Technique maximizes converter efficiency
Roger C Whipple, Hazeltine Corp, Braintree, MA - January 15, 2013
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Originally published in the January 5, 1989, issue of EDN
For a designer wishing to use a Linear Technology LT1072 switching regulator in the buck mode and being forced to deal with high input voltages, achieving the highest efficiency possible poses a problem. If, for example, you need to convert 20V to 5V at a relatively low power level of 1.25W, the quiescent current of the device itself (typically 6 mA) will become an important part of the circuit’s power consumption.
Because the quiescent current is relatively unaffected by the input voltage, the power that the IC consumes is directly proportional to its applied supply voltage. If your system has an external low-voltage supply available, you could run the IC from it—the LT1052 operates down to 2.6V. If such an auxiliary supply is absent, you can operate the IC from its own output by incorporating a switch-over circuit (Figure 1). Adding this feature boosts the supply’s overall efficiency from 77% to 83%.
Figure 1 This switch-over circuit powers the regulator IC from the supply’s output after the supply starts up. Running the IC from the lower-voltage output instead of the input raises the supply’s efficiency from 77% to 83%.
When you first apply power to the supply, the regulator has no output: R8 and D7 hold C6 discharged and the gate of MOSFET Q4 at ground. Because Q4 is turned off, the rising supply voltage pulls the gate of Q3 up via R5. As the supply voltage rises, Q3 turns on, applying the full input voltage to the IC and allowing the regulator to begin operation.
Once the regulator starts and the output voltage rises, C6 begins charging through R8. When the voltage on the gate of Q4 reaches about 2.5V, Q4 turns on, pulling the gate of Q3 to ground and shutting it off. This shutoff removes the input voltage from the IC. As C5 discharges into the IC, D5 becomes forward-biased and supplies voltage from the output to the IC.
If a power glitch or a momentary short circuit causes the output voltage to drop below the minimum that the LT1072 needs to operate, diode D7 will rapidly discharge C6, allowing the input voltage again to be applied to the IC. When the voltage rises again, normal operation will resume.