3/10/16 Day 6: More Mesh Analysis, Diodes and Transistors

Today in class we first took a quiz applying mesh analysis to a circuit to find some currents, which was similar to the usual problems we see involving mesh analysis. After the quiz we learned about supermeshes, or a current source isolated between two nodes, and how to recognize and take them into account when analyzing the circuit. We then did a problem involving a supermesh and solved for a current. We then compared nodal analysis and mesh analysis in order to determine which one to use. We then performed the lab called "Mesh Analysis 2", which is simple and straight-forward. After the lab, we learned a little about diodes and how they only allow current to flow in one direction. They contain a positive and negative terminal, and depending on the direction of the source current will or will not flow. We also learned more about transistors and how they are split into NPN and PNP types. All transistors allow current to flow between the collector and emitter only when a voltage is applied to the base. We then learned more about BJTs, which are types of transistors. BJTs amplify the current entering the base by a constant labeled beta, which ranges from 50 to 1000. Another constant, termed alpha, denotes the amount of electrons entering the emitter from the collector (1 being all electrons). Then, a problem involving a transistor in a circuit was analyzed. Lastly, the lab titled "A BJT Curve Tracer" was performed. The lab is explained in more detail in the lab section.


LECTURE:

Above is a problem involving a circuit containing a supermesh. The supermesh was a current source and a resistor in series, attached to the labeled two nodes between i2 and i3. The problem asks to solve for the current io. The constraint equation of the supermesh using KCL was obtained, followed by applying KVL by mesh analysis in order to solve for the three unknown currents i1, i2 and i3. Then, once the three currents were solved for, io was obtained by subtracting i1 and i3 to get a value of -1.73 A.

The above table provides information on the situations in which nodal analysis or mesh analysis is more powerful to use. On an efficient basis, a good rule of thumb is to use the method that results in the least amount of equations. Because of this, nodal analysis tends to be better used in circuits that have many parallel connected elements, whereas mesh analysis is better used in circuits with many series connected elements. Lastly, by a logical approach it is better to use nodal analysis when solving for voltages, whereas it is better to use mesh analysis when solving for currents.

This circuit includes a transistor, and the objective was to analyze the circuit to find the voltage out of the emitter. This was done by solving for current through the base in order to solve for the dependent current source in the collector. KVL was then used to solve for the voltage of the emitter.


LAB:

Mesh Analysis 2:

Purpose:

The purpose of this experiment was to analyze a circuit using mesh analysis to obtain values for wanted voltages and currents. These calculated values were then checked for accuracy using EveryCircuit. Then the circuit was built and those voltages and currents were tested. The calculated and measured values  were then compared.



Prelab:

The prelab of this experiment was to use mesh analysis to solve for the voltage across the 6.8 kohm resistor, labeled V1, the voltage across the 10 kohm resistor, labeled V2, and the current through the 10 kohm resistor, labeled I1. The value of I1 was found to be -0.370 mA using mesh analysis. The value of V1 was calculated as 5 V. Lastly, the value of V2 was found to be 3.22 V.

The accuracy of the calculated values for V1, V2 and I1 was checked using EveryCircuit, as shown below:

As can be seen in the EveryCircuit schematic, the calculated values for voltages V1 and V2 and current I1 were correct. This also shows that mesh analysis and nodal analysis, which EveryCircuit uses, do end up with the same answers, which shows that both are correct methods to use in any circuit.



Apparatus:

The apparatus of this experiment consisted of the usual materials, such as a digital multimeter, an analog discovery tool kit, a laptop with the Waveforms application, a breadboard, some resistors, wires and alligator clips.



Procedure:

The schematic of the circuit seen in the prelab above was built. Next, the true resistances of the resistors was measured using an ohmmeter. Then, the voltages across the 10 kohm and 6.8 kohm resistors were tested, along with the current across the 10 kohm resistor. The only difference is that the ground was attached to the 6.8 kohm resistor instead of the 4.7 kohm resistor, but that does not practically affect anything. As in the nodal analysis lab, the 3 V voltage source was supplied using the yellow W1 wire (the waveform generator on the analog discovery). To test the voltages, the DMM was put in parallel to the branches, whereas to test current, the DMM had to be put in series with the 10 kohm resistor, requiring to break the circuit. Testing the voltages is seen below:

As can be seen, the voltage across the resistor is tested by connecting the voltmeter in parallel.

