ilovefuzz.com

Would a voltage divider circuit where the V out is less than V in cause a 9v battery to drain quicker than usual? This is speaking of some DIY circuits I have found whose operating voltage is 6v. I have found that even in breadboarding and prototyping I am going through 9v batteries quickly (testing their remaining voltage with a multimeter). I apologize in advance for my layperson terminology as I starting out but learning fast. I wonder if my focus should be using wall warts to power the circuits and move away from the batteries. Thought? Suggestion? All perspectives will be helpful.

Doesn't seem like fuzz circuits would draw that much in use unless you are breadboarding huge fuzz circuits that are more complicated. But if you are doing testing for prolonged intervals of time and it's plugged in/active it could cause faster drainage.

Best best would be to get a boss or onespot to power your testing circuits. Another thing that you could be doing is leaving the battery connected to where it is draining all of the time. Using a power supply, you won't need to worry about this. And in the end, much cheaper...

A voltage divider shouldn't draw much current, not unless you're using really low value resistors. For example a pretty standard one using two 10k resistors to generate 4.5V would only use about 500uA. A 9V battery at a guess might have about 200mAh so could last up to about 400 hours - this is just a rough guess. You should read up on Ohms law, Kirchoffs laws and Thevenins theory.

It kind of sounds like you could be trying to power a whole circuit through the voltage divider by using the voltage divider to lower the voltage ? If you're trying to do this, it's just going to work like a voltage starve(it's at a high impedance - ie. it can't supply any current and is load dependent). You can use a voltage regulator instead.

I agree with MysteriousJ. If you need 6V for a circuit, the best way to handle it is with a voltage regulator. You can go with a LM317 with adjustable output voltage, or you can get a fixed 6V regulator. Or, if you have a 5V regulator (and a lot of people do because of PT2399s and other digital ICs), you can add a diode between the ground terminal and ground to "push up" the output voltage. The forward voltage rating of the diode will be the amount of voltage added. So if you used a 1N4148 on a 5V regulator, for example, the final output voltage would be ~5.7V.

nice!