Improve power density with a baby boost converter in a PFC circuit

1 Introduction

In the year 2000, server front end power-supply units (PSUs) – with an AC input to a 12-V/48-V DC rail – achieved around 10 W/in3 of power density, with about 85% peak efficiency [1]. Today, so many server PSUs can meet 80 Plus Platinum (94% peak) and 80 Plus Titanium (96% peak) requirements [2] that the latter requirement, along with very high power density (>90 W/in³), are both now becoming minimum requirements.

One reason why server PSUs can achieve higher power density levels is because of technological innovations in the semiconductor industry. New semiconductor fabrication processes enable devices to have lower parasitics and better figures of merit [3], significantly improving power dissipation and facilitating higher power density.

Topology and architecture innovations are also behind PSUs with high power density. Applying a totem-pole bridgeless power factor correction (PFC) circuit to the AC/DC rectifier stage of new server PSUs – along with wide bandgap devices such as gallium nitride (GaN) and silicon carbide (SiC) (Figure 1-1) – achieves the best converter efficiency over other bridge or bridgeless PFC topologies [4] [5]. Although higher efficiency does minimize the area needed for heat dissipation, a bulk capacitor (C_BULK in Figure 1-1) is still required to hold the output voltage in regulation after AC dropout. In order to accomplish this for more than 10 mS, a 3-kW server PSU would need a total capacitance of over 1.3 mF, which would consume at least 30% of the overall space. To further improve power density, you must reduce the bulk capacitance.

Figure 1-1 Server PSU block diagram.

In this article, the concept and operational principles of a “baby” boost converter (a compact boost converter that only operates during AC dropout events) is introduced to reduce the bulk capacitance. Test results on a PFC reference design [6] with a baby boost converter show that a 910-µF bulk capacitor (as opposed to a 1.3-mF capacitor) is enough to hold the output voltage above 320 V for more than 10 mS after AC dropout with a 3-kW load.

2 Selecting C_Bulk capacitance

The server front end PSU shown in Figure 1-1 generally consists of two stages: an AC/DC rectifier stage and an isolated DC/DC converter stage, with peak efficiency targets of >98.5% and >97.5%, respectively. In order for the isolated DC/DC stage to achieve an efficiency target >97.5%, the operational input voltage range of the isolated DC/DC converter (V_Bulk) generally has to be limited inside 320 V to 410 V (V_Bulk,max). Assuming a nominal bulk capacitor voltage (V_Bulk,nom) of 390 V, Equation 1 calculates the capacitance required for holding up 3 kW for 10 mS as:

Equation 1.

$C_{B u l k} = \frac{2 P_{O U T} \times t_{h o l d u p}}{V_{B u l k, n o m}^{2} - V_{B u l k, m i n}^{2}} = \frac{2 \times 3 k W \times 10 m S}{{(390 V)}^{2} - {(320 V)}^{2}} = 1.207 m F$

Considering V_Bulk voltage ripple and capacitance tolerance, the system shown in Figure 1-1 would require a capacitor with over 1.3 mF of capacitance. It is notable that the capacitor energy used to hold up the output voltage after AC dropout is only 32.6% of the total energy stored in the bulk capacitor during normal operation.

Inserting a baby boost converter stage in between the AC/DC rectifier stage and the isolated DC/DC converter stage (as shown in Figure 2-1) makes it possible to turn off the bypass field-effect transistor (FET) and enable the baby boost converter to allow the charging of C_BB to above 320 V from C_BULK after AC dropout. V_Bulk can then go much lower than 320 V, thus requiring less capacitance on the bulk capacitor to hold the output voltage for the same amount of time.

Assuming that V_Bulk can go down to 240 V (V_Bulk,min) with the baby boost converter during the AC dropout period, using Equation 1 equates to a required C_BULK of 635 μF, using 62% of the total capacitor energy.

Figure 2-1 Server PSU block diagram with a baby boost converter.

