Design Equations for Impulse Current Generators

Definitions

• $V_p$: Peak voltage. $I_p$: Peak current.

• $T_{fv}$: Front-time of the open-circuit voltage waveform, defined as its 30%-90% rise time multiplied by 1.67. For a waveform with a front-time of 1.2 μs, its 30%-90% rise time is 0.718 μs.

• $T_{dv}$: Duration of the open-circuit voltage waveform, defined as duration between the 50% point of rising edge and the 50% of the falling edge.

• $T_{fi}$: Front-time of the short-circuit current waveform, defined as its 10%-90% rise time multiplied by 1.25. For a waveform with a front-time of 8 μs, its 10%-90% rise time is 6.4 μs.

• $T_{di}$: Duration of the short-circuit current waveform, defined as duration between the 50% point of rising edge and the 50% point of the falling edge.

Step 1

Given:

$$y(x)=\exp (-x)-\exp (-\frac{\beta }{\alpha }x)$$

Numerically find the constant $k=\frac{\beta }{\alpha }$ to make $y(x)$’s waveform satisfy the ratio of $T_{fv}/T_{dv}$. For example, find a $k$ to make $y(x)$’s $T_{fv}/T_{dv}=50/1.2$.

Step 2

Given:

$$v(t)=\exp (-\alpha t)-\exp (-\beta t)$$

Numerically find $\alpha$ and $\beta$ under the constraint of $k=\frac{\beta }{\alpha }$, to make $v(t)$’s waveform satisfy the requirement of $T_{dv}$ (then $T_{fi}$ is also automatically satisfied). For example, find $\alpha$ and $\beta$ to make the time difference between the 50% point of falling edge and the 50% of the rising edge to be $50\cdot {10}^{-6}\text{ s}$.

Step 3

Given:

$$y(x)=\exp (-\frac{z}{2Q})\cdot \sin (\sqrt{1-{(\frac{1}{2Q})}^2}\cdot z)$$

Numerically find the constant $Q$ to make $y(x)$’s waveform satisfy the ratio of $T_{fi}/T_{di}$. For example, find a $Q$ to make $T_{fi}/T_{di}=20/8$.

Step 4

Given:

$${\omega }_n={\omega }_0\cdot \sqrt{1-{(\frac{1}{2Q})}^2}$$

And:

$$i(t)=\exp (-\frac{{\omega }_0}{2Q}\cdot t)\cdot \sin ({\omega }_{n}t)$$

Numerically find the constant ${\omega }_0$ under the constraint of $Q$ to make $i(t)$’s waveform satisfy the requirements of $T_{di}$ (then $T_{fi}$ is also automatically satisfied). For example, find ${\omega }_0$ to make the time difference between the 50% point of falling edge and the 50% of the rising edge to be $20\cdot {10}^{-6}\text{ s}$.