Add an auxiliary voltage to a buck regulator

John Betten, Texas Instruments, Dallas, TX -- 10/31/2002

You often need more than one regulated output voltage in a system. A frequently used and reasonably simple way to create this auxiliary output voltage is to add a second winding to the output inductor, creating a coupled inductor or a transformer, followed by a diode to rectify (peak-detect) this output voltage. The biggest drawback of this approach is that the diode's voltage drop varies with temperature and load current and can have a 2-to-1 variation, resulting in poor output-voltage regulation. This problem be-comes more critical as output voltages decrease and may require the addition of a linear regulator. The circuit shown in Figure 1 is an alternative approach that replaces this diode with Q₂, a p-channel FET. The circuit works as follows:

During the conduction time of FET Q₁, the voltage across the primary winding of transformer T₁ clamps to the voltage, V_OUT+V_F1, where V_F1 is the voltage drop across FET Q₁. Through transformer action, the voltage on the secondary winding of inductor L₁ is equal to the turns ratio between the windings times the voltage across the primary winding. The output capacitor on the auxiliary output, V₀₂, then charges to the peak of the secondary-winding voltage. FET Q₂ turns off when Q₃ turns back on to prevent the output capacitor from discharging. The secondary voltage floats; you can add it to the main output voltage by tying one end of the secondary winding to the main output. You can also tie it to ground for an output voltage lower than V₀₁, if desired. The equation that defines the auxiliary-output voltage for the circuit in Figure 1 is:

The second half of this equation represents a voltage-error term between FETs Q₁ and Q₂. To cancel out the error attributable to the FET voltage drops, you need to make the voltage drop of FET Q₂ equal to V_F2=V_F1×(N_S/N_P), where N_S/N_P is the transformer's turns ratio. Because these FET voltages are a function of the output currents and the on-resistance of the FETs, you can select the on-resistance of FET Q₂ by using the following equation:

In Figure 2, the main output voltage is 3.3V, yielding an inductor primary voltage when Q₁ is conducting equal to only 3.44V, because of the low voltage drop across FET Q₁. Thus, if you wanted a 5V output, the secondary winding would need to develop an additional 1.7V, necessitating a 2-to-1 step-down turns ratio. The desired on-resistance of the FET internal to IC₁ from the above equation should be 0.16Ω to cancel the voltage drop across Q₁ at maximum loads and while operating from a 5V input voltage. This example uses a 0.20Ω FET with a voltage drop equal to only 88 mV. This choice allows for good voltage matching between FETs Q₁ and the FET internal to IC₁, resulting in excellent error cancellation, less power loss, and better overall output-voltage regulation than diode rectification provide. An added benefit of this approach is that you can use it with controllers that have integrated switching FETs, because you don't need access to Q₁ and Q₃ gate drives. Measured results, although varying both outputs' loads over their full operational range, showed less than a ±3% variation in the 5V output.

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Add an auxiliary voltage to a buck regulator

John Betten, Texas Instruments, Dallas, TX -- 10/31/2002 document.write(get_publication('EDN'));

John Betten, Texas Instruments, Dallas, TX -- 10/31/2002