March 24 Non-Ideal Power Sources and Max power

Non-Ideal Power Sources

Pre lab

At first, we do a pre-lab finding the power of a source when it is ideal and not ideal.

a),We find that when we assume it is ideal, there is no internal resistance. We find that the current across it is 0.0455 A. And the voltage across it is being set to be 1 V. Thus, the power is just P=VI= 0.0455 W. 

B),Then, we consider there is an internal resistance. We calculate the current to be 1/(22+Rs). We calculate the output Voltage is 22/(22+Rs). So the Power of it is 22/(22+Rs)^2 W. 

We choose a 22 ohm resistor which is actually 22.3 ohm, we manually calibrating the output voltage from Waveform to 1V and measure the voltage across the 22.3 ohm resistor.Then we calculated the internal resistance of power source is 0.502 ohm which make sense cause it should around 0.5 ohm.

Lab data

Circuit

Surprising oil usage 

plot of function of max power

Max power lab

prelab

 We calculate the theoretical power of it to be 2.84*10^-3 W. 

Measured Voltage 

The percentage difference is 0.3% we consider the predicted value is same with the experimental value.

Summary:

We also that power sources contain an internal resistance and how to calculate the resistance.

It is important to transfer maximum power to the load. In order to do that, the load resistance should equal the Thevenin resistance. 

March 26 inverting Voltage Amplifier

Class practice with opam.

Pre lab
We do a pre-lab to see that the relationship between V in and V out. We find that theoretically the relationship would be Vout= -2 Vin approximately. 

Data table

Plot between Vin and Vout

Circuit

Discussion:

We find that that the output voltage start as a horizontal line when the input voltage is higher than 1.5V or lower than-2V. Ideally the output voltage of the op-amp should be 5V (or -5V); since the op amp is cheap, the realm is -3.45V to 4.22V. The upper range is larger since the offset of the OPAM.

 

Summary:

Today, we talk about operational amplifiers, and explain the internal of it. Most time, we can treat op amp as ideal op amps so that it is easy to analysis. We also learn how useful the opam is and examine the physical property of opam. 

March 31 summing amplifier

Class practice
We done some practices in class.

Summing amplifier Prelab
By doing this lab, we can know and prove how a Summing Amplifiers combines two input signals. For any amplifier shown in the circuit, the Vout = - R3/R1 * (Va+Vb). 
We fix R1 and R2 equal to 1K and need to find out the value for R3. We don't want the resulting voltage be greater than 4V so that the OP will saturate. We need R3/R1 smaller than 4/6, so we choose R3 to be 220. 

Lab data
The results are pretty close to what we expected. The percentage difference is quit small. 

Real circuit 

Difference amplifier 
The purpose of this lab is to find and prove output voltage to be the difference between two input voltages. the pre-lab section, we figure out that if we use four identical resistors, the out put voltage is simply Vout = Va - Vb. we use four 10K resistors We get our four resistor to have true value around 9.7 k ohm. 

Ciucuit

Lab results for Va=1V

The Graph of Vin and Vout, we can see the relationship between vin and Vout is almost linear. The non-linear part may due to the saturation in high voltage input.

Now we changes the Va to -1V and adjust different Vb again. The results are pretty good since the percentage difference is quite small. 

as we see the graph reaches negative saturation at high voltage, the rest part of graph is linear which fits our expectations. 

Summery 

Today, we learn more about two different types of operational amplifiers. summing Op add up the input signals while the difference OP subtract one signal from another. In the lab we did good job because the experimental results are very close to the predicted values. The difference may due to the non-ideal OP and the inaccurate resistance value.

March 19,Thevenin's Theorem

Use of every circuit

Prelab result

Thevenin's resistance is 7.1K

Thevenin's voltage is 0.458V

Thevenin's circuit

If R=1k, I should be 54.4uA

Actual Circuit used in lab

Measured Vth=0.461V, compare to pre-lab value 0.458V, the percentage difference is 0.6%

Measured Rth=7.10K,compare to pre-lab value 7.1K, the percentage difference is 0%.Those two values are exact same. 

we put a 1K resistor between a and b then measured voltage across 1K load resistor, V=0.055V

Then we change the load resistor to 4.7K and measure the voltage across this resistor.V=0.17V

Result table.

For 1K load the theoretical value is 0.457*(1/(1+7.1))=0.056V while the measurement is 0.055V, percentage difference is 1.9% we can say they are same.

For 4.7K load the theoretical value is 0.457*(4.7/(4.7+7.1))=0.182V while the measurement is 0.17V, percentage difference is 3.6%, they are also could be same.

The slight difference may due to the difference in resistance we use.

Then we build the Thevenin's circuit. We use two 3.5K resistors and a 100 resistor connect in series to make a 7.1K resistor. 

Compare original circuit and Thevnin's equivalent circuit

For 1K load the original is 0.055V while the THEQ circuit is 0.054V, the difference is 1.9%, we could say they are same.

For 4.7K load, the original is 0.17V while the THEQ circuit is 0.17V, the difference is 0%, we could say they are same.From the result we examine the Thevenin's equivalent circuit is equivalent to the original circuit.

Data table for the potentiometer. 

The maximum power is when Rl=Rth=7.1K ohm.

P=(V(Rl/(Rl+Rth)))^2/Rl

Theoretically the Pmax=7.48*10^(-6)W. 

From the table the maximum power is 7.58*10^(-6)W when Rl=5600 ohms, which is a little bit off from the expectation. That may due to the inaccurate measurement, since from the graph the point on 5600 seems a outlier. possible we make a mistake when we take the resistance measurement. 

IN the future I could suggest that take more data from potentiometer to create a better graph to see the trend of power change with load resistance change. 

Summary:
In today’s lab, we practice more on Thevenin’s Theorem, and examine that Thevenin's theorem actually work. Also we try to find the max power deliver to the load resistance, based on our experiment, we can see that the power of a load resistance is about constant. It will only change a little bit when the resistance changes. 

March 17, Time-varying Signals, A BJT Curve Tracer

Time varying signals

Pre lab

Vout is half of the Vin since the two resistors are same. the result wave should be same shape but half amplitude.

R1=R2 in this sircuit.

Lab circuit

Original Sin wave

Result sin wave

original square wave

result square wave

original triangle wave

result triangle wave

build stair wave

In all there pattern of waves the result wave has same frequency and half the amplitude which fit our predictions.

BJT curve tracer circuit

result BJT wave which has 5 levels.  Summary:

Today we learn some new methods of circuit analysis. We talk about linear circuit.  We do a Time-Varying signal lab to see how a voltage divider actually works, and we introduce BJT, and figure out how BJT works.So that we know how to build different wave in WaveForm which could be very useful in future.