Analog Circuit Design Laboratory Report - MyWeb at WIT ...

The next configuration, with a 10-kΩ load resistor connected to pin 6, illustrates
sourcing current. Sourcing current used the positive supply to feed the output and an
additional source is connected to the non-inverting pin 3. Isupply(+) is equal to the sum of
the negative supply current and the current; see Equation 2. This satisfied Kirchoff’s
Current Law.
Isupply(+) enters into the op-amp, and I supply(-) and the load current, IL, exit the op-
amp. The circuit was broken so that current measurements could be taken for Isupply(-), Iout
and Isupply(+). The sum of Isupply(-) equal to 0.6mA and IL equal to 1.44mA was within 1.4%
of the direct measurement of Isupply(+) equal to 2.13mA. The polarity of the source current
proved that the op-amp was sourcing current to the load as it was positive and equal to
+Vsat.
Next, the circuit was configured to sink current. With the 5V source at the
connected to the inverting pin 2, the output became -Vsat, since voltage follows from
higher to lower potential, the load current is drawn into the op-amp. The op-amp was
then sinking an IL equal to 1.37mA and using the measured value Isupply(+) at 0.6mA from
this circuit, to calculate an Isupply(-) equal to 2.31mA from Equation 2.
Based on the sum of Isupply(+) and Iout when compared to the direct current reading
of Isupply(-) provided a percent difference of 12.26%, which is not within the excepted
10%. Sinking current to the load did provide a –Vsat.
This exercise was also simulated on PSpice, see Figure 10. This was devised with
simulated components and adjacent voltage supplies connected by assigning the same
name of the source at the inputs. The results of PSpice were close to those measured in
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