Keithley 2182

Ratio and Delta 5-7
Figure 5-1
Test circuit using constant current source
Figure 5-1B shows what happens when the current is reversed. The measurement by the
Model 2182 still includes the 10µV of thermal EMF, but the voltage across the DUT is now
negative. Therefore, the Model 2182 will measure 90µV:
V
2182
= V
THERM
+ V
DUT
= 10µV - 100µV
= –90µV
As demonstrated in Figure 5-1, neither measurement by the Model 2182 accurately measured
the voltage across the DUT. However, if you take a simple average of the magnitudes of the two
readings (110µV and 90µV), the result is 100µV, which is the actual voltage drop across the
DUT. This is what the calculation for Delta does.
DUT
0.1Ω
V
THERM
V
10µV
HI
LO
+
–
2182
CH 1
1mA
A. Positive Current Source
DUT
0.1Ω
V
THERM
V
10µV
HI
LO
–
+
2182
CH 1
1mA
B. Negative Current Source
V
DUT
= 100µV
V
2182
= 10µV + 100µV
= 110µV
V
DUT
= –100µV
V
2182
= 10µV – 100µV
= –90µV