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The drive to increase efficiency and reduce the size of switching power supplies has proliferated into just about every aspect of industrial, consumer, and lighting electronics. The simple step-down buck converter has largely been used in non-isolated dc-to-dc applications where a voltage conversion ratio of 5 to 1 or less is usually required (e.g. 24V DC to 5V DC) and where galvanic input to output isolation is not required. The buck is now finding its way into certain off-line applications where logic level voltages are required without the need for isolation from the AC mains. Since a typical 120V AC, off-line rectified "bulk" voltage is around 165V DC, stepping down to 5V DC or even 12V DC results in a greater than 10 to 1 voltage step-down ratio. The ratio is double for European applications where the mains are 230V AC. This large step-down ratio results in a high peak-to-average current ratio and poor conversion efficiency in the buck, particularly if the power required is greater than a few watts. This article illustrates a technique that will help overcome the inefficiencies of high conversion ratios by utilizing a tap on the buck inductor which not only effectively lowers the conversion ratio, and can provide a significant boost in the output current also.
Buck basics
Let's analyze the typical off-line buck converter shown in Figure 1. This is a 12V, 300-mA output buck supply using ON Semiconductor's monolithic NCP1014 current mode controller with internal 700V MOSFET. Voltage regulation and feedback is accomplished by the simple circuit of zener diode Z1, optocoupler U2, and the associated resistors. Referring to the NCP1014 datasheet, we see that the maximum peak current in the MOSFET at the nominal current limit point is 450 mA. The maximum output load current of the buck will have to be less than this value due to the additive magnetizing current component of the output choke. For a maximum load current of 300 mA, a safe limit to the magnetizing current should be 450 mA " 300 mA = 150 mA. The minimum acceptable inductance of the choke can be calculated from this once we establish a couple of other parameters.
Figure 1: Conventional off-line Buck converter.
(Click this image to view a larger, more detailed version)
The input-to-output DC transfer function for the buck converter is Vdc out = D x Vdc in where D is the on-time of the MOSFET's switching period. This relationship shows that the buck is a true linear integrator in which the area under the switching waveform determines the output voltage. This relationship holds true only if the current in the inductor is continuous throughout the entire switching period, or stated otherwise, it never goes to zero during a switching cycle under normal loading conditions. For the buck example of Figure 1, the relationship can be rearranged to determine the required duty cycle D for a typical AC input voltage and 12V output. For 120V AV input D will be as follows:
D = Vdc out/Vdc in = 12 Vdc/170 Vdc = 0.07 or 0.7%
In the above equation, Vdc in = 1.414 x 120 Vac = 170 Vdc due to peak charging of the off-line rectifier capacitive filter of C2 and C3. Note that for a 100-kHz switching frequency where the switching period T is 10 μs, the MOSFET on-time with D = 0.07 will be:
Ton = 0.07 x 10 μs = 0.7 μs
In other words, out of a switching period of 10 μs, the MOSFET needs to be on for only 0.7 μs out of the entire switching period to produce the desired output voltage. Also note that for a 5V output, the necessary on-time would be about 0.3 μs. This latter value approaches the typical internal propagation delay of many controllers and could result in erratic or burst mode operation since the controller can't smoothly decrease the duty cycle to a value less than the chip's propagation delay. From this we can see that a high input-to-output conversion ratio results in very low duty cycle operation for the MOSFET with a consequential high peak to average current ratio because the peak load current must be delivered during this period in addition to the buck inductor's magnetizing current which we must now determine based on the choke's inductance.
During the MOSFET off-time, or for 9.3 μs, the output inductor must maintain an output load current which "freewheels" through fast recovery diode D5. The voltage across the inductor is equal to the output voltage plus the forward drop of the freewheeling diode, or about 13V. In order for the current to remain continuous during this period (even though it is falling linearly), the choke's minimum inductance must be determined by the familiar relationship E = L x di/dt. Rearranging this for L we get:
L = (E x dt)/di = (13 V x 9.3 us)/ 0.15 A = 800 μH.
An 820 μH inductor rated at 500 mA should suffice here. This will appear to be a rather large buck inductance for designers that are used to lower conversion ratios. As mentioned before, the short duty cycle and the fact that the buck MOSFET must carry the maximum load current plus the inductor's inherent magnetizing current usually results in less than acceptable efficiencies with respect to the new Energy Star requirements for power supplies in this power range. The same calculations can be carried out for the 230V AC European mains and an even less desirable result will be obtained.
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