Lab 9 - Capacitor Charging/Discharging

Objective:
- To test a charge/discharge system that uses a 12 V DC power supply
- Verify if the system charges within 20s with a  resulting stored energy of 2.5 mJ
- Verify if the system will discharge the 2.5 mJ in 2s.

Pre-Circuit Tasks:
The circuit below shows the simple schematic of the charging/discharging system on the capacitor.  To charge/discharge the capacitor, the connected to the positive side of the capacitor is placed in its respective positions.  Meaning the connected cable will be attached to the CHARGE position when the capacitor is to be charged and changed to the DISCHARGE position to discharge.  The rate at the which the energy is transferred is controlled by C, Rcharge, and Rdischarge.

Before determining the values of the components, the expressions for the Thevenin voltage and resistance for the charging and discharging circuits were calculated.

Calculation for Thevenin Expressions

Calculations for Charging/Discharging Circuit Components:

Charging Circuit Component 

Discharging Circuit Component 

The results for the component values:

C = 34.7 µF 

Rcharge = 121212 Ω

Rdischarge = 12121.2 Ω

The Circuit:

Schematic of the Actual Circuit

For each circuit, an oscilloscope was used to show the movement of energy within the circuit.  

Charging Circuit

Charging Circuit

The charging process for the capacitor was about 20 seconds.

The Vfinal obtained after 20secs was about 11.96V.

With the value of Vfinal, RLeak was determined to be 0.145 µΩ.

Discharging Circuit

Discharging Circuit

The discharging circuit did not completely dissipate all its energy within 2 secs. 

Follow up Question

Practical Question:

Lab 8 - Practical Ingegrator

Objective:

Sketch the input and output waveforms for 1kHz sine wave, triangle wave, and square wave inputs.

Explain the purpose of the 10 MΩ resistor's usage in the circuit

Explain the effects of removing the 10 MΩ  from the circuit.

The circuit schematic below was constructed by Professor Mason.

It contains a 100k Ω resistor, a 10MΩ resistor, a 0.01µF capacitor, and an Op-Amp

Circuit Schematic (Constructed by Professor Mason)

The images of the wave forms are displayed on an Oscilloscope.
Red waves represent the Input waves.
Yellow waves represent the Output waves.

Output: Sine Wave

Output: Square Wave

Output: Cosine Wave

In the circuit, the 10MΩ resistor is used to counter any feedback from the DC voltage.  This means that if the resistor is not present within the circuit, the output waves would not be "nice" wave forms but a bunch of irregular looking waves that may contain a lot of noise and spike at random points. 

Lab 7 - Practical Signal Conditioning

Objective:
Use an Op-Amp in a scaling and level-shifting circuit to change the output signal of an electronic temperature sensor to produce a temperature reading in degrees Fahrenheit.

Pre-Circuit Information/Calculations:
For the lab, the national semiconductor LM35 is used to represent the temperature in degrees Celsius.  The conversion rate for the LM35 to degree Celsius is 10 mV/°C.  The LM35 is tested to see if it is working properly by following the schematic below.

Schematic for to check LM35

Once the LM35 has been tested, the following circuit is to be constructed after determining the resistors needed to satisfy the equation to convert the temperature from Celsius to Fahrenheit.

Schematic for the Circuit

The equation to convert from Celsius to Fahrenheit:

TF = 1.8TC + 32 (Original)

TF = 1.8TC + .32 (LM35 equvialent)

The temperature conversion equation in terms of circuit elements:

VF = (1 + R2/R1)VC - R2/R1*Vref

Calculation for Circuit Components

In the circuit element equation, the 1.8 from the temperature equation is represented by (1 + R2/R1).  This yields the ratio value of R2/R1 which is 0.8.  With the ratio obtained, the value of Vref can be determined by setting R2/R1*Vref = 0.32.  This yields the value of Vref to be 0.4V.  The values of R2 and R1 are determined by making sure that the ratio of the two resistors is about 0.8. 

Values of Components:

Vref = 0.4 V

R1 = 2144 Ω

R2 = 1765 Ω

The Circuit:

The Circuit

For the circuit, three power supplies were used.  Two were set up at around 9V to power the op-amp and the LM35.  The other power supply was to supply the 0.4 V for Vref.  The voltmeter read about 77mV for the temperature of the room in Fahrenheit.  

LM35 value

The value of the LM35 was about .237 mV, which is about 24°C. Plugging this value into the Celsius to Fahrenheit conversion equation:

TF = 1.8(23.7) + 32 = 74.66 °F

The Fahrenheit temperature obtained by the voltmeter for the circuit is about 77.0 mV, which is about 77°F.

The percent error of TF:

Percent Error = (77-74.66)/ 74.66 *100 ≈ 3.13% 

Lab 6 - Op Amps 1

Objective:
The purpose of the lab understanding how an inverting op-amp functions in a circuit.

Pre-Circuit Information:
Before the circuit can be assembled, the values of resistors connected to the op-amp and sensor must be determined.  The resistors connected to the op-amp are Ri and Rf.  The sensor has resistors RX and RY (the potentiometer/variable resistance).

Given :
Output range: 0 to +1V
Gain between 0 and -10V
Current through sensor cannot exceed 1 mA
Power Supply current into op-amp cannot exceed 30 mW
For the sensor, the Vsupply = 12V & VNode between Rx & Ry = 1V

Calculation:

Op-Amp

Schematic without sensor &  Ri

Rf

Sensor

Schematic with sensor &  RX

 RX  with a 1/8W Power Rating

RY & RTH

Ri = 1000Ω

Rf = 10000Ω

RX = 110000Ω

RX @ 1/8W = 1152Ω   ≈ 1300Ω 

RYmax = 108.33Ω   ≈ 130Ω

RTH = 118.182Ω

Measurements of the Components:

Note: Used RX = 10000Ω instead of calculated value since the calculated value did not allow the circuit to function correctly

The Circuit Model:

The Data collected by adjusting the potentiometer (RY):

After these measurements, the current being supplied to the op-amp was measured for each power supply:

IV1 = 1 mA

IV2 = 1 mA

Using the measurements of the current through each power supply, the power being supplied was by the power supply was calculated. 

The current through Rf 

The power supplied into the op-amp from the power supplies were PV1 = 12.08 mW  and PV2 = 12.13 mW. Since the power supplied are below 30 mW, the power supply constraint was satisfied.  A way to reduce the power drawn by the power supply without changing the gain of the op-amp is to increase the values of the resistors.