Keithley 6485 Instruction Manual page 51

Model 6485 Picoammeter Instruction Manual
To prevent accidental damage, a series resistor should be added to the test setup. The min-
imum value of this series resistor depends on the lowest current range to be used in the
measurement. If it will not be necessary to use the lower measurement ranges, a smaller
series resistor can be used, reducing the effect it will have on measurement accuracy. The
lowest necessary measurement range can be determined from the measurement range
accuracy specs, the applied voltage and largest resistance desired to measure. If using auto
range, program the Model 6485 to not use its lowest ranges when autoranging.
To set the auto range lower limit from the front panel:
1.
2.
3.
4.
To set the auto range lower limit over the bus, use [CURRent]:RANGe:AUTO:LLIMit
(Section
Use the following formula to determine the minimum resistance for proper current limit-
ing resistors:
The series limiting resistor should have a minimum power rating of:
Example: If measuring 100GΩ resistances using an external voltage source of 500V, and
thus, a lowest necessary current range of 20nA, the minimum series resistance that will
prevent damage in the case of a shorted resistor would be:
minimum R
minimum power rating = (500V)
NOTE
The 6485 can be programmed to calculate the resistance and subtract the series resistance.
Using the M/X+B function, in the example above, one would set M to 500, B to -14e6, and
the units character to "omega". For more details on the M/X+B function, see
Press the CONFIG key.
Press the down range key ( ).
Use the and range keys to scroll through the available lower limit settings.
Press ENTER to save the displayed value as the lower limit. Press EXIT to return
to the previous setting.
4).
MinR
series
Lowest range to be used
2nA or 20nA
200nA or 2µA
20µA or 200µA
2mA or 20mA
MinPowerRating
= (500V - 220V)/220V
series
The 14MΩ in series will increase the measured resistance to 100.014GΩ
 –
SourceVoltage 220V
= ---------------------------------------------------------- -

220V
=
SourceVoltage
×
11 MΩ = 14MΩ
2
/14MΩ =18mW
Measurement Concepts
 R

in
R
in
11MΩ
3.5MΩ
50kΩ
500Ω
2
⁄
R
series
Section
2-7
5.