Wayne Rewinkel, National Semiconductor, Schaumburg, IL
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 100mV 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 VOUT2. Inductor L1's ripple current flows through R5 and C5 and produces a certain amount of ripple voltage at VOUT1. 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, 50mV of ripple at the pin may correspond to excessive ripple voltage at VOUT1. 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 VOUT2 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 VOUT1, 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, VOUT2 falls below VOUT1 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 VOUT1 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 400mA, switching frequency is 480kHz, and desired minimum feedback ripple is 50mVp-p. Select L1 to operate at a ripple current of 200mA. Solving for L1, you obtain:
Select a Coilcraft DO1813P-473HC with ESR of 0.47¥Ø based on its small pc-board footprint. For C5, choose a ceramic capacitor that's large enough to limit the ripple voltage on VOUT to less than 10mVp-p. Given the maximum ripple voltage and a known trianglewave current drive, calculate a value for C5:
For C5, you can use TDK's 10¥ìF C3216X7R0J106 ceramic capacitor that presents an ESR of 3m¥Ø or less. Because the internal reference voltage for the LM5007 or LM5008 is 2.5V, set feedback resistors R3 and R4 to 1k¥Ø 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 100mV of ripple at VOUT1 and 50mV ripple at the feedback pin in the circuit of Figure 1, the design requires a value of 0.25¥Ø for R5. Adding C4 changes the value of R5 to 0.125¥Ø to provide 50mV ripple at VOUT1 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, VOUT2 droops to 100mV at full load but exhibits only 6mV of ripple without C4 in the circuit. Losses in R5 amount to 40mW. Adding C4 delivers 50mV of ripple to the feedback pin and VOUT1; setting R5 to 0.125¥Ø reduces R5's power loss to 20mW. In the third scenario, R6 is an open circuit, and the design in Figure 2 requires an R5 value of 0.125¥Ø to provide 50mV ripple at FB. VOUT1 exhibits no voltage droop at full load current and has only 6mV of ripple voltage. R5's power loss is 20mW.
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 50mV of ripple voltage at FB. Use the following equations to calculate the value:
With R6 in the circuit, VOUT drops slightly because R6 and R4 effectively connect in parallel. To compensate, you can slightly increase R6 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 VOUT by only 85mV. Adding R6 produces 6mV of ripple at VOUT and little or no loss in R5, but load regulation will not be perfect.
Inductor L1's ESR of 0.47¥Ø introduces a voltage drop of about 200mV at full load, which increases the voltage at the junction of L1 and R6 and also reduces VOUT to maintain 2.5V at FB. You can calculate the magnitude of the change by multiplying L1's voltage drop of 200mV 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).
