The control circuit in Figure 1 presents a relatively low input resistance and thus imposes a low value on external pulldown resistor R₁ to provide the desired low-level input voltage. In turn, R₁ wastes power by drawing a relatively high current through switch S₁. For example, suppose that the control circuit presents an input resistance of 2.2 kΩ and requires a logic-low input of 5V or less. At a V_CC of 24V, R₁ must not exceed 500Ω for a current drain of 24/500=48 mA. The power dissipated in R₁ is thus 24²/0.5=1152 mW, which requires a 2W resistor for reliable operation.

In this application, the system includes three controls with three input circuits, which present a total current of 432 mA and adds approximately 10W to the power budget. To reduce wasted power, it uses an active pulldown circuit (inside the dashed line in Figure 2). As long as switch S₁ remains closed, PNP transistor Q₁'s base voltage exceeds its emitter voltage due to diode D₁'s forward voltage drop. Thus, Q₁ doesn't conduct, and the control circuit's input voltage rests at V_CC–0.7V (D₁'s forward voltage drop).

Opening S₁ reverse-biases D₁, and the base current flowing through resistor R₁ turns on Q₁, which saturates and pulls the circuit's input to V_CE(SAT). Closed-circuit current through S₁ is thus V_CC/R₁. For example, with V_CC of 24V and R₁ having a value of 10kΩ, I=24/10=2.4 mA, and R₁ dissipates 0.058W, or approximately 20 times less power than in Figure 1. In this application, current demand of the nine control-circuit inputs decreases from 432 to 22 mA and saves pc-board space by eliminating the need for using 2W. As a variant, Figure 3 shows an active-pullup version of the circuit.

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