Buck regulator operates without a dedicated clock
Most switching regulators rely on a dedicated clock oscillator to determine the switching frequency of operation. A dedicated oscillator circuit within the power controller usually generates the clock signal. A class of hysteretic switching regulators can actually operate at a relatively fixed frequency without a clock, even with changing input-line and output-loading conditions. Figure 1 shows a simplified buck regulator operating in continuous-conduction mode. (The inductor current always remains positive.) The output voltage, VOUT, is equal to DVIN, where D is the duty-cycle ratio of buck switch Q1 and VIN is the input voltage. The duty cycle, D, at fixed-frequency operation is TON/TS, where TON is the on-time of Q1 and TS is the switching-frequency period, 1/FS. Some rearranging and substitution leads to the expression D=VOUT/VIN=TON/(1/FS)=TONFS.
Now, look at a regulator circuit, which, rather than using a fixed clock and a PWM, uses a circuit that turns on Q1 for a time, TON, that's inversely proportional to the input voltage, VIN. Figure 2 shows a regulator based on this principle. This regulator does not contain a clock oscillator, yet it remains at a fixed operating frequency even while the input voltage varies from 14 to 75V. The two main regulation blocks within this regulator are the on-timer and the regulation comparator. The comparator monitors the output voltage. If the output voltage is lower than the target value, the comparator enables the output switch, Q1, for a period of time that the on-timer determines. The time period of the on-timer is TON=KRON/VIN, where K is a constant (1.3×10–10), R is a configuration resistor, and VIN is the input voltage. If you now substitute TON in the previous buck-regulator equations, an interesting result occurs: VOUT/VIN=FSKRON/VIN. If you solve for FS, you obtain FS=VOUT/KRON. Because VOUT remains regulated and the K and RON terms are constants, the switching frequency also remains constant.
The constant-frequency relationship holds true provided that the inductor current remains continuous. At lighter loading, the current in the inductor becomes discontinuous. (The inductor current is zero for some portion of the switching cycle.) At the onset of discontinuous operation, the switching frequency begins to decrease. This reduction is a desirable feature to maintain high efficiency as the load decreases, because switching losses greatly decrease at lower switching frequencies. You derive the switching frequency in discontinuous mode as follows: The peak inductor current IP=VINTON/L=VINKRON/LVIN=KRON/L, where L is the output-inductor value. The output power is POUT=VOUT2/ROUT=LIP2F/2=FK2RON2/L. Solving for F: F=(VOUT2L)/(ROUTK2RON2). As you can see, the switching frequency varies inversely with the output resistance, ROUT.
Fixed-frequency operation without an oscillator offers a low-cost, easy-to-implement step-down regulator. You needn't worry about any loop-compensation or stability issues. The transient response is fast because the circuit has no bandwidth-limiting feedback components. Depending on the inductor value and loading, the operating frequency remains constant for most of the output-power range. A desirable reduction in operating frequency occurs at low loading levels.