Operational Amplifiers

In this lab we had the problems of wanting a voltage source varying  0-1V while having 0-10V of output. In this case we used an inverting amplifier to obtain a gain of -10. 6V battery supplies powered the op amp.

We first calc. the smallest Ri value for input resistance using Ohm's law (V = IR)         Ri = 1000Ω.

Afterwards, nodal analysis was used to calculate a value of Rf, the resistor between the inverting input and output of the op amp., that would satisfy a gain of -10.          Rf = 10000Ω.

For the  variable source (Rx) to vary 0-1V, a resistor Rx was placed, acting as a voltage divider. Assuming worst case scenario, where Ry = 0Ω, we calc. Rx = 283Ω. However, closest resistor available was 360Ω.

We determined Ry using KCL, measuring  5.95V from power supply and asuming V=1V for Ry.    
Ry = 72.72Ω.   Actual R used: 75Ω.

After calculations, we built the breadboard, switching the +/- terminals of one battery to have both
+ and - 5.95V power supplies. We then measured different Voltage values at 0-1Vin, as followed:

VIN          Vout  (V            Gain              VRi (V)            IRi  (mA)            VRf  (V)
                                      (Calculated)                              (Calculated) 
0.0V            0.02                  0                       0                       0.00                  0.03       
0.25V          2.62                  10.08                0.26                  0.26                  2.62
0.50V          4.06                  9.90                  0.41                  0.41                  4.17
0.75V          4.06                  9.44                  0.43                  0.43                  4.39
1.00V          4.06                  9.02                  0.45                  0.45                  4.62

With VIN = 1.00V, we measured the current from the power supplies:

(+Vcc) IV1 = 0.03 mA
(-Vcc)  IV2 = 0.76 mA

The gain was consistent, although it decreased as the voltage increased, and the current did not pass 1mA.

Webassign Blog

        Wanting the Thevenin Eq. of the circuit external of R3 (= 2kΩ), the R3 is removed from the circuit. VTh becomes the open circuit voltage at the terminals, which is 75V. 
        For RTh, we can treat current sources as open circuits, voltage sources as short circuits, and add Rs in parallel and series. Adding R1 and R2 in parallel gives RTh = 5.33kΩ.

        Alternatively, we could use Norton's theory to treat the open circuit as a short circuit and find the current through it. Using Kirchoff''s current law, I3= 11.81 + 2.25 ~ 14mA. Using IN(I3) = VTh/RTh, RTh = 5.33Ω.

              VTH and RTH is found the same way as part 1. The voltage across the open circuit approximates to 61.2V, and adding R1 and R2 in parallel gives the RTh = 42.4Ω.
              If the R of the load is to dissipate as much power as possible, it must be equivalent to RTh; therefore, RL = 42.4. Using PL = VTh/4RTh, PL = 22.1W.

 DC sweep was introduced in this Pspice lab, where node voltages and currents can be obtained with source that changes through a range of values, rather than staying constant as in nodal analysis.

 

In this circuit V1 had a DC sweep from 0-20V in 1-V increments. The graph above has its traces with the voltages with respect to current.

For more complicated circuits, Thevenin and Norton equivalents can be found for elements. In this case, a DC sweep graph was done at branch I2. With it being Voltage over Current, the resistance is the slope of the graph (around 6 ohms) and VTh is the zero intercept, which is 20V.

 The current source is then replaced with a voltage source to find the Norton equivalent. With the graph being Current vs Voltage, the slope of graph represent the inverse of resistance, conductance, with is around 0.17S. Zero y-intercept gives IN = 3.34A.

In the last circuit we graphed a chart of power. As expected, the power is maximum when the load resistance = the Thevenin Resistance, or in this case, when R is 1k.