Capacitor Charging/Discharging

Purpose
The purpose of this lab is to understand the concept of RC charging and discharging circuit and then see how a real capacitor works in the circuit.

Background
In the RC charging circuit, the voltage and current across capacitor can be express as V(t)=Vs(1-e^(-t/tau)) and I(t)=Vs/R*e^(-t/tau), where tau is time constant which is Rcharge*C. In addition, in the discharging circuit the voltage and current across the capacitor are V(t)=Vic*e^(-t/taud) and I(t)=-Vic/Rdiscahrge*e^(-t/taud), where taud is time constant in discharging circuit. However, in real capacitor model the capacitor connected to resistor Rleak in parallel, so we must apply Thevenin's theorem to find out the real value of resistance.

Problem
We use a 9V power supply to employ a charging interval of about 20 s with a result stored energy of 2.5mJ, and then discharges that 2.5mJ in 2s.

Approach
So, the value of capacitance can be determined by applying w=0.5*cV^2, which is 61.7 uF. For R(charging), we can use 5 R(charging)*C=20, so R(charging)=64.8k ohms. From the values above, we can determine peak current and power in the charging circuit which are 0.139 mA and 1.25mW. For discharging circuit, 5 tau = 2, so R(discharging)=6.48k ohms. The peak discharge current and power are 1.39 mA and 12.5mW respectively.

Assemble
R(charging) = 65k ohms
R(discharging) = 6.47k ohms
C=63.6 uF
All values above are measured.

Figure1. Build-up Circuit on Breadboard

Figure 2. Build-up Circuit Connected Logger Pro and MAC

Graphs

• Charing

Figure 3. Vc v.s Time When Charging

• Discharging

Figure 3. Vc v.s Time When Discharging

Data Analysis

Vs = 9.01 V

Vfinal = 8.562 V = Vs*Rleak/(Rleak+65k), so R(leak)=1.24 Mohms

The Thevenin equivalent voltage and resistance seen by the capacitor during charging are

Vth=9.01*1.24M/(1.24M+65k)=8.55V and Rth=1.24M and 65k in parallel which is 61.8k ohms, and the Thevenin equivalent voltage and resistance seen by the capacitor during discharging are

Vth=9.01*1.24M/(1.24M+65k)=8.55V and Rth=1.24M and 6.47k in parallel which is 6.44k ohms 

Op Amp Application

Purpose
   The purpose of this lab is to use a scaling and level-shifting op-amp circuit which is LM741to process the output signal from an electronic temperature sensor so that the output can show reading in degrees Fahrenheit converting from Centigrade reading.

Background
   The temperature sensor we have used is AS35 which produces an output voltage that is proportional to the air temperature based on a scale factor of 10mV/degree C. Also, AS35 requires an 4 to 20V to operate.

Pin Connection

Figure 1. Front View of AS35

Figure 2. Top View of AS35

Design
   The converting formula between Centigrade and Fahrenheit readings is TF = 1.8TC + 32. So, we can use a single 741 op-amp to scale and shift a signal. our design is shown as following,

Figure  3. Designed Circuit

By assuming the op amp is ideal, so V+ = V-= Vc. Using nodal analysis on the node which is just above R1, we can write (Vc-Vref)/R1 = -(Vc-Vf)/R2 

==> Vf/R2 = (Vc/R2+Vc/R1)-Vref/R1 

==> Vf = R2*(Vc/R2+Vc/R1)-R2/R1*Vref

==> Vf = (1+R2/R1)Vc-R2/R1*Vref. Comparing the equation to TF = 1.8TC + 32, we can get (1+R2/R1)=1.8 and 0.8*Vref = 0.32, which means Vref = 0.4V; however, Vref has to be negative if the overall shift is to be positive. The design also has constraint which the output current from the op-amp is much less than the maximum allowable output current specified in the data sheet which is 25 mA for LM741.

Assemble

R1=0.999 M ohms

R2=0.812 M ohms (both readings from multimeter), so that the ratio of R2/R1=0.813 which is close to the ideal value, 0.80. Two power supplies, one used as positive 9V and the other used as negative 9V, must connect to AS35 and LM 741 properly. In addition, we use a variable power supply to provide 0.4V as Vref instead of using a power supply and two resistors as a voltage divider. 

Figure 4. Build-up Circuit Connected to Multimeter and Power Supply

Figure 5. The Build-up Circuit

Data Analysis

Vc=0.228V

Vf=0.711V(readings from multimeter)

Theoretical Value of Tf = (1.8*0.228+0.32)*100=73.04 degree Fahrenheit

% error = (73.04-71.1)/73.04*100% = 2.66%

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