Electronic Power supply

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zaky

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I have build a simple power suply that works very well, as follow:.
9 volts in (9 volt battery) one voltage regulator (1.2 to 33 volts DC),2-150 ohms resistors and one 500 ohms potentiometer, I can adjusted from 1.2v to about 8.8 vdc.
I use it to drive a small portable cassette player(recorder) that requires 3 volt only.
At 3vdc it draws from 120 to 170 mA of current.
The 9 volt battery last abut 20 minutes.
I connected 2-9 volt batteries(in parallel) and the battery last it about 42 minutes.
My question is: Should I add other components to the circuit to make the battery last longer?
If any one has an idea please let me know.
Thanks to all of you for reading this messege.
zaky.
 
Stop using 9 volt batteries and use an 8 cell NiMH pack, 9 volt batteries have HORRIBLE energy density. If an 8 cell pack is too big for your taste use 3 Lipoly cells in series instead.
 
If you're driving a 3 volt cassette player, you might consider using a pair of 1.5 volt AA cells without the regulator. About the same size as the 9V, over 1000 mAH instead of 50 mAH.
 
AA size alkaline battery cells are 1.5V each for only a short time when they supply hundreds of mA. Their average voltage is 1.2V which is too low to drive many circuits.
A 9V alkaline battery is 625mAh. AA alkaline cells are 2850mAh.
 
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I assumed (perhaps wrongly so) that the "3V" was intended for a player/recorder designed for a pair of AA or AAA cells. It would be my guess that the OP was hoping to get more total energy from a 9V, which we know from our experience isn't the way it works.

A 9V battery operating at 150 mA delivers a lot less than it would at its ideal rating.
 
audioguru said:
AA size alkaline battery cells are 1.5V each for only a short time when they supply hundreds of mA. Their average voltage is 1.2V which is too low to drive many circuits.
A 9V alkaline battery is 625mAh. AA alkaline cells are 2850mAh.
Click to expand...

What is meant by mAH?
 
miliamp hours.
100maH's is 100ma's for 1 hour. Or 1 amp for 1/10th of an hour. when current draw increases much past the battery packs rateing in ma's the capacity decreases though.
 
For a battery, 100 mAH is generally measured for 10 hours at 10 mA load (it can be specified differently). Since the energy delivery in an alkaline cell at high currents is extremely inefficient, a 100 mAH Alkaline battery can't deliver 100 mA for a full hour.

Alkalines are poor in this respect (Old carbon cells are terrible); NiCd, NiMH, and Lead/acid deliver fairly high peak currents without losing as much capacity.
 
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Three NiMH cells in series will deliver 3.6V most of their life time. Probably won't do any harm. But a "3V' device is probably designed for 2.4 to 3.0V and could run from two of them.
 
Thanks to all of you for the information provided, now I can select one that work better for my project.
Thanks again
zaky
 
zaky said:
I have build a simple power suply that works very well, as follow:.
9 volts in (9 volt battery) one voltage regulator (1.2 to 33 volts DC),
Click to expand...
Is this a switching regulator or an LM317?

How did you get 33V

I can adjusted from 1.2v to about 8.8 vdc.
Click to expand...
I thought it was designed for 33VDC?

You won't even get 8.8V from an LM317 with a 9V battery, you'll be lucky to get 7V and that's on a brand new battery!
 
To: Hero999,The voltage regulator is a LM350 it can handle up to 33 vdc, my circuit is designed for 9 volts (or less) adjustable with a potentiometer, with a power supply input at 9 vdc I can adj the output to about 8.8vdc.
zaky
 
zaky said:
To: Hero999,The voltage regulator is a LM350 it can handle up to 33 vdc, my circuit is designed for 9 volts (or less) adjustable with a potentiometer, with a power supply input at 9 vdc I can adj the output to about 8.8vdc.
zaky
Click to expand...
No. An LM350 regulator has a dropout voltage of about 2V (more at high current) so with a 9V input its max output voltage is 7V or less. It might be 7.5V if there is no load.
 
audioguru.Yes, you'r problably right I did not test the power supply with loads above 3.5 vdc (but I will) I agree heavy loads more voltage drop.
thanks
zaky
 
The very small voltage drop from a regulator is different from its dropout voltage. When its input voltage is high enough for it to regulate properly, its output voltage drops only about 10mV with a load.
When it drops-out because its input voltage is too low then it doesn't regulate, its output could drop volts with a load.
 
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