Oscilloscope

Statement of Purpose
To learn the utility of a oscilloscope in electronic circuits.

Procedure

Period: 200 micro seconds

  Peak-to-Peak Amplitude: 11.4V

  zero-to-peak amplitude: 5.8V

 

  expected RMS value: 5.336V

 

So we measure to confirm,

Here a DC offset of 2.5V was set in the function generator.

The following measurements with the DMM were made.
VDC = 2.51 V
VAC = 3.37 V

Next, a squarewave function was generated.

 

 The following measurements with the DMM were made.
 VDC = 10mV
 VAC = 5.33 V

With a mystery signal, adjusting the oscilloscope properly is important to seeing the correct signal.

Conclusion
Oscilloscope allows us to graphically view an electric signal. Without such a device we can only make quantitative measurements which does not always give us a full picture of an electronic circuit.

Capacitor

Statement of Purpose
To see the affect of a discharging capacitor.

Procedure

First, we know that it takes 5 time constants to discharge a capacitor to near 0 levels, which depends on the resistance and capacitance of the RC circuit.

So to charge a capacitor in 20 seconds of 2.5mJ with a 9V power supply we make the following calculations.
The energy of a capacitor is U=1/2CV^2, so C=0.0617 mF
The time constant is 5T =RC -> R=64.8 kΩ
The power generated in the 1W resistor box is 1.25 mW, which will not fry the resistor box. 
The following charge curve is generated by LoggerPro,

To discharge a capacitor in 2 seconds of 2.5mJ, we make the following calculations.
The time constant is 5T =RC -> R=6.48 kΩ
The power generated in the 1W resistor box is 12.42 mW, which will not fry the resistor box. 
The following charge curve is generated by LoggerPro,

Questions:
Using the Thevenin equations,
1. R_cth = 59.4 kΩ, V_cth = 8.25V
 2. R_dth = 6.42 kΩ, V_dth = 0V

3. 0.6321*V_f = 5.215 V
5.215 V is around 4.6s
tau_c = RC = 4.6s
R = 4.6/C = 74.55 kΩ
Error = 15%

Practical Question:
1. U = (1/2)CV^2
C_eq = 2*U/V^2 = 2*160*10^6/(15*10^3)^2 = 1.42 F

2. 1/2C + 1/2C + 1/2C + 1/2C = 2C = C_eq
C = 1/2C_eq = 0.71 F


Conclusion
The time to charge and discharge a capacitor is fairly simple. What matters practically is the time constant, resistors, and power loading.

Operational Amplifier II

Statement of Purpose
To observe the signal produced by an inverting operational amplifier on a given load.

Procedure
Part 1.

Set up the operational amplifier with a sensor as the input, which is emulated by a variable resistor.

Although there is only two power supplies, we know that hooking the same potential can be attained by hooking the variable resistor in parallel. With this setup the following data is attained.

V_in (Desired)	V_in (Actual)	V_out (Measured)	V_RF (Measured)	I_op (Calculated)
0.25 V	0.24 V	-2.41 V	2.46 V	-0.0246 mA
0.5 V	0.50 V	-4.90 V	4.87 V	-0.0502 mA
1.0 V	1.00 V	-10.04 V	9.86 V	-0.1028 mA

I_cc = 0.874 mA
I_ee = -0.981 mA
I_cc + I_ee = -0.107 mA
P_cc = V*I = 10.488 mW
P_ee = 11.772 mW

Part 2.

This time a load resistor was connected to the output of the operational amplifier.

V_in (Desired)	V_out (Measured)	V_RF (Measured)	I_op (Calculated)	I_cc (Measured)	I_ee (Measured)
1.0 V	-9.99 V	9.72 V	-0.102 mA	0.887 mA	-0.987 mA

I_ee + I_cc = -0.1 mA
P_cc = 10.644 mW
P_ee = 11.844 mW


Part bonus

A useful purpose for an operational amplifier is its ability to change the amplitude of a signal by adjusting the values of the input and feedback resistors. We can take advantage of this fact that the gain is represented by the ratio of input and feedback resistors by making one of those resistors adjustable, instead of a fixed value.



The results are as shown below,

V_in (Desired)	V_out (Measured)	V_RF (Measured)	I_op (Calculated)	I_cc (Measured)	I_ee (Measured)
1.0 V	-5.03 V	4.99 V	-0.101 mA	0.885 mA	-0.985 mA

I_ee + I_cc = -0.1 mA

Conclusion 
We see that the current goes unchanged when adding a resistor to the output of the operational amplifier, since changes to the output of the operational amplifier have no affect to the input side, we see the many benefits of an operational amplifier. The use of this can be seen most obviously in the buffer operational amplifier.

Temperature Probe-Op Amplifier

Statement of Purpose
To amplify a signal produced by a sensor.

Procedure
The sensor being considered is a LM35 temperature sensor which will produce 10mV per each degree centigrade. The idea is that the signal is too low for purpose, so we amplify the signal to a desirable one.

The desired voltage range in this case is 0-5V. Since the LM35 produces a constant signal at 350mV and a second input of 150mV is required, the gain required for this circuit is 25 times so that the difference between the voltages is 25(150mV) and 25(350mV) = 5V. This can be handled appropriately with a 150k and 6k Ohm resistor.

Data

Conclusion
The operational amplifier handles simple multiplication and subtraction quite simply and a useful signal can always be produced with this handy tool.

Operational Amplifier I

Statement of Purpose
To amplify a given signal using an operational amplifier. The amplification is determined by the gain resistors. The math is pretty simple.

Procedure

Half of the power rating of a 1/4 W resistor, 
Rx = (V^2)*8 = (12^2)*8 = 1152 Ω.  
Ry = R_x/11 = 104.7 Ω.

Rth = Rx*Ry/(Rx + Ry) = 96 Ω. Vth = 1 V.
Ri = 20*96 = 1.92 kΩ.Rf = 19.2 kΩ.

Data

Component	Nominal Value	Measured Value	Power or Current Rating
Ri	2000 Ω	1957 Ω	1/8 W
Rf	20000 Ω	19760 Ω	1/8 W
Rx	1152 Ω	1786 Ω	1/4 W
Ry	104.7 Ω	7020 Ω	1/4 W
V1 = Vcc	12 V	12.12 Ω	2 A
V2 = Vee	12 V	12.08 Ω	2 A

Vin	Vout (Measured)	GAIN (Calculated)	VRi (Measured)	IRi (Calculated)	VRf (Measured)
0.0 V	0.00 V	0.00	0.00 V	0.00 A	0.00 V
0.25 V	-2.55 V	-10.20	0.25 V	0.125 mA	2.57 V
0.50 V	-5.07 V	-10.14	0.50 V	0.250 mA	5.11 V
0.75 V	-7.62 V	-10.16	0.75 V	0.375 mA	7.65 V
1.00 V	-10.16 V	-10.06	1.01 V	0.505 mA	10.18 V

I_v1 = 2.31 mA

I_v2 = -1.645 mA

Pv1 = V*I = 28.00 mW

Pv2 = -19.87 mW

Iv1 + I_v2 = 0.665 mA

If = 0.509 mA

Error = 30.6%

Conclusion
To emulate a sensor's signal we use potentiometer to adjust the voltage to the amplifier. The result will amplify the signal ten times the potential input.