AN-1197 dcdc inductorsAN-1197 dcdc inductors, zasilacze
[ Pobierz całość w formacie PDF ]
Selecting Inductors for Buck Converters National Semiconductor Application Note 1197 Sanjaya Maniktala July 2002 Introduction This Application Note provides design information to help select an off-the-shelf inductor for any continuous-mode buck converter application. The first part shows how the designer should estimate his requirements, specifically the required inductance. The next part takes an off-the-shelf inductor and shows how to interpret the specs provided by the vendor in greater detail. A step-by-step procedure is provided. Finally, all the previous steps are consolidated in a single design table, which answers the question: “How will the selected inductor actually perform in a specific application?” The important point to note here is that though every induc- tor is designed assuming certain specific ‘design conditions’, that does not imply that these conditions cannot be varied. In fact every inductor can be satisfactorily used for many appli- cations. But to be able to do this, the designer must know how to be able to accurately predict, or extrapolate, the performance of the inductor to a new set of conditions, which are his specific ‘application conditions’. It will be shown that ‘intuition’ can be rather misleading. A detailed procedure is required and is presented in the form of the Design Table ( Table 2 ) and the Selection Flow Chart ( Figure 2 ). I O . Note that by definition ’r’ is a constant for a given converter/application (as it is calculated only at maxi- mum load), and it is also defined only for continuous conduction mode. A high inductance reduces I and results in lower ‘r’ (and lower RMS current in the output capacitor), but may result in a very large and impractical inductor. So typically, for most buck regulators, ‘r’ is chosen to be in the range of 0.25–0.5 (at the maximum rated load). See Appendix to this Applica- tion Note. Once the inductance is selected, as we decrease the load on the converter (keeping input voltage constant), ∆ I remains fixed but the DC level decreases and so the current ripple ratio increases. Ultimately, at the point of transition to discontinuous mode of operation, the DC level is ∆ ∆ I/2 as shown in Figure 1 .So • The current ripple ratio at the point of transition to discon- tinuous mode is 2. Therefore, the upper limit for ’r’ is also 2. • The load at which this happens can be shown by simple geometry to be r/2 times I O . So for example, if the induc- tance is chosen to be such that ‘r’ is 0.3 at a load of 2A, the transition to discontinuous mode of operation will occur at 0.15 times 2A, which is 300 mA. Note: If the inductor is a ‘swinging’ inductor, its inductance normally in- creases as load current decreases and the point of transition to discontinuous mode may be significantly lower. We do not consider such inductors in this Application Note. Background: The Inductor Current Waveform Refer to Figure 1 , which shows the current through an in- ductor in continuous mode operation (bold line). Consider its main elements: 1. I DC — is the geometrical center of the AC/ramp component — is the average value of the total inductor current waveform — is the current into the load, since the average current through the output capacitor, as for any capacitor in steady state, is zero Estimating Requirements for the Application There are two equivalent ways to go about calculating the required inductance and the designer should be aware of both. BASIC METHOD TO CALCULATE L From the general rule V = L * dl/dt we get during the ON time of the converter: 2. I PEAK is I DC + I/2, and it determines the peak energy in the core (e = 1 ⁄ 2 * L * I 2 ), which in turn is directly related to the peak field the core must withstand without saturat- ing. ∆ 3. I TROUGH is I DC - I/2 and determines the constant re- sidual level of current/energy in the inductor. Note that it depends on the load, even though it is not itself trans- ferred to the load. ∆ where V IN is the applied DC input voltage, V SW is the voltage across the switch when it is ON, D is the duty cycle and f is the switching frequency in Hz. Solving for 4. The AC component of the current is I AC = ∆ I we can write ‘r’ I=I PEAK −I TROUGH ∆ as: 5. The DC component is the load current for the