Charging and Discharging Capacitros

The purpose of the experiment is to learn how to control capacitor charge and discharge times using a nonideal case.

Prelab:
We first build the Thevenin expressions for the charge and discharge circuits below to use later.

Charging: Rth = Rleak*Rcharge/(Rleak + Rcharge), Vth = Rleak*Vs/(Rcharge + Rleak)
Discharge: Rth = Rleak*Rdischarge/(Rleak + Rdischarge), Vth = 0.

The setup is to use 9V to charge for 20s with 2.5 mJ, and discharge the 2.5 mJ in 2s.
Doing some math, we find:
C = 2*U/V^2 = 0.0617 mF
Charging: 5τc = 20s
Rc = 4/C = 64.8 kΩ
The power Pc = V^2/R_c =1.25 mW which is under the limit of 1W.

Discharging: 5tau_d = 2s
Rd = 2/(5*C) = 6.48 kΩ
The power Pd = 12.42 mW which is under the limit of 1W.

Procedure:

 Build the circuit according to the pre-calculated values in the prelab questions.

Capacitors combined in parallel to reach desired value

Entire circuit with resistor boxes

Analysis:
Logger pro was used to measure the change in voltage over time.

Capacitor charging

The peak voltage V_f = 8.25 V due logger pro's limit of 8Vs and the effect of the Rleak.
The time is estimated to be around 18s.
Rleak = Rcharge /(V_s/V_f - 1) = 712.8 kΩ

Capacitor discharge

Starting from Vf, the discharge time is around 2s.

Questions:
Using the Thevenin equations,
1. Rcth = 59.4 kΩ, Vcth = 8.25V
 2. Rdth = 6.42 kΩ, Vdth = 0V

3. 0.6321*V_f = 5.215 V
5.215 V is around 4.6s
 

τc = RC = 4.6s
R = 4.6/C = 74.55 kΩ
Error = 15%

Practical Question:
1)       U = (1/2)CV^2
     Ceq = 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 = Ceq
        C = 1/2Ceq = 0.71 F

Conclusion:
 We had successfully reach a charging time of about 20s, and a discharging time of about 2s. The experiment was successful, and the pre-lab calculations were correct. The result in this lab proved that it was possible to control the times.

Temperature sensitive OP-AMP circuit

The purpose of this lab is to amplify a thermosensor to a specific output range.

Prelab:
We start with the LM35 temperature sensitive electronic device which produces 10 mV/°C.
We want to measure between 15 °C to 35 °C.
Using a difference amplifier, we shift and amplify the range of 150 mV - 350 mV to 0V - 5V.

To do so, we have V_2 = LM35, V_1 = 150 mV.
Our gain is 25, so R_f = 150 kΩ, and R_i = 6 kΩ.

Procedure: 
We start by building the circuit.

V_1 = Left pot, V_2 = Right pot

However to test the op amp, we made V_2 = 350 mV to see if the analysis was correct.

We then measure V_out = 5.05 V

Conclusion:

The lab was only a minor success. We were able to use our opamp to amplify 350 mV correctly into 5V by using power supply and voltage dividers to represent the LM35 at 35 degree C. However, when the LM35 was placed into the circuit, we failed to get the expected 5V output. Although there was not much evidence to suggest the LM35 was broken, it was the only possible conclusion for why the circuit failed when the LM35 was placed into it.

Lab 8 OpAmpII

          In this experiment, we will be exploring the relationship between the output voltage, input voltage, and the ratio of resistors (Rfeedback and Rin). Meanwhile, we will probe the output current to testify whether Kirchoff's current law holds true in OpAmp.

We will be using an inverted op-amp with the rails' voltages at -12V to 12V.

 

A seperate voltage divider is to replace Vsen to divide a 1V input voltage out of a 12V power supply.

 

 

How to set up the voltage divider:

 



1) Obtain a potentiometer, and some wire.

2) Connect 1 of the potentiometer side pin to +12V, another side pin to ground, and the middle pin in series to the input resistor.

3) Measure the potential drop between ground and the middle pin, adjust the potentiometer untill you obtain the voltage you for Vsen.

Since we want to get a gain of -10, we will need the following resistor according to the op-amp's equations:

Resistors	Nominal Values, kΩ	Measured Values, kΩ
R­f­	10kΩ	9.91kΩ
R­in­	100kΩ	97.7kΩ

 

Finding Iop:

 

If we say Vsen is 1V, and our gain is -10, the current Iop = -10V/Rf since the feedback branch is the only way that Iop can go through.

 

                    VRf=Vout=-10V=Iop(Rf)     ===>>          Iop= -10/100kΩ = -0.1mA

 

 

After measuring the voltage across different parts of the circuits, we obtained the following results:

V­in­ desired	V­in  measured­	V­out measured	V­Rf  measured	I­op  calculated
0.25V	0.24V	-2.46V	2.46V	0.0246mA
0.5V	0.5V	-4.90V	4.87V	0.0490mA
1.0V	1.00V	-10.04V	9.86V	0.1004mA

 

The current in the rails (@ Vin =1V) are the following:

 

ICC = 0.874mA                 IEE=-0.981mA

 

The sum of the the rail currents = ICC+IEE

                                                                                 =(0.874-0.981)mA

                                                   =-0.107mA.

