Sunday, June 7, 2015

May 28

Signals with Multiple Frequency Components

In this lab, we will calculate the magnitude response of an electrical circuit and use this information to infer the effect of the circuit on some relatively complex input signals.

 There is an 90 degree angle delay of the output signal.

As the frequency increases, the output power decreases. 

May 26

Apparent Power & Power Factor

In this lab, we learned the use of apparent power and power factor to quantify the AC power delivered to a load and the power dissipated by the process of transmitting this power.
for  three different cases:
  1. When transmission line resistance is 10 Ω (actual = 10.3 Ω)
  2. When transmission line resistance is 47 Ω (actual = 47.2 Ω)
  3. When transmission line resistance is 100 Ω (actual = 98.6 Ω)

This is how the circuit was connected.

After all the calculations, these are how the power dissipated when R = 10 , 47, and 100 ohms.

 case 1 R = 10ohms
 Case 2 R = 47ohms
 Case 3 R = 100 ohms

 When resistance of the transmission line RT = 47.2 Ω, with 1uF capacitor in parallel with 1mH inductor
When resistance of the transmission line RT = 98.6 Ω, with 1uF capacitor in parallel with 1mH inductor

When R increases, Apparent Power decreases.

Thursday, May 21, 2015

May 14

Inverting Voltage Amplifier

In today's lab, we will be concerned with the steady-state response of an inverting voltage amplifier to sinusoidal inputs.

 At f = 100, 1000, and 5000 we calculated the gains and angles for the output.

This is how the circuit is connected.



This is the input/output data at f = 100
The angle is 27.7 and the gain is 0.87
This is the input/output data at f = 1000
The angle is 79.2 and the gain is 0.191


This is the input/output data at f = 5000
The angle is 90 and the gain is 0.05

As the frequency increases, the angle increases, and the gain decreases. 

May 12


Phasors: Passive RL Circuit Response

In today's lab, we will be concerned with the steady-state response of electrical circuits to sinusoidal inputs. We will measure the gain and phase responses of a passive RL circuit and compare these measurements with the calculations. 

 For w = 4, we found Zr = 10, Z1H = 4j, Z0.5H = 2j, Z 0.1F = -2.5j
We found Ix = 7.5<108.43


The picture shows the calculation is correct, at this frequency.
w = wc.

May 7


Impedance

In today's lab, we measure impedances of resistors, capacitors, and inductors. The measured values will be compared with the expectations values.

Zc = 1/(jwc)
Zl = jwlL
Zr = R

Zseries = Z1 + Z2 + Z3 .....
Zparallel = 1/ ( 1/Z1 + 1/Z2  + 1/Z3 + .......)


We setup three circuits each with a 47 ohm resistor in series with a resistor, an inductor, or a capacitor.
 From left to right: RC, RL, RR circuit.

There is no phase shift for the RR circuit.

 For the inductor the current lead the input voltage by around 90 degree. The frequency have affect on the angle.

For the capacitor, the input voltage lead the voltage across the capacitor by around 90 degree. The frequency have affect on the angle.



April 30

RLC Circuit Response

This lab will emphasize modeling and testing of a second order circuit containing two resistors, a capacitor, and an inductor.
a = R/2L = 550000,  Wo = 1/sqrt(LC) = 316228.
Since a > Wo the circuit is over damped


 With a square wave input, the voltage across the resistor R2 shot over the input voltage. then drop back down. It matches the prediction of an over damped circuit.



The pictures show how the circuit is connected.

Wednesday, May 20, 2015

April 28

Series RLC Circuit Step Response

In this lab, we will build and test a RLC second order circuit in series.

 C = 460nF, L = 1uF, R = 1.3ohm.
a = R/2L = 5.5*10^-5 Wo = 1/sqrt(LC) = 1.459*10^-6. a > Wo.
The circuit is under damped.

To make a = Wo which is critically damped, we need to use a 3.3uF capacitor.

The three elements are connected in series in RLC order.


As expected, the voltage across the capacitor shot beyond the input voltage then oscillate down to meet the input voltage.

