Class Activity (week 7)

Charging and Discharging a Capacitor

Charging and discharging of a capacitor. Voltage in a capacitor is 90 degree lag of its current.

I = C dV(t)/dt  or V(t) = ∫I*dt + V(0).

Figure 1: Relation between The Current and Voltage of a Capacitor

Figure 2: Sinusoidal and Triangular wave input Voltage of a  Capacitor and its Current

According to the figure 2, current of a capacitor is derivative of its voltage.

Figure 3: A series-parallel circuit

When a voltage source at t = 0 connects to a empty capacitor, capacitor will act as a short circuit

(wire). Also, a completely charged capacitor acts as a open circuit.

Figure 4: Mathematical Relation of  Discharging a Capacitor

Figure 5: Schematic of Voltage an Ideal Capacitor

An Ideal capacitor keeps its voltage for a long time, but an ideal capacitor lose its voltage gradually.

Thus, an ideal capacitor is modeled similar to a capacitor parallel with a resistor. When resistance of

the resistor is equal with infinity, the capacitor is an ideal capacitor.

Figure 6: Schematic of Charging and Discharging a Capacitor

Voltage-Current Inductor Relations

         A nonideal inductor has a significant resistance, as shown in Fig. 1 because a inductor is

made of a conducting material such as copper, which has some resistance. This resistance is called 

the winding resistance (Rw), and it appears in series with the inductance of the inductor. Winding 

resistance is usually very small. A nonideal inductor also has a winding capacitance due to the

capacitive coupling (Cw) between the conducting coils. A capacitive coupling is very small and can 

be ignored in most cases, except at high frequencies. 

Figure 1: A Schematic of a non Ideal Inductor

Figure 1: Equivalent of Series- parallel Inductors

Inductor voltage difference is proportional with derivative of its current. VL(t) = L di(t)/dt ,


VL(t) = Vmax * e^(-Rt/L) , Time Constant = L/R

A inductor acts the same open circuit at t = 0s, and also it acts the same short circuit at t = ∞.

Passive RC Circuit Natural response & Inductor Voltage-Current Relations

     In this lab time constant of a capacitor is determined. R2 = 2.2 KΩ is parallel with C = 22 μF and

R1 = 1 KΩ is series with this two elements. Also, a 5V square voltage with a 1 KHz frequency is

connected to the circuit.

Figure 1: Calculation of Time Constant 

Figure 2: Calculation of a Real Time Constant

Figure 3: A Discharge Graph of a Capacitor

After 1 time constant, voltage a capacitor became Vmax/e = 0.37*V(max). According to the y-axis

in the above graph, 1 volt is equal with 27 mm so 94 mm is equal with 3.48 v = V(max).

V(max) = 3.48 v so V = 0.37*(3.48) = 1.3 v. Also, on the x-axis every 50 ms is equal with 50 mm.

1.3 Volts is correspond with 35 mm on the x-axis. This imply that time is equal with 16 ms. This

time is time constant.  Percent Error = [16 - 14.9 / 14.9 ]*100% = 7.4%

Figure 4: A Schematic of Calculation of Time Constant

Video 1: Discharging of a Capacitor

Video 2: Discharging of a capacitor 

 Inductor Voltage-Current Relations

This lab is similar to above lab. Instead of a capacitor, a inductor is used. A triangular input voltage

with frequency = 1 KHz, amplitude = 1v, and offset = 0v is connected to circuit.

Figure 1: Calculation of Time Constant

R1 = 0.983 KΩ , R2 = 2.17 KΩ , L = 1 mH , Time Constant = L/Req , Time Constant = 10^(-3)/0.677

Time Constant (Theory) = 1.48 μs     (of course, resistance of the inductor is ignored.)

Figure 3: Schematic of a Inductor circuit

 All of quantity such as resistor (R1) voltage difference, total current, and inductor current have

triangular wave. According to, VL(t) = L dI(t)/dt , inductor voltage difference is square.

Figure 3: A Real Graph of a Part of a Real Inductor

After 1 time constant, voltage a inductor grow up to [1-(1/e)]*Vmax = 0.63*V(max). According to

the y-axis in the above graph, 2 mv is equal with 15 mm so 45 mm is equal with 6.0 mv = V(max).

V(max) = 6.0 mv so V = 0.63*(6.0) = 3.8 mv. Also, on the x-axis every 1 μs is equal with 27 mm.

