want independent currents for separately controlled motors.
Quote


Also, now I have the 10xAA case for 1.5V batteries, but that would give 15V which I'm afraid is too much, even if the IC drops 3V from the motors, that'll be 12V. Is that too much then? Because when I supplied them with a 9V battery, so -3V means 6V, they worked fine.


You can use PWM to reduce the effective voltage, say perhaps limiting the PWM from 0 to 50%...  However
it would be better to only use fewer cells in series and be able to drive up to 100%.
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delodephius

Well, until I find a casing with less cells I'll have to try PWM.

If I calculated correctly, the motors use 4.5V, the IC has a drop of 3V, so I would need 7.5V, and if I'm providing 12V (not 15V, the batteries turned out to be 1.2V), that would leave PWM at 255/12=21.25 multiplied by 7.5V ~ 160. Correct?

jackrae

You can still use the 10 cell casing but with less working cells in it.  Simply use some wood dowels the size of the cells, neatly wrap each in some aluminium cooking foil (including both ends) then insert them into the casing as substitute cells.  2 "dummy" cells will reduce the system voltage down to 9 volts. 

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