case shown in the figure. I DC =I O where I O is the maximum rated load. 6. and ‘r’ is defined as the ratio of the AC to DC compo- nents (current ripple ratio) evaluated at maximum load, Now, for a buck regulator, we can show that the duty cycle is © 2002 National Semiconductor Corporation AN200212 www.national.com Estimating Requirements for the Application (Continued) Note: During a hard power-up (no soft start) or abnormal conditions like a short circuit on the output, the feedback loop is not effective in limiting the current to the value used above for calculating the energy handling capability. The current is actually going to hit the internal current limit of the device, I CLIM in Figure 1 , and this could be much higher than the steady state value calculated above. If the inductor has saturated, and if the input DC voltage is higher than 40V the current could slew up at a rate so high that the controller may not be able to limit the current at all, leading to destruction of the switch. Luckily, most off-the-shelf inductors are designed with large inherent air gaps and do not saturate very sharply even under overload conditions. However we strongly recommend that at least when the input voltage is above 40V, the inductor should be sized to handle the worst case energy e CLIM : where V D is the forward drop across the catch diode ( ≅ 0.5V for a Schottky diode). ‘r’ can be finally written as: where L is in µH and I CLIM is the internal limit of the regulator in amps. and L is therefore VOLTSECONDS METHOD TO CALCULATE L Talking in terms of voltseconds allows very general equa- tions and curves to be generated. Here we talk of voltsµsecs or ‘Et’ which is simply the voltage across the winding of the inductor times the duration in µsecs for which it is applied. Note: • where f is in Hz. EXAMPLE 1 The input DC voltage is 24V into an LM2593HV buck con- verter. The output is 12V at a maximum load of 1A. We require an output voltage ripple of 30 mV peak-to-peak ( ± 15 mV). We assume V SW = 1.5V, V D = 0.5V and f = 150,000 Hz. Since, for loop stability reasons, we should not use any output capacitor of less than 100 m Ω , and since we do not wish to use an LC post filter, our Current ramps up to the same peak value whether V (the applied voltage across inductor) is large but t (the time for which V is applied) is small, or whether V is small but t is large. So an infinite number of regulators with different combinations of input and output voltages but having the same voltseconds are actually the same regulator from the viewpoint of basic magnetics design. Et is what really counts. (The only exception to this is the Core Loss term since this depends directly on the absolute value of the frequency too, not just the Et). ∆ I must be • Also, Et can be calculated during the ON-time, (Vµsecs gained), or during the OFF-time (Vµsecs lost). Both will give the same result since there is no net change in Vµsecs per cycle in steady state. So ‘r’ is • Also, remember that though Vµsecs is related to the energy in the core, it does not tell us the total energy. The Vµsecs gives information only about the AC component of ‘r’, i.e., I. Combined with the DC component I DC ,it determines the peak current and energy of the inductor. So both I O and Et are the variables on which our design procedure and tables are based upon. But a given appli- cation is completely defined by a certain I O and Et (and frequency for the core loss term), and so these cannot be changed. Our only degree of freedom is L (or ‘r’) and we fix it according to the guidelines in the Appendix. From the general equation V = L * dl/dt we can write that V * dt =L * dl. Here V * dt is the applied voltseconds. So by definition Et=V ∆ t=L ∆ I Vµsecs where L is in µH. ‘r’ can therefore be written as ∆ The required inductance is L = 127 µH The required energy handling capability is next calculated. Every cycle, the peak current is I PEAK = 1.15A The required energy handling capability ‘e’ is Solving for L where L is in µH. So which gives us an alternate and more general way of calcu- lating L. www.national.com 2 Estimating Requirements for the Application (Continued) • The core loss equation for the core is 6.11 x 10 −18 xB 2.7 xf 2.04 mW where f is in Hz and B is in Gauss. • The inductor was designed for a frequency of 250 kHz. • Et 100 is