Transistor Lab

The purpose of today's lab was to use a transistor to observe the linear relationships between the emitter current and the base current. Additionally we are trying to find Beta.

Our data: where Ib = base current, Ic = emitter current

Ib (mA)	Ic (mA)
0.41	4.10
0.66	4.40
1.29	5.00
1.87	5.50
1.96	5.60

Nodal Analysis

PROBLEM
We want to construct a "reliable" power system - a system that has the ability to survive a casualty or cascade of damage. We model a circuit consisting of 2 power supplies, 2 loads, and cables with cable resistances.

Given the circuit above, we use nodal analysis to find the two unknown voltages V2, V3.


CALCULATIONS
  Nodal equations:
  (V2-Vbatt1)/Rc1 + (V2-V3)/Rc2 + (V2)/RL1 = 0
  (V3-V2)/Rc2 + (V3-V4)/Rc3 + V3/RL2 = 0
where: Rc1 = 100 ohms, Rc2=Rc3= 220 ohms, RL1=RL2=1 kohm, Vbatt1 = 12V, Vbatt2= 9V.
Solving the two equations:  V2 = 10.2 V,  V3 = 8.7 V

Next, we find the current leaving each battery & the power supplied by each battery.

Ibatt1= (Vbatt1-V2)/Rc1 = 0.018 A = 18mA
Ibatt2 = (Vbatt2-V3)/Rc3= 0.014 A = 14mA

Psupp = Vi
Pbatt1= 0.216 W
Pbatt2= 0.126 W

Data
Components Data Table

Component	Nominal Val	Measured Val	Power/Current Rating
Rc1	100 ohms	98.2 ohms	0.25 W
Rc2	220 ohms	217 ohms	.0125 W
Rc3	220 ohms	217 ohms	.0125 W
RL1	1 kohm	976 ohms	.0125 W
RL2	1 kohm	979 ohms	.0125 W
Vbatt1	12 V	12.07 V	2A
Vbatt2	9 V	9.08 V	2A

Experimental Values Data Table

Variable	Theoretical Val	Measured Val	Percent Error
Ibatt1	0.017 A	17.4 mA
Ibatt2	0.0014 A	12 mA
V2	10.2 V	10.32
V3	8.7 V	8.7

Next, we calculate the power delivered by the two batteries:
Pbatt1 - 17.4mA*(12.07V) = 0.210 W
Pbatt2 - 12mA*(9.08V) = 0.109 W

Finally, assume we change the problem: suppose V2=V3=9V. Use the nodal equations to find the required battery voltages Vbatt1, Vbatt2.

Vbatt1 = 9.9 V
Vbatt2 = 10.8 V

Implement the circuit using the new battery voltages and record the achieved node voltages and battery currents.

V2 = 8.96 V
V3 = 8.79 V
Ibatt1 = 9.48 mA
Ibatt2 = 8.24 mA

FREEMAT funtimes

Today's lab was done on FREEMAT. Free tools, gotta love it.

Given the following circuit, use FREEMAT to find the current i3 through R3:

Done. How valuable is that???

Introduction to Biasing

PROBLEM
    Suppose you want to use a 9V Alkaline battery to light to LEDs. The LEDS may be part of an instrument and are being used to indicate whether the device is ON or OFF. LED1 is rated for 5 V and 22.75 mA and LED2 is rated for 2V and 20 mA.

METHOD
    A circuit is to be constructed:

Voltage across LED1 is 5 V w/ current of 22.75 mA , across LED2 is 2 V with current of 20 mA

 We determine the theoretical resistances of LED1 and LED2 using Ohm's Law and the rated current and voltage values for each LED respectively.  RLED1 = 5/.02275=219.8 ohms, RLED2 = 2/.02=100 ohms. Next we determine the values for the bias resistors, R1 and R2 using the current values across each LED. R1=4/.02275=176 ohms, R2=7/.02=350 ohms. Finally, we determine the power dissipated by the resistors - PR1 - 91 W, PR2 - 140 W.


