Design of a fixed frequency buck regulator

Most switching regulators rely on a dedicated clock oscillator to determine the switching frequency. A dedicated oscillator circuit within the power supply controller usually generates this clock signal. There is a class of hysteretic switching regulators that do not require a clock and can actually operate at a relatively fixed frequency even with input line and output load variations. 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 of the buck switch Q1 and VIN is the input voltage. The duty factor D for fixed frequency operation is TON/TS, where TON is the on-time of Q1 and TS is the switching frequency period 1/FS. Rearranging the circuit and replacing components yields the following formula: D=VOUT/VIN=TON/(1/FS)=TONFS.

Figure 1. In a typical buck regulator, the output voltage is equal to the switch duty cycle times the input voltage.

Now let's look at a regulator circuit that uses a circuit that makes Q1's on-time TON inversely proportional to the input voltage VIN, rather than a fixed clock and a PWM. Figure 2 shows a regulator based on this principle. This regulator does not contain a clock oscillator, but it maintains a fixed operating frequency even when the input voltage varies from 14V to 75V. The two main voltage regulation components in 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 turns on the output switch for a time determined by the on-timer. The time period of the on-timer is TON = KRON / VIN, where K is a constant (1.3 × 10-10), R is the configuration resistor, and VIN is the input voltage. If we now substitute TON into the previous buck regulator formula, we get an interesting result: VOUT / VIN = FSKRON / VIN. If we solve for FS, we get FS = VOUT / KRON. Because VOUT is still regulated and the K and RON terms are constant, the switching frequency remains constant.
Assuming the inductor current is still continuous, the constant frequency relationship is still valid. At light loads, the current in the inductor becomes discontinuous. (The inductor current is zero for part of the switching cycle.) Once discontinuous operation begins, the switching frequency begins to decrease. This decrease is a desirable characteristic that maintains high efficiency as the load decreases because the switching losses are greatly reduced at the lower switching frequency. The switching frequency in discontinuous mode is derived as follows: 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 changes inversely with the output resistance ROUT.
Fixed frequency operation without an oscillator provides a low-cost, easy-to-implement buck regulator. There is no need to worry about any loop compensation or stability issues. Transient response is very fast because the circuit has no bandwidth-limiting feedback components. The operating frequency remains constant over most of the output power range, depending on the inductor value and load, with a desired drop in operating frequency at light loads.

Figure 2 In this step-down regulator, the switching frequency remains constant over a wide input voltage range.