the Vµsecs at which ‘B’ is 100 Gauss. Note: For core loss equations it is conventional to use half the peak-to-peak flux swing. So, like most vendors, the ‘B’ above actually refers to EXAMPLE 2 We repeat Example 1 from the viewpoint of Et. The ON-time is ∆ B/2. This must be kept in mind in the calculations that follow. The step-by-step calculations are: a) AC Component of Current: This can be easily calculated from Et=L I Vµsecs where L is in µH. So ∆ t ON = 3.62 µsecs So Et is Et=(V IN –V SW –V O )xt ON = (24–1.5–12) x 3.62 Vµsecs Et = 38.0 Vµsecs L is therefore b) ‘r’: So this inductor has been designed for the following ‘r’ r = 0.438 at a load current of 0.99A. c) Peak Current: L = 127 µH which gives us the same result as in Example 1 as expected. SUMMARY OF REQUIREMENTS • An inductance of 127 µH (or greater, based on maximum ‘r’ of 0.3) • DC load of 1A (to ensure acceptable temperature rise, specify I PEAK = 1.21A d) RMS Current: ∆ T) OR steady state Energy handling capability of 84 µJ • Peak load of 4.0A (to rule out core saturation if DC input voltage 40V) OR peak energy handling capability of 1016 µJ. (Max Current Limit of LM2593HV is 4.0A) • Et of 38 Vµsecs • Frequency 150 kHz These can be communicated directly to a vendor for a custom-built design. ≥ Characterizing an Off-the-Shelf Inductor With reference to our design flow chart in Figure 2 , the first pass selection is based upon inductance and DC current rating. We tentatively select a part from Pulse Engineering because its L and I DC are close to our requirements, even though the rest does not seem to fit our application (see Table 1 and bullets below). In particular the frequency for which the inductor was designed is 250 kHz, but our appli- cation is 150 kHz. We are intuitively lead to believe that since we are decreasing the frequency our core losses will go up, and so will the peak flux density. In fact the reverse happens in our case, and that is why it is important to follow the full procedure presented below. ‘Intuition’ can be very mislead- ing. The vendor also states that: • I RMS = 0.998A e) Copper Loss: This is P CU =I RMS 2 x DCR mW where DCR is in m Ω . In most cases, to a close approximation, we can simply use I DC instead of I RMS in the above equation. Also sometimes, the vendor may have directly given the RMS current rating of the inductor. P CU = 0.998 2 x 387 mW P CU = 385 mW f) The AC Component of the B-Field: This is proportional to the AC component of the inductor current. The inductor is such that 380 mW dissipation corre- sponds to 50˚C rise in temperature. 3 www.national.com Characterizing an Off-the-Shelf Inductor (Continued) The vendor has provided the information that an Et of Et 100 = 10.12 Vµsecs produces 100 Gauss (B). So since the inductor is designed for an Et = 59.4 Vµsecs, we get P CORE = 6.11 x 10 −18 x 587 2.7 x 250000 2.04 mW P CORE = 18.7 mW j) Total Inductor Loss: P=P CU +P CORE mW P = 385 + 18.7 mW P = 404 mW k) Thermal Resistance of Inductor: The vendor has stated that 380 mW dissipation corresponds to a 50˚C rise in temperature. So thermal resistance of the inductor is B = 587 Gauss But this is half the peak-to-peak swing by convention. So ∆ B=2 • B = 1174 Gauss CHECK: We can use the alternative form for B as given in Table 2 . We asked the vendor for more details than he had provided on the datasheet and we learned that the effective area of the core, A e , is 0.0602 cm 2 and the number of turns is N = 84. So ∆ R TH = 131.6˚C/W l) Estimated Temperature Rise of Inductor: which is what we expected. g) The DC Component of the B-Field: This is proportional to the DC component of the inductor current. In fact the instantaneous value of B can always be considered proportional to the instantaneous value of the current (for a given inductor). The proportionality constant is known from f) above, i.e., a ∆ T = 53˚C Here the temperature rise ‘ T’ is the temperature of the core, ‘T CORE ’ minus the worst case ambient temperature ‘T AMBIENT ’. The ‘ambient’ is the local ambient around the inductor. m) Energy Handling Capability of Core: ∆ ∆ I of 0.434A produces a B of 1174 Gauss. So the DC compo- nent of the B-field must be ∆ where I DC =I O = 0.99A where L is in µH e = 100 µJ As before, we warn that the energy in the core during hard power-up or a short circuit on the outputs, may be