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Author Topic: Making a 12vdc to 9vdc regulator  (Read 4380 times)
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pyrolater
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« on: January 01, 2009, 07:55:34 PM »

I want to use a 9vdc Makita battery to run a chronograph.  The 9v dc makita battery meters at 12vdc when its fully charged.  The chrono takes a regular 9volt battery.  I'm not sure how much extra voltage the chrono can take so I want to feed the chrono a steady 9vdc so I don't smoke it

I'm sure this cant be to tough but I don't know how to do it.  Can you help me out??

Thanks
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Hak Foo
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« Reply #1 on: January 01, 2009, 08:08:06 PM »

Look into a part called a 7809.  Costs like 50 cents.
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pyrolater
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« Reply #2 on: January 01, 2009, 10:16:00 PM »

I looked at that and I also found a LM2940T-9.0-ND.  It drops out a a lower voltage.  They show 2 capacitors in the picture for the lm2940.  Does it make any difference on the type of capacitor used in this application??
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linear
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« Reply #3 on: January 01, 2009, 11:28:48 PM »

You can skip the bypass caps in a battery-supplied circuit. They are there to shunt any AC to ground, but you won't run into much of that.

What's your goal in doing this though? You may be throwing away runtime because the regulator dissipates quite a bit of energy to do its gig.

Did you measure the Makita battery under any kind of load? (I know the answer) It would be a good idea to figure out what kind of load your device presents, and pick a dummy resistor to simulate it, then measure the voltage at your battery. You may not need to do a damn thing to your battery.

Dummy load example: say your dingus draws 50mA current. The resistance that would give you that load current with an ideal 9V supply is R = V/I = 9V/0.050A = 180 ohms. The power dissipated in the load is going to be 9V*0.050A = 0.45W or 450 mW. So you'd want a 1/2 watt resistor for your dummy load (preferably a 1 watt).

The load current you (likely) measured under was effectively zero into a infinity ohm load, so the battery isn't behaving the same way as it will under load. The batteries have a series resistance of their own that will cause the voltage seen at the terminals to sag under load (more under larger loads). The way you (likely) measured won't account for that effect.
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pyrolater
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« Reply #4 on: January 02, 2009, 01:08:56 AM »

My goal is to quit buying 9v alkaline batteries.  I'm pretty sure I could get away with out doing any mods to this whole deal but I got a bug up my butt to make a little dohicky for my Chrono.  I'm not to worried about loosing any run time.    I got a bunch of makita batteries and no matter what they will last linger than those puny little alkaline batteries.

I did figure the makita battery voltage would drop but I didn't think it would be much. And your right I didn't load it the first time I metered it.  I don't have a 1 watt resistor handy but I do know if I jump a 12vdc auto tail lamp ( more than a watt) across the battery it will drop to 9.8 volts and hold,  for how long I don't know. 

My Dingus will run a 9v alkaline batt all day long so I don't think it draws much.  I just checked and pulling a full load its .034 amps.  So it won't draw much from the makita.

So do you think its worth the trouble to use the lm2940 or do you have a better way?





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