Simple scheme keeps current drain constant
Peter Güttler, APS Software Engineering GmbH, Cologne, Germany -- EDN, April 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=ILOAD+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 ISUP=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, ISUP, 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.
