Datasheet LT1316 (Analog Devices) - 7
| Manufacturer | Analog Devices |
| Description | Micropower DC/DC Converter with Programmable Peak Current Limit |
| Pages / Page | 16 / 7 — APPLICATIONS INFORMATION. Example:. Calculating Duty Cycle. Figure 4. … |
| File Format / Size | PDF / 313 Kb |
| Document Language | English |
APPLICATIONS INFORMATION. Example:. Calculating Duty Cycle. Figure 4. Discontinuous Mode Operation
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LT1316
U U W U APPLICATIONS INFORMATION
then adding in the amount of overshoot that will occur due V L = OUT – VIN + VD (t to turn-off delay of the power transistor. This turn-off OFF) 0.4(IPEAK) (2) delay is approximately 300ns. where t Peak switch current = DC current limit from graph + OFF = 2µs and VD = 0.4V. VIN/L(turn-off delay) As a result of equations 1 and 2, ripple current during switching will be 40% of the peak current (see Figure 2).
Example:
Using these equations at the specified IOUT, the part is Set peak switch current to 100mA for: V delivering approximately 60% of its maximum output IN = 2V, L = 33µH power. In other words, the part is operating on a 40% reserve. This is a safe margin to use and can be decreased Overshoot = VIN/L(turn-off delay) = (2/33µH)(300ns) if input voltage and output current are tightly controlled. = 18.2mA Refer to R For some applications, this recommended inductor size SET graph and locate (100mA – 18.2mA) ≈ 82mA may be too large. Inductance can be reduced but available output power will decrease. Also, ripple current during RSET ≈ 33k switching will increase and may cause discontinuous operation. Discontinuous operation occurs when
Calculating Duty Cycle
inductor current ramps down to zero at the end of each For a boost converter running in continuous conduction switch cycle (see Figure 4). Shown in Figure 5 is minimum mode, duty cycle is constrained by VIN and VOUT according inductance vs peak current for the part to remain in to the equation: continuous mode. V DC = OUT – VIN + VD 0mA INDUCTOR VOUT – VSAT + VD CURRENT 100mA/DIV where VD = diode voltage drop ≈ 0.4V and VSAT = switch saturation voltage ≈ 0.2V. SW PIN If the duty cycle exceeds the LT1316’s minimum specified 5V/DIV duty cycle of 0.73, the converter cannot operate in con- 2µs/DIV 1316 F04 tinuous conduction mode and must be designed for discontinuous mode operation.
Figure 4. Discontinuous Mode Operation
1000
Inductor Selection and Peak Current Limit for Continuous Conduction Mode
µH) 5V TO 18V 5V TO 12V Peak current and inductance determine available output power. Both must be chosen properly. If peak current or 2V TO 5V inductance is increased, output power increases. Once 100 output power or current and duty cycle are known, peak current can be set by the following equation, assuming continuous mode operation: MINIMUM INDUCTANCE FOR CONTINOUS MODE OPERATION ( 2(I 10 10 100 1000 I OUT) PEAK = (1) 1 – DC PEAK CURRENT (mA) 1316 F05
Figure 5. Minimum Inductance vs Peak Current
Inductance can now be calculated using the peak current:
for Continuous Mode Operation
7
U U W U APPLICATIONS INFORMATION
then adding in the amount of overshoot that will occur due V L = OUT – VIN + VD (t to turn-off delay of the power transistor. This turn-off OFF) 0.4(IPEAK) (2) delay is approximately 300ns. where t Peak switch current = DC current limit from graph + OFF = 2µs and VD = 0.4V. VIN/L(turn-off delay) As a result of equations 1 and 2, ripple current during switching will be 40% of the peak current (see Figure 2).
Example:
Using these equations at the specified IOUT, the part is Set peak switch current to 100mA for: V delivering approximately 60% of its maximum output IN = 2V, L = 33µH power. In other words, the part is operating on a 40% reserve. This is a safe margin to use and can be decreased Overshoot = VIN/L(turn-off delay) = (2/33µH)(300ns) if input voltage and output current are tightly controlled. = 18.2mA Refer to R For some applications, this recommended inductor size SET graph and locate (100mA – 18.2mA) ≈ 82mA may be too large. Inductance can be reduced but available output power will decrease. Also, ripple current during RSET ≈ 33k switching will increase and may cause discontinuous operation. Discontinuous operation occurs when
Calculating Duty Cycle
inductor current ramps down to zero at the end of each For a boost converter running in continuous conduction switch cycle (see Figure 4). Shown in Figure 5 is minimum mode, duty cycle is constrained by VIN and VOUT according inductance vs peak current for the part to remain in to the equation: continuous mode. V DC = OUT – VIN + VD 0mA INDUCTOR VOUT – VSAT + VD CURRENT 100mA/DIV where VD = diode voltage drop ≈ 0.4V and VSAT = switch saturation voltage ≈ 0.2V. SW PIN If the duty cycle exceeds the LT1316’s minimum specified 5V/DIV duty cycle of 0.73, the converter cannot operate in con- 2µs/DIV 1316 F04 tinuous conduction mode and must be designed for discontinuous mode operation.
Figure 4. Discontinuous Mode Operation
1000
Inductor Selection and Peak Current Limit for Continuous Conduction Mode
µH) 5V TO 18V 5V TO 12V Peak current and inductance determine available output power. Both must be chosen properly. If peak current or 2V TO 5V inductance is increased, output power increases. Once 100 output power or current and duty cycle are known, peak current can be set by the following equation, assuming continuous mode operation: MINIMUM INDUCTANCE FOR CONTINOUS MODE OPERATION ( 2(I 10 10 100 1000 I OUT) PEAK = (1) 1 – DC PEAK CURRENT (mA) 1316 F05
Figure 5. Minimum Inductance vs Peak Current
Inductance can now be calculated using the peak current:
for Continuous Mode Operation
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