Figure 1 below shows the circuit we'll be using to explore the step response of an RLC circuit. The voltage source produces a 1V step at t=0. Initially the resistor has been set to 0Ω. We'll be probing the voltage across the capacitor, which will indicate the amount of charge on the plates of the capacitor (q=Cv), and the current through the inductor, which will indicate the flux linkage of the magnetic field of the inductor (λ=Li).
TRANTransient Analysis
Figure 1. RLC Circuit driven by a voltage step
Run a 50μs TRAN simulation on the circuit. You'll see that the energy imparted by the voltage step oscillates back and forth between the flux of the inductor and the charge of the capacitor (as indicated by the oscillating voltage across the capacitor and the oscillating current through the inductor).
Leaving the inductance L as-is, adjust the capacitance C so that the frequency of oscillation is 100 kHz with a period of 10μs. You can do this by experimentation, but a more effective approach would be to use Equation 12.45 to compute the correct value for C given L and the desired ω0. Note that the units of ω0 are in radians/sec, so you'll have to convert Hz to radians/sec using 1Hz=2π rad/s. Your plots should like those shown to the right.
Please enter the adjusted capacitance C below:
Adjusted capacitance C (in farads):
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Keeping the adjusted C, now adjust the resistance R so that the system is just operating in the over-damped region, i.e., so that the voltage across the capacitor makes a single 0→1 transition, never exceeding 1V. Again you can use the analysis in Section 12.2.2 to compute the R necessary to produce a plot like that to the right. Hint: if there's any voltage sample that is greater than 1V, the voltage scale will have a maximum of 2V, so one quick way to tell if the voltage is staying less than or equal to 1V is when the maximum value on the voltage scale is 1V.
Please enter the adjusted resistance R below:
Adjusted resistance R (in ohms):
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Now let's use the properties of second-order systems to build a boost converter, a DC-to-DC power supply useful where high voltages are required but not directly available. Powering the flash bulb in a camera is one such example. A boost converter circuit is shown in Figure 2. In this case the supply voltage is 3V and the goal is produce a relatively stable supply of 6V to drive a load, here represented as a 1kΩ resistor. Your task to is adjust L, C and the duty cycle D of the square wave controlling the MOSFET switch so that the output voltage falls between 5.9V and 6.1V (6V with a maximum of 0.1V ripple).
DCDC AnalysisTRANTransient Analysis
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Figure 2. Boost converter
The operation of the boost converter is analyzed in detail in Section 12.10 of the text. The circuit above differs from the example in how the switches are implemented. S1 in the Example is implemented by a MOSFET switch, where a square-wave voltage is used to turn the switch on and off. S2 in the Example is implemented using a diode, which conducts when the voltage on the inductor side sufficiently exceeds the voltage on the capacitor side. In the steady state this happens when S1 is off, just what we wanted for the correct operation of the circuit.
The boost converter operates in a two-state cyclic manner. During the first state, the MOSFET is turned on. This connects the inductor across the power supply, so the inductor current iL increases. During this time, the diode is reverse biased and so behaves as an open circuit, disconnecting the capacitor and load resistor from the remainder of the circuit. In this state, the capacitor powers the load resistor, discharging in the process.
During the second state, the MOSFET is turned off but the inductor current cannot go to zero instaneously. So the voltage on the diode increases making the diode forward biased and the decreasing current from the inductor flows through the diode to power the load and simultaneously replenish the charge in the capacitor. The inductor flux diminishes in the process.
This cycle repeats itself indefinitely. Let the duration of the first state be DT, and the duration of the second state be (1−D)T , where D is the fraction of a cycle for which the square wave controlling the MOSFET is high (sometimes called the duty cycle) and T is the switching period (set to 1/30kHz in the circuit above).
Try a 2ms TRAN simulation to see how the circuit works, paying attention to the current through the inductor and the output voltage. Start by adjusting the capacitance C so that the ripple on the output voltage is 0.1V or less using the equation iΔt=CΔv to help choose the appropriate value. Here i is the desired load current 6V/1kΩ, Δt is the discharge time in one cycle DT and Δv is the desired ripple 0.1V.
Now choose values for the inductance L and duty cycle D to produce the desired output voltage. As explained in the text, the average output voltage is roughly VIN/(1−D), so increasing the duty cycle will raise the output voltage. One wants to choose D and the cycle time T so that the inductor current reaches zero just as the next cycle begins.
Experiment with the various parameters until the output voltage is 6V with a maximum 0.1V ripple. As a final check, perform a 10ms TRAN simulation and then click CHECK. The on-line system will be veryifying that the output voltage meets the specification for 9.5ms<t<10ms.
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To adjust the capacitance C in the RLC circuit, we can use Equation 12.45, which relates the frequency of oscillation ω0 to the inductance L and capacitance C:
ω0 = 1 / sqrt(LC)
Given that the desired frequency of oscillation is 100 kHz, we can convert this to radians/second by using the conversion factor 1Hz = 2π rad/s. Thus, ω0 = 100,000 Hz * 2π rad/s = 200,000π rad/s.
Rearranging Equation 12.45, we can solve for C:
C = 1 / (L * ω0^2)
Plugging in the values of L and ω0, we can calculate the adjusted capacitance C.
For the adjusted resistance R, we want the system to operate in the over-damped region, where the voltage across the capacitor makes a single 0→1 transition, never exceeding 1V. We can use the analysis in Section 12.2.2 to determine the necessary value of R to achieve this.
To build a boost converter circuit that produces a stable 6V output with a maximum ripple of 0.1V, we need to adjust the values of the inductance L, capacitance C, and the duty cycle D of the square wave controlling the MOSFET switch. The ripple on the output voltage can be determined using the equation iΔt = CΔv, where i is the load current, Δt is the discharge time in one cycle, and Δv is the desired ripple.
The average output voltage of the boost converter is roughly equal to VIN/(1-D), where VIN is the supply voltage. By adjusting the duty cycle D and the switching period T, we can choose values that ensure the inductor current reaches zero just as the next cycle begins.
It is advised to perform simulations and experimentation to fine-tune the values of C, R, L, and D to meet the desired specifications of the circuits.