Electronic – the max current from a 9 V battery

batteriescurrentohms-law

Let's take, for example, a 9 V battery. Forgetting about internal resistance or any temperature restrictions, what is the maximum current I can draw from this?

Using Ohm's law with a 1 Ω load, this should give us:

  • V = I/R
  • I = 9 V * 1 Ω
  • Current = 9 A

According to my calculations, this would give us ≈3.5 min of battery life

I also thought of it like this:

  • 9 V battery, 550 mAh battery life
  • 550 mA for 1 hour
  • 550 mA/h * 3600 secs = 1980 A for 1 sec

Drawing this much current at 9 V would require around 5 milliohms according to my calculations. I know this isn't possible in the real world, but theoretically maybe?

Even theoretically ignoring the temperature restrictions and internal resistance, this seems obviously impossible. Where am I going wrong?

Best Answer

If you "forget about" internal resistance, then the maximum current is infinite. An "ideal" component, non-existent in the real world, can provide mathematically "pure" infinite or zero amounts of resistance, voltage, current, and all the rest.

Different battery compositions will have different amounts of real-world "impure" limitations. Internal resistance, temperature versus performance characteristics, "memory" and recovery effects, and so on.

One of the difficult times I had learning about electronics was doing calculations and then wondering why the physical components on the breadboard were different. The figures on paper say I should measure 9 volts. I'm actually measuring 8.654 volts. What gives?

A short length of wire might well be only 5 mΩ, but when you connect the battery using only the wire, it doesn't vaporize the wire with a massive surge of almost 2000 amperes. Why? Because the battery is limited by real-world physics.

Some batteries are capable of some extremely high current. Consider automotive "wet cell" lead batteries. You'll find that they're capable of 1000 amperes or more, especially for turning over huge engines during start. In electronics and physics, many things are a trade off. If you want super high current, you may have to accept lower voltage, lower battery life, or extremely high cost.

A capacitor, as another example, can supply extremely high currents (compared to batteries), but they store charge, and are not a charge pump, as a battery is. As such, they're sort of like super-high-speed batteries with extremely limited capacity.

It was the biggest eye-opener for me as a kid in school to realize that applying Ohm's law to components was not exactly straightforward. You have to take the physics into consideration, and it's messy. A capacitor isn't just a capacitor: it has some resistance and inductance as well. The best way to think about components and batteries, I think, is that any component is a mixture of a bunch of other components, but imagine a control panel with sliders. A resistor might have its "resistance" slider at a large amount, but the "capacitance" and "inductance" sliders can't be at zero. A wirewound resistor, for example, will have more inductance than say a carbon composition resistor.

Your math isn't wrong, but it's for ideal components. Check out a battery datasheet; it'll provide you with some figures that show where it isn't exactly ideal.

(If you happen to have a 2000A-capable 9V battery, I know some electric vehicle engineers that would like to chat with you!)