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Simple Buck-Boost Converter Shines In USB Applications


Liu Yang

March 29, 2004

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In USB applications, a constant 5 V is often required. But according to the USB standard, the input voltage can vary from 4.5 to 5.25 V. So converters must be used to stabilize the voltage. However, widely available buck (step-down) or boost (step-up) converters can't fulfill the requirement. Instead, a buck-boost converter is needed. It permits stable operation when the input voltage is either higher or lower than the desired output voltage.

The buck-boost converter in Figure 1 uses an Elantec EL7515 pulse-width modulation (PWM) step-up dc-dc regulator that incorporates an internal NFET switch, with the drain connected to the LX pin and the source connected to the PGND pin. By simply adding a PFET (Q1) and a diode (D2) to the typical operating circuit, the converter becomes a buck-boost converter.

When the internal NFET is turned on, the LX pin is pulled to low, as is Q1's source. Q1, L1, and the internal NFET form a current loop with the input voltage source (VIN), which charges L1. As a result, L1's current increases. When the NFET is turned off, LX's voltage rises to VOUT plus a diode drop; Q1 is off; and the current in L1 dumps to output capacitor C5. L1, D1, C5, and D2 form a current loop. As L1's current decreases, the output voltage is set by:

VOUT = 1.3\[1 + (R2/R1)]

The converter's voltage gain is:

(VOUT/VIN) = D/(1−D)

where D is the FET switching duty ratio (on-time/off-time).

Figures 2 and 3 show the converter's efficiency and regulation curves. For most of the load range, efficiency is 70% to75%, and load/line regulation is better than 2%.

This converter can also be used in higher-voltage applications, up to a maximum VOUT voltage of 17 V. The limitation is that VIN can't be too much higher (1 V) than VOUT, because Q1 isn't independently controlled. However, experimental results show that the converter works quite well for USB applications that require a 5-V output voltage.

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