Is there current limiting on NI's (PXI-5402/5406) function generators? I am trying to replace an old piece of lab equipment (Agilent 3245A) which does not have current limiting. The application is a standalone system generating a 5V p-p square & sinewaves @ 100Hz.
Thank you for using NI Forums. To answer your question there is overload protection implemented on the PXI-5402 and PXI-5406. The Specifications Page for these two items details: "Maximum Overload Protection - The CH 0 output terminal can be connected to a 50 Ω, ±12 V source without sustaining any damage. No damage occurs if the CH 0 output is shorted to ground indefinitely." Also as far as how much current the device can source, with a max voltage of 5V and an output impedance set at 50 ohms the max current would be 100 mA. Hope this helps. Let me know if you have any other questions.
Thanks for the response. My test is a rather special condition for which I need a current limiting (not overload protection) of ~ 20mA on this AC signal. I could end up putting another 200ohm resistor in line with the 50 ohm (5V/ 250 ohm). This might end up being OK if the part has a high impedance. Any other "clean" options? I might have to look at the ..........errr Agilent rack 😞
It would be possible to use another resistor to put in line with the device to get your current at the certain level, but I am not sure how exact the current level would be. The function generators are designed to keep a constant voltage and I have not personally tried to limit the current with a FGEN. Another option would be to use one of our programmable power supplies that would be able to keep a constant current. They would be able to provide a 100 Hz Square Wave but it would be software timed, so it depends on how exact you need that square wave to be. Another option would be to look into using a Current Buffer with the FGEN, once again I have never done this but it could be something that would be worth looking into.
Also, my math earlier was a little off, the answer was right, but the methodolgy to get there was incorrect. The math is as such: The max voltage is 10V over a voltage divider, where it is R(load) / (R(load) + R(source)). If you look at the detailed specs for the device it shows this configuration.