Datasheet LT1107 (Analog Devices) - 7

LT1107
O U U W U APPLICATI S I FOR ATIO Inductor Selection –– Step-Up Converter
As an example, suppose 12V at 60mA is to be generated from a 3V to 6V input. Recalling equation (1), In a step-up, or boost converter (Figure 1), power gener- ated by the inductor makes up the difference between P V 12 0. V 5 V 3 6 m 0 A = m 570 W ( )( ) L = + − 6 ( ) input and output. Power required from the inductor is determined by: Energy required from the inductor is: P = V + V − V  I ( ) 1 P mW L OUT D L IN MIN OUT  ( ) ( ) = 570 = 9 0 . 5 J µ 7 ( ) f kHz OSC 63 where VD is the diode drop (0.5V for a 1N5818 Schottky). Energy required by the inductor per cycle must be equal or Picking an inductor value of 33µH with 0.2Ω DCR results greater than: in a peak switch current of: P  −1Ω•11µ  L / f OSC ( ) 2 s V 3 I 1 e 33 H = − µ  m 850 A 8 ( ) in order for the converter to regulate the output. PEAK 1Ω   =   When the switch is closed, current in the inductor builds according to: Substituting IPEAK into Equation 4 results in:  − 1 R′t  2 µ . µ V E 33 H 0 8 A 5 11 91 J L = ( )( ) = 9 ( ) I t IN ( ) =  − 1 e  2 L L ( ) 3 R′   Since 11.9µJ > 9.05µJ, the 33µH inductor will work. This where R′ is the sum of the switch equivalent resistance trial-and-error approach can be used to select the opti- (0.8Ω typical at 25°C) and the inductor DC resistance. mum inductor. When the drop across the switch is small compared to VIN, A resistor can be added in series with the ILIM pin to invoke the simple lossless equation: switch current limit. The resistor should be picked so the calculated I V PEAK at minimum VIN is equal to the Maximum I t IN = t ( ) L ( ) 4 Switch Current (from Typical Performance Characteristic L curves). Then, as VIN increases, peak switch current is can be used. These equations assume that at t = 0, held constant, resulting in increasing efficiency. inductor current is zero. This situation is called “discon- tinuous mode operation” in switching regulator parlance.
Inductor Selection –– Step-Down Converter
Setting “t” to the switch ON time from the LT1107 speci- The step-down case (Figure 2) differs from the step-up in fication table (typically 11µs) will yield IPEAK for a specific that the inductor current flows through the load during “L” and VIN. Once IPEAK is known, energy in the inductor both the charge and discharge periods of the inductor. at the end of the switch ON time can be calculated as: Current through the switch should be limited to ~650mA 2 in this mode. Higher current can be obtained by using an E = 1 LI L 5 ( ) PEAK 2 external switch (see LT1111 and LT1110 data sheets). The E I L must be greater than PL/fOSC for the converter to deliver LIM pin is the key to successful operation over varying the required power. For best efficiency I inputs. PEAK should be kept to 1A or less. Higher switch currents will cause After establishing output voltage, output current and input excessive drop across the switch resulting in reduced voltage range, peak switch current can be calculated by the efficiency. In general, switch current should be held to as formula: low a value as possible in order to keep switch, diode and 2I  V + V  inductor losses at a minimum. I OUT OUT D PEAK =   10 ( ) DC V − V + V  IN SW D  1107fa 7