Optimizing Voltage Selection in Buck Converters Understanding the impact of input voltage and gate-drive voltage on MOSFET power losses enables designers to achieve maximum efficiency in the design of synchronous buck converters. F or several years, the dc-dc converters used to power microprocessor core supplies in PCs have been fed by a 12-V source. This voltage was chosen to limit the current carried through the harness of the ATX power supply as well as limit the maximum current in the ATX output rectifiers. Although this rationale makes sense with regard to the ATX supply, it is contrary to the requirements of the core supply—typically a synchronous buck converter—on the motherboard. That’s because dynamic losses in the buck converter rise in proportion to input voltage. Hence, the higher the input voltage, the lower the buck converter’s efficiency. Laboratory experiments show that both the input voltage and the gate-drive voltage play a major role in power loss in both the control or high-side (HS) MOSFET and the synchronous rectifier or low-side (LS) MOSFET, although to varying degrees. In this article, we will explore these effects and formulate the governing equations, and investigate the effects of different parameters on the converter’s power losses. A brief examination of the synchronous buck converter’s By Alan Elbanhawy Elbanhawy, Director, Computing and Telecommunications Segments, Advanced Power Systems Center, Fairchild Semiconductor, San Jose, Calif. circuit reveals the dynamic losses may be calculated from: 1 PDYNAMIC = ID VIN FS (T (TR + +T TF ) 2 From this equation, we can see that this loss mechanism is directly proportional to input voltage (VIN), the load current, which is approximately the MOSFET drain current (ID), and the switching frequency (FS). Therefore, the smaller any one of these parameters, the smaller the dynamic losses. The other major loss mechanism is conduction losses, which may be calculated for the HS MOSFET from the equation: PCONDUCTION = ID2 R DS(O DS(ON) DS (ON) N) ∆ where ∆ is the duty cycle: ∆≈ Output_Voltage Input_Voltage Efficiency (%) The smaller the input voltage, the larger the conduction losses for the same MOSFET on-resistance (RDS(ON)). Clearly, there is a point when the input voltage has a value where the combined dynamic and conduction losses are at a minimum, and this point is of particular interest to us because it represents the point of highest efficiency for the core supply. The effect of the gate-drive voltage is more complex because it involves the nonlinear relationship between the gate-drive voltage and the MOSFET on-resistance. By driving the MOSFET at the right drive level, RDS(ON) may be reduced by more than 50% from that of the nonoptimum gate-drive conditions. We conducted two sets of tests: one to verify the effect of the input voltage on losses and efficiency and the other to verify the effect of the gate-drive voltage on the same efficiency and losses. Fig. 1 depicts a voltage regulator module (VRM) efficiency as a function of the input Fig. 1. The measured converter efficiency of a VRM switching at 1 MHz is plotted as a voltage at different load currents. Given function of the input voltage over the 3.5-V to 12-V range. Power Electronics Technology June 2005 24 www.powerelectronics.com OPTIMIZING VOLTAGE Efficiency (%) an efficiency increase of about 3% may be gained when operating at the optimal gate-drive voltage as compared to driving at the conventional 12 V. Fig. 3 shows the effective loss resistance (ROL)[9] of the same converter in Fig. 2 as a function of the gate drive at different load currents. At full load, from 4.7-mΩ down to 3.9-mΩ, a gain of 17% can be achieved. Fig. 4 shows the gains that