Thevenin Equivalents

   Let assume that we have a circuit that contains multiple sources and loads
We are able to use Thevenin's Theorem to reduce a complicated into a simple circuit such as below with a Voltage and a resistor that represent the complicated circuit.














We are given  a circuit and we need to measure the Thevenin equivalent voltage and resistance as shown below




The Thevenin Resistance can be detremined from R_TH=V_oc/I_sc  where I is the current measured as short circuit at the two points

The given element has the theoratical value of

R_c1	R_c2	R_c3	R_L1	Vs1	Vs2
100Ohm	39Ohm	39Ohm	680Ohm	9V	9V

First we used nodal analysis to find the open-circuit voltage

we got a equation:
(Vx-Vs2)/Rc2 + (Vx-Vs1)/Rc1+Vx/Rcl=0

Vx has the value of 8.64V, which is the Thevenin Voltage


Next, we calculate the voltage V_y in the fugure. Then find the short-circuit current use it to compute the Thevenin resistance
(Vx-Vs2)/Rc2 +(Vx-Vs1)/Rc1+Vx/Rc1+Vy/Rc3=0
Vy=5.11V

Isc=5.11/39=0.131A
Rth=66Ohm

After knowing the R_th and V_th. We can know draw another circuit that contains two resistors in series, which are R_th and R_load. We can use a voltage divider since we know the voltage drop across the R_load is 8V to find the theoretical value of R_load
R_L2=825Ohm


After that we just device the experiment and measure all the components and the value of voltage across different components
 

Component	Nominal Value	Measured Value	Power Rating
R_th(Ohm)	66	66.6	0.3W
R_L2,min(Ohm)	825	822	0.3W
V_TH	8.64	8.64	x

Now, we can build the THevenin Equivalent Circuit on the Breadboard and record data

Configuration	Theoretical Value	Measured Value	Percent Error
RL2=RL2,min	VLoad2=8V	7.99	0.12%
R=L2=infinite	VLoad2=8.64	8.64	0%

Measured Value

Component	Nominal Value	Measured Value	Power Rating
R_c1	100Ohm	99.6Ohm	0.25W
R_c2	39Ohm	39.0Ohm	0.25W
R_c3	339Ohm	4.02Ohm	0.25W
R_L1	680Ohm	679 Ohm	0.5W
V_s1	9V	9.07V	18W
V_s2	9V	8.97V	18W

Building the Original Circuit

 

We obtained the following data


 

Configuration	Theoretical Value	Measured Value	Percent Error
R_L2=R_L2min	Vload2=8.64	8.31V	3.90%
R_L2=infinite	Vload2=8.64	8.62V	0.23%

    We can caculate the power that will be supplied to (absorbed by) R_L2 when its equal to R_th P=V^2/4R=0.28W   Next, we set the R_L2 to different value then measured the load voltage to find the power
 

Configuration	V_load2 (V)	P_load2 (W)
0.5R_th	2.87	0.2496
R_th	8.31	1.0463
2R_th	5.74	0.2496

Day 6- Beta of a BJT

In this experiment, we will figure out the Beta(gain) of the npn bipolar junction transistor
the basic schematic of the transistor is sown

With the relationship of
Ic=a*Ie
Ic=B*Ib
B=a/(1-a)

In this experiment, we set up the circuit as shown below

with a NPN type 2N3904 transistor

we will R1 (variable resistance box) and R2 of 100ohm(measured 98.2ohm) in the circuit

We will measure the change in R1 respect to the change in Ib and Ie

Measured V         Rb(Theoretical)         Rb(actual)            Ib                   Ie

6.05 V                       530 k ohm           529 k ohm          0.011 mA      1.36 mA

6.05 V                       265 k ohm           263 k ohm          0.021 mA      2.81 mA

6.05 V                       176 k ohm           176.6 k ohm       0.033 mA      4.18 mA

6.05 V                       132 k ohm           132.5 k ohm       0.046 mA      5.56 mA

6.05 V                       106 k ohm           106.4 k ohm       0.058 mA      6.98 mA

we can plot the graph with logger pro with Ie vs. Ib graph




Since we know that Ie=(1+beta)Ib
and we know the slope of the line is 116.2
116.2-1 will be the beta value of 115.2

According to the textbook, transistor has some saturated value that after the value, the transistor will acts like open and close switch. Since the kind of transistor we are using(BJT) is usually use as parts in amplifier, we do not want it function as switch. Instead, we want it to act like flow controller that control the desire current we want.  Thus, we are only interesting in the linear portion of the graph of the transistor.