Index
1. Definition
2. Basic Topologies
2.1. Buck Converter
2.2. Boost Converter
2.3. Buck-Boost Converter1. Definition
LDOs can have poor power efficiency when input and output voltages differ by a lot. For example, the power efficiency for a 12V-to-5V LDO is at most 5/12 = 41.7%. We seek high efficiency solutions by utilizing inductors and capacitors to transfer energy with low loss regardless of the difference between input and output voltages. Several topologies that serve this purpose are introduced. In each of these topologies, switches are used to put the circuit at alternating states, hence these types of DC-DC regulators are called switching regulators.
2. Basic Topologies
2.1 Buck Converter
Buck converter, or step-down converter, has a output DC voltage level that is lower than the input. Below is an asynchronous buck converter.
Operation Principle
At on-state, the switch S closes, and the power supply charges the inductor. Assume the output voltage is constant due to a large-enough capacitor, then the current through the inductor ramps up since VL,on = L*ΔiL,on/Δton = positive constant. At off-state, inductor current ramps down since VL,off is a negative constant. Note that the total increase and decrease of inductor current needs to be the same, i.e., ΔiL,on = -ΔiL,off as shown below, in order for output to stay constant; this in term requires ton/toff to be constant. We define “duty cycle” to be ton/ttotal = ton/(ton + toff) and denote it with D. D is sometimes expressed as percentage.
Example
Suppose VIN = 12 V, ton = 0.5 us, toff = 0.7 us, L = 10 uH, C is large to ensure constant output, R = 5 Ω, the diode is ideal, i.e., Vd = 0 V. At on-state, VL = VIN – VOUT = L*ΔiL,on/ton. At off-state, VL = 0 – VOUT = L*ΔiL,off/toff. For total energy transfer, ΔiL,on = -ΔiL,off, hence -(VIN – VOUT)/VOUT = -toff/ton = -7/5 => VOUT = 5 V. In general, VIN and VOUT in a buck converter follows the formula: VOUT/VIN = D = ton/ttotal
The above example assumes constant output voltage, but with a simple analysis we know it cannot be true. For a constant output, the average inductor current is supplied to the output, while the differential current goes through the capacitor. But the capacitor v/i formula iC = C*dVOUT/dt shows that when iC ≠ 0, the output voltage is not constant, which alters the inductor current analysis and recursively brings higher-order effects, let alone the parasitic component of the passive devices. The good news is that the variation is very small comparing to the output level, and as the order gets higher, the effect decays. For educational purpose, we can evaluate the second-order output ripple when C = 5 uF, assuming inductor current is still perfectly linear.
The total differential current in the inductor is ΔiL,on = (VIN – VOUT)*ton/L = 0.35 A. This 0.35 A differential current nearly all goes into the capacitor. Use iC = C*dVOUT/dt, we obtain ΔVOUT = 0.6 uV, less than 0.01% of the output. In reality, the energy loss from the parasitics or non-ideal components has a much greater impact on output ripple, but overall, the ripple is typically required to not exceed 5% or 2% of the output level.
As described, inductor current ramps up and down, and the average current is supplied to the load, while the differential current goes through capacitor. For normal operation the inductor current is always non-negative, i.e., iload – ΔiL,on/2 ≥ 0 because there is no back flowing path. What if the load is changed to 5 kΩ? In this case, iload – ΔiL,on/2 = 1 mA – 0.125A < 0, the mode of operation has to change.
