Electronics Lab - Day 2

For our second day learning about electronics, we learned about voltage dividers. I'm finding the concept to be simple and really hard at the same time. From what I can figure out, it sounds like a voltage divider tells us the voltage across one of the resistors when there are two resistors in series. We can use the formula Vout = (R2/R1+R2)Vp to figure out the output voltage.

Now the key concept that we learned today was Thevenin's Idea. He told the world that if there is a complicated circuit in some black box, we could essentially just treat everything in that circuit as if it were just one voltage source and one resistor. This is really important because whenever we're creating circuits, we're trying to figure out whats going to happen when we connect one complicated thing with another complicated thing.

What we did today was power a motor directly with a 4.5v battery pack and then using the output pins on a logochip. Using an oscilloscope we monitored what happened to the voltage across the power and ground buses under different conditions. When the motor connected directly to the battery pack, where was only around a 0.2v drop in voltage when we stalled the motor whereas there was almost a 3v drop in voltage when we stalled the motor connected to the logochip.

In order to better understand why this happens, according to Thevanin, we can treat the power source as just a power source and a resistor with a specific thevanin resistance and we can treat the motor as just a resistor with some other thevanin resistance (as long as we keep the motor stalled). So the challenge is to figure out these resistances.

Here's how we did it:

In the equation for figuring out the output voltage, R1 is essentially the thevanin resistance of the power source. If we put in a known resistance as the second resistor(R2=47 ohm) then all we need in order to calculate the thevanin resistance are the input(Vp) and output(Vout) voltages. For the input voltage, we just have to measure the voltage across just the power source and to find the output voltage, we can measure the voltage across the second known resistor like in a voltage divider circuit.

R1 = (R2*Vp - Vout*R2)/Vout

We also replaced the 47 ohm resistor with the motor and measured the voltage across a stalled motor powered by the battery pack and the logochip in order to calculate the thevanin resistor of the motor.

R2 = R1*Vout/(Vp-Vout)

*we should calculate the same thevanin resistance no matter whats powering the motor.

Results:

(1) Battery Pack:

Vp = Vth = 4.4v

Vout=4.3v

Rth = 1.1 ohm

(2) Logochip:

Vp = Vth =4.3v

Vout=1.8v

Rth = 65.3 ohm

(3) Motor:

Vout(battery pack) = 4.2 v

Rth = 23.1 ohm

Vout(logochip) = 1v

Rth = 19.8 ohm

Although the two calculated thevanin resistances for the motor are different, they still verify that the resistance of the motor is much larger than the Rth of the batter pack and much smaller than that of the logochip. In a voltage divider, when R1<>R2, Vout

Lab 1-4

First we measured the open circuit voltage across a power source and a 10 k resistor and found that Voc= 2.2v. We then added another 10k resistor and measured the short circuit current across the second resistor and found Isc = 0.439 mA. Using these two values, and the equation R=V/I we calculate the equivalent thevanin resistance to be Rth = 5 kohm.

Then, leaving the input voltage at 2.2v, we replaced the 10 k resistor with a 5k one and then measured the current across the power and ground buses. We found the current to be .45 mA which is the just about the same as the current in the previous circuit, thus proving that thevanin is, in fact, correct.

Lab 1-5

We connected a function generator to the oscilloscope and did various things with both instruments to become more familiar with the both of them. First we output a square wave from the function generator and measured the rise time using the scope which we found to be 20 ns.

We also made use of the sync function which makes it easy to see where the peaks and troughs of an oscillating waveform are. This makes it much easier to trigger on the peak if you want to.

Lab 1-6

When we input a 1 kHz sine wave into the voltage divider, we found that the output voltage also oscillated at the same frequency as the input voltage. The only difference that we found was that the ouput voltage was shifted to always be positive where as the input voltage oscillated between a positive and negative voltage.

The voltage divider must have an oscillating voltage because

Vout = (R2/R1+R2)Vp and if Vp is changing with time, so must Vout.