few more questions.

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  #1  
Old 03-09-05, 06:52 PM
jaxx751
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few more questions.

ok, so i'm wondering, since i only SORTA get this ohms/watts/amperage/volts

so... i have:

**all 12Vdc**

1A 12Vdc transformer.
6W pump
2W fan

my question, i am deciding to put in a SPST switch to turn on the whole thing, and have the fan connected after the 1st switch, then a SPDT switch for 7V and 12V... or just 7V...

any suggestions at all? im having a really hard time.

-jaxx
 
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  #2  
Old 03-09-05, 07:11 PM
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You plan on putting the fan and pump in series or in parallel? Of course if the loads are both 12-volt DC loads, you'll have to put them in parallel if you want them to work. So that will be a 8-watt load on a transformer capable of 12 watts.

What exactly do you expect these two switches to do? What will be running with S1-off/S2-off, S1-on/S2-off, S1-off/S2-on, and S1-on/S2-on?
 
  #3  
Old 03-09-05, 07:50 PM
jaxx751
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never mind, im pretty sure that i want to use linear voltage regulator... i am just going to want to run the pump at 7V, no variation.

the pump is rated at 6W, 12Vdc, .5A...

any way that i can do this? i will still have a SPST switch to turn everthing on.

thanks again,
-jaxx
 
  #4  
Old 03-09-05, 08:55 PM
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"the pump is rated at 6W, 12Vdc, .5A..."

12 volts - 7 volts = 5 volts
you want to drop 5 volts
5 volts / .5 amps = 10 ohms
5 volts x .5 amps = 2.5 watts

Use a 10 ohm resistor, larger then 2-1/2 watts for a 0.5 amp load.
Get a10 ohm resistor 5 watts.
The resistor may get between warm and hot. thats ok.
 
  #5  
Old 03-13-05, 01:26 AM
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I made an error in my calculations.
I did not take in account that your pump will use less current with less voltage.

12 volt pump / .5 A = 24 ohms (pump resistance)
7 volts / 24 ohms = 0.29 A (current your pump will pull at 7 volts)

12 volts - 7 volts = 5 volts
5 volts / .29 A = 17.24 ohm resistor.
5 volts x .29 A = 1.45 watt resistor.
you need a 17 ohm 2 watt resistor.
 
  #6  
Old 03-13-05, 03:37 AM
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Location: Oregon
Posts: 1,219
GWIZ,

A pump is _not_ a resistive load, so your calculations don't apply very well. The current consumed by a motor depends upon the applied voltage, the resistance of the coils and the _speed_ of the rotor; since the speed of the rotor depends upon the connected load, you almost can't know what you are going to get. It is even possible that by reducing the voltage to the motor that the current would _increase_.

Jaax,

I don't know where the 7V number comes from. If the pump is rated for 12V, then it should be able to take 12V. I presume that you want to be able to slow the pump down. You are just going to have to try out different resistors to see what works.

-Jon
 
  #7  
Old 03-13-05, 03:35 PM
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Join Date: Oct 2004
Location: USA
Posts: 719
WINNIE, is right.
But now you need a tachometer, dynamometer, amp meter, volt meter, constant voltage power supply and a flow meter.

The small DC motor, question and application. Did not warrant the technical explanation to get that last percent of accuracy.
The calculations will work for 95% of most peoples DC requirements.
 
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