may be attained at different load currents by optimizing the gate drive individually at each point. � Fig. 2. The measured converter efficiency of a VRM switching at 200 kHz is plotted as function of gate-drive voltage over the 5-V to 11-V range. Mathematical Representations of the Losses First, we will derive formula to calculate the input voltage that delivers the highest power efficiency of a buck converter. To begin, consider the power dissipation relationship with input voltage for the top MOSFET. The power dissipation equation for the top MOSFET is as follows: OL PD = ID2 R DS(ON) Fig. 3. The effective output loss resistance for the VRM of Fig. 2 is plotted as a function of load current and gate-drive voltage. VO 1 + VIN ID FS ((T TR +T +TF ) VIN 2 where TR and TF are the rise and fall times, respectively; VIN is the input voltage; ID is the load current; FS is the switching frequency; RDS(ON) is the MOSFET on-resistance; VO is the output voltage; and VO/VIN is the duty cycle. Assuming that TR=TF , we get: Efficiency (%) PD = ID2 R DS(ON) (Note that we have ignored the losses due to the MOSFET output capacitance (COSS) because this term plays a secondary role in the power dissipation and it would complicate the solution immensely.) Taking the first derivative of PD with respect to VIN, we get: V 0 = -ID2 R DS(ON) O2 + ID FS TR (Eq. 1) VIN Optimum Drive Gate Drive = 12 V Fig. 4. A plot of efficiency versus load current reveals the efficiency improvement obtained with the optimized gate drive as compared with the standard 12-V gate drive. Then, taking the second derivative with respect to VIN yields: V 0 = 22IID2 R DS(ON) O3 VIN The second derivative is positive, indicating a minimum for power dissipation. Solving Eq. 1 for the optimum input voltage (VINOPT), we get: the typical 1.5-V VRM output voltage, optimum efficiency is achieved at an input voltage of 7 V to 8 V, not at the current 12 V. In this particular situation, an entire four percentage points in efficiency may be gained at full load, and considering that at this point the total losses are only 16 percentage points (100%–84%), this gain represents a 25% reduction in the converter los ses, a very sobering fact. Fig. 2 depicts the efficiency of a different 1.5-V output VRM as a function of the gate-drive voltage showing that Power Electronics Technology June 2005 VO + VIN ID FS TR VIN VINOPT = ID R DS(ON)T VO FS TR where RDS(ON)T is the top MOSFET’s on-resistance. For an example calculation of VINOPT , assume the following parameters: VO = 1.7 V, V IN = 12 V, T R = 15 ns, and 26 www.powerelectronics.com OPTIMIZING VOLTAGE RDS(ON)T = 0.012 . Plotting the optimal input voltage (VINOPT) versus drain current for switching frequencies of 300 kHz, 500 kHz and 1 MHz, we get the results graphed in Fig. 5. Now, consider the power dissipation relationship with input voltage for the bottom MOSFET. The power dissipation equation for the bottom MOSFET is as follows: V PD = ID2 R DS(ON)B 1 − O + VD ID FS TR VIN where RDS(ON)B = bottom MOSFET on-resistance. Taking the first derivative with respect to VIN: 0 = ID2 R DS(ON)B VO VIN 2 Now, taking the second derivative: 0 = -22IID2 R DSON(B) VO VIN 3 Fig. 5. Optimum input-source voltage for the top MOSFET alone is plotted as a function of drain current and switching frequency. between RDS(ON) and VG is linear for simplicity: RDS(ON)VG = RDS(ON) – BVG where VO is the amplitude of the gate-driver output voltage and B is a constant. Assume that VO=1.7 V, FS=1106 Hz, TR=1510-9 sec, and CIN=510-9 F. Then, substituting in the equation for PwrElec Ignition Coils1/4p 5/12/05 12:44 PM Page two points at VG=5 V and 10 V. The second derivative is negative, indicating a maximum as VIN approaches