3 Baby boost converter design considerations

Although a baby boost converter can reduce a bulk capacitor’s size and capacitance, minimizing the converter’s footprint will help preserve the original high power density goal. Since a baby boost converter operates in a very short amount of time (an AC dropout event), the peak operational current and voltage stresses will determine your power-stage component selections, instead of continuous power dissipation. At V_Bulk,min, current stress should be at maximum. Select the boost diode and metal-oxide semiconductor FET (MOSFET) to handle current stress at V_Bulk,min while rated for V_Bulk,max. The baby boost inductor needs to handle peak current at V_Bulk,min.

Equation 2 determines the baby boost inductor inductance:

Equation 2.

L_{B B} = \frac{V_{B u l k, m i n} \times (V_{B B} - V_{B u l k, m i n})}{{∆ i}_{L B B} \times F_{s, B B} \times V_{B B}}

where V_BB is the voltage at C_BB, Δi_LBB is the baby boost inductor peak-to-peak ripple current and F_s,BB is the baby boost converter switching frequency.

Since the goal is to minimize the footprint of the inductor, Equation 3 assumes that the peak-to-peak ripple current equals twice the input current at V_Bulk,min and the maximum output power:

Equation 3.

{∆ i}_{L B B} = \frac{2 P_{O U T}}{V_{B u l k, m i n}} = \frac{2 \times 3 k W}{240 V} = 25 A

With V_BB regulating to 390 V and assuming F_s,BB = 500 kHz, Equation 2 calculates LBB as 7.385 µH.

Because the footprint is a higher design priority than the power dissipation, an inductor core with a higher saturation point is preferable; in the case of the baby boost converter, a powder iron core is a better choice than a ferrite core. The soft-saturation characteristic of the powder iron core [7] makes the design of the baby boost inductor design a bit tricky, however. With the core permeability dropping (inductance dropping) as the current increases, you must ensure the LBB calculated in Equation 2 is the inductance at i_LBB peak. Equation 4 estimates the inductance at a given magnetizing field:

Equation 4.

L = A_{L} \times μ_{i} % \times N^{2}

where A_L is the inductance factor in henry/turns², µ_i% is the remaining percentage of the initial permeability at a given magnetizing field, and N is the number of turns applied to the inductor.

Equation 5 expresses the relationship between µ_i% and the magnetizing field, according to the core manufacturer:

Equation 5.

μ_{i} % = \frac{1}{(a + {b H}^{c})}

where a, b and c are constant coefficients and H is the magnetizing field.

Assuming the application of a Magnetics 0076381A7 [8] (a Kool Mμ Hƒ core [9]) for the baby boost inductor, the constant coefficients a, b and c would be 0.01, 4.064∙10-7 and 2.131, respectively.

According to Ampere’s law, Equation 6 expresses the relationship between H and N:

Equation 6.

H = \frac{N \times I}{l_{e}}

where I is the current through the winding and l_e is the effective magnetic path length in centimeters.

With Equation 2 and Equation 3 calculating L_BB, Equation 4, Equation 5 and Equation 6 determine the N needed to achieve the inductance at a given magnetizing field.

It is also possible to estimate N iteratively. Assuming that a given inductor has an inductance that operates at a certain H with a given current, you can use Equation 4, Equation 5 and Equation 6 to evaluate whether the calculated H is close to the assumed H.

For example, if your initial guess is that H = 140 Oe with I = 25 A and the inductor has an inductance of 7.385 µH, Equation 4 calculates µ_i% at 39.65%. Then, taking the L_BB calculated from Equation 2 and Equation 3, along with the µ_i% calculated into Equation 4, N is then 20.8.

To verify H using Equation 6 with calculated N, you have H = 125.67 Oe. Since there is still error between the guessed H and the calculated H, you can make a second guess about H and calculate H again until the error becomes negligible. You will find the correct turns (operation point) after a couple of iterations. Using the iterative method, H is 108.75 Oe, with N = 18.009. The inductance is 7.385 µH at 25 A.