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I am new to electronics and trying to produce 850mA out of a 9V battery. I am just wondering if this would be possible and if so, how to do it.

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marked as duplicate by Finbarr, Sean Houlihane, Voltage Spike, W5VO Apr 26 at 4:50

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    \$\begingroup\$ That's way over the specification of most alkaline 9 V batteries. You can get AA battery packs that will deliver 9 V, though. And AA batteries can probably handle it (though that's still a lot.) A C cell based battery pack would be fine. \$\endgroup\$ – jonk Apr 21 at 16:45
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    \$\begingroup\$ What are you trying to do? Typical 9V batteries are designed for low current use (100 to 200mA.) If you put a resistor of about 10 ohms across the poles of a 9V battery, then about 900mA will flow - but not for long. A typical 9V battery has a capacity of maybe 200mAh. So, maybe 10 minutes (probably less) if you put that heavy a load on it. What are you really up to? There's probably a better way to do it than torturing underrated batteries. \$\endgroup\$ – JRE Apr 21 at 16:49
  • \$\begingroup\$ @JRE, I don't think the 9V can realistically produce 900 mA@9V, even for a short time. The internal resistance alone is several ohms. I second that the OP probably has some application in mind, and probably a much better alternative to 9V battery exist for that application. \$\endgroup\$ – anrieff Apr 21 at 16:51
  • \$\begingroup\$ A 12V battery and a dc to dc converter. \$\endgroup\$ – Solar Mike Apr 21 at 16:58
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    \$\begingroup\$ powerstream.com/9V-Alkaline-tests.htm \$\endgroup\$ – Bruce Abbott Apr 21 at 23:55
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Unlikely to be possible. 9V batteries cannot supply a lot of current; they're intended more for low-draw applications, like smoke detectors or alarm clock battery backups.

The specifications for a typical 9V battery only go out to a discharge current of 100 mA, and show an effective capacity of under 400 mAh at that current. If a 9V battery were able to supply 900 mA at all, its lifetime would likely be measured in minutes or less.

Consider using another type of battery for this application. You can get 9V from a pack of six 1.5V cells, for instance.

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  • \$\begingroup\$ Thanks for helping me understand, I'll try to figure something out with the 1.5V cells. \$\endgroup\$ – coopw Apr 21 at 23:25
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    \$\begingroup\$ A 9V battery is a pack of six 1.5V cells, after all. \$\endgroup\$ – Hearth Apr 21 at 23:49
  • \$\begingroup\$ @Hearth, true, a 9V battery is a pack of six 1.5V cells, but they are 6 tiny cells, not like as implied AA cells :-) \$\endgroup\$ – Ale..chenski Apr 22 at 4:14
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Internal impedance of modern and FRESH alkaline-based 9-V battery is about 1.5 Ω per this home-made experiment, and per this episodic datasheet is down to 0.040-0.17 Ω. Likely there is more statistics on the Internet, I am lazy to do a better search. The ESR is obviously non-linear and depends on battery load, quality, technology details, and how fresh it is.

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Taking the first example and assuming ESR = 1.5 Ω, if loaded with 850 mA, the voltage on battery terminals will drop by ~1.3V, to 9.6 - 1.3 = 8.3 V. Thus, if a load resistor is about 10 Ω you will get about 850 mA of current (at ~8 V). Assuming again that typical capacity of a 9-V battery is 500 mAh, the battery will last maybe 0.5 hour.

Given no information about your particular battery, your mileage may vary.

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