Introduction to Biasing

Purpose: To light two LED lights with a 9V battery without burning them out. We will use biasing, establishing the correct voltage across and current through a component so it functions properly.



Introduction:

• Using a bench-top power supply as our 9V
• LED1: 5V and 22.75 mA
• LED2: 2V and 20 MA

Using V=IR we find the values of the resistance in the LEDs:

• LED1: 219.78 Ohms
• LED2: 100.0 Ohms

In order to find R1 and R2 required we use the concept that parallel branches have the same voltage. To find the power we use the power equation from the previous blog. (Values shown later)

We choose the available resistors closest to the calculated value. and set up the following circuit:

The currents and voltages were measured in 3 loops

1. all components active
2. LED 1 active
3. LED 2 active

Data:

Calculations;

Introduction to DC Circuits

Objective: The  objective is to see how long the cables for a circuit can be while still supplying enough voltage to a load to be functional. This can apply to real applications such as a remotely operated vehicle that may be powered remotely via wire.

Introduction:

• Our "load" is rated to consume 0.144W when supplied with 12V
• "Load" will operate properly as long as the voltage is greater than 11V
• Battery voltage is a constant  12V (approximately) with a capacity of 0.8Ahr.

                   
Experiment:
       We simplify out circuit by modeling the incoming and outgoing wires as resistors is series and combine them and we use a resistor box as our variable resistor in order to solve our problem.



Then add a couple meters to measure the voltage and current to make sure our load is getting enough voltage.

                   




Our load is represented with a resistor of 1000 Ohms. This was calculated using the rated values and the equation:

Power= (voltage^2)/(resistance)

The circuit was set up and the power supply set close to 12V. Both this voltage and the current in the circuit were measured.

The voltage across the "load" was measured and the resistor box was varied until we got to our minimum voltage.

Data:

• Our resistor actually measures out 973 Ohms for our "load"
• The voltage across the battery is 12.18 V
• Voltage across load = 11.01± 0.05 V
• Current out of battery = 11.65 ± 0.10 mA
• Resistance in "cable" (resistor box) = 73 Ohms

Data Analysis: 

a,b) Calculation of the time to discharge and efficiency:

c) We are not exeeding the power capability of the resistor box because the power through it did not exceed 1W. This can be proven by applying the same power equation from the above calculations.

d)

e)If we used 60 foot tether of AWG #28 wire with signals at 20 mA and 5V:
       The nominal rating for below 2 V (as indicated by picture) we have 5-0.4V= 4.6V.
       AWG # 28 wire has a 0.0764 Ω/ft resistance density
           so, the longest possible length would be   (73)/(0.0764Ω/ft) = 2810 ft

f) If we were sending 48 volts at 10 amps while the sub required 36 volts minimum, we would have 12 volts that would be able to dissipate in the wire. If we use the length from the previous problem the minimum cable gauge we can use would be what?
48V = IR + 36V
12V = 10R
R = 1.2Ω
12/955.5=0.012559   (this is the maximum resistance)
According to this the minimum gauge we can use is a 10 gauge wire.