LT1070 Design Manual (Linear Technology) - 4

Application Note 19
PREFACE
This formula assumes continuous mode operation. If ΔI, as calculated by this formula, exceeds I
Smaller Versions of the LT1070
P, it may be possible to go to discontinuous mode operation, with Since this application note was written, several new ver- further reductions in inductance. Discontinuous mode sions of the LT1070 have been developed. The LT1071 and requires higher switch currents and not all the AN19 LT1072 are identical to the LT1070 except for switch topologies show design equations for this mode, but it current ratings, 2.5A and 1.25A, respectively. Designs should definitely be considered for very low output powers which result in lower switch currents can take advantage or where inductor/transformer size is critical. All topolo- of the cost savings of these smaller chips. Design equa- gies work well in discontinuous mode with the exception tions for the LT1071 and LT1072 are identical to the of fully isolated flyback. Drawbacks of discontinuous LT1070 with the following exceptions: mode include higher output ripple and slightly lower Peak Switch Current (I efficiency. P) = 5A LT 1070 = 2.5A LT1071
Example 1:
Negative buck converter with VIN = –24V, = 1.25A LT1072 VOUT = – 5V and IOUT = 1.5A, Switch “On” Resistance (R) ≈0.2Ω LT1070 V ( −V V)( ) ≈0.4Ω LT1071 IN OUT OUT I Equation ( 37) = I I OUT + 1.5A P OUT ≈0.8Ω LT1072 • V • f • L ( ) = = 2 IN ≈ ∞ V Δ C Pin to Switch Current ≈8A/V LT1070 IMAX = 2(ISW – IP) = 2(5 – 1.5) = 7A (LT1070) Transconductance ≈4A/V LT1071 = 2(2.5 – 1.5) = 2A 9(LT1071) ≈2A/V LT1072 = 2(1.25 – 1.5) - N.A. (LT1072) Also available in the 2nd quarter of 1989 will be 100kHz The LT1072 is too small (IP > ISW), so select the LT1071, versions of the LT1070/LT1071/LT1072. which yields a maximum ΔI of 2A. A conservative value of actual ΔI is selected at 1A. This allows room for efficiency
Inductance Calculations
losses and variations in component values. Using Equa- Feedback from readers of AN19 shows that there is tion 37: confusion about the use of ΔI to calculate inductance values. ΔI is the change in inductor or primary current V ( −V V)( ) (24 )5( ) IN OUT OUT − 5 L = = = 99 H μ during switch “on” time, and the suggested value is V ΔI • f 24 1 • 4 k ( ) ( ) IN 0 approximately 20% of the peak current rating of the LT1070 switch (5A), or in some cases, 20% of the average
Example 2:
Flyback converter with VIN = 6V, VOUT = ±15V inductor current. This 20% rule-of-thumb is designed to at 35mA and 5V at 0.2A, N = 0.4 (primary to 5V secondary). give near maximum output power for a given switch For calculations, the entire output power of 2.05W is current rating. If maximum output power is not needed, referred to the 5V secondary, yielding one value for N(0.4), much smaller inductors/transformers may be used by VOUT (5V) and IOUT = 0.41A. allowing ΔI to increase. The design approach is to calcu- late peak inductor/switch current (IP) using the formulas provided in AN19, with L = ∞. Then compare this current to the peak switch current. The difference is the “room” allowable for ΔI; ΔIMAX = 2(ISWITCH(PEAK) – IP) an19fc AN19-4