April 21

Inverting Differentiator

In this lab, we will build and test a inverting differentiating circuit with an op-amp and a capacitor.

Vo = - RC dVin/dt
After analyzing, we found Vo = 2(pi)RCAfsin(2pi*f*t)

f = 500 Hz a = 1V,
Vo = 0.69sin(1000pi*t)

f = 1000Hz a = 1V
Vo = 1.39sin(2000pi*t)

f = 2kHz a = 1V
Vo = 2.78sin(4000pi*t)

Comparing the calculated values with the experimental values. The calculations are correct. Picture is missing.

April 16

Passive RC Circuit Natural Response

In this lab, we examine the natural response of a simple RC circuit. 


For part a, we disconnected the power supply manually. The capacitor starts discharging within the circuit only containing C and R2. 
The time for Vc to drop to 0.368V0 is 0.0484s. t = RC.

For part b, as soon as the power supply is disconnected, the wire is connected back to another loop. The capacitor starts discharging through two resistors in parallel. t = (R1R2)/(R1+R2) * C = 0.015125s.

The experiment agrees with the calculations. But a picture is missing. 

April 14

Capacitor Voltage - Current Relations

In this lab, we measure the relationship between the voltage across and the current passing a capacitor.

This is what we predicted for Ic and Vc.
The current through a capacitor is i=Cdv/dt.
The voltage across a capacitor is v = 1/c * int(idt)

When passing sine wave through the capacitor, the current lacks behind the input voltage for around 90 degree. It's not 90 degree because the frequency of the input can have effect on the angle of both Vc and Ic.

When passing triangle wave through the capacitor, the current also lacks behind the input voltage for around 90 degree. It's not a square wave as we predicted because there is latency inside the capacitor. Also, it's not 90 degree because the frequency have effect on the angle.

April 9

Temperature Measurement System Design

In this lab, we will design a simple temperature measurement system which uses a difference amplifier to increase the overall sensitivity of the thermistor.



The change of resistance for the thermistor is 3Kohms. 
With the 10Kohms resistors set up for the wheatstone bridge.
We are able to boot the sensitivity of the thermistor to beyond 0.2V/degree C.

This is how the circuit is setup.


The video shows the measurement system is very sensitive.

Thursday, April 2, 2015

March 31st

Summing Amplifier
In this lab, we implemented a simple op-amp circuit to perform summation.
 The picture shown how the op-amp was connected to the circuit and all the resistances for the resistors. By connecting two voltage supply into the same terminal, we get a summation effect.
Vo = -R3(V1/R1 + V2/R2)
The op-amp has an offset of -12 mV.

 This was how the circuit looks like.


Difference Amplifier
In this lab, we implemented a simple op-amp circuit to perform subtraction.
 The picture shown how the op-amp was connected to the circuit and all the resistances for the resistors. By connecting two voltage supplies into both positive and negative terminals, we get a subtraction effect.
Vo = (R2/R1 +1)R4/(R3+R4)*V2-R2/R1*V1
The outputs which has an arrow pointing to it have reached the saturation voltage.
This is a picture of the circuit.

Monday, March 30, 2015

March 17

Time-varying Signals
In this lab, we learned how to use the arbitrary waveform generator to generate time-varying signals and use an oscilloscope to measure the signals.

The picture shows how the circuit was connected. We set that R1=R2, so we predicted that Vout is 1/2 Vin for all 3 different waveforms.

This is a picture of the circuit. 

This is the input waveform for the sin wave. Vmax = 2V.
This is the output waveform for the sin wave. Vmax =1V.



This is the output waveform for the square wave.

This is the output waveform for the triangle wave.

All the output waveforms have  their maximum voltage output as 1/2 the input.



A BJT Curve Tracer.
The purpose of this lab is to investigate the collector current vs collector voltage of the BJT. 

This is the schematic circuit diagram.

 We plot channel 1(Vce) on the X axis, and channel 2 (Ic) on the y axis. Current Ic can be obtained from dividing C2/100 ohm
There is a linear region in between the threshold voltage and saturated voltage.