Does a 9V battery supply more current than a 6V battery?

Discussion in 'General Electronics Chat' started by Vincenzo1309, Dec 28, 2008.

  1. Vincenzo1309

    Thread Starter Active Member

    Dec 28, 2008
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    Hi, I am curious, does higher voltage rated batteries supply more current than a 6V battery ( a pack of 4 AA batteries)?

    I have a solenoid which operate from a range of 6 - 9 Volts. Surprisingly a 9V battery cannot actuate the solenoid, but when I use the 6V battery, the solenoid can be actuated.

    Why is it so??

    I thought the 9V battery should supply more current than the 6V battery.

    Pls advise.
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    Yes some batteries can supply more current due to their lower internal resistance. Maybe your 9V battery is empty.
     
  3. Vincenzo1309

    Thread Starter Active Member

    Dec 28, 2008
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    Thks alot for your prompt reply.

    That's the correct word, internal resistance. It should be higher internal resistance right? Not lower. Someone told me that too!!
    I can confirm that the 9V battery is not empty, just that it has depreciate to 7.8V. But still it should be also to actuate the solenoid.

    So the theory of higher voltage-rated batteries supplying more current is not true afterall?

    But how do we know which batteries has lower internal resistance?
     
  4. mik3

    Senior Member

    Feb 4, 2008
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    If you measure the voltage of the 9V battery to be 7.8V when nothing is connected to it then it is empty.

    The fact that higher voltage batteries can supply more current is not always true.

    Usually large batteries (in size) and with higher Ah ratings can supply more current and without being destroyed.
     
  5. Vincenzo1309

    Thread Starter Active Member

    Dec 28, 2008
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    So do you mean that if we measure it to be 9V with nothing connected, means that it is full. And when we measure it to be , let's say, 8.2V, it is empty.

    When you say empty, what do you mean?
    The current is empty??

    But the battery is still capable of starting a DC motor.

    Higher Ah means ampere per hour right?
     
  6. whale

    Active Member

    Dec 21, 2008
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    it's all because of internal resistance,
    you first measure the resistance of two battries and compare them you will get it.
     
  7. mik3

    Senior Member

    Feb 4, 2008
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    A 9V battery to be full its voltage has to be greater than 9V. If you measure it below 9V then it starts to get empty.

    By empty, it is meant that the chemicals in the battery can't produce an EMF (voltage) anymore.

    Yes, Ah means ampere-hour.
     
  8. Vincenzo1309

    Thread Starter Active Member

    Dec 28, 2008
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    I see, thanks alot for all of your guidance.

    As you see, I am now working on my final year project, and I need to control a solenoid, and 2 DC motors normally found in RF controlled cars.
    The scope is to build a Bluetooth controlled wheeled robot which can "kick" a ball by means of the solenoid.

    Any advice on what kind of battery is able to supply enough current to drive these 3 actuators?

    Iam now currently using a 9V Eveready and a pack of 4 AA Eveready batteries bought over the counter.
     
  9. mik3

    Senior Member

    Feb 4, 2008
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    What are the voltage and current requirements of the motors and the solenoid?
     
  10. Vincenzo1309

    Thread Starter Active Member

    Dec 28, 2008
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    I got the DC motors from a cheap RF controlled toy car. There are no voltage and current requirement available. All I know is that the car is powered up by a 3.6V battery pack.

    The voltage requirement of the solenoid is 9V, but I don't know about the current requirements.

    With all these info, will it be tough to select a suitable power supply?
     
  11. mik3

    Senior Member

    Feb 4, 2008
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    I guess the motors draw about 1A. Measure the coils resistance to see what will be the current at DC voltage.
     
  12. Vincenzo1309

    Thread Starter Active Member

    Dec 28, 2008
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    Hi, sorry, for the 3.6V battery pack, the current is 300mAh. So I think the DC motors' requirements is this?
     
  13. mik3

    Senior Member

    Feb 4, 2008
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    No, this does not mean the motors draw 300mA each but I think they do not draw more than 1A.
    What is the resistance of the coil?
     
  14. Vincenzo1309

    Thread Starter Active Member

    Dec 28, 2008
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    I am sorry, but I really can't find the specification of the motors.
    All I know is that the battery pack for the car is 3.6V, 300mAh. So 300mAh is enough to drive the 2 motors right?
     
  15. mik3

    Senior Member

    Feb 4, 2008
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    300mAh means that if you draw 300mA from the battery it will get empty in one hour. This does not say that the motor draws 300mA.
     
  16. Vincenzo1309

    Thread Starter Active Member

    Dec 28, 2008
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    Oh I see, sorry as I am a beginner in this kind of electronics project, please tolerate me.

    So does that mean that if one motor draw 500mA for example, and the solenoid draw 1A, for example, I need to find a battery source of current, say 850mA?

    How do we derive what current specs that our power supply needs?

    Kindly advise

    Thanks alot!!
     
  17. mik3

    Senior Member

    Feb 4, 2008
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    If the motor and the solenoid will work the same time the power supply will need to be able to supply 1.5 A (500mA +1A) without get destroyed.
     
  18. Vincenzo1309

    Thread Starter Active Member

    Dec 28, 2008
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    But since there are 2 DC motors, so add up to 2A. Does that mean that I have to purchase a power supply of current 2Ah?? Then what will be the voltage??

    If thats the case, the power supply will be big man!

    That will not be very appropriate for my project.

    What do you think?

    The 3 actuators will not always work together, maybe sometimes.

    What do you mean by "power supply will need to be able to supply 1.5 A (500mA +1A) without get destroyed."
    You mean the power supply will be destroyed if the actuators draw much more current than the power supply can give?
     
  19. mik3

    Senior Member

    Feb 4, 2008
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    I a power supply is rated to provide 2A at 9V and you draw more than 2A it will get destroyed unless it has overload protection.

    In your case you will need to purchase a 9V 2A power supply. You will power the coil directly from it but you will need to use a simple transistor voltage regulator (to drop the 9V down to 3.6V) or limit the current for the motors via a resistor as not to destroy them.
     
  20. Vincenzo1309

    Thread Starter Active Member

    Dec 28, 2008
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    "you will need to use a simple transistor voltage regulator (to drop the 9V down to 3.6V) or limit the current for the motors via a resistor as not to destroy them."

    So you mean that we just need to limit one of the parameters, be it the voltage or current? If I limit the voltage, I do not have to limit the current?

    But I have tried using 9V battery to power up the motors, they are working ok. So do I actually really need to limit the voltage?
     
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