How Long Will My Battery Last with a Boost?

Batteries are ubiquitous in everyday portable items; yet for designers, they're still frustratingly complicated nonlinear devices. And the more the effort is to understand batteries, it seems the more complicated they become. In mathematical terms, you can describe them by using “stiff” equations. This means they have a large range of time constants involved in their description, resulting in analysis problems similar to those a simulator has trying to make a Bode plot of a switching power supply using a switching model.

At high frequencies (that is, high for batteries, i.e. 1 KHz and up) a battery appears inductive ^[1]. This can be due to the internal construction of the cells and lead lengths to the system. This means that for many switch-mode power supply applications, it's necessary to parallel the battery with a capacitor to ensure the Middlebrook criterion is satisfied: The source impedance must be lower than the load impedance to prevent oscillations.

At intermediate frequencies, say down to 1 Hz, nonlinearities appear in the I-V characteristic of batteries. The output voltage of the battery depends on how much current is pulled from it. Manufacturers' datasheets show this as an internal resistance. This is a reasonable approximation at low currents, but the real characteristic is a hyperbolic sine. This approximation works because for small values of the argument, the sinh function is about linear.

At the lowest frequencies, a battery has a finite amount of charge stored in it, and its output voltage goes down as this charge is removed. This time, scale determines how long a battery can run a device. The amount of charge recoverable and the voltage as a function of time depend on the current being drawn from the battery. At light rates (currents that would drain the battery in a month or more), the time-voltage curve is approximately linear. But at high rates (currents that would drain the battery in a day or less), a notable change is obvious in the slope of the time-voltage curve after just a few hours operation. Figs. 1 and 2 show typical V-t curves for an Energizer D-cell alkaline battery.

Looking at the low rate curve of Fig. 1 first, let's approximate the time-voltage curve at 30mA continuous current as linear. The voltage drops from 1.5V to 900mV in 600 hr, corresponding to a slope of -1mV/hr. Because the curve is linear, let's suppose that discharging the battery at half the current will double the time; and so for low rates, let's approximate the battery's voltage as V= 1.5 - 0.0001V/hr × I × t/30mA with t in hours, or more generally

Looking next at the high rate curve of Fig. 2, we can estimate that the break between the two slopes occurs at about 2 hr with a voltage of 1.3V. Using this, and discharging down to 800mV at 27.5 hr results in the two-part expression for high rate discharge

LEDs as a Battery Load

A common load for a battery is an LED. These two are used together in many applications, such as cell phone backlighting, flashlights, and so on. Since LEDs are semiconductor diodes, they have the typical nonlinear I-V characteristic: their forward current is exponential in their forward voltage. Fig. 3 shows measured characteristics of a super-bright white LED (green line) along with a best fit curve (red line) given by I = 106nA × exp (3.402 × V). The exponential function matches the data closely. Normal brightness at the nominal current of 20mA occurs with a V_f close to 3.6V.

In the simplest circuit, a battery is fed directly into an LED in series with a current limiting resistor. As shown in Fig. 4, to get enough voltage to forward bias the LED requires three of the 1.5V cells in series. Seven LEDs are paralleled to get adequate brightness. Selecting the value of the resistors sets the maximum brightness with a fresh battery. This must be a tradeoff against the battery operational life.

As a first step, we can estimate LED brightness as a function of time for this circuit using the previous equations. Because we will be having something on the order of 20mA through each LED, seven LEDs will pull 140mA, and we'll probably be in the high rate regime. The current (divided by 7) is set by the voltage across the resistor R, which in turn is equal to the battery voltage minus LED voltage:

This equation being valid until the battery voltage reaches 3.9V (1.3V per battery). This ugly integral equation can be changed to a differential equation by differentiating both sides by t, with the initial condition that at t = 0, the current is approximately

With R = 20Ω, the initial current is approximately 45mA per LED. The solution can be given in terms of Lambert's W-function ^[2]:

The voltage reaches 3.9V at 4.3 hr, at which time the current is 24mA per LED. The same equation with the 1.5V replaced by 1.3V and the 0.2 replaced by 0.04 gives the rest of the solution. As you can see in Fig. 5, the current per LED reaches 10mA (at which point the LEDs are noticeably much dimmer) at 45.8 hr, which is the run time of this battery in this application. The cells are down to 1.088V at this point, so this application is not fully using their capacity.

Using a Boost Driver

The other common method of driving LEDs is to use a boost converter such as the RYC9901, to drive all of the LEDs in series. As shown in Fig. 6, the current in the LEDs is set by a current sense resistor in series with the string. The distinguishing difference from the simple resistor ballasting is that the current through the LEDs is a constant; this implies that the power drawn from the battery is also constant, so the current goes up as the voltage goes down. It might be supposed, then, that the run time with a boost driver would be less than that with the simple resistor, especially considering efficiency loss in the switching. Surprisingly, the run time is actually longer.

Let's select 17mA of LED current, since this is the time average current produced by the resistive circuit

over the battery's operational life. This gives an LED voltage of 3.52V, so the power output is 17mA × 7 × 3.52V = 419mW; assume a typical efficiency of 80%, so the power required from the battery is 524mW. This is again, of course, high rate for the battery.

The equations for the case of a switching converter are conceptually similar to those for the resistive ballasting. In this case, we have a constant output power from the battery, which is I × V

and with the initial condition that initially, I(0) = 0.524W / 4.5V = 116mA. We can again differentiate with respect to time, and get a differential equation

which can be solved exactly with the initial conditions,

The original equation is good until V = 3.9V, and it is easy to find that this occurs at 8.1 hr. The same equation again holds for the rest of the discharge, with the solution

In the case of the RYC9901 boost converter, the circuit runs down to 2V, and this voltage is reached at 77.3 hr. The battery voltage during operation is shown in Fig. 7.

Running with a Boost

A more thorough analysis would include the effects of the internal resistance of the cells. Although this changes the details, the overall conclusion remains the same. The resistive circuit runs the LEDs for only 46 hr, while the intensity of the light decreases by a factor of 4. The boost circuit, despite its switching losses, manages to run the LEDs at constant illumination for 77 hr, in part because of its ability to get all of the energy out of the battery.

Because the batteries are considerably more expensive than the IC circuitry, overall cost of ownership is reduced by using a boost converter, quite aside from user convenience in having to change batteries less frequently and having more constant illumination. The same holds true for other types of batteries and other operating conditions.

References

Lenk, Ron, “Practical Design of Power Supplies,” IEEE Press / McGraw-Hill, 1998.
All calculations and plots were created with Maple 8 by Waterloo Maple.

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