Why A 9V Battery (Usually) Isn't Good For Circuit Power Jim Wagner, Oregon Research Electronics!

November, 2009 Major Revision: August, 2013

There have been a number of queries in various on-line forums about using a 9V battery to power digital circuits or motors. The simple fact is that they are not a very good match for these circuits and here is why. 1. Total Stored Energy: Many users of 9V batteries are confused about the amount of energy stored in batteries, generally. The perception is often that higher voltage batteries store more energy. Sadly, that is not so. Before digging into this, lets first discuss what we mean by “power” and “energy”. In electrical circuits, power is the product of the voltage being provided to some device and the current flowing through that device. For example, if we have a circuit that drives a motor with 9V and the motor draws 100mA (same as 0.1A) of current, that motor is consuming 9V * 0.1A = 0.9W of power. Power is important because some power sources cannot deliver as much power as other sources can. With most power sources, the voltage will drop as you try to increase the current. This is independent of how long the power has been drawn from the source. Other sources, notably batteries, provided a lower and lower voltage, the longer the power is used. Such sources provide a fixed amount of energy, and, as the energy is used up, the voltage drops. Energy is a measure of the total work done. In an electrical system, it is the product of power and time and is measured in watt-seconds (Joules) or watt-hours. For example, if a battery delivers 0.5W for 10 seconds, it has delivered, from its finite storage, 5 Watt-seconds (5 Joules). If it was carrying 20 Watt-seconds of energy, there is only 15 Watt-seconds remaining. The same amount of energy would have been used if the battery delivered 0.05W for 100 seconds.

Batteries are only indirectly rated for energy. The usual number you see on small batteries is “milliampere-hours”. The total energy stored in the battery is the product of its milliampere-hour (mA-Hr) rating and the average voltage over the life of the battery. The table, below, shows the pertinent numbers for a variety of common commercial “alkaline” batteries at low discharge currents. Type

Rated Voltage

mA-Hr

Avg Volt

Energy

AAA

1.5V

1500

1.1V

1.65 W-Hr

AA

1.5V

2900

1.1V

3.2 W-Hr

C

1.5V

8000

1.1V

8.8 W-Hr

D

1.5V

20000

1.1V

22 W-Hr

9V

9V

600

6.75

4.0 W-Hr

This table would appear to tell us that a 9V battery has a lot more energy than either an AA or AAA type battery. For a single battery, that is correct. However, if we are after a particular voltage, say, 9V, then it takes 6 of the 1.5V batteries to make 9V. Now, what is the available energy? A 9V battery made up of 6 of the AAA type cells has 9.9 W-Hr of energy available! Using AA cells in the same way would provide 19.2 WHr of energy (almost 5 times the energy of a standard 9V battery). To be fair, the table, above, assumes a discharge to a lower voltage than many of us would accept. If you cannot allow such a deep discharge, perhaps because your circuit stops working, then the available mA-Hr drops and the available stored energy drops, though the average voltage will be higher. 2. Volume Energy Delivery: The next point is a fairly subtle one. 9V batteries simply are not very efficient providers of energy, in terms of stored energy per volume. To make this point, I will use data on consumer-grade "Alkaline" batteries from http://www.duracell.com/en-US/products.jspx (Click "Technical Library", then select "Product Data Sheets"). The same data was used for the preceding table.

A typical 9V alkaline battery has a volume of 21.1 cubic centimeters and has a capacity of 600mA-hr (when discharged all the way down to 4.8V). The average voltage is around 6.75V over that discharge cycle. The actual energy is about 4000 milliWatt-Hr or 4.0W-Hr. Lets compare this to a pair of the worst 1.5V cells, the "AAA". One cell has a current capacity of 1500mA-Hr (when discharged down to 0.8V) and occupies about 3.8 cubic centimeters. This package is cylindrical and but occupies a rectangular space; the effective volume is about 4.4 cc. With an average discharge voltage of about 1.1V, the energy from one AAA cell is about 1.65W-Hr. Outwardly, the 9V battery seems to be almost three times as good. But, when you look at the volume required to provide that energy, a 9V battery occupies 21.1 cc compared to 4.4 cc for a AAA battery. So, 9V batteries deliver about 0.2W-Hr for each cc of volume while AAA batteries deliver about 0.38W-Hr for each cc of volume. The result of this comparison is especially important in limited volume applications, you get almost 75% more energy for every cubic centimeter of battery volume from AAA batteries. And, even more from AA batteries. 3. Peak Current Delivery: In this area, the differences between the cylindrical 1.5V batteries and the 9V battery is even more stark. For a 9V battery, the life graphs only show loads up to 50mA. An AA battery is characterized at loads up to 1A (1000mA). And it goes up from there, way up. D type batteries are characterized for loads up to 2A! Thus, if you have something, such as a motor, that takes a lot of current, then a 9V battery may not provide what you need. 4. Regulators: What if you don't have an application that is constrained by volume? You just want something that lasts a long time. Why not use a 9V battery since it holds almost twice as much energy as a pair of AAA batteries? The answer to this is "regulator".

