Keithley 6514 Instruction Manual page 78

4-6 Amps Measurements
Floating current measurements — As discussed in Section 3 for volts measurements,
guarding uses a conductor at essentially the same potential as input HI to drastically reduce leak-
age currents in high-impedance test circuits. No current can flow when there is a 0V drop across
a leakage resistance.
For floating current measurements, ammeter input LO can be used as the guard since it totally
surrounds input HI (via the input triax cable), and is at nearly the same potential as input HI. The
actual voltage drop, known as voltage burden, depends on which measurement range is being
used. The voltage burden values are listed in the specifications (Appendix A).
Figure 4-3A shows an unguarded floating current measurement in a high impedance circuit.
The goal is to measure the current (I
from ammeter input LO to test circuit common. Since the ammeter drops essentially 0V, approx-
imately 10V is dropped by R
(10V/1GΩ = 10nA). Therefore, the current that is measured by Model 6514 is the sum of the
two currents (I = I
corrupt the measurement.
Figure 4-3B shows the guarded version of the same circuit. Notice that the only difference is
that the connections to the electrometer are reversed. Resistor R
from ammeter input HI to ammeter input LO, and resistor R
ter input LO (guard) to test circuit common. As previously mentioned, the ammeter drops almost
0V. If the actual voltage drop across the ammeter is <2mV, it then follows that there is a <2mV
drop across R
that is being measured by Model 6514 is the sum of the two currents (I = I
of guarding reduced the leakage current from 10nA to <2pA. Note that the 10nA leakage current
(I ) from ammeter input LO to test circuit common still exists, but it is of no consequence since
G
it is not measured by Model 6514.
) through resistor R. However, a leakage path (R
R
. The current through R
L
+10nA). Obviously, if I
R
. Therefore, the current through R
L
will be approximately 10nA
L
is a low level current, then the 10nA leakage will
R
now represents the leakage
L
represents the leakage from amme-
G
is <2pA (<2mV/1GΩ = <2pA). The current
L
) exists
L
+ <2pA). The use
R