infinity. This clearly indicates that losses in the synchronous rectifier do not have a minimum as a function of VIN. Next, let’s consider both the top and the bottom MOSFETs’ losses together and attempt to find the optimum input voltage that would result in minimum losses, and hence, highest efficiency for the buck converter. The equation for the combined losses in the top and bottom MOSFETs is: V V PD = ID2 R DS(ON)B 1 − O + R DS(ON)T O + VIN VIN VD ID FS TR + VIN ID FS TR (Eq. 2) Taking the first derivative of Eq. 2, we get: R DS(ON)B VO R DS(ON)T VO A = ID2 + ID FS TR V 2 V 2 IN Transformers, Inductors, Coils & Ignition Coils UL/CSA/CE Standards IN Solving for VIN yields two solutions: -FS TR ID VO (R DS(ON)BB - R DS(ON)T ) FS TR − and -FS TR ID VO (R DS(ON)B - R DDSS(ON)T ) FS TR and taking the positive solution leaves: FS TR ID VO (R DS(ON)BB - R DS(ON)T ) FS TR For an example using this equation, assume that VO =1.7 V, V IN =12 V, T R =15 ns, R DS(ON)T =0.01 , and RDS(ON)B=0.006 . Once again, let us represent this equation in a graph form at switching frequencies of 300 kHz , 500 kHz and 1 MHz to derive the data plotted in Fig. 6. Next, let us consider the dependency of power dissipation on the gate-drive voltage. Assume that the relationship Featuring Standard & Custom SMT Products Capacitors • Varistors • Transformers • Plug-In Power Adapters 346 Monroe Ave., Kenilworth, NJ 07033 Tel: (908) 272-9262 • Fax: (908) 272-7630 www.ventronicsinc.com • e-mail: ventronics@prodigy.net CIRCLE 218 on Reader Service Card or freeproductinfo.net/pet www.powerelectronics.com 27 Power Electronics Technology June 2005 1 VINJ (V) VINJ (V) OPTIMIZING VOLTAGE G1 IDJ (A) IDJ (A) Fig. 6. Optimum power-source voltage is plotted for the combined top and bottom MOSFETs. 0.100 = RDS(ON) – B•5 0.008 = RDS(ON) – B•10 Solving Eq. 3 and Eq. 4 for RDS(ON) and B: -5 A = (11 -10 -10 ) (RB DS(ON) DS(O N) ) = A-1 i(0.010 0.008 ) Fig. 7. Optimum gate-drive voltage is plotted as a function of switching frequency and load current. I D 2 B∆ (Eq. 9) 2C IN FS Next, let’s consider the situation where we determine one optimum gate-drive voltage for both the top and bottom MOSFETs: 0.012=RDS(ON)T – B•5 0.010=RDS(ON)T – B•10 (Eq. 3) (Eq. 4) VG = (Eq. 5) Solving Eq. 5, we get: RDS(ON)=0.012 and B=4 10-4 The duty cycle () may be calculated according to this equation: A = (11 A = (11 Let us consider the losses in the top MOSFET: PD = ID (R DS(ON) - BVG )∆ + VG C IN FS + PDYN where CIN=total input capacitance measured at the gate of the MOSFET, including the Miller capacitance. Taking the first derivative of PD with respect to VG, (Eq. 6) -5 -10 ) (RB DS(ON)B B ) = A-1 i(0.008 0.006 ) 0 = -ID2 B T ∆ - ID2 BB ((11 - ∆) + 2VG (C INT )F FS INT + C INB ) (Eq. 7) Then, taking the second derivative: 0 = 2(C 2(C INT )F FS INT + C INB ) Now, let’s consider the bottom MOSFET given that VIN=12 V and =1-(V =1-(VO/VIN). In this case: Now, solving the first derivative equation yields: PD = ID (R DS(ON) (ON) − BVG )∆ + VG C IN FS + PDYN (ON) 2 VG = Taking the first derivative of PD with respect to VG yields: ID2 B T ∆ + ID2 BB (1 - ∆) 2(C C INT + C INB )FS Assume that CINT = 2.2 x10-9 F and CINB = 4x10-9 F. Then once again we may represent the equation for optimum gatedrive voltage in graphic form for switching frequencies of 300 kHz, 500 kHz and 1 MHz, as shown in Fig. 7. Finally, using the equations derived previously, Fig. 8 0 = -ID2 B∆ + 2VG C IN FS (Eq. 8) Now, solving for Eq. 6 the optimum gate-drive voltage VG, we get: Power Electronics Technology June 2005 ) = A-1 i (0.012 0.010 ) where BT and BB are the coefficients of the equation for RDS(ON) as a function of VG, and RDS(ON)T and RDS(ON)B are the other coefficients of the top and bottom MOSFET, respectively. Taking the first derivative with respect to VG: Now, solving Eq. 6 for the optimum gate-drive voltage VG, we get: 2 DS(ON)T T VG 2 ((C C INTT + C INB ))F FS + PDYN 2 I 2 B∆ VG = D 2C IN FS (RB The total power dissipation of both the top and bottom MOSFETs is: 2 PD = ID2 (R DS(ON)T (R DS(ON)B - BB VG )(1 − ∆) + (ON)T − B T VG )∆ + I D (R (ON) 2 0 = -ID B∆ + 2VG C IN FS ) 0.008=RDS(ON)B – BB•5 0.006=RDS(ON)B – BB•10 V ∆= O VIN 2 -5 -10 28 www.powerelectronics.com OPTIMIZING VOLTAGE Intel Technology Symposium 2003. 