Simple Circuit Generates +3.3 V At 1 A From –5 V

May 24, 2004
Figure 1 depicts a circuit that uses pulse-frequency modulation (PFM) to generate +3.3 V from a −5-V power supply without the need for a transformer. It's ideal for applications that have a stable...

Figure 1 depicts a circuit that uses pulse-frequency modulation (PFM) to generate +3.3 V from a −5-V power supply without the need for a transformer. It's ideal for applications that have a stable −5-V power supply and don't require isolation.

The switch-mode regulator (IC1) operates in the classic boost configuration in non-bootstrapped mode. Conversion with respect to the IC is +5 to +8.3 V. But the connections shown (GND and AGND to −5 V and V+ to system ground) let the circuit produce an output of 3.3 V relative to the system ground. Its conversion efficiency is over 90% for 1-A loads and about 84% for light loads of approximately 10 mA (Fig. 2).

Figure 3 shows the output voltage (set by external resistors) versus load current. For a 3.3-V output, load regulation is within 3 mV for load currents up to 500 mA and within 3% of 3.3 V for load currents up to 2 A. You adjust the circuit output by selecting the values of R1 and R2:

where the feedback resistor (R1) is between 10 kΩ and 500 kΩ, and VREF = 1.5 V.

The input and output capacitors (C1 and C4) should have very low effective series resistance (ESR) to ensure high efficiency and minimum output-voltage ripple. As shown, C1 is a 100-µF OS-CON capacitor with 25-mΩ ESR, and C4-C5 are 150-µF OS-CON capacitors with 25-mΩ ESR. For light loads or in applications that can tolerate higher output ripple, capacitors with smaller values or higher ESR (or both) are acceptable.

Because the −5-V supply voltage serves as a ground reference for IC1, that voltage should be very stable. Any voltage change in the −5-V rail appears with unchanged magnitude at the +3.3-V output.

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