Introduction to Biasing

In this lab we need to find the correct values of resistors in order to bias 2 LEDs so that they will not burn out when we put them in series and in parallel.

LED 1 is rated for 5V and 22.75mA

LED 2 is rated for 2V and 20mA

We model our system with laboratory equipment.

Our 9V battery will be our power supply. LED1 is rated to be at roughly 220 ohms and LED2 at 100 ohms. Using nodal analysis we find the following data:

I_R1 = 22.75 mA

I_R2 = 20 mA

VR1 = 4

VR2 = 7

R1 = 175.8 ohms

R2 = 350 ohms

P_R1 = .09W

P_R2 = .14W

We build our circuit and both LEDs are biased correctly.

Measurements from the circuit.

Configuration 1:

ILED1 = 14.20+_0.005mA

VLED1 = 5.39+_0.005V

ILED2 = 32.0+_0.05mA

VLED2= 2.41+_0.005V

IBATT = 47.0+_0.05mA

Configuration 2:

ILED1 = 10.0+_0.05mA

VLED1 = 5.41+_0.005V

IBATT = 10.0+_0.05mA

LED1 removed:

ILED2 = 37.0+_0.05mA

VLED2 = 2.42+_0.005V

IBATT = 37.0+_0.05mA

Questions:

a. The circuit can operate for .2Ahr/.047= 4.3 hours

b. LED1 had a percent error of ~38%. LED2 had a percent error of ~60%. The LEDs are not rated to be within strict tolerances as their function is to produce light. they are not good resistors.

c. The efficiency = p_out/(p_out+p_lost) = ~36%

d. The efficiency would increase as the voltage would be closer to LED1, 5V. The optimal efficiency would increase until you get to a point where the voltage becomes too much of a difference from LED2. I have not done the calculations to find this voltage.

Introduction to DC Circuits

In this lab we are given a theoretical problem and asked to solve it, both analytically and experimentally. We are given a few assumptions:

Our load is rated to consume .144W at 12V.

The load will operate as long as the voltage is greater than 11V.

Battery voltage remains constant at approximately 12V and the battery has a capacity of .8Ahr.

We must determine the maximum cable resistance, the maximum distance the battery can be seperated if we use AWG #30 cable, the distribution efficiency, and the approximate time before the battery dies.

We begin by modeling the system with laboratory equipment.

From the given constraints we can find that the load resistance to be approximately 1000 ohms.

The resistor box is rated to be at 0.3W. Our power supply is rated to be at 12V and 2A max voltage and current, respectively. Our 1000 ohm resistor is rated to 1/8 W.

We find the resistance of our cable to be approximately 101 ohms and is within tolerance. our load resistance, which is being modeled by a 1000 ohm resistor is also within tolerance.

We measure our voltage battery to be 12.05 V.

We then build our circuit with all the circuit elements using alligator clips. Our final readings for load voltage is 10.99V, current across the battery is 11.10A, and equivalent resistance of the cable being 101 ohms.

Questions:

a. Our time to discharge is calculated to be Ahr/current = 72.1 hours.

b. The power to the load is current*voltage = .125W. The power to the cable is .013W. The efficiency is given by p_out/(p_out+p_lost) and results in an efficiency of roughly 90%.

c. No, the resistor box is rated to .3W and we are only absorbing .013W through the cable.

d. 101 ohm/(2*0.345 ohm/m)= 146m.

Using a Multimeter

This review lab consisted of getting reacquainted with the multimeter from what we learned in Physics 4B.

We set up the multimeter to measure resistance and measure resistance of air, the probes, and 4 different resistors. For my resistors I chose 100, 220, 39, and 6700 ohm resistors.

The color coding for these resistors were brown/black/brown, red/red/brown, orange/white/black, and blue/grey/red, respectively.

We then set up the multimeter to measure voltage. Since we are measuring voltage from a transformer, which outputs DC voltage, it does not alternate. We use the DC voltage setting to measure DC voltage. I measured my transformer to be at 36.71 V.


For the next section we set up the described circuit on our breadboard. Our measured voltage across the resistor is 2.41 V and across the LED is 2.13 V. Together they sum up to roughly the voltage drop across the battery, which is 4.55 V.

We then measure the current across the entire circuit, which ends up being around 15.6 mA.