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Last Modified: November 30, 2020

Determine the extended range pulsing parameters, assuming the following operating point.

Output function | Pulse Current |

Pulse voltage limit, V_{pulse} |
40 V |

Pulse current level, I_{pulse} |
6 A |

Bias voltage limit, V_{bias} |
0.1 V |

Bias current level, I_{bias} |
0 A |

Pulse on time, t_{on} |
1.5 ms |

Chassis' slot cooling capacity | ≥58 W |

Solution

Begin by calculating the pulse power using the following equation.

$\text{Pulse power}={\text{V}}_{\text{pulse}}*\text{\hspace{0.17em}}{\text{I}}_{\text{pulse}}$ $=\text{40 V}*\text{\hspace{0.17em}6 A}$ $=2\text{40 W}$

For PXIe-4139 (40W), refer to the following figures to identify next steps. First, verify the the region of operation using Figure 1, which shows 240 W is in the extended range pulsing region.

Next, refer to Figure 1, which shows the maximum pulse on time, t_{on}, is limited by the maximum
pulse energy, E_{pulseMax}. Use the pulse energy equation *(Equation 2)* from Table 1 to calculate the maximum pulse on time, t_{onMax}
*(Equation 6)*.

${\text{t}}_{\text{onMax}}=\left|\frac{{\text{E}}_{\text{pulseMax}}}{{\text{V}}_{\text{pulse}}*\text{\hspace{0.17em}}\text{}{\text{I}}_{\text{pulse}}}\right|\text{}(Eq\mathrm{.}6)$ $=\left|\frac{\text{0.4 J}}{\text{40 V}*\text{\hspace{0.17em}6 A}}\right|$ $=\text{1.67 ms}$

Next, refer to Figure 2, which shows the maximum duty cycle, D, is limited by the cycle average power,
P_{CA}.If the required pulse on time is 1.5 ms and the module is
installed in a chassis with slot cooling capacity ≥58 W, use the cycle average power
equation *(Equation 5)* from Table 1 to calculate the minimum pulse off time, t_{offMin}
*(Equation 7)*.

${\text{t}}_{\text{offMin}}=\left|\frac{{\text{P}}_{\text{CA}}*\text{\hspace{0.17em}}{\text{t}}_{\text{on}}-{\text{V}}_{\text{pulse}}*\text{\hspace{0.17em}}{\text{I}}_{\text{pulse}}*\text{\hspace{0.17em}}{\text{t}}_{\text{on}}}{{\text{P}}_{\text{CA}}-{\text{V}}_{\text{bias}}*\text{\hspace{0.17em}}{\text{I}}_{\text{bias}}}\right|\text{}\text{}(Eq\mathrm{.}7)$ $=\left|\frac{\text{40 W}*\text{\hspace{0.17em}1.5 ms}-\text{40 V}*\text{\hspace{0.17em}6 A}*\text{\hspace{0.17em}1.5 ms}}{\text{40 W}-\text{0.1 V}*\text{\hspace{0.17em}0 A}}\right|$ $=\text{7.5 ms}$

Finally, verify that the pulse cycle time, t_{cycle}, is greater than or equal to the
minimum pulse cycle time, t_{cycleMin} (5 ms). To calculate the pulse cycle
time, use the following equation:

${\text{t}}_{\text{cycle}}={\text{t}}_{\text{on}}+{\text{t}}_{\text{off}}\text{}\text{(Eq. 4)}$ $=\text{1.5 ms}+\text{7.5 ms}$ $=\text{9 ms}$

In this case, the pulse cycle time meets the minimum pulse cycle time specification.

Therefore, a 40 V, 6 A pulse with an on time of 1.5 ms and a pulse off time of 7.5 ms is supported, since it fulfills the following criteria:

- Greater than the minimum pulse on time of 10 μs
- Equal to the minimum pulse off time of 7.5 ms to meet maximum cycle average power
- Greater than the minimum pulse cycle time of 5 ms

Determine the extended range pulsing parameters, assuming the following operating point.

