Simple scheme keeps current drain constant

Peter Güttler, APS Software Engineering GmbH, Cologne, Germany -- EDN, 4/12/2001

It is sometimes advantageous to keep the overall current consumption of an electronic device constant. A large, seven-segment display, for example, draws nearly zero current when no segment is on to hundreds of milliamps when fully lit. This heavily varying current can cause EMI problems when a device receives its power through long cables from a remote power supply. The low-parts-count circuit in Figure 1 keeps current consumption constant. IC₂ is an ordinary three-terminal regulator that supplies 5V to the load, R₂. IC₂ draws a total current I₃=I_LOAD+I₄. (I₄ is approximately 8 mA, the quiescent current of IC₂). The negative three-terminal voltage regulator, IC₁, maintains 5V across R₁. The current through R₁ is I₂+I₃. So, I₂=5V/R₁–I₃, and total supply current I_SUP=I₁+5V/R₁. I₁ is approximately 2 mA, the quiescent current of IC₁. If the load draws more current, IC₁ reduces I₂ and vice versa.

This regulation works well as long as I₃ is smaller than 5V/R₁. If the load draws more current, IC₁ stops regulating and the voltage drop across R₁ rises above 5V. This example sets R₁ at 50W , setting the supply current, I_SUP, to approximately 102 mA. C₁ and C₄ are input-filter capacitors, C₂ improves ripple rejection, and C₃ provides stability. Note that R₁ dissipates (5V)²/R₁ and must have an adequate power rating. IC₁ and IC₂ may require heat-sinking. The minimum supply voltage for this circuit is 12V. (The minimum input voltage for IC₂=7V+IC₁'s reference voltage.) If your application cannot tolerate the 5V drop across R₁, try using an LM337 with a 1.25V reference voltage for IC₁.