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5-to-1.8V converter works without magnetics

-April 13, 2000

To derive 1.8V from 5V, you might think of using a switch-mode regulator. Switchers are highly efficient but also complicated and expensive. Linear regulators, too, are out of the question unless your design can tolerate 40% efficiencies. The circuit in Figure 1, on the other hand, is more than 70% efficient (Figure 2a), sources as much as 100 mA, costs less than a switch-mode regulator, and requires less space.

IC1 is a CMOS charge-pump voltage converter that the circuit configures as a voltage inverter. With its output grounded and 5V at its V+ pin, IC1 generates V+/2, or approximately 2.5V at Pin 3. This nominal 2.5V output, which sags as the device sources current, drives linear-regulator IC2, which regulates the 2.5V input to 1.8V. IC2 is can source 100 mA before its sagging input voltage falls below the dropout level (Figure 2b). Using larger values for C1 and C2 enables IC1 to maintain its output voltage with heavier load currents. (DI #2511)


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