signifi- cantly higher. In case of soft-start it should also be remembered that there are several ways to implement this feature, and not all lead to a reduction in switch or inductor current at start-up. The worst condition is start-up with a short already present on the output. The inductor waveforms should therefore be moni- tored on the bench during all conditions to check this out. Also it will be seen that all inductors of a ‘family’, i.e., using the same core will typically have the same rated energy capability. So if this core is found to be inadequate, normally the only way out is to move to a physically larger core/inductor. Other options include the use of improved and more expensive core materials. B DC = 2678 Gauss h) Peak B-Field: Since B is proportional to I, we can write for the peak B-field B PEAK = 3265 Gauss i) Core Loss: The vendor has stated that core loss (in mW) is 6.11 x 10 −18 xB 2.7 xf 2.04 watts where f is in Hz and B is in Gauss. www.national.com 4 Characterizing an Off-the-Shelf Inductor (Continued) • DCR (m Ω ) • f (Hz) • The form for core losses (mW) as a * B b * f c , where B is in Gauss, f in Hz. Note that B is half the peak-to-peak flux swing. • Thermal resistance of inductor in free air (˚C/W) If any of these are unknown, the vendor should be con- tacted. Table 2 condenses the step-by-step procedure given earlier and also shows how to ‘extrapolate’ the performance of the inductor. SUMMARY OF INDUCTOR PARAMETERS • The inductor is designed for about 50˚C rise in tempera- ture over ambient at a load of 1A. • The copper losses (385 mW) predominate (as is usual for such inductors/core materials) and the core losses are relatively small. • The peak flux density is about 3200 Gauss, which occurs at a peak instantaneous current of 1.2A. • The rated energy handling capability of the core is 100 µJ. Note: Most vendors do not explicitly provide the material used, though an astute designer can figure this out by looking at the exponents of B and f in the core loss equation provided, or of course simply by asking the vendor. In this case we know that the material is ferrite and can typically handle a peak flux density of over 3000–4000 Gauss before it starts to saturate. (Caution: not all ferrite grades are similar in this regard and also that the saturation flux density B SAT falls as the core heats up.) EXAMPLE 3 This shows the complete selection procedure. Refer to Table 2 and Figure 2 . We have seen that the ‘Design Conditions’ of the inductor are: • Et = 59.4 Vµsecs • f = 250,000 Hz • I DC = 0.99A Our ‘Application Conditions’ are • Et' = 38 Vµsecs • f' = 150,000 Hz • I' DC =1A (We assume that T AMBIENT is unchanged so we can ignore it above). We need to verify that using the inductor in the given appli- cation: a) current ripple ratio ‘r’ is close to desired b) peak flux density/current are within bounds c) temperature rise is acceptable Using Table 2 : a) ‘r’: Design Value: Evaluating the Inductor for the Actual Application Above we have the limits of the inductor operating under its design conditions. We will now extrapolate its performance to our specific application conditions. Unprimed parameters are the original ‘design values’, and the corresponding primed parameters are the extrapolated ‘application values’. The following are the design conditions (these may be al- lowed to change): • I DC • Et • f • T AMBIENT The ‘Application Conditions’ are: • I' DC • Et' • f' • T' AMBIENT In going from the ‘Design Conditions’ to the ‘Application Conditions’ the following are considered constant • L • DCR • Rth • The core loss equation And, finally, to ‘approve’ the inductor for the given application we need to certify • r = 0.438 Extrapolated to our Application: ‘r’ is acceptable (choice of L). • B PEAK OK. I PEAK < I CLIM . • T OK (evaluate P CU +P CORE ). • B CLIM < B SAT (if DC input voltage is ≥ 40V). We assume the vendor has provided all the following inputs: • • ∆ r' = 0.277 We expected ‘r'’ to be slightly lower than 0.3 since the chosen inductor has a higher inductance than we required (137 µH instead of 127 µH). This is acceptable however as the output voltage ripple will be less than demanded. b) Peak Flux Density Design Value: Et (Vµsecs) • Et 100 (Vµsecs per 100 Gauss) • L (µH) • I DC (Amps) 5 www.national.com [ Pobierz całość w formacie PDF ] |
Podobne
|