Switching regulator forms constant-current source
Edited by Bill Travis
Stefan Strozecki, Institute of Telecommunications, Bydgoszcz, Poland -- EDN, May 30, 2002
Many applications require current sources rather than voltage sources. When you need a high-current source, using a linear regulator is inadvisable, because of the high power dissipation in the series resistor. To solve the wasted-power problem, you can use a switch-mode regulator. The circuit of Figure 1 uses IC1, an LM2576 adjustable regulator. It needs only a few external elements and has an adjustable sensing input, which you use for controlling the output current. Resistor RSC is a current sensor. IC2A, one-half of a TL082 op amp, operates as a difference amplifier. When R1=R2=R3=R4, the output voltage is proportional to the current flowing in RSC. Good common-mode rejection and a wide common-mode voltage range are important, because the amplifier works with large, changing common-mode signals.
The second half of the TL082 op amp, IC2B, operates as a noninverting amplifier. The required gain depends on the output current you need: G=VREF/VSC, where G is gain, VREF is the voltage on the sensing input of the LM2576, and VSC is the voltage across RSC. Note that VSC=IOUTRSC, where IOUT is the output current. For example, if IOUT=2A and RSC=0.12Ω, then VSC=0.24V. Typically, for the LM2576, VREF=1.237V. So, you can obtain the gain of the noninverting amplifier from the gain equation: G=5.15V/V. The overall gain of the noninverting amplifier is G=1+R7/R6. If R7=100 kΩ and G=5.15, you can solve for R6 (24.1 kΩ). When you need a precise output current, you can replace the fixed resistor, R6,with a series connection of a fixed resistor and a potentiometer. Tests showed that the output current is practically constant with varying loads. For example, the 2A output current changed less than 1% for an output-voltage range of 0.3 to 15V.
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