Datasheet LT1107 (Analog Devices) - 8
| Manufacturer | Analog Devices |
| Description | Micropower DC/DC Converter Adjustable and Fixed 5V, 12V |
| Pages / Page | 16 / 8 — APPLICATI. S I FOR ATIO. Inductor Selection –– Positive-to-Negative … |
| File Format / Size | PDF / 217 Kb |
| Document Language | English |
APPLICATI. S I FOR ATIO. Inductor Selection –– Positive-to-Negative Converter. Step-Up (Boost Mode) Operation
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LT1107
O U U W U APPLICATI S I FOR ATIO
where DC = duty cycle (0.50 in step-down mode) In this mode the switch is arranged in common collector V or step-down mode. The switch drop can be modeled as SW = switch drop in step-down mode a 0.75V source in series with a 0.65Ω resistor. When the VD = diode drop (0.5V for a 1N5818) switch closes, current in the inductor builds according to: IOUT = output current VOUT = output voltage −R′t V V I t L 1− e IN = minimum input voltage L L ( )= 1 ( 5) R′ VSW is actually a function of switch current which is in turn a function of VIN, L, time, and VOUT. To simplify, 1.5V can where R′ = 0.65Ω + DCRL be used for VSW as a very conservative value. VL = VIN – 0.75V Once IPEAK is known, inductor value can be derived from: As an example, suppose –5V at 50mA is to be generated from a 4.5V to 5.5V input. Recalling Equation (14), V − V − V IN MIN SW OUT ( ) L = × tON ( ) 11 I P 5V 0.5V 50mA = 275mW (16 ) L = − + ( )( ) PEAK where tON = switch ON time (7µs). Energy required from the inductor is: Next, the current limit resistor RLIM is selected to give PL mW = 275 = 4 4 . J µ 1 ( 7) IPEAK from the Maximum Switch Current vs RLIM curve. f kHz OSC 63 The addition of this resistor keeps maximum switch cur- rent constant as the input voltage is increased. Picking an inductor value of 100µH with 0.2Ω DCR results in a peak switch current of: As an example, suppose 5V at 300mA is to be generated from a 12V to 24V input. Recalling Equation (10): −0 8 . 5Ω • 9 s ( µ 4. V 5 − 0 7 . V 5 ) I = 1 e 100 H µ PEAK 2 mA (300 ) 5+0 5. (0 6.5Ω +0 2.Ω) − I = mA PEAK = 600 12 ( ) 0 5 . 0 12− 1 5 . + 0 5 . = mA 325 18 ( ) Next, inductor value is calculated using Equation (11): Substituting IPEAK into Equation (04) results in: 12 − 1 5 . − 5 1 L = 7 s µ = 64 H µ 1 ( 3) E 100 H 0. A 325 2 µ 5 2 . 8 J L = ( )( ) = µ 1 ( 9) mA 600 2 Since 5.28 Use the next lowest standard value (56µH). µJ > 3.82µJ, the 100µH inductor will work. With this relatively small input range, R Then pick R LIM is not usually LIM from the curve. For IPEAK = 600mA, RLIM necessary and the I = 56Ω. LIM pin can be tied directly to VIN. As in the step-down case, peak switch current should be limited
Inductor Selection –– Positive-to-Negative Converter
to ~650mA. Figure 4 shows hookup for positive-to-negative conver-
Step-Up (Boost Mode) Operation
sion. All of the output power must come from the inductor. In this case, A step-up DC/DC converter delivers an output voltage higher than the input voltage. Step-up converters are not P V V I (14) short-circuit protected since there is a DC path from input L = OUT + ( D)( OUT ) to output. 1107fa 8
O U U W U APPLICATI S I FOR ATIO
where DC = duty cycle (0.50 in step-down mode) In this mode the switch is arranged in common collector V or step-down mode. The switch drop can be modeled as SW = switch drop in step-down mode a 0.75V source in series with a 0.65Ω resistor. When the VD = diode drop (0.5V for a 1N5818) switch closes, current in the inductor builds according to: IOUT = output current VOUT = output voltage −R′t V V I t L 1− e IN = minimum input voltage L L ( )= 1 ( 5) R′ VSW is actually a function of switch current which is in turn a function of VIN, L, time, and VOUT. To simplify, 1.5V can where R′ = 0.65Ω + DCRL be used for VSW as a very conservative value. VL = VIN – 0.75V Once IPEAK is known, inductor value can be derived from: As an example, suppose –5V at 50mA is to be generated from a 4.5V to 5.5V input. Recalling Equation (14), V − V − V IN MIN SW OUT ( ) L = × tON ( ) 11 I P 5V 0.5V 50mA = 275mW (16 ) L = − + ( )( ) PEAK where tON = switch ON time (7µs). Energy required from the inductor is: Next, the current limit resistor RLIM is selected to give PL mW = 275 = 4 4 . J µ 1 ( 7) IPEAK from the Maximum Switch Current vs RLIM curve. f kHz OSC 63 The addition of this resistor keeps maximum switch cur- rent constant as the input voltage is increased. Picking an inductor value of 100µH with 0.2Ω DCR results in a peak switch current of: As an example, suppose 5V at 300mA is to be generated from a 12V to 24V input. Recalling Equation (10): −0 8 . 5Ω • 9 s ( µ 4. V 5 − 0 7 . V 5 ) I = 1 e 100 H µ PEAK 2 mA (300 ) 5+0 5. (0 6.5Ω +0 2.Ω) − I = mA PEAK = 600 12 ( ) 0 5 . 0 12− 1 5 . + 0 5 . = mA 325 18 ( ) Next, inductor value is calculated using Equation (11): Substituting IPEAK into Equation (04) results in: 12 − 1 5 . − 5 1 L = 7 s µ = 64 H µ 1 ( 3) E 100 H 0. A 325 2 µ 5 2 . 8 J L = ( )( ) = µ 1 ( 9) mA 600 2 Since 5.28 Use the next lowest standard value (56µH). µJ > 3.82µJ, the 100µH inductor will work. With this relatively small input range, R Then pick R LIM is not usually LIM from the curve. For IPEAK = 600mA, RLIM necessary and the I = 56Ω. LIM pin can be tied directly to VIN. As in the step-down case, peak switch current should be limited
Inductor Selection –– Positive-to-Negative Converter
to ~650mA. Figure 4 shows hookup for positive-to-negative conver-
Step-Up (Boost Mode) Operation
sion. All of the output power must come from the inductor. In this case, A step-up DC/DC converter delivers an output voltage higher than the input voltage. Step-up converters are not P V V I (14) short-circuit protected since there is a DC path from input L = OUT + ( D)( OUT ) to output. 1107fa 8
