IEC 61000-4-5

It's worth noting that both "1.2/50 μs" voltage and "8/20 μs" current impulses are classic waveforms with a well-established history of use in high-voltage testing for electric power transmission.[2] Thus, these waveforms are also defined by IEC 60060-1 "High-Voltage Test Techniques" and other standards in this context. In fact, the waveform definitions in IEC 61000-4-5 were originally based on IEC 60060-1.[3][4]

Nevertheless, there are important differences. In traditional high-voltage testing, voltage and current impulses are tested separately, not in combination. The "1.2/50 μs" generator is designed for insulation testing, and produces a high-voltage, low-current impulse into a high-impedance load. The output current of this generator is on the milliampere scale.[2][5] The "8/20 μs" generator is designed for surge arrester testing, and produces a high-current surge into a low-impedance load.[2] On the other hand, modern electronic devices can be high and low-impedance loads simultaneously due to non-linear devices, protection circuits, and arcing in a dielectric breakdown. As a result, it motivated the creation of the Combination Wave Generator with the ability to generate a high-voltage, high-current output during the same surge.[2] In addition, both standards have different waveform tolerances[6] and other technical requirements. Thus, IEC 61000-4-5 is not to be confused with IEC 60060-1 and other high-voltage tests that also use a "1.2/50 μs" or "8/20 μs" impulse.[5]

When a Coupling Network is used, past experience has shown inconsistent waveforms between different generators. Thus, an important change in IEC 61000-4-5 Ed. 3 is that a Combination Wave Generator must be verified only with a 18 μF capacitor attached at the output. This causes a significant impact to the short-circuit current waveform. If the generator is to be designed without the coupling capacitor in mind, the output would no longer be standard compliant.[7]

Third Edition also simplified waveform definitions. The earlier standard contained two definitions of "1.2/50-8/20 μs" waveform parameters, based on either IEC 60060-1 or IEC 60469-1, and two definitions of "10/700-5/320 μs" waveform parameters, based on either IEC 60060-1 or ITU-T K series. Ed. 3 removed references to these standards and gives standalone definitions.[3][4] Especially, front time has been redefined in terms of rise time, rather than a time interval from an extrapolated "virtual origin" using IEC 60060-1's approach. This allows one to use the built-in measurement feature on an oscilloscope, simplifying test procedures. For practical purposes, the differences between both definitions are negligible.[7] However, because the new definition was created using IEC 60060-1 as its basis, a generator calibrated according to IEC 60469-1's definitions may no longer be standard compliant.[4]

Simplified Schematic of the 1.2/50-8/20 μs Combination Wave Generator

The Combination Wave Generator is essentially a capacitor discharge circuit. Initially, the switch is open, a high voltage source charges the energy-storage capacitor ${\displaystyle C_{c}}$ through a current-limiting resistor ${\displaystyle R_{c}}$. The switch is then closed to deliver an impulse to the load through a pulse-forming network, which consists of a rise time shaping inductor ${\displaystyle L_{r}}$, two impulse duration shaping resistors ${\displaystyle R_{s1}}$ and ${\displaystyle R_{s2}}$, and an impedance matching resistor ${\displaystyle R_{m}}$.

The standard does not specify component values or practical circuits, any suitable design that conforms to the standard requirements can be used.

A complete circuit analysis of the ideal surge generator, including design equations and component values, is available in the presentation Introduction To Voltage Surge Immunity Testing by Hesterman et, al.[8] An updated derivation for the Third Edition is given in the paper Elementary and ideal equivalent circuit model of the 1,2/50-8/20 μs combination wave generator by Carobbi et, al.[7]

The following design equations are derived by Carobbi et, al. In these equations, the charging voltage is ${\displaystyle E}$, and the components are ${\displaystyle C=C_{c}}$, ${\displaystyle R_{1}=R_{s1}}$, ${\displaystyle R_{2}=R_{m}}$, ${\displaystyle R_{3}=R_{s2}}$, and ${\displaystyle L=L_{r}}$.[7]

For open-circuit voltage, its Laplace transform is:

${\displaystyle V(s)={\frac {E\cdot {\frac {R_{3}}{L}}}{s^{2}+s(\alpha +\beta )+\alpha \beta }}}$

(1)

Where:

${\displaystyle \alpha +\beta ={\frac {R_{2}+R_{3}}{L}}+{\frac {1}{R_{1}C}}}$

(2)

${\displaystyle \alpha \beta ={\frac {R_{1}+R_{2}+R_{3}}{R_{1}LC}}}$

(3)

Thus, open-circuit voltage is a double exponential waveform:

${\displaystyle v(t)={\frac {E\cdot {\frac {R_{3}}{L}}}{\beta -\alpha }}(e^{-\alpha t}-e^{-\beta t})}$

(4)

The voltage reaches its peak value at:

${\displaystyle t_{pe}={\frac {\ln {(\beta /\alpha )}}{\beta -\alpha }}}$

(5)

And the peak voltage is:

${\displaystyle V_{p}={\frac {E\cdot {\frac {R_{3}}{L}}}{\beta -\alpha }}(e^{-\alpha t_{pe}}-e^{-\beta t_{pe}})}$

(6)

When the output is shorted, note that the last resistor ${\displaystyle R_{3}}$ (${\displaystyle R_{s2}}$ in the schematic) is effectively removed.

