04-14-2015 10:29 AM
Hello
I'm trying to measure current going through the capacitor shown in the diagram below. The NI 9215 is connected to a shunt resistor (R1) in order to measure the current. The concern here is that the voltage supply is in the range of 7kV and so my understanding is that the voltage drop accross the shunt resistor will be around 7kV for a short amount of time. Do you think it will damage the NI 9215? The capacitance is between 40 to 80 pF.
Thanks
04-15-2015 01:12 PM
Depending no whether you have BNC or screw terminal connections, the 9215 will have CAT I or CAT II isolation. Either way, your 7kV range is way out of the isolation specifications. You can find more information on the product's web page or in the specifications for the 9215.
04-15-2015 01:37 PM
04-15-2015 09:05 PM - edited 04-15-2015 09:12 PM
So the connectors are screw terminals, the voltage supply is limited to 0.4mA and the bandwidth is pretty low no more than 10 Hz.
04-16-2015 01:09 AM
The current limit of the source doesn't really tell something IF there is a 'small' 100nF buffer capacitor at the output and someone plug in the cable 😉
My crystal ball has a EMC damage, so please provide more information.
04-19-2015 11:38 AM
Your circuit is so incomplete that it is impossible to predict what would actually happen.
If an ideal 7 kV source is connected instantaneously to the capacitor (40 pf) and R1 (220 ohms) the instantaneous initial current is 31.82 A and the voltage is 7 kV across R1. Since your source is limited to 0.4 mA, the details of the source circuit including current limiter and any energy storage (capacitors or inductors) are required to analyze the actual behavior.
If the instantaneous peak current from the source is limited to 0.4 mA, the largest voltage across R1 will be 88 mV, assuming that the current in the other resistor is negligible.
The time constant of the capacitor and R1 is a few nanoseconds. Most high voltage power supplies do not react nearly that fast, so the transient repsonse is dominated by stray reactances. Charging a 40 pF capacitor to 7 kV at 0.4 mA requires 700 microseconds.
A source with 7 kV and 0.4 mA could be a photomultiplier tube power source. Those are often small inverters and may have some output capacitance. That output capacitance can make a huge difference in the behavior of the circuit.
Please tell us what the actual circuit is and what the purpose of the current measurement is. There may be a better way to get the data you need.
Lynn