PSpice Tutorial

The first image which i did in the class is just what i followed by the handout of the pspice tutorial. The thing that what i learned is the software need to run as administrator, otherwise people would not find the running system. Also, make sure the nodes are connected correctly. 

#6 Thevenin Equivalents

In this lab we attempted to verify Thevenin's theorem.  The idea is that in a system of multiple power sources and loads we want to be able to easily calculate the affect of replacing one element so we reduce it into it's Thevenin equivalent (a source in series with a resistor).   We took a complex circuit and solved for it's Thevenin equivalent values(Vth =8.643V, Rth = 66ohms) and also for the value of the resistor necessary to produce an 8V drop across the load(R load min = 820 ohm).  Then we set up two circuits to verify - the first was the actual Thevenin Eq circuit and the second was the original circuit.  Our results were very accurate for the Thevenin Eq circuit but not for the original circuit.  This may have been due to a faulty breadboard and the additional wire resistance required for the full circuit.  

Component   Nominal Value           Measured Value Power or Current Rating
RTH                   66 Ω                         66.6 Ω                  0.3 W
RL2,min           825 Ω                         822 Ω                  0.3 W
VTH                   8.64 V                         8.64 V                


Config                 Theoretical Value Measured Value Percent Error
RL2 = RL2,min VLoad2 =8 V         7.98 V                  0.25%
RL2 = ∞Ω          VLoad2 =8.64 V 8.63V                  0.12%

#3 Introduction to Biasing

We have two LEDs in parallel, each with unique voltage ceilings. Since we have only one power supply (also in parallel), we have to find out how to differ the voltage through each of the LEDs. This is called biasing. We can limit voltage across the LEDs by adding resistors to the circuit. These resistors "soak up" some of the parallel voltage so that not all of it is going through our sensitive LEDs.

Given a 9V source and two parallel LEDs with voltage limits of 5V and 2V, respectively, we could calculate that a 220 ohm resistor is needed in series with the 5V LED, and a 100 ohm resistor with the 2V LED. So we set up our circuit as such, and connected a voltmeter in parallel + ammeter in series with the LEDs to take some measurements. We also set up configurations with the individual LEDs.

#1 Model the system on breadboard

Here are our measurements:

Config 1
LED1 Current = 14.5mA
LED1 Voltage = 6.86V
LED2 Current = 21.0mA
LED2 Voltage = 1.65V
Supply Current = 35.4mA


Config 2

LED1 Current = 14.6mA
LED1 Voltage = 6.81V
Supply Current = 14.4mA


Config 3
LED2 Current = 20.7mA
LED2 Voltage = 1.65V
Supply Current = 20.7mA

Based on these results and a hypothetical 9V, 0.2 amp-hour battery, we calculated that the LEDs could be powered for 4.68 hours.

We then calculated the actual vs. theoretical LED current. Our actual current was lower due to the fact that we were using a non-ideal voltmeter in parallel.

Finally, we calculated the efficiency of our setup, which was quite low due to the resistors stealing a lot of our power.

Power out:  0.13412 W

Power in:    0.3186 W

Power lost:  0.18448 W

Efficiency:  42.1%

We were asked about what would happen to the efficiency if a 6V power source had been used instead of a 9V. I reasoned that it would increase, because less power is now going to the resistors (which aren't doing any work). 5V would actually correspond to the most efficient setup, since we're now using a value that allows us to eliminate the most resistance possible (one of our LEDs has a 5V max) and run a resistor only alongside the 2V max LED.

#5 Nodal Analysis

In this lab, we are modeling a redundant power system that has two voltage sources, in case one fails. This system should have breakers (in order to isolate bad components), which we are modeling with 100 ohm and 220 ohm resistors. Finally, we model two loads, each with 1K ohm resistors. Before we set up the circuit, we were able to calculate the voltage across the loads using nodal analysis, solving for the resulting system of equations. Substituting back into our equations, we could calculate current through each of the power sources and the power they supply.

V2	V3	Ibat.1	Ibat.2	Pbat1	Pbat2
10.26 V	8.67 V	17.4 mA	1.50 mA	0.209 W	0.0135 W

Next, we measured the actual values for the resistors, and set everything up on a breadboard. Here is our circuit:

 #1: Built the circuit.

We measured the actual values for the currents through source 1 and source 2 and voltage across the two loads:

 #2: Built the circuit.

 #3 set up the power

Variable	Theroetical	Measured	% Error
I bat 1	17.4 mA	17.9 mA	2.87%
I bat 2	1.50 mA	1.47 mA	-2.00%
V2	10.26 V	10.35 V	0.88%
V3	8.67 V	8.75 V	0.92%

Then, we calculated power delivered by the batteries:

P bat 1 = 217 mW

P bat 2 = 13.3 mW

For the last step, we assumed that V2 and V3 were 9V, and used our previous equations to solve for the corresponding source voltages. To achieve this source voltage drop, we hooked up resistor boxes in series with the two sources and tweaked the resistances until the desired 9V was achieved across the loads. These resistances ended up being R1s = 203 ohms and R2s = 103 ohms.

These were the resulting values for this scenario:

V2	V3	Ibat.1	Ibat.2
9.04V	9.00V	10.0 mA	8.61 mA

#4: step 4 :set both Voltage =9V

#5: Adjust power,Reading required battery voltage=10V

#6 the other one reading 2.9V

Freemat Assignment 4

 

#1: Using the KCL & KVL the equition are given.

#2: Solving Simultaneous Equations with MATLAB

Lab #2: "Introduction to DC circuits"

#1: Build the circuit.

#2: The Power supply(max Voltage 12V; max Current 2A)

#1: Using a multmeter

#1: Measure the voltage across the LED