# How to get high current from 9 volt batteries

I am planning to make a solenoid.. Which should be so powerful as to lift up about 600N force. I have thought about it, and I have a doubt. How can I get DC currents up to 20-30 A using a 9 V battery? Please suggest a way I can set them up in a way that would give enough current.

Another question, I thought about stepping up the voltage for a larger current. Would that work?

• Increasing the voltage decreases the current available. Jun 18 '13 at 5:35
• google.com/search?q=flash+circuit Jun 18 '13 at 5:36
• Only way to get high current from 9 V batteries is to connect large number of them in parallel, but that would have it's own down-sides. Really, 9 V batteries are extremely poor source of power. If you need current, get rechargeable 12 V battery or some lithium-polymer batteries. They'll be much cheapr in the long run. Jun 18 '13 at 6:01
• What is the dc resistance of your solenoid and what is the capacity of your 9V battery and type. Jun 18 '13 at 7:15
• You would need a really big 9V battery. Stepping up the voltage makes the problem worse, not better, if it's the current you are about. Aug 21 '19 at 14:58

A 9V battery has approximately a stored energy of:

$$\require{cancel} \frac{560\cancel{m}A\cancel{h} \cdot 9V}{1} \frac{3600s}{\cancel{h}} \frac{1}{1000\cancel{m}} \approx 18144VAs \approx 18 kJ$$

A joule is a watt-second, or a newton-meter. Under the most ideal conditions, with perfectly efficient machines everywhere, there is enough stored energy in a 9V battery to apply your specified 600N force over a distance of:

$$\frac{18\cancel{k}\cancel{J}}{1} \frac{\cancel{N}m}{\cancel{J}} \frac{1}{600\cancel{N}} \frac{1000}{\cancel{k}} = 30m$$

Your proposed solenoid, requiring maybe $25A$ at $9V$, consumes electrical power at the rate of:

$$25A \cdot 9V = 225W$$

Applying your specified $600N$ force, and given that power, we can solve for the velocity your solenoid, were it 100% efficient, could provide:

$$\frac{225\cancel{W}}{1} \frac{\cancel{J}}{\cancel{W}s} \frac{\cancel{N}m}{\cancel{J}} \frac{1}{600\cancel{N}} = 0.375m/s$$

So you see, even if we can extract all the stored energy of the 9V battery with 100% efficiency, there isn't a whole ton of it. Knowing that your ideal solenoid is moving at $0.375m/s$, and that the battery has enough energy for it to move $30m$, the runtime is:

$$\frac{30\cancel{m}}{1} \frac{s}{0.375\cancel{m}} = 80s$$

Or we could calculate it from the battery energy and solenoid power:

$$\frac{18000\cancel{W}s}{1} \frac{1}{225\cancel{W}} = 80s$$

But maybe it's enough. The question is how to do it efficiently. Electrical power in a resistance is given by:

$$P = I^2 R$$

The internal resistance of a 9V battery is maybe $1.5\Omega$, when fresh. It goes up as the battery drains. Your solenoid is probably at least another $1\Omega$. So at $25A$, your resistive losses alone would be:

$$(25A)^2 (1.5\Omega + 1\Omega) = 1562.5W$$

Compare this to the power used by the ideal solenoid considered above ($225W$) and you can see this is an absurdly inefficient system. Just dealing with the heat from these losses will be a challenge. Of course, you can't actually get this out of a 9V battery, because the voltage lost over its internal resistance at $25A$ is:

$$25A \cdot 1.5\Omega = 37.5V$$

...which is more than the 9V supplied by the battery.

Besides the battery, or the solenoid, transferring $225W$ of electrical power is a problem itself. Because power is the product of voltage and current ($P=IE$), to move a lot of power you can have high current, or high voltage. But, even wires have resistance, and since power lost to this resistance is proportional to the square of current, it's more practical to move high amounts of electrical power at high voltage than it is at high current. This is why the electric utility transmits power over long distances at very high voltage.

So, if you want to move $225W$ at $9V$, you must keep the resistance very low, to avoid resistive losses from being very high. That means fat wire (including the wire in your solenoid, which accounts for most of the wire in the circuit) and batteries with low internal resistance. You can also trade current for voltage, or voltage for current, in the design of your solenoid, as supercat's answer describes.

• Thank you for answering this from an ESR standpoint. Too many people look at batteries as ideal voltage sources, and miss important effects of actual batteries, particularly highly compromised types like the common 9V and coin cells. Jun 18 '13 at 16:39

Since you haven't specified, I'd assume that you mean a household, commercially available 9V battery. A standard 9V battery has about 400-600 mAh capacity. In the most basic terms, these batteries can supply about 500 milliamps for one hour before being "dead". You could, in theory, draw the current you are looking for, but even for several 9V batteries in parallel (sums the capacity), you'd get around 1-2 minutes from each set of batteries. With standard consumer batteries, that's pretty unrealistic, and pretty inefficient. I can imagine that you probably don't want to be changing batteries every minute.

If size and weight aren't huge factors, I'd look into large, high capacity, fast-discharge lead acid batteries. They come in 12V, which is close to what you're looking for, and can supply hundreds of amps of current for varying amounts of time (depending on how much you spend/which battery you get).

• The OP hasn't said (yet) what the battery is and therefore your answer presupposes that it is a "standard" type. You may be correct in your assumption but you may not be. Jun 18 '13 at 7:17
• Excellent point. Edited my answer a bit. Jun 18 '13 at 14:39

The design of a solenoid is a trade-off between using more turns versus using more current. I really doubt that any solenoid small which is small enough that one could realistically operate it off a 9-volt battery would achieve optimal behavior with a current as high as you suggest. Efficiency would most likely be better using more turns.

That having been said, the maximum solenoid current that can be gotten from a 9-volt battery would be obtained by connecting a capacitor in parallel with the battery, and then using a couple of efficient switches to alternately connect the solenoid to the battery and short it out (one must avoid ever having both switches closed, and should minimize the amount of time both are open; use a flyback diode to safely dissipate energy during the "both open" time). To absolutely maximize solenoid current without regard for battery life, set ratio of battery time to "shorting time" so that the battery is drawn down to about 4.5 volts. That will draw maximum power out of the battery, even though it will cause the battery to waste about half of its energy heating itself. Setting the ratio a bit lower, so it only draws the battery down to 6 volts, will reduce the coil current slightly (by about 12%) but improve battery life by 50%, since only a third of the battery's energy will be wasted in self-heating.

Using a 9V battery imples using one, single, 9V battery.

If that is not a real constraint you are working under, I recommend serially connecting several to achieve a higher voltage. If you can only use one single 9V battery, I recommend acquiring a high-capacitance capacitor and charging it through a current-limiting circuit, then switching it to power your solenoid. This will not give a sustained output, but depending on circuit and solenoid, it should give you 600 N

Use two IMR high drain lithium batteries in series. Sizes 18650 to 26650 or more can deliver currents as high as 60 amps. If voltage is too low (8.4) you can use LiFePo batteries for 9.6 volts, (3 times 3.2 volts).