Operational Amplifiers I

Purpose
  Sometimes, output of the circuit which we have designed is not optimal for directional application to what we have expected if there is an unwanted offset, or noise. So, we can use an Op, LM741, to eliminate the problem.


Op AMP 741

Figure 1. Circuit Symbol

Figure 2. S.O. Package

Designed Circuit

Figure 3. Designed Circuit

Assumption
  We determine a circuit that includes an inverting amplifier with a gain of -10.It must draw no more than 1mA of current from the sensor, and power supplies should supply no more than 30mW each; moreover, the circuit should be designed with standard 1/4 W resistors. 

Figure 4. Build-up Circuit on Breadboard

Figure 5. The Circuit connected to Multimeter

Calculations
  We calculated for safe values for the resistor, built the circuit, and measured the V_{IN and} 

V_{OUT for the gain.}

• Given

  • V_{1 =} 12.13V

  • V_{2 =  12.06V}                                                                           

   _Result
  

  • Ri = 1K Ω 

  • Rf = 10KΩ 

  • Rx = 1152 Ω 

  • Ry= 104.7 Ω 

  
  

  
  

  Vin (V)	Vout (V)	GAIN	VRi (V)	IRi (uA)	VRf (V)
  0	-0.05	0	0.007	7.00	0.06
  0.25	-2.54	-10.16	0.269	269	2.57
  0.50	-5.08	-10.17	0.536	536	5.12
  0.75	-7.56	-10.08	0.797	797	7.62
  1.00	-9.91	-9.91	0.946	946	9.91

  The power supply current

  Iv1=2.32mA

  Iv2=1.49mA

  
  

  So, Pv1= 12.13*2.32m=28.14mw; Pv2= 12.06*1.49m=17.97mw which satisfy the power supply constraint to supply no more than 30mW each.

Pert Chart of Ping-Pong Shooter Project

PSpice Review

Given problem

Place a DC current source between terminals and sweep the current from 0 to 3 amps. We can find Vth=10V when I terminal = 0A. Rth is the slope which is equal to 10/3 =3.33 ohms.

Then, assume RL=R4=10 ohms. The Thevenin Equivalent circuit is shown as following

Norton Equivalent circuit is 

Set RL is from 1 to 10 as following

Then graph power (V6*V6)/RL vs RL, and then RL should be 3.33 ohms in order to get the maximum power which is 7.5 W.

PSPICE Tutorial

Purpose


The purpose of this lab is to learn how to PSpice to simulate a complex circuit.

Given circuit

By using PSpice, the voltage at each node and the current on each branch are easy obtained as following,

Thevenin Equivalents

Purpose

       The purpose of this lab is to simplify a complex circuit by using Thevenin Equivalent method. We were given the engineering problem of finding the smallest equivalent load resistance that can be used in a system of multiple sources and loads. Using Thevenin Equivalent, only VTh, RTh, and Rload 2. Therefore, using nodal analysis, we calculated Vth = 8.64V and I load 2 = 0.131 A, and found RTh with RTh = VTh/ I load 2 = 66 Ω,  RL2 = 825 Ω.

Results

Config          Theoretical Value          Measured Value           % error

RL2                   Vload2 = 8V                   7.99V                       0.13%

RL2 infi.            Vload2 =8.64V               8.63V                       0.12%

Extra

Pmax = 8.64^2/(4*66)=0.28W

Overall, values were fairly close with both circuits. Percent error may have been the result of using 9.07V for power sources instead of an ideal 8V.

Beta of a BJT

Purpose

The purpose of this lab is to figure our the beta(gain) of a BJT by finding the linear relationship between Ie and Ib. For a NPN typed BJT, Ie = (1+beta)*Ib

The built circuit looks like as following

Figure 1.

Measured V         Rb(Theoretical)         Rb(actual)            Ib                   Ie

6.05 V                       530 k ohm           529 k ohm          0.011 mA      1.36 mA

6.05 V                       265 k ohm           263 k ohm          0.021 mA      2.81 mA

6.05 V                       176 k ohm           176.6 k ohm       0.033 mA      4.18 mA

6.05 V                       132 k ohm           132.5 k ohm       0.046 mA      5.56 mA

6.05 V                       106 k ohm           106.4 k ohm       0.058 mA      6.98 mA

Using EXCEL to graph Ie v.s Ib 

Therefore, 1+beta=117.3 ========> beta=116.3

Nodal Analysis

Purpose

The purpose of this lab to analyze a circuit with nodal analysis since the given circuit combines elements in series or parallel. Therefore, choosing a node and identifying a ground are important steps before solving the circuit.

Given Problem

Figure 1. 

We can write down equation on V2 and V3 as following

(V1-V2)/100=V2/1000+(V2-V3)/220------->1

(V2-V3)/220+(V4-V3)/220=V3/1000------->2

Solve for equation 1&2 we can get V2=10.26V and V3=8.67V

Ibat1=(V1-V2)/100=0.0174A

Ibat2=(V4-V3)/220=0.0015A

Pbat1=-0.0174*12=-0.2088W

Pbat2=-0.0015*9=-0.0135W

Next we build the circuit up 

Figure 2. 

Connect it with two voltage supplies, and measure the battery current

Figure 3.

The results come out as following

Variable    Theoretical Value     Measured Value         Percent Error 

Ibat1         17.4  mA                 17.12  mA                  1.61 %

Ibat2         1.5    mA                 1.77   mA                    18  %

V2            10.26  V                  10.34  V                     0.78 %

V3            8.67  V                    8.77    V                    1.15 %

Next, we design that we want V2=V3=9V, we need to solve the desired battery voltages

Set the equations up

(V1-9)/100=9/1000+9/220======>V1=9.9 V

(V4-9)/220=9/1000======>V2=10.98 V

We implement the new battery voltage in the circuit and measure the new current again

V2=8.92 V, V3=8.80 , I bat1 = 9.29 mA, I bat2 = 8.03 mA

P.S. This guy allows us to adjust our voltage we want in a range between 0V to 20 V