Data and Analysis:

The experimental voltage across the 6.8 kohm resistor was found to be 4.97 V, which was found to be very similar to the calculated voltage of 5 V (a percent difference of -0.6 %). The experimental current across the 10 kohm resistor was measured to be 0.340 mA, which is reasonably close to the theoretical value of 0.322 mA (a percent difference of 5.59%). The only reason the percent difference is relatively high is because the values are very low, so the small changes resulting from uncertainty and measurement errors increase the percent difference relatively large. Lastly, the experimental voltage across the 10 kohm resistor was found to be 3.57 V, which is fairly close to the theoretical value but still has a relatively significant difference (a percent difference of 10.9%). The reason the percent difference is fairly large is because of the uncertainties and errors in measurements of the resistors and DMM. In addition, it is also due to the resistance of the resistors increasing as their temperature increases over time of use, resulting in a larger voltage drop than expected.

A BJT Curve Tracer:

Purpose:

The purpose of this experiment was to experimentally find some characteristics of an NPN BJT, such as the collector current and voltage, the current gain and the base current.

Apparatus:

The equipment of this experiment included an analog discovery box, a laptop with Waveforms software, a breadboard, a digitial multimeter, wires, a 100 kohm resistor, a 100 ohm resistor and a 2N3904 NPN bipolar junction transistor.

Procedure:

Source: dmercer "Activity 4. A BJT Curve Tracer" Analog Devices Wiki

The above circuit was first created using the NPN transistor, a breadboard, the 100k ohm resistor, the 100 ohm resistor and the analog discovery. The built circuit is shown below:

The circuit is an exact representation of the schematic above. Channel 2+ and 2- (orange wires) were put across the 100 ohm resistor, and so on. Next, the true resistance values of the resistor were obtained by measurement using the digital multimeter as an ohmmeter, as shown below:

Using Waveforms AWG tool, a stair step function with 5 steps was made. The channel was placed at a frequency of 40 Hz, an amplitude of 2 V, and an offset of 2.6 V. The first step started at 0.6 V and then each step increased by a 1 V amount. Therefore, step 2 was at 1.6 V, step 3 at 2.6 V, step 4 at 3.6 V and step 5 at 4.6 V. Each step was prolonged for 5 milliseconds, therefore having the wave function last for a total of 25 milliseconds. This function was placed in waveform 2, as seen in the schematic. Then, for channel 1 a triangular wave was made using Waveforms AWG tool, with an amplitude of 2.5 V and an offset of 2.5 V. The frequency of channel 2 was then placed at 200 Hz. The waveform in channel 1 went from 0 V to 5 V and then back to 0 V in a triangular fashion for the same amount of time of one step in channel 2's waveform (5 milliseconds). The graphs of these functions are shown below:

 Then, the waveforms graphs were run in the circuit and data was collected. The data was then used to graph the current of channel 2 on the y axis and the voltage of channel 1 on the x axis. Five curves were obtained, each separated by an inputted constant of 2 mA. These graphs show the maximum current reaching and running through the transistor, independent on changes in voltage after the threshold voltage. The graph is seen below;

The Gummel graph above can be used to find the current gain (beta) of the transistor; only the domain before the graph becomes linear is applicable in this analysis. Analyzing the nonlinear portions of the curve, beta, taken from the slope, is found to be 40. This shows that the current exposed to the base of the transistor is amplified 40 times as much.

Analyzing the graph further, the early voltage can be found by using the highest base current step. The early voltage was found to be roughly 0.02 V. Then, using beta and early voltage the beta early voltage was found by multiplying the two values, obtaining a beta early voltage of around 0.8 V. The value is 0.1 V off from the theoretical value (0.7 V) due to rounding errors from the amount of significant figures of the graph. Overall, though, our obtained data and values are very similar to the manufacturer specifications for the BJT.