Many microcontrollers have a maximum working voltage of (about) 5.5V with a maximum recommended working voltage of about 5V. So, how do you get 5V from a 9V battery? You need SOME kind of regulator. You have only two fundamental choices: a linear regulator or a switching regulator. Linear regulators are no different, in terms of efficiency, if they are just a resistor, a series zener diode, or an active (integrated circuit) regulator. They ALL dissipate significant power. The dissipated power is the product of the current THROUGH times the end-end voltage DROP. Thus, if you have a linear regulator that supplies 5V to a load that draws 10ma and the primary source is 9V, the end-end drop is 9V-5V = 4V. The power dissipated inside that regulator is 10mA * 4V = 40mW. The load, itself, consumes 5V * 10ma or 50mW. Thus, the whole system takes 90mW and the regulator burns 40mW/90mW = 44% of all the power in the battery. This power is converted to heat and is lost to practical use. This power loss gets smaller as the battery discharges. Over the life of the battery, however, you will loose about 30% of the battery energy in such a regulator. Switching regulators are "constant power" devices. Except for small circuit losses, the power into the regulator is the same as the power delivered to the load. Well designed switching regulators CAN achieve 80% to 90% efficiency. But, switching regulators take a lot of parts, are not inexpensive, and use precious board area. You can add to this dilemma the fact that the energy delivery of the 9V battery was measured for a battery discharged down to 4.8V. Even good linear regulators (and most switchers) will stop operating properly at about 5.2V input when delivering 5V. Below 5.2V, they may still deliver some output voltage, but it will no longer be regulated. Thus, you might not even be able to extract all of the specified energy from that 9V battery! Thus, given the choice between the poor efficiency of a linear regulator and the cost and complexity of a switching regulator, neither choice

looks very good. If these choices are not very good, what are some better choices? 5. Other Power Sources: Since many microcontrollers will operate down to 2.8V, it is attractive to connect three 1.5V cells in series, and simply NOT regulate. You will get nearly 100% of the available energy. This option, however, is not so good when you have support other hardware with limited operating voltage; 3.3V is quite common with a recommended range of 3.1V to 3.5V. Also, lower supply voltages may not allow the clock frequency you need. Then, one option is to use a linear regulator of the "low dropout" type (LDO). You can get them, fairly easily, with the ability to operate with as little as a 50mV difference between input and output. But this raises the cost and you are back with the inefficiency of a linear regulator. Another option is to use a "step-up" switcher, starting, perhaps, from a nominal 1.5V to generate 3.3V. There are also "charge pump" voltage doublers and triplers (with regulators, in some cases) that could be useful. Some designers opt for Lithium-Ion (Li-Ion) batteries. These are typically 4.3V fully charged and can be as low as 2.8V at maximum discharge. They carry a lot of energy per unit volume. The down-side is that they are more expensive and are complex to charge correctly. These batteries may be useful when performance is important enough to support the higher cost. There may be other battery options worth considering. Lithium polymer (LiPo) are easier to charge but don't hold quite as much energy as a LiIon battery. There are also other battery types that are available from distributors (but, rarely, from hobby shops). Each has its own set of constraints and advantages. 5. Conclusions: On almost all counts, the common 9V battery is a poor choice for microcontroller circuits and for applications that have high peak current requirements (such as motors).

A number of alternatives have been suggested. But, all are physically larger than a 9V battery. This points out the importance of addressing battery power early in a project so that you wonʼt be “surprised” when things do not operate the way you want.