3. Elbanhawy, Alan. Mathematical Treatment for HS MOSFET Turn Off. Proc. PEDS 2003. 4. Elbanhawy, Alan. A Quantum Leap in Semiconductor Packaging. Proc. PCIM China, pp. 60-64. 5. Alan Elbanhawy. The Road to 200 Amps at one Volt VRM. Proc. PCIM Europe 2004, pp. 54-58. 6. Elbanhawy, Alan and W. New berr y. Packag ing Parasitic Fig. 8. Optimum input voltage as a function Fig. 9. Optimum gate-drive voltage as a Resistance Frequency Effect Effects. Power of the switching frequency FS and the load function of the switching frequency FS and Electronics Conference USA 2004 current ID . load current ID . PCIM San Francisco. depicts the optimum input voltage as a function of the load 7. Elbanhawy, Alan, and J. Ejury. Investigations of the current ID and the switching frequency FS. Examination of Influence of PCB Layout Parasitic Inductances in DC/DC the graph shows that the optimum input voltage for high Converters on the Efficiency. Proc. PCIM Europe 2004, frequency high current has a value of about 4 V to 6 V, a pp. 31-36. far cry from the current 12 V. Fig. 9 depicts the gate-drive 8. Elbanhawy, Alan. Are Traditional Packages Suitable for voltage again as a function of ID and FS and clearly shows the New Generation of DC-DC Converter? Proc. IPEMC China that higher currents require higher drive voltage, while 2004. higher frequency requires lower drive voltage for optimum 9. Elbanhawy, Alan. Is the Power Conversion Efficiency power loss. Running Out of Steam as a Comparison Tool? Proc. Portable Several conclusions can be drawn from the experiments Power Developer Conference USA 2005. PETech described here. We have shown in Figs. 5 and 6 that the optimal input-source voltage is not 12 V, but rather in the neighborhood of 3 V to 5 V, depending on the load current and the switching frequency. The larger current drawn from a 5-V source compared to a 12-V one can easily be dealt with through proper motherboard layout. Unfortunately, the off-line power-supply (silver-box) manufacturers have championed the push for higher source voltage because this allows them to continue using cheaper rectifiers instead of synchronous rectifiers. This savings results in lower-efficiency dc-dc converters and aggravates the thermal management problem in PCs. We also have shown that the gate-drive voltage has an optimal value. This could be accommodated by PWM controller manufacturers giving the end customer the choice of gate-drive voltage. The optimal gate-drive voltage, as seen from Eq. 7 and Eq. 9, is inversely proportional to the switching frequency and directly proportional to the square of the load current. There is no optimal source voltage for the synchronous rectifier on its own since a larger input voltage means longer on-time for the synchronous rectifier and hence more power dissipation. References 1. Elbanhawy, Alan. Effect of Parasitic Inductance on switching performance. Proc. PCIM Europe 2003, pp. 251-255. 2. Elbanhawy, Alan. Effect of Parasitic Inductance on Switching Performance of Synchronous Buck Converter. Proc. www.powerelectronics.com CIRCLE 219 on Reader Service Card or freeproductinfo.net/pet 29 Power Electronics Technology June 2005