3.8 mv is correspond with 40 mm on the x-axis. This imply that time is equal with 1.48 μs. This

time is time constant.  Percent Error  is zero.

Capacitor Voltage-Current Relations

      A series RC circuit is designed with a variable voltage in its input. In the series circuit total

current is equal with resistor current, and also it is equal with the capacitor current.

Part a: a sinusoidal wave, V(t), with frequency = 1 KHz, and amplitude  = 2v, and offset = 0v is

connected to the input of a RC circuit. R = 100 Ω  , C = 1 μF

Figure 1: An sinusoidal Input with f = 1 KHz , Vin = 2V

According to the figure 1, one cycle is equal with 1 ms, this is period of sinusoidal wave. So f = 1/T,

f = 1 KHz. Channel 1 of oscilloscope (C1) shows the resistor voltage difference, and channel 2 (C2)

shows the capacitor voltage difference. Of course, the circuit current is in phase with the resistor

voltage difference, but the capacitor voltage difference is 90 degree lag of the circuit current.

Therefore, according to the figure 1, C2 is started at maximum and C1 is stared at zero. This show 90

degree difference between current and voltage of the capacitor.

Figure 2: Schematic of  a Series RC circuit

Figure 3: An sinusoidal Input with f = 2 KHz , Vin = 2V

Part b: a sinusoidal wave, V(t), with frequency = 2 KHz, and amplitude  = 2v, and offset = 0v is

connected to the input of a RC circuit. R = 100 Ω  , C = 1 μF

Figure 4: A Triangle Input Voltage f = 100 Hz, Amplitude = 4v

According to the figure 4, one cycle is equal to 10 ms, so f = 1/T, f = 1/ 0.01 s , f = 100 Hz

A triangular input voltage with frequency = 100 Hz, and amplitude = 4v, and offset 0v is connected

to the series RC circuit. The resistor voltage and current is triangular (channel 2, C2), and also the

total current is triangular shape. The capacitor voltage is square (channel 1, C1) because derivative

of a square function is a triangle function.

Temperature Measurement System Design

     Temperature Measurement System Design

         A simple temperature measurement system is designed by a Wheatstone bridge circuit and a

difference amplifier. A thermistor as a variable resistor is used in the Wheatstone bridge circuit in

order to show temperature. In other words, the Wheatstone bridge converts the resistance variation of

the thermistor to a voltage variation. Also the op-amp is able to show the temperature by a DC

voltage in its output.

Figure 1: Wheatstone Bridge & an Differential Op-Amp

 The output of the Wheatstone bridge circuit, Vab = Va - Vb, is very small and the differential

amplifier in order to increase this voltage is chosen. The key point, ratio of R2 to R1 is chosen less

than one because output voltage (Vo = 5V) does not go in saturation region.  Vo = (R2/R1)*Vab 

Figure 2: A Schematic of Real Temperature Measurement System

The output voltage from the system is between zero and +/- 20 mV at room temperature (25 C).

Movie 1

When the thermistor becomes hot, the output voltage of the differential amplifier will change.

Class Activity (week 6)

In this week many different kinds of Op-amps are introduced by the Professor Mason.

Figure 1: Voltage as a Time Function

When g(t) = 2.5 +2.5*sin(wt)  , this means first two parts are added and then those are multiplied.

When g(t) = 2.5*(1 + sin(wt) ) , this means first two parts are multiplied and then those are added.

Mathematically there is no different between situation 1 and 2, but electrically those have different 

meaning.

Figure 2: The Sum of Two voltages

The summation, subtraction, and multiplication of two voltages or currents are as the mathematical 

functions. 

Figure 3: The Input and Output Voltages an Inverting Op-amp

The gain is bigger than one, it is negative. When Vin = - 100 mV, Vo = (-3)*(-100) = 300 mV.

When Vin = 100 mV, Vo = -300 mV, but the lowest saturation voltage of op-amp is zero, so Vo = 0.

Figure 4: The Input and Output Voltages an Non-Inverting Op-amp

Vo = (33/133)*(1.33)*V2 , Vo = (0.33)*( V2)

The gain is less than one, but it is positive. When Vin = 5V, Vo = (0.33)*(5) = 1.65 V

When Vin = -5 V, Vo = (0.33)*(-5) = -1.65 V, but the lowest saturation voltage of  the non-inverting

op-amp is zero, so Vo = 0. 

