2.2 Model circuit analysis using Laplace transform
In Fig. 3, the voltage can be assumed since $i_{L}(0)=0$, $i_{R}(0)=0$ and $i_{C}(0)=0$. Therefore,
it has an the initial condition of $v_{2}(0)=0$, $v_{2}'(0)=0$.
Equation(1) is established by using the Laplace transform of Fig. 3 considering the initial valuesof $v_{2}$.
Equation(1) is transformed into
equation(2).
Fig. 1. Gas discharge lamp circuit
Fig. 2. Voltage source wave form
Fig. 3. Voltage and current of model circuit
The voltage source waveform in
Fig. 2 is expressed as
equation(3).
(4)
The Laplace transform of one period of the voltage source waveform is obtained by
equation(4).
Using
equation(4), the Laplace transform of the voltage source waveform is represented by
equation(5).
By substituting
equation(5) into
equation(2), it is summarized as equation.(6)
In order for $v_{2}$ to oscillate,
equation(7) must have an imaginary solution, so the condition of
equation(8) must be met.
(5)
The solution of
equation(7) is represented by equation.(9)
In the case of $s=\alpha +- j\beta$ ($\alpha$ and $\beta$ are real numbers, $\beta\ne
0$), $\alpha$ is calculated by
equation(10), and $\beta$ is calculated by
equation(11).
Using $\alpha$ and $\beta$,
equation(11) becomes
equation(12).
Equation(13) is used to analyze the damped oscillating waveform.
The inverse Laplace transform of
equation(13) becomes equation.(14)
Where, $S_{a}=\dfrac{\alpha}{\alpha^{2}+\beta^{2}}$, $T_{a}=-\dfrac{\beta}{\alpha^{2}+\beta^{2}}$,
$A=s H_{1}(s)|_{s=0}=LC$.
Equation(14) is transformed into equation(15).
Substituting
equation(15) into
equation(12) to obtain equation.(16)
In
equation(16), the Laplace transform of the first half-period waveform is represented by equation.(17)
Equation(18) can be obtained by inverse Laplace transform of equation.(17)
In
equation(18), $\alpha$ is the damping coefficient and $\beta$ is the angular frequency of the
oscillation wave.
2.3 Calculationofimpedanceof discharge lamp
The impedance of the discharge lamp is calculated in the following order:
① voltage waveform measurement
② area calculation
③ calculate DC average value
④ calculate the AC value (waveform value minus DC average value)
⑤ absolute value calculation of AC value
⑥ trend line calculation
⑦ damping rate calculation
⑧ average period calculation of oscillation wave
⑨ calculation of effective resistance and effective capacitance
In this paper, a voltage v is 60kHz, 310V square wave, and inductance L is 20μH is
applied for the circuit presented in Fig. 1.
Fig. 4 is a measurement of the current and voltage wave forms of the discharge lamp in the
Fig. 1 circuit.
Fig. 4. Discharge lamp current and voltage wave forms
In
Fig. 4, the time and voltage of the peaks and valleys are measured on the half cycle of
the voltage waveform. The area of the graph is obtained by using the trapezoidal
formula, and the area is divided by time to obtain the average value of voltage.
Fig. 5. Measured crest, valley and average value of voltage wave form
Fig. 6. Measured value minus average value of voltage wave form
Fig. 7. Trend line of selected data
Fig. 8. LTspice simulation circuit
Fig. 9. Simulated voltage and current wave forms
Fig. 5 shows the peaks and valleys of the measured voltage waveform, and the average voltage
value, 310V, of the measured voltage waveform. The calculated average value of the
voltage should be the same as the voltage of the applied square wave as shown in equation.(18)
Fig. 6 shows the waveform with the average value (DC component) removed from the measured
voltage waveform.
The waveform in Fig. 6 shows an exponential multiply sine wave function form ($v=A e^{\alpha t}\sin\omega
t$). In order to obtain the attenuation coefficient α, the absolute value of the waveform
in Fig. 6 is taken, and the trend line is calculated by using the first three values. Fig. 7 shows this process.(6)
It is $\alpha =-5.365\times 10^{5}$ calculated from the trend line.
In Fig. 6 the average period is calculated by using the time from the first peak to the second
valley. The calculated average period is 1.37μs. From the above relation, it is calculated
as $\beta =4.59\times 10^{6}$ by equation.(19)
From
equation(10) and
equation(11), the circuit capacitance C is calculated as
equation(20), and the circuit resistance R is calculated as
equation(21).
In the circuit used in this paper, the circuit inductance L is 20μH. Therefore the
circuit capacitance C is calculated as 2.35nF and the circuit resistance R is calculated
as 397Ω.
In order to verify the calculated resistance and capacitance values of the discharge
lamp, Fig. 1(b) circuit was simulated by using LTspice.
Fig. 9 shows the simulation result of Figure 8 circuit. It can be seen that the simulation
waveform has a similar trend to the actual waveform in Fig. 4.
Fig. 10. Measured and simulated voltage
When the measured voltage waveform and the simulated voltage waveform are compared
quantitatively, a difference between the two can be seen.