Boost converter works with wide-range negative-input supply

Assume that a design requires positive voltage, but only a negative-voltage power source is available. Using a standard boost-converter IC in the circuit of Figure 1, you can efficiently generate a positive voltage from a negative source. The boost converter generates an output voltage that's higher than the input voltage. Because the output voltage—5V in this example—is higher than the negative-input-voltage ground level, the circuit does not violate the boost-converter principle. The circuit in Figure 1 uses the EL7515, a standard boost converter. The ground pins of the converter IC connect to a negative-voltage input source. Ground becomes the "positive" input source. V_OUT is as follows: V_OUT=V_FB(R₂/R₁)=1.33V(37.5 kΩ/10 kΩ)=5V. The Q₁ and Q₂ pnp transistors form a translator that scales the 5V output voltage (referred to ground) to a feedback voltage referred to the negative input. The transistor pair also eliminates temperature-change and voltage-drop effects. As the negative input voltage decreases, Q₂ runs at an increasingly higher current than Q₁, causing additional transistor-offset mismatch.

For optimal line regulation, you should set Q₁ and Q₂ to operate at the same currents with the nominal input-voltage applied. Figure 2 shows the line-regulation results. The maximum output-to-input voltage difference must be within the boost converter's internal power FET drain-to-source breakdown voltage (V_DS). For the EL7515, the maximum V_DS is 18V. For the 5V output, the minimum (most negative) input voltage is –12V. A 1V safety margin compensates for the D₁ diode drop and any voltage spikes on the drain of the power FET. Figure 3 shows the load-regulation test results. The maximum output current is a function of the input-to-output voltage ratio and current-limit setting of the boost converter. As Figure 4 shows, the circuit yields greater than 80% efficiency at 200-mA output.

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