Added components improve switching-regulator stability

Wayne Rewinkel, National Semiconductor, Schaumburg, IL; Edited by Brad Thompson and Fran Granville -- 9/29/2005

This Design Idea shows how adding one or two passive parts can reduce a hysteretic constant-on-time switched-mode voltage regulator's output voltage ripple and reduce its susceptibility to variations in external load capacitance and ESR (equivalent series resistance). The regulator operates much like a pure hysteretic switcher. The device's internal one-shot sets its pass transistor's on-time, making it less prone to frequency runaway but still susceptible to noise injected at the regulator's FB (feedback) pin. To switch cleanly with predictable frequency and duty cycle, the switcher requires application of approximately 50 to 100 mV of ripple voltage to the feedback pin. This Design Idea shows how four circuit implementations using National's LM5007 and LM5008 regulators satisfy the important feedback-ripple requirements by adjusting on-resistance and maintaining a nearly constant switching frequency as input voltage varies.

Figure 1 shows a basic buck regulator whose output capacitor, C₅, presents a high internal ESR, R₅. Note that the designer cannot access R₅ and V_OUT2. Inductor L₁'s ripple current flows through R₅ and C₅ and produces a certain amount of ripple voltage at V_OUT1. Although common, this simple design presents two problems: First, feedback resistors R₄ and R₃ form a voltage divider that reduces the output ripple presented to IC₅'s FB pin. Thus, 50 mV of ripple at the pin may correspond to excessive ripple voltage at V_OUT1. Adding a compensation capacitor, C₄, forces the output-ripple voltage to appear at the feedback pin. Second, a typical pc board may include many low-ESR ceramic bypass capacitors that attenuate ripple voltage to a level that destabilizes the circuit.

Suppose that you replace C₅ in Figure 1 with a low-ESR capacitor and add R₅ as a discrete resistor. You can connect the external load to V_OUT2 to reduce output-ripple voltage at the load and increase the circuit's immunity to added load capacitance. The regulator's feedback voltage derives from V_OUT1, and C₄ reduces output-ripple losses in the feedback divider, R₄ and R₃. Unfortunately, this frequently used design introduces new problems: As output current increases, V_OUT2 falls below V_OUT1 and degrades load-voltage regulation. Second, R₅ carries full load current and dissipates power, reducing overall efficiency. Using a large, high-wattage fractional-ohm resistor at R₅ increases product cost and regulator-package dimensions.

Figure 2 shows a rearranged buck-mode-switching regulator with two additional components. Assume that C₅'s ESR is negligible and that R₆ is open. Now, R₅ remains inside the feedback loop, and C₄ couples voltage ripple from the inductor side of R₅ to feedback. This action stabilizes the regulator's operation and requires almost no output-ripple voltage, and R₅ introduces only a small reduction in efficiency. Taking dc feedback at the load preserves the circuit's excellent load regulation.

In another scenario, suppose that you replace R₅ with a short circuit and select values for R₆ and C₄ that provide the desired amount of ripple voltage at the feedback pin. This configuration produces almost no output-voltage ripple and eliminates R₅'s power losses. Load regulation suffers because, as load current increases, inductor L₁'s ESR introduces voltage droop at V_OUT1 and forces the voltage at switching node SW slightly higher. However, designers can select L₁ for a low ESR that minimizes its effects on load regulation and can make R₆'s resistance larger than that of R₄.

The following examples compare output ripple, circuit losses, and component count for the design scenarios. Assume that input voltage is 50V, output voltage is 5V, output current is 400 mA, switching frequency is 480 kHz, and desired minimum feedback ripple is 50 mV p-p. Select L₁ to operate at a ripple current of 200 mA. Solving for L₁, you obtain:

Substituting on-time into the above equation yields

Select a Coilcraft DO1813P-473HC with ESR of 0.47Ω based on its small pc-board footprint. For C₅, choose a ceramic capacitor that's large enough to limit the ripple voltage on V_OUT to less than 10 mV p-p. Given the maximum ripple voltage and a known triangle-wave current drive, calculate a value for C₅:

For C₅, you can use TDK's 10-μF C3216X7R0J106 ceramic capacitor that presents an ESR of 3 mΩ or less. Because the internal reference voltage for the LM5007 or LM5008 is 2.5V, set feedback resistors R₃ and R₄ to 1 kΩ to divide the regulator's 5V output to 2.5V. Next, select values for R₅, C₄, and R₆ to compare results for each design. In the first scenario, to provide 100 mV of ripple at V_OUT1 and 50-mV ripple at the feedback pin in the circuit of Figure 1, the design requires a value of 0.25Ω for R₅. Adding C₄ changes the value of R₅ to 0.125Ω to provide 50-mV ripple at V_OUT1 and the feedback pin. You calculate a value for C₄ that passes the ripple current:

where R₄|| R₃ represents the value of the parallel combination of R₄ and R₃.

In the second scenario, V_OUT2 droops to 100 mV at full load but exhibits only 6 mV of ripple without C₄ in the circuit. Losses in R₅ amount to 40 mW. Adding C₄ delivers 50 mV of ripple to the feedback pin and V_OUT1; setting R₅ to 0.125Ω reduces R₅'s power loss to 20 mW. In the third scenario, R₆ is an open circuit, and the design in Figure 2 requires an R₅ value of 0.125Ω to provide 50-mV ripple at FB. V_OUT1 exhibits no voltage droop at full load current and has only 6 mV of ripple voltage. R₅'s power loss is 20 mW.

In the fourth case, a short circuit replaces R₅, and the design in Figure 2 requires a value for R₆ that increases the voltage across C₄ to provide 50 mV of ripple voltage at FB. Use the following equations to calculate the value:

With R₆ in the circuit, V_OUT drops slightly because R₆ and R₄ effectively connect in parallel. To compensate, you can slightly increase R₄ so that the new value of R₄ in parallel with R₆ equals R₄'s original value. Thus,

 

In this instance, you may decide not to use the new value of R₄ because adding R₆ raises V_OUT by only 85 mV. Adding R₆ produces 6 mV of ripple at V_OUT and little or no loss in R₅, but load regulation will not be perfect.

Inductor L₁'s ESR of 0.47Ω introduces a voltage drop of about 200 mV at full load, which increases the voltage at the junction of L₁ and R₆ and also reduces V_OUT to maintain 2.5V at FB. You can calculate the magnitude of the change by multiplying L₁'s voltage drop of 200 mV by the ratio of R₄ to R₆:

Note that output-voltage droop and power dissipation become more significant in designs that deliver higher output current or lower output voltage (Table 1).

 

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