 

From circuit analysis, Iop =ICC+IEE ; Where Iop = -0.1004mA and ICC+EE=-0.107mA.

Since Iop is about the same as ICC+EE, we conclude that KCL holds true in op-amps.

 

 

Power supplied by each 12V power source at the operating point:

 

P-12=-0.981mA*-12V =11.77mW

P12=0.874mA*12V=10.49mW

 

Now we replace the open circuit with a 1kΩ resistor and measure the following :

 

­V­in desired	V­­out measured	­V­Rf measured­	­I­op calculated­	­I­cc­ ­measured	I­EE measured
1.0V	-9.99V	9.72V	-0.102 mA	0.887mA	-0.987mA

Sum of rail's current:

(0.887-0.987)mA= -0.100mA ≈ -0.102mA ≈Iop   -----> KCL still holds true.

 

P-12=-0.987mA*-12V =11.84mW

P12=0.887mA*12V=10.64mW

 

 

Bouns:

 

Use the resistance box to find Rf so that the gain becomes -5.

 

By Theory, we need a Rf of 50kΩ to obtain a gain of -5.

As we measure the actual resistance in the resistor box after setting it to 50kΩ, the actual reading gives us 49.9kΩ.

 

 

­V­in desired	V­­out measured	­V­Rf measured­	­I­op calculated­	­I­cc­ ­measured	I­EE measured
1.0V	-5.03V	4.99V	-0.101mA	0.884mA	-0.985mA



 

 

 

Conclusion:

 

The gain of an inverted op-amo is always the negative ratio of -Rf/Rin. So, if the gain is -5, -Rf/Rin has to be -5. At the same time, the current at the operation point is the sum of the current from the two rails into the op-amp (KCL is always true).

 

 



 

Lab 8 OpAmpII

          In this experiment, we will be exploring the relationship between the output voltage, input voltage, and the ratio of resistors (Rfeedback and Rin). Meanwhile, we will probe the output current to testify whether Kirchoff's current law holds true in OpAmp.

We will be using an inverted op-amp with the rails' voltages at -12V to 12V.

 

A seperate voltage divider is to replace Vsen to divide a 1V input voltage out of a 12V power supply.

 

 

How to set up the voltage divider:

 



1) Obtain a potentiometer, and some wire.

2) Connect 1 of the potentiometer side pin to +12V, another side pin to ground, and the middle pin in series to the input resistor.

3) Measure the potential drop between ground and the middle pin, adjust the potentiometer untill you obtain the voltage you for Vsen.

Since we want to get a gain of -10, we will need the following resistor according to the op-amp's equations:

Resistors	Nominal Values, kΩ	Measured Values, kΩ
R­f­	10kΩ	9.91kΩ
R­in­	100kΩ	97.7kΩ

 

Finding Iop:

 

If we say Vsen is 1V, and our gain is -10, the current Iop = -10V/Rf since the feedback branch is the only way that Iop can go through.

 

                    VRf=Vout=-10V=Iop(Rf)     ===>>          Iop= -10/100kΩ = -0.1mA

 

 

After measuring the voltage across different parts of the circuits, we obtained the following results:

V­in­ desired	V­in  measured­	V­out measured	V­Rf  measured	I­op  calculated
0.25V	0.24V	-2.46V	2.46V	0.0246mA
0.5V	0.5V	-4.90V	4.87V	0.0490mA
1.0V	1.00V	-10.04V	9.86V	0.1004mA

 

The current in the rails (@ Vin =1V) are the following:

 

ICC = 0.874mA                 IEE=-0.981mA

 

The sum of the the rail currents = ICC+IEE

                                                                                 =(0.874-0.981)mA

                                                   =-0.107mA.

 

From circuit analysis, Iop =ICC+IEE ; Where Iop = -0.1004mA and ICC+EE=-0.107mA.

Since Iop is about the same as ICC+EE, we conclude that KCL holds true in op-amps.

 

 

Power supplied by each 12V power source at the operating point:

 

P-12=-0.981mA*-12V =11.77mW

P12=0.874mA*12V=10.49mW

 

Now we replace the open circuit with a 1kΩ resistor and measure the following :

 

­V­in desired	V­­out measured	­V­Rf measured­	­I­op calculated­	­I­cc­ ­measured	I­EE measured
1.0V	-9.99V	9.72V	-0.102 mA	0.887mA	-0.987mA

Sum of rail's current:

(0.887-0.987)mA= -0.100mA ≈ -0.102mA ≈Iop   -----> KCL still holds true.

 

P-12=-0.987mA*-12V =11.84mW

P12=0.887mA*12V=10.64mW

 

 

Bouns:

 

Use the resistance box to find Rf so that the gain becomes -5.

 

By Theory, we need a Rf of 50kΩ to obtain a gain of -5.

As we measure the actual resistance in the resistor box after setting it to 50kΩ, the actual reading gives us 49.9kΩ.

 

 

­V­in desired	V­­out measured	­V­Rf measured­	­I­op calculated­	­I­cc­ ­measured	I­EE measured
1.0V	-5.03V	4.99V	-0.101mA	0.884mA	-0.985mA



 

 

 

Conclusion:

 

The gain of an inverted op-amo is always the negative ratio of -Rf/Rin. So, if the gain is -5, -Rf/Rin has to be -5. At the same time, the current at the operation point is the sum of the current from the two rails into the op-amp (KCL is always true).