Resistor combinations: R1 = 176 ohms = (1)130 & (1)47 ohm resistors - measure - 176.3 ohms - 1/4 W
                                      R2 = 350 ohms = (1)360 ohm resistor - measure 355.7 ohms - 1/4 W

Data was recorded for three configurations: 1 - both LEDs in the circuit, 2- remove LED2, 3 remove LED1

DATA

Configuration	I LED 1	V LED 1	I LED 2	V LED 2	I Supply
1	14.67 mA	6.8 V	19.62 mA	2.26 V	26.4 mA
2	14.63 mA	6.83 V	x	x	73.3 mA
3	x	x	19.79 mA	2.28 V	64.1 mA.

CALCULATIONS

 a) Capacity of 9V Alkaline batter is approx. 0.6A-hr; however, voltage drops significantly as the battery gets to the "end" of it's life. So let's assume the "useful" life of the battery is 0.2A-hr for a 9V. With both LEDs in the circuit, how long can the circuit operate before the battery voltage goes too low?

Soln: 0.2A-hr = 02.A(3600s)=720  As= 720 C. Since, W=qV , we know there is initally 720(9) J= 6480 J of "useful" energy left in the battery. Additionally, P=W/t >> t= W/P   where P is the total power absorbed (Vi) by the LEDs. Pled1=0.0998 W, Pled2=0.04434 >> P=sum(p1+p2)=0.1441 W. Accordingly, t= 6480/0.1441 J/(J/s)=44956 s*(1hr/3600s)=12.48 hrs.

Answer: Approximately 12.5 hours


b)What was percent error between Achieved LED current and desired value with both LEDs in the circuit? What was the cause of the error?

%err=absval(act. - theo)/(theo) * 100% >>
LED1= 14.67-22.75/22.75*100 = 35.6%
LED2= 19.62-20/20*100 = 1.9%

The error is rather unreasonable. If we analyze our dataset using Ohm's law, we can see that the resistance of the system was different than what we expected. R1 is 463 ohms, which is greater than our Rled1 resistance of 219.8 ohms. Additionally, R2 is 115 ohms, which is again greater than our expected Rled2 resistance of 100 ohms. The error is likely due to the resistance of the cables, which reduced the current flow.

c) Determine the efficiency of the circuit with both LEDs in the circuit.
Pout is power consumed by LEDs, Pin is power supplied by battery
%Eff = Pout/Pin*100% >> 0.1441/0.238*100 = 60.5% efficiency

d) If we repeat the design with a 6V battery - changing the biasing resistors R1, R2, but WITHOUT CHANGING LED parameters - would efficiency go: up, down, or stay the same? What theoretical value of battery voltage would provide the "best" efficiency and what is the efficiency?

Intro to DC Circuits

PROBLEM
Today in lab, we constructed a circuit similar to that above.
The problem is to determine:
   - maximum cable resistance required for load to function normally
   - maximum distance batter and load can be separated using AWG #30 cable
   - distribution efficiency
   - approximate time before the battery discharges

METHOD
We first determined the theoretical value for Rload, given that the load consumes p = 0.144W when supplied   v = 12 V, where R = (v^2)/p = 1 kohm. We took measurements of all components before assembling the circuit, and recorded the data. To measure the current, we placed an ammeter in the circuit; a voltmeter was used to measure the voltage across the load. The circuit was modified:



The RcableTot resistance was adjusted until the voltage across RcableTot read 11V; this value was recorded. Values for voltage across the load, Vload, and current, Iamm.
  - Vload = 10.8 V
  - Ibatt = 11.4 mA
  - RcableTot = 97 ohms

QUESTIONS/CALCULATIONS
A)  Given an Amp-hour rating of 0.8Ahr for the battery, and that Ahr=amp*time >> time=Ahr/amp.
  - Time to Discharge: t=70 hrs

B) The power to the load(power out) and the power to the cable(power lost) need to be determined to find the distribution efficiency.
  - Pload = Pout = 0.1254 W
  - Pcable = Plost = 0.013 W
Efficiency = (Pout/(Pout+Plost))*100 = 91% EFFICIENCY

C) We are not exceeding the power capability of the resistor because our efficiency is reasonable. Accordingly, I can infer that the power capability of the resistor is not exceeded.

D) Resistance of AWG#30 wire is 0.3451 ohm/m. Max distance between battery and load,           dMax=R/p=97/0.3451= 281.1 m

E)

F)