Op-Amps 1

 PROBLEM: We need to couple a sensor to a processing agent. Oftentimes the output of the sensor is not optimal for direct application to your processing agent, so we insert a signal conditioning circuit. A signal conditioning circuit is required when the range of Vsensor(output sensor voltage) is too small/large, or if there is unwanted offset, or if we wish to filter out unwanted noise, or protect the micro-controller input from an unwanted over-voltage.

METHODS:
We want to increase the voltage range of the sensor. The range of Vsensor is 0-1V and 0-10V range for the micro-controller. The operation amplifier (op-amp) offers an ideal solution to the problem. We will use the LM741 op-amp (pin diagram shown in Figure 1). The current into the LM741 must be less than 1mA; additionally, the op-amp power supplies should supply no more than 30mW of power each. The circuit should be designed with standard 1/4 W resistors. The circuit is an inverting amplifier with a gain of -10 as shown in Figure 2.

Figure 1: LM741 Pin Diagram (NC= NO CONNECTION)

We represent the sensor that produces a voltage which varies between 0V and 1V. We create a voltage divider to synthesize Vin. If Rx and Ry are much smaller than Ri, then the inverting amplifier will not appreciably "load" the divider circuit.

Calculations:
We calculate Ri using Ohm's law and the conditions for the sensor voltage and current into op-amp. 1V=Ri(1mA) >> Ri = 1kOhm
We can determine the feedback resistor Rf using the proportionality of the input and output voltages of the op-amp; we assume the op-amp to be ideal. Thus,
Vout=Rf/Ri*(Vin) >> Rf = Vout/Vin*(Ri) -- Vout=10V, Vin=1V, Ri=1kOhm >> Rf= 10kOhm

To determine Rx, we assume that we want it to be a standard 1/4 W resistor operating at half its rated power. The worst case across Rx occurs when Ry is 0. Thus,
PRx=(Vs^2)/Rx >> Rx=(Vs^2)/PRx >> PRx(Max power dissiapted by Rx)= 1/4W*1/2=1/8 W, Vs = 6V and 12 V

Accordingly, Rx = 288 Ohm, 1152 Ohm for Vs = 6V, 12 V
For Ry we use a voltage divider assuming that the op-amp does not load the sensor circuit.

Vin=(Ry*Vs)/(Rx+Ry) >> Ry = 57.6 Ohm, 26.2 Ohm for Vs = 6V, 12V

We determine the Thevenin Equivalent of the divider looking back to the left of Ri.
Where Rth = Rx + Ry and Vs = 6 V, 12 V respectively.

We see that the the Thevenin resistance is not at least 20 times less than the resistance value for Ri = 1kOhm. To achieve the output of 1V across Ri, we use a potentiometer for Ry and adjust it until we reach the desired voltage value.

DATA: 


Component Data Table

Component	Nominal Value	Measured Value	Power/Current Rating
Ri	1 kOhm	0.96 kOhms	1/8 W
Rf	10 kOhm	10.09 kOhms	1/8 W
Rx	288/1152 Ohm	287/1152 Ohms	1/8 W
Ry	57.6/26.2 Ohm	58/27 Ohms	1 W
V1	6/12 V	6.06/12.02 V	2 Amp
V2	6/12 V	6.06/12.02 V	2 Amp

6 V Power Supplies Data Table

Vin	Vout	Gain	Vri	Iri	VRf
0.00 V	0.07 V	0.00	8.9 mV	9.27 uA	90.1 mV
0.25 V	2.70 V	10.80	264 mV	0.275 mA	2.66 V
0.50 V	4.02 V	8.04 V	0.39 V	0.406 mA	4.12 V
0.75 V	4.02 V	5.36 V	0.42 V	0.438 mA	4.36 V
1.00 V	4.02 V	4.02 V	0.44 V	0.458 mA	4.62 V

12 V Power Supplies Data Table

Vin	Vout	Gain	Vri	Iri	VRf
0.00 V	0.00 V	0	4.40 mV	4.58 uA	52 mV
0.25 V	2.60 V	10.4	0.25 V	0.26 mA	2.73 V
0.50 V	5.20 V	10.4	0.54 V	0.56 mA	5.61 V
0.75 V	7.70 V	10.27	0.75 V	0.78 mA	7.76 V
1.00 V	9.80 V	9.8	0.95 V	0.99 mA	9.85 V

We can see the saturation of the op-amp from the output voltages of the 6V Power Supply Data Table; hence,  the reasoning for the switching the experiment to use a 12 V power supply.

