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Low-cost active load draws constant battery power

-April 13, 2000

Unlike nickel-metal-hydride and lithium-ion rechargeable batteries, the discharge voltage of alkaline batteries is not constant and tends to vary from 1.5 to 0.8V per cell. Alkaline-battery manufacturers specify the discharge life of their batteries under constant powerloads. Because of the varying discharge voltage, you can't simply connect a load resistor to the cells to verify or measure power profiles without incurring a lot of error in your measurements. The circuit in Figure 1draws a constant power from the battery pack. Thus, as the battery voltage decreases, the load current increases. A single potentiometer allows a user to set the desired wattage draw. The circuit is simple and inexpensive to build.

The heart of the load is an analog multiplier/divider circuit based on a cheap transistor array, IC1. The five transistors have closely matching VBEs, which is crucial to the accuracy of the circuit. The circuit does not use the one transistor, Q1E, whose emitter ties to the chip's substrate.

The circuit applies the battery voltage between V1 and circuit ground. Op amp IC2A forces current I1 to flow through R1; thus, I1=VBATTERY/R1. Likewise, current I2 flows through R2 and is equal to V2/R2. The setting of potentiometer RSET determines I2. Q1A and Q1B, which the circuit wires as diodes, are within the feedback paths of the op amps, so their currents equal I1 and I2, respectively. Because of the logarithmic nature of VBEs, the voltage difference between the two diodes is as follows:

The circuit applies this voltage differential to the bases of Q1C and Q1D. Because all four transistors closely match one another, the ratio of currents I3/I4 equal the ratio I2/I1, or

I3, like I1 and I2, is set by an op-amp feedback network: I3=V5/R3. Because I3 must flow through Q1D, IC2C will supply enough current to the differential pair Q1C/Q1D to force I4=V6/R5. IC2D will then force its output voltage, which drives current sink Q2, such that V6 equals I4×R4. All you need to do now if finish the math:

or

Note that the circuit forces V6 across the current-setting resistor, R6, so that the current forced through transistor Q2 is ILOAD=V6/R6. Thus,

The load current is inversely proportional to the applied battery voltage, which is precisely what you want. V2 and V5 are scaling voltages, allowing you to trim the current source value. In this case, the reference diode, D1, sets V5 at 1.225V, and a potentiometer sets V2. The resistor ratios in Figure 1 allow DMM measurements of V2 to correspond to a power-dissipation level of 1V =1W from a four-cell battery pack. You should try to keep I1 through I4 at 10 to 500 µA for best linearity. Using an op amp with a lower input offset voltage also helps improve accuracy.

R1 draws a small amount of power from the battery as well, adding to the preceding equation. However, this error term is small, and you can ignore it. The current-source bias voltage, namely the load current times the sum of R6 and Q2's on-resistance, or RFET, limits the minimum battery voltage. Doing the math, you'll find that
VIN(MIN)=  where V2 is the desired power draw.

A small power supply that outputs approximately 6.3V ac powers the circuit. Two 1N4001 diodes half-wave-rectify the positive and negative voltages, and two 100-µF capacitors provide filtering. The extremely light load of the circuit means that postfilter regulation is unnecessary. If portability is necessary, you could power the circuit from a pair of series-wired 9V batteries. You can also use asymmetrical power supplies, but you need a positive voltage of around 8V to allow Q2 to turn fully on. If you substitute a different op amp, make sure that it can withstand the power-supply voltages. Q2 dissipates most of the power, so it needs an attached heat sink. (DI #2512)


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