For short-circuit current, its Laplace transform is:

${\displaystyle I(s)={\frac {\frac {E}{L}}{\left(s+{\frac {\omega _{0}}{2Q}}\right)^{2}+\omega _{n}^{2}}}}$

(7)

Where:

${\displaystyle \omega _{0}^{2}={\frac {R_{1}+R_{2}}{LCR_{1}}}}$

(8)

${\displaystyle {\frac {w_{0}}{Q}}={\frac {R_{2}}{L}}+{\frac {1}{R_{1}C}}}$

(9)

${\displaystyle \omega _{n}=\omega _{0}{\sqrt {1-(1/2Q)^{2}}}}$

(10)

Thus, short-circuit current is a damped sine wave (from an underdamped RLC circuit):

${\displaystyle i(t)={\frac {E}{\omega _{n}L}}e^{-{\frac {\omega _{0}}{2Q}}t}\sin {(\omega _{n}t)}}$

(11)

The current reaches its peak value at:

${\displaystyle t_{ps}={\frac {1}{\omega _{n}}}\arctan {\left({\sqrt {(2Q)^{2}-1}}\right)}}$

(12)

And the peak current is:

${\displaystyle I_{p}={\frac {E}{\omega _{0}L}}e^{-{\frac {\omega _{0}}{2Q}}t_{ps}}}$

(13)

Ignore the amplitude in 4, it becomes:

${\displaystyle v'(t)=e^{-\alpha t}-e^{-\beta t}}$

(14)

By substituting ${\displaystyle x=\alpha t}$:

${\displaystyle y(x)=e^{-x}-e^{-{\frac {\beta }{\alpha }}x}}$

(15)

The ratio ${\displaystyle {\frac {\beta }{\alpha }}}$ should be selected to make ${\displaystyle y(x)}$'s waveform have a duration over front-time ratio of ${\displaystyle 50/1.2\approx 41.7}$. By numerically evaluating ${\displaystyle y(x)}$'s waveform (including its front time and duration) while varying this ratio, the solution is found to be ${\displaystyle {\frac {\beta }{\alpha }}=168}$. Next, ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are computed by numerically varying ${\displaystyle \alpha }$ until 14's waveform has a front time of 1.2 μs. The solution is ${\displaystyle \alpha ^{-1}}$ = 68.2 μs. Therefore, ${\displaystyle \beta ^{-1}}$ = 0.4 μs.

Ignore the amplitude in 11, it becomes:

${\displaystyle i'(t)=e^{-{\frac {\omega _{0}}{2Q}}t}\sin {(\omega _{n}t)}}$

(16)

By substituting ${\displaystyle z=\omega _{0}t}$:

${\displaystyle y'(z)=e^{\frac {-z}{2Q}}\sin {\left({\sqrt {1-(1/2Q)^{2}}}z\right)}}$

(17)

The value ${\displaystyle Q}$ should be selected to make ${\displaystyle y'(z)}$'s waveform have a duration over front time ratio of ${\displaystyle 20/8=2.5}$. By numerically evaluating ${\displaystyle y'(z)}$'s waveform (including its front time and duration) while varying ${\displaystyle Q}$, the solution is found to be ${\displaystyle Q=1.46}$. Next, ${\displaystyle \omega _{0}}$ is computed by varying it numerically until 16's waveform has a duration of 20 μs. With the correct duration, front time is also automatically satisfied. The solution is ${\displaystyle {\frac {\omega _{0}}{2\pi }}=f_{0}=20.03{\text{ kHz}}}$.

Once ${\displaystyle \alpha }$, ${\displaystyle \beta }$, ${\displaystyle \omega _{0}}$ and ${\displaystyle Q}$ are solved, the circuit component values can be obtained, ${\displaystyle R_{3}}$ is derived first.

Note that the effective output impedance is (by dividing 6 by 13):

${\displaystyle R={\frac {V_{p}}{I_{p}}}={\frac {\omega _{0}}{\beta -\alpha }}(e^{-\alpha t_{pe}}-e^{-\beta t_{pe}})e^{{\frac {\omega _{0}}{2Q}}t_{ps}}R_{3}}$

(18)

And can be rearranged as:

${\displaystyle R_{3}={\frac {(\beta -\alpha )e^{-{\frac {\omega _{0}}{2Q}}t_{ps}}}{{\omega _{0}(e^{-\alpha t_{pe}}}-e^{-\beta t_{pe}})}}R}$

(19)

Set output impedance ${\displaystyle R}$ = 2 Ω, the solution is ${\displaystyle R_{3}}$ = 26.1 Ω.

