||||| Boost Converter Design The Design of Boost DC to DC Converters Before reading this section, please read the. All of the circuits in this tutorial can be simulated in LTspice ®. If you are new to LTspice, please have a look at my Consider the architecture of a boost converter, shown in FIG 1. FIG 1 Ignore components C1, C2, R3. The MOSFET, Q1, switches on creating a short circuit between the right hand side of the inductor, L1, and 0V. Thus a fixed voltage of 3.3V is applied across the inductor, so its current will ramp up according to or or 1.5 million amps per second. Thus if the MOSFET switches off after 1us, the current through the inductor will have ramped up by 1.5A.
When the MOSFET switches off, the inductor tries to maintain its current flow. It does this by generating a voltage across its terminals very similar to a battery, where the current flows from the negative terminal, through the battery, to the positive terminal. New super mario bros 3 rom hack - download free apps on my iphone. In the circuit of FIG 1, we can see that to maintain current flow, the right hand side of the inductor has to increase in voltage with respect to the left hand side. The left hand side is connected to the input voltage (so cannot change), thus the right hand side voltage increases above the input voltage and continues to do so until something conducts. Theoretically, this voltage will rise to an infinite value, making the inductor very good at generating high voltages from low voltages. In FIG 1, the inductor voltage increases until diode D1 conducts after which the energy in the inductor flows into the output capacitor C3, causing the voltage across C3 to increase slightly. It is worth noting that even before the MOSFET has started to switch, there is a dc path flowing from the input, through L1 and diode D1 into C3, so at startup C3 will have a voltage across it (equal to Vin – Vdiode).
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When the MOSFET switches off and the inductor discharges, the inductor still behaves according to except this time, the voltage across the inductor is equal to Vout – Vin (ignoring the diode drop). When the inductor has discharged, the MOSFET switches on and the process starts again. Repeating this process produces pulses of energy from the inductor into the output capacitor making the voltage across the output capacitor rise. In FIG1, resistors R1 and R2 monitor the output voltage and when the voltage at the FB pin reaches a certain point, the chip terminates the drive to the MOSFET until the voltage on the output capacitor droops. The LTspice model of this circuit can be downloaded here (right click over the link and save as a '.asc' file). The LTspice simulation results are shown in FIG 2.
Here we are looking at the part operating once the output voltage has ramped up to 5V. FIG 2 The blue waveform is the Gate voltage to the MOSFET. When the FET turns on, the inductor current (in red) ramps up from 1.09A to 2.18A in 739ns (this can be measured in LTspice), or at a rate of 1.474 million Amps per second, close to what we calculated above. The discrepancy is due to the fact that the FET does not provide a true short circuit to ground and actually has a voltage across it of approximately 50mV when switched on, thus reducing the voltage across the inductor. Likewise when the FET switches off, the current ramps from 2.18A to 1.09A in 1.083us. From the equation once the output has reached regulation the voltage across the inductor is [(5+Vd) - 3.3], where Vd is the voltage across the diode (approx. 0.5V), so we can calculate the current ramp to be or 1 million amps per second.
Over a period of 1.083us, the current ramps down by 1.083A, so again our LTspice simulation is very close to the calculated value. It is interesting to note that the value of di/dt is determined ONLY by the inductance value and the voltage across the inductor.
The controller IC has nothing to do with setting the inductor ramp current. It is also useful to calculate the duty cycle (the ratio of the ON time of the FET to the total period of oscillation). Again from the equation we can see that during ramp up, the inductor current di is represented by where dt 1 is the ON time of the FET and Vin is the input voltage.
During ramp down, the inductor current is represented by where Vout is the output voltage and dt 2 is the OFF time of the FET. To make life easier, we have neglected the diode voltage drop. For a fixed input voltage and a fixed output voltage, di is the same when ramping up as ramping down. Thus, equating di gives From this we can calculate Now, our duty cycle, DC is represented by Hence So So from there we can work out that Again, the duty cycle is set by the input and output voltages only. The inductor value does not feature in setting the duty cycle, nor does the controller IC. The above is true as long as the current in the inductor does not fall to zero.