Next we insert an ammeter in series with the +/-Vcc inputs on the op-amp; we measure the current leaving each power supply, Iv1, Iv2 respectively.
Iv1 = 0.503 mA
Iv2 = 1.48 mA

Lastly we calculate the power deliver by each of the power supplies, V1 & V2. Where, V1=V2=12.02V,   Pv1= Iv1*(V1),   Pv2 =Iv2*(V2)

Pv1 = 6.05 mW
Pv2 = 17.7 mW

QUESTIONS
-If we assume the currents into the non-inverting and inverting terminals are zero, based on power supply currents and calculated current through Rf, show that KCL is satisfied for condition, Vin = 1.00V.
        -Solution:

-Do we satisfy the power supply constraint to supply no more than 30mW each? If we wanted to reduce the power drawn from the power supplies even more, how could we modify our design to accomplish this without changing the amplifier gain?

Pspice

Today we did a Pspice lab. We needed to find the Thevenin and Norton Equivalent for a circuit. Graphs were used to determine the values.

Rth = Rn = 3.33 ohms
Vth = 10 V
iN = 3 A
Pmax occurs when RL is less than 0.5 kohms

Thevinin Equivalents

PROBLEM
 Suppose you have two power supplies, and multiple loads. Given a minimum acceptable voltage across a load, determine the smallest equivalent load resistance that can be successfully used, and what is the power consumed. What voltage will exist across the terminals of the load if we remove it (open circuit/highest voltage/Voc) and what short-circuit current will flow if we replace the load by a short (isc/highest current)?

METHOD
 We model a system similar to Figure 1:

Figure 1

We want to reduce the circuit to it's Thevenin Equivalent consisting of a voltage source and a resistor -- we want to "thevenize" the left-hand side of the circuit. We remove RL2 from the circuit to find Voc=Vx=Vth.

Vs1=Vs2= 9 V

We use nodal analysis at node x to find Vx, which results in the following nodal equation:
Vx/680 + (Vx-9)/100 + (Vx-9)/39 = 0  >> Vx = 8.659 V = Voc = Vth

We find Rth two ways:

We first find Rth using equivalent resistances: Rth = Rc3 + 1/(1/Rc1 +1/Rc2) = 66 ohms

Additionally, we can find Vy using nodal analysis. Next we find isc = Vy/Rc3; finally, we can find Rth = Voc/isc= 66ohms

The nodal equation for Vy:

(Vy-9)/39 + (Vy-9)/100 + Vy/680 + Vy/39 = 0 >> Vy = 5.11 V >> isc = 5.11/39 = 0.131 A
Thus, Rth = 8.659/0.131 = 66 ohms

Finally, we have our Thevenin equivalent circuit:

Vth = 8.659 V      Rth = 66 ohms

Given that the minimum load across Vload is 8 V, we can find the smallest permissible value for RL2 using a voltage divider. We also find the isc using Ohm's Law, and the open circuit voltage by inspection.

Vload=Vth*RL2/(RL2+Rth) >> RL2 = Rth*Vload/(Vth-Vload) = 66*8/(8.659-8) = 801 ohms

isc = vload/RL2 = 8/801= 9.99 mA

Voc = Vth = 8.659 V

DATA
 Next, we build the Thevenin equivalent circuit to find the measured voltage across the load, Vload and the open current voltage Voc. Percent error is calculated.
Components Data Table

Component	Nominal Value	Measured Value	Power/Current Rating
Rth	66 ohms	66.5 ohms	0.3 W
RL2,min	825 ohms	826 ohms	0.3 W
Vth	8.659	8.71	10.0 A

Thevenin Equivalent Experimental Data Table

Configuration	Theoretical Value	Measured Value	Percent Error
RL2=RL2,min	8.00 V	8.04 V	0.50%
RL2=infinity	8.569 V	8.54 V	0.34%

Next, we build the system in Figure 1 on the breadboard and again measure the voltage across the load, Vload and the open current voltage Voc. We also calculate the power supplied to/absorbed by RL2 when RL2=Rth. Percent error is calculated.
Components Data Table

Component	Nominal Value	Measured Value	Power/Current Rating
Rc1	100 ohms	100. 3 ohms	1/8 W
Rc2	39 ohms	38.1 ohms	1/8 W
Rc3	39 ohms	39.1 ohms	1/8 W
RL1	680 ohms	679 ohms	1/4 W
Vs1	9 V	8.96 V	10.0 A
Vs2	9 V	9.05 V	10.0 A

Figure 1 Circuit Experimental Data Table

Configuration	Theoretical Value	Measured Value	Percent Error
RL2=RL2,min	8.00 V	7.94 V	0.75%
RL2=infinity	8.569 V	8.64 V	0.83%

Maximum power to RL2 when RL2=Rth:
P=((Vth)^2)/4Rth >> (8.659)^2/4*66 = .284 W
Thus, Pmax = 0.284

Finally, we verify the maximum power by adjusting values of RL2 and measuring the load voltage each time and calculating power, where Pload = ((Vload^2))/R

Power Data Table

Configuration	Vload	Pload
RL2=0.5Rth	2.75 V	0.229 W
RL2=Rth	4.14 V	0.259 W
RL2=2Rth	5.57 V	0.235 W