Output function | Pulse Current |

Pulse voltage limit, V_{pulse} |
40 V |

Pulse current level, I_{pulse} |
6 A |

Bias voltage limit, V_{bias} |
0.1 V |

Bias current level, I_{bias} |
0 A |

Pulse on time, t_{on} |
1.5 ms |

Chassis' slot cooling capacity | ≥58 W |

Solution

Begin by calculating the pulse power using the following equation.

$\text{Pulse power}={\text{V}}_{\text{pulse}}*\text{\hspace{0.17em}}{\text{I}}_{\text{pulse}}$ $=\text{40 V}*\text{\hspace{0.17em}6 A}$ $=\text{240 W}$

Since the pulse power of 240 W is within the 500 W region of Figure 2, the maximum configurable on time is 400 μs and maximum duty cycle is 2%.

For example, if the required pulse on time is 100 μs, and the required pulse cycle time is 10 ms, calculate the pulse off time and verify the duty cycle using the following equations.

${\text{t}}_{\text{off}}={\text{t}}_{\text{cycle}}-{\text{t}}_{\text{on}}$ $=\text{10 ms}-\text{100}\mu \text{s}$ $=9\text{.9 ms}$

$\text{Duty cycle}=\frac{{\text{t}}_{\text{on}}}{{\text{t}}_{\text{cycle}}}*\text{\hspace{0.17em}100\%}$ $=1\%$

Therefore, a pulse with an on time of 100 μs and 1% duty cycle would be supported, since it fulfills the following criteria:

- Greater than the minimum pulse on time of 50 μs
- Less than the maximum pulse on time of 400 μs and duty cycle of 2%
- Greater than the minimum pulse cycle time of 5 ms

Determine the appropriate operating parameters and custom transient response settings, assuming the following example parameters.

Output function | Pulse Current |

Pulse voltage limit, V_{pulse} |
50 V |

Pulse current level, I_{pulse} |
5 A |

Bias voltage limit, V_{bias} |
0.1 V |

Bias current level, I_{bias} |
0 A |

Transient response | Fast |

Load, cable impedance | 4.5 Ω, 40 μH |

Pulse on time, t_{on} |
10 μs |

Pulse off time, t_{off} |
4.99 ms |

The SMU Transient Response can be configured to three predefined
settings, Slow, Normal, and Fast. If these settings do not provide the desired pulse
response, a fourth setting, Custom, enables NI SourceAdapt^{[1]} technology which provides the
ability to customize the SMU response to any load, and achieve an ideal response
with minimum rise times and no overshoots or oscillations.

Solution

SourceAdapt allows users to set the desired gain bandwidth, compensation frequency, and pole-zero ratio through custom transient response to obtain the desired pulse waveform. To use SourceAdapt, first set the Transient Response to Custom.

To achieve the resulting waveform in the following figure, use the parameters in the following table.

Transient response | Custom |

Current: Gain bandwidth | 260 kHz |

Current: Compensation frequency | 140 kHz |

Current: Pole-zero ratio | 0.75 |

Gain bandwidth is directly proportional to the step response slew rate. The higher the gain bandwidth, the higher the slew rate. It is worth noting that increasing the gain bandwidth will likely increase ringing. However, this can likely be removed by appropriately setting the compensation frequency and the pole-zero ratio.

Compensation frequency and pole-zero ratio are used to determine the frequencies of the SMU control loop pole and zero, which can be used to optimize the system transient response by increasing phase margin and reducing ringing. To reduce the overshoot, it is recommended to set the compensation frequency close to the overshoot ringing frequency, see Fc in Figure 3, and set the pole-zero ratio to be greater than 1.

For reference, the pole frequency and zero frequency are derived by the following equations.

$\text{Pole frequency}=\text{Compensation frequency}*\text{\hspace{0.17em}}\sqrt{\text{Pole-zero ratio}}$

$\text{Zero frequency}=\frac{\text{Compensation frequency}}{\text{Pole-zero ratio}}$

These settings can be accessed through the Transient Response set to Custom: Voltage or Current.

^{1}Visit ni.com for more information about NI SourceAdapt Next-Generation SMU Technology.