Switching regulator forms constant-current source

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 R_SC is a current sensor. IC_2A, 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 R_SC. 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, IC_2B, operates as a noninverting amplifier. The required gain depends on the output current you need: G=V_REF/V_SC, where G is gain, V_REF is the voltage on the sensing input of the LM2576, and V_SC is the voltage across R_SC. Note that V_SC=I_OUTR_SC, where I_OUT is the output current. For example, if I_OUT=2A and R_SC=0.12Ω, then V_SC=0.24V. Typically, for the LM2576, V_REF=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.