Figure 5: A Non-Inverting Op-amp

Gain = Vo/Vin = 1 + (Rf/Rs) , If Rf << Rs → Gain = 1. (This is a buffer)

Figure 6: A Inverting Op-amp

Figure 7: A Summing op-amp

The output voltage is function of two input voltages. Also, output voltage is weighted by R3/R1 and 

R3/R2. In other words, the output voltage is a weight function.

Figure 8: A Summing Amplifier

If R1 = R2 = 2*R3  →  Vo = - (Va + Vb)/2. The output voltage is average of its input voltages.

Figure 9: A Real Circuit of an Summing Amplifier

Figure 10: A Summing Amplifier with Three input 

Figure 11: A Non-Inverting Summing  Amplifier

The gain of inverting amplifiers are A1 and A2. Also, the output voltages of these amplifies are the 

input voltages a summing amplifier. 

Vo1 = - (6/2)*V1 , Vo2 = -(8/4)*V2 , Vo = (-10)*[(Vo1/5) + (Vo2/15)] , Vo = 6*V1 + (4/3)*V2

Figure 12: A Cascade Amplifier

Vin = 20 mV  →  Amplifier → Vo = 350 mV

Gain 1 = 1 + (12/3) , Gain 2 = 1 + (10/4) , Total Gain = A1*A2 , Vo = (A1*A2)*Vin

Vo = [1 + (12/3)]*[(1 + (10/4)]* Vin , Vo = 350 mV

Operational Amplifiers

An OP27 operational amplifier is used in this lab in order to show an inverting op-amp.

Gain = V(out) / V(in) = - (R2/R1). If R1 is equal R2, output voltage V(out) is equal negative of

input voltage V(in). An op-amp with gain = -2 is provided. Also a DMM in order to measure the

input and output voltage is used.

The measured resistance of R1 and R2 are: R1 = 2.1 k , R2 = 4.3 k So Gain = - 4.30/2.10 = -2.05

Figure 1: The schematic of a Real inverting Op-Amp

V(in)	V(out)
-3.49	3.46
-2.99	3.46
-2.49	3.46
-1.99	3.46
-1.49	3.01
-0.99	2
0	0
0.49	-1
0.996	-2
1.50	-3
1.99	-4.01
2.49	-4.23
2.99	-4.23
3.49	-4.23
3.99	-4.23

                        Data Table: 1

Figure 2: The Graph of Output Voltage vs Input Voltage

The saturated voltages are equal to the negative and positive power supplies.

V(in)	V(out)
-1.49	3.01
-0.99	2
0	0
0.49	-1
0.996	-2
1.50	-3
1.99	-4.01

                Data Table 2: Linear part 

According to the data table 2, gain is slope of the linear part of the graph output voltage vs input

voltage.           V(out) = - 2.0143 V(in) + 0.0039      ,     Gain = - 2.0143

percent error of gain = [(2.05 - 2.01)/(2.05)]* 100% = 1.95%

Figure 3: A Schematic of the inverting op-amp

Class Activity (week 5)

             An operational amplifiers (Op-Amp or Opamp) is a DC-coupled high-gain electronic voltage amplifier. The op-amps are used to do mathematical operations in many linear, non-linear and frequency-dependent circuits. An op-amp output voltage is typically hundreds of thousands of times larger than the voltage difference between its input voltages. An op-amp consists of five terminals. Two terminals for input consists a non-inverting input (+) with voltage V+ and an inverting input (–) with voltage V−. One output terminal, and two terminals for power supply consists of positive power supply (VS+) and negative power supply (VS−). An ideal op-amp has infinite input resistance (R_i) and gain (A). Also it has zero input current (i_in) and output resistance (R_o). A realistic op-amp has finite input resistance and gain. Also it has a non-zero input current and output resistance. An op-amp has a linear region and two saturated regions. In saturate region output voltage is equal negative power supply or positive power supply. When input voltage increases in saturate region, output voltage will be constant.

Figure 1: A Real Inverting Op-Amp

Figure 2: A Real Inverting Op-Amp

The inverting op-amp gain is negative. Gain = V(out) / V(in) = - (R2/R1) 

Figure 3: A Real Op-Amp