Basic concepts of linear regulator and switching mode power supplies, Part one

-August 28, 2013

Applications Where Linear Regulators Are Preferable


There are many applications in which linear regulators or LDOs provide superior solutions to switching supplies, including:


1. Simple/low cost solutions. Linear regulator or LDO solutions are simple and easy to use, especially for low power applications with low output current where thermal stress is not critical. No external power inductor is required.


2. Low noise/low ripple applications. For noise-sensitive applications, such as communication and radio devices, minimizing the supply noise is very critical. Linear regulators have very low output voltage ripple because there are no elements switching on and off frequently and linear regulators can have very high bandwidth. So there is little EMI problem. Some special LDOs, such as Linear Technology’s LT1761 LDO family, have as low as 20μVRMS noise voltage on the output. It is almost impossible for an SMPS to achieve this low noise level. An SMPS usually has mV of output ripple even with very low ESR capacitors.


3. Fast transient applications. The linear regulator feedback loop is usually internal, so no external compensation is required. Typically, linear regulators have wider control loop bandwidth and faster transient response than that of SMPS.


4. Low dropout applications. For applications where output voltage is close to the input voltage, LDOs may be more efficient than an SMPS. There are very low dropout LDOs (VLDO) such as Linear’s LTC1844, LT3020 and LTC3025 with from 20mV to 90mV dropout voltage and up to 150mA current. The minimum input voltage can be as low as 0.9V. Because there is no AC switching loss in an LR, the light load efficiency of an LR or an LDO is similar to its full load efficiency. An SMPS usually has lower light load efficiency because of its AC switching losses. In battery powered applications in which light load efficiency is also critical, an LDO can provide a better solution than an SMPS.


In summary, designers use linear regulators or LDOs because they are simple, low noise, low cost, easy to use and provide fast transient response. If VO is close to VIN, an LDO may be more efficient than an SMPS.




Why Use a Switching Mode Supply? A quick answer is high efficiency. In an SMPS, the transistors are operated in switching mode instead of linear mode. This means that when the transistor is on and conducting current, the voltage drop across its power path is minimal. When the transistor is off and blocking high voltage, there is almost no current through its power path. So the semiconductor transistor is like an ideal switch. The power loss in the transistor is therefore minimized. High efficiency, low power dissipation and high power density (small size) are the main reasons for designers to use SMPS instead of linear regulators or LDOs, especially in high current applications. For example, nowadays a 12VIN, 3.3VOUT switching mode synchronous buck step-down supply can usually achieve >90% efficiency vs. less than 27.5% from a linear regulator. This means a power loss or size reduction of at least eight times.


The Most Popular Switching Supply—the Buck Converter


Figure 8 shows the simplest and most popular switching regulator, the buck DC/DC converter. It has two operating modes, depending on if the transistor Q1 is turned on or off. To simplify the discussion, all the power devices are assumed to be ideal. When switch (transistor) Q1 is turned on, the switching node voltage VSW = VIN and inductor L current is being charged up by (VIN – VO). Figure 8(a) shows the equivalent circuit in this inductor-charging mode. When switch Q1 is turned off, inductor current goes through the freewheeling diode D1, as shown in Figure 8(b). The switching node voltage VSW = 0V and inductor L current is discharged by the VO load. Since the ideal inductor cannot have DC voltage in the steady state, the average output voltage VO can be given as: 


VO(DC) = AVG [VSW] = (TON/TS ) • VIN               (2)


Where TON is the on-time interval within the switching period TS. If the ratio of TON/TS is defined as duty cycle D, the output voltage VO is:


 VO(DC) = (TON/TS ) • VIN = D • VIN                  (3)


When the filter inductor L and output capacitor CO values are sufficiently high, the output voltage VO is a DC voltage with only mV ripple. In this case, for a 12V input buck supply, conceptually, a 27.5% duty cycle provides a 3.3V output voltage.



Figure 8. Buck Converter Operating Modes and Typical Waveforms


Other than the above averaging approach, there is another way to derive the duty cycle equation. The ideal inductor cannot have DC voltage in steady state. So it must maintain inductor volt-second balance within a switching period. According to the inductor voltage waveform in Figure 8, volt-second balance requires:


(VIN – VO) • D • TS = VO • (1 – D) • TS           (4) 


Hence, VO = VIN • D              (5)


Equation (5) is the same as equation (3). The same volt-second balance approach can be used for other DC/DC topologies to derive the duty cycle vs. VIN and VO equations.


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