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I have one Keithley 6487 at lab. when I use Labview control the unit and let it output voltage, it behaves normally until the setting is beyond 265V. if you set it output more than 265V, for example 300V, you will get 265V no matter what the setting is. I measured the voltage using Fluke 83V DMM. At the same time the current meter of the 6487 is also included in the measurement loop because I want to know how much is the background current noise; the current reading is about 1E-9A or less. I wonder if this is normal.
BTW, I noticed one thing, if I did not include the - ampere meter in my voltage measurement loop, for example, directly from the 6487's voltage output to Fluke DMM's input, I got 500V reading.
I tried again this morning, if I directly link the output of the 6487 to a DMM, I got the voltage value whatever I set up either manually or remotely. for example, I set 411V, I got 411V on the DMM. but if I include the current meter of the same 6487 unit into my measurement loop, I got a leakage current. for example, if I set the 6487 to output 50V, I could get 50V on DMM, but the current meter of 6487 shows there is a 5uA current reading. and if I set the voltage higher, for exmple,300V, it ends up that the frond panel of 6487 flashes "COMP" (I assume for "compliance"), and the current reading is 26.5uA while the voltage reading on DMM is 265V. I used 2 DMMs and got the same results. I wonder if anyone could shine some light on it.
I found out if I set the current limit ( accomplice level) on the 6487 higher, I will get the voltage I set for. for example, if I set the current limit to 2mA and let 6487 output 400V, I measured with DMM and got a reading of 400V, but at the same time the current meter of 6487 reads about 40uA, which means a resistance of 10Mohm. so I assume this 10Mohm is the internal resistance of the DMM. since DMM has a limited value of input resistance like 10Mohm here, when I measure voltage, DMM acts as a resistor and a small current flows. if I set the current limit too low, then the overall voltage would equal to V=resistance*maximum current/current limit. I guess that is the story in my case.
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