Finally, the closed-form solution of other component values is:

${\displaystyle L={\frac {R_{3}}{\alpha +\beta -\omega _{0}/Q}}}$

(20)

${\displaystyle R_{2}=\left({\frac {\omega _{0}}{Q}}-{\frac {\alpha \beta -\omega _{0}^{2}}{\alpha +\beta -\omega _{0}/Q}}\right)L}$

(21)

${\displaystyle R_{1}={\frac {\omega _{0}^{2}}{\alpha \beta -\omega _{0}^{2}}}R_{3}-R_{2}}$

(22)

${\displaystyle C={\frac {1}{R_{1}}}{\frac {\alpha +\beta -\omega _{0}/Q}{\alpha \beta -\omega _{0}^{2}}}}$

(23)

The solution is ${\displaystyle C}$ = 5.93 μF, ${\displaystyle L}$ = 10.9 μH, ${\displaystyle R_{1}}$ = 20.2 Ω, and ${\displaystyle R_{2}}$ = 0.814 Ω.

Output peak voltage is slightly lower than the charging voltage. To scale the voltage, use the amplitude in 4 and set E = 1, this yields ${\displaystyle {\frac {1\cdot {\frac {R_{3}}{L}}}{\beta -\alpha }}=0.943}$. Thus, the capacitor charging voltage is ${\displaystyle {\frac {1}{0.943}}=1.06}$ times the output peak voltage.

Note that this solution doesn't consider the coupling capacitor, and also has an undershoot of ${\displaystyle e^{-{\frac {\pi }{2Q}}}=0.34}$. The solution to both problems are discussed in the following sections.

An extra 18 μF series coupling capacitor has almost no effect on the open-circuit voltage, but affects short-circuit current significantly.

Carobbi et, al. suggested the following iterative, trial-and-error design procedure to take the effect of the series coupling capacitor into account. First, without considering the capacitor, the original circuit analysis is reused, and circuit components values are obtained through a numerical solver. Next, the capacitor is added and the change of short-circuit waveform is noted. Then, the target waveform parameters for the numerical solver are "pre-distorted", obtaining a new set of component values (by changing front time, duration, and effective output impedance). For example, if the peak current becomes too low, component values are recalculated for a higher peak current by adjusting the effective output impedance target. These steps are repeated until the desired waveform is obtained. The result given here is accurate within 1.5% after two iterations, more iterations are required for higher accuracy. [7]

Ideal Component Values (Numerical Approximation)
18 μF Series Capacitor	Charging Voltage (V)	${\displaystyle C_{c}}$ (μF)	${\displaystyle L_{r}}$ (μH)	${\displaystyle R_{m}}$ (Ω)	${\displaystyle R_{s1}}$ (Ω)	${\displaystyle R_{s2}}$ (Ω)	Current Undershoot (%)	By
No	1060	5.93	10.9	0.814	20.2	26.1	34	Carobbi et, al.[7]
No	1082	6.04	10.4	0.941	25.1	19.8	27.4	Hesterman, et. al.[8]
Yes	1063	9.98	10.7	0.832	9.39	25.5	39	Carobbi et, al.[7]

Both sources showed that it's not possible to exactly meet the waveform requirements without violating the 30% short-circuit current overshoot limit. Nevertheless, Hesterman, et. al. presented an approximate solution by adjusting the waveform parameters within tolerance.[8] The derivation by Carobbi et, al. ignored the undershoot requirement, pointing out that a practical circuit may reduce overshoot to even practically zero in some cases if an unidirectional switch is used.[7] Also, IEC 61000-4-5 states that there's no overshoot or undershoot requirement at the output of a coupling network.

These solutions are only valid for an ideal generator, suitable for circuit simulation. It can be used as a starting point of practical generator design, but component values have to be adjusted further due to switch non-idealities. In an ideal circuit, open-circuit voltage rise time is governed by the time constant ${\displaystyle {\frac {L_{r}}{R_{m}+R_{3}}}}$, but a practical switch may cause rise time degradation. Further, due to the use of different switch types, a real generator may produce either a bidirectional impulse with undershoot, or an unidirectional impulse without undershoot. An ideal circuit model cannot predict these non-linear effects, and should not be treated as a complete circuit model of practical generators.[7]

A different Combination Wave Generator is used for the 10/700-5/320 μs surge.

Simplified Schematic of the 10/700-5/320 μs Combination Wave Generator