In this lab, we will determine the function and application of the transitor. We will use a BJT large signal model to demonstrate the different currents and voltage differences running through different parts of the transitor. We will get deep down and use top thinking in finding how the tranistor truly operates. We will also uses a LED and the three basic relationships of Ohm’s Law, KCL, and KVL in our transitor tests. We will make a circuit so the transitor in our test can operate properly.

1. Purpose: use BJT large signal model to find currents and voltages across all elements when input voltage is 0 V.

Analysis: when input voltage is at 0 V,

IB = 0 µA and VB = 0 V,

since there is no voltage from the input power supply, no voltage goes through the

gates of the diode. under Ohm’s Law V = IR, no voltage means no current through all

of the elements on the base branch of the transitor.

IC = 0 mA,

since the active region equation IC = ß IB , there will be no current in the collector if

there is no current in the base.

IE = 0 mA,

since the emitter current is the total of the base and collector currents IE = IC + IB ,

there is no emitter current since there is no base or collector currents.

VBE = 0 V,

since the Vin = 0 V, no voltage flows from the base to the emitter (at least ideally).

VCE = - 1.8 V,

since we solve for the circuit:

0 V = 1.8 V + VCE + IC RS , with IC = 0 mA and RS = 330 ohms,

VCE = 0 V - ( 0 mA X 330 ohms ) - 1.8 V = - 1.8 V

VCB = VBE - VCE = 0 V - (- 1.8 V) = 1.8 V

2. Purpose: use BJT large signal model to find current and voltages across all elements when input voltage is 5 V.

Analysis: when input voltage is at 5 V, and the ß = 200, R8 = 20 k ohm, R24 = 330 ohm

IB = 0 µA and VB = 0 V,

VB = IB R8 = ( 5 V - 0.7 V - 0.7 V ) = IB (20 k ohm),

IB = 180 µA

IC = 9.1 mA,

IC = ( 5 V - 1.8 V - 0.2 ) / 330 ohms

IC = 9.1 mA

IE = 9.28 mA,

IE = IC + IB = 9.1 mA + 180 µA

IE = 9.28 mA

VBE = 0.7 V,

since the voltage on a transitor from the base to the emittor acts as a diode (0.7 V).

VCE = 0.197 V,

5 V = 1.8 V + VCE + IC RS , with IC = 9.1 mA and RS = 330 ohms,

VCE = 5 V - ( 9.1 mA X 330 ohms ) - 1.8 V = 0.197 V

VCB = VBE - VCE = 0.7 V - ( 0.197 V ) = 0.503 V

PLED = IC VLED = ( 9.1 mA )( 1.8 V ) = 16.2 mW

3. Analysis: the input voltage that turns the LED on,

VCE = 5 V - ( 9.1 mA X 330 ohm ) = 1.997 V

VCB = VBE - VCE = 0.503 V - 1.997 V = - 0.5 V

Vin = -0.5 V + 0.7 V + 0.7 V = 0.1 V

4. The R8 resistor was chosen to be 20 k ohm so a little amount of current is supplied to the transitor. A small base current is needed to control the transitor’s voltage for it’s collector branch and the LED. Hence, the R24 resistor has chosen to draw more current so the circuit can operate the LED. It regulates the current supplied to the LED. The transitor is like an LED, but it controls and regulates voltage and current at two places instead of one like in the diode. The transitor regulates the voltage through the LED and functions the LED in when the light will turn on. The D6 functions as a voltage regulator for the base branch of the transitor. It justs supplies a lower voltage to the transitor. Remember, the base currents needs to be very low, and the R8 resistor and diode helps control the current to a low value.

5. On graph, please see next page.

6. Measurements: when input voltage = 0 V

R1 = 19.74 k ohms

R2 = 333 ohms

LED = 1.788 V

IB = 0 mA

IC = 0 mA

IE = 0 mA

VBE = 0.01 V

VCE = 3.42V

VCB = 2.640 V

IE = IC + IB = 0 mA + 0 mA = 0 mA,

VCE = VCB + VBE = 2.640 V + 0.010 V = 2.65 V

Comparisons: KVL does not apply to the circuit when the current is 0 V. This may be because that the transitor is in saturation and cutoff regions and do not operate functional normally. The voltage is has a larger value than expected between the collector and emitter. However, KCL still applies through the transitor.

Our analysis and the actual measurements don’t match accordingly. It appears that the saturation and cutoffs have alter our data.

7. Measurements: when input voltage = 5 V

IB = 0.2 mA

IC = 8.12 mA

IE = 8.33 mA

VBE = 0.727 V

VCE = 0.096 V

VCB = 0.630 V

IE = IC + IB = 8.12 mA + 0.2 mA = 8.32 mA

VCE = VCB + VBE = 0.727 - 0.630 = 0.097

Comparisons: This circuit does indeed follow KCL AND KVL. We see that the emitter current is the sum of the collector and the base currents. We see that the voltage and currents are regulated by the transitor at the collector and base branches, and flow to the emitter branch.

Unlike the input voltage of zero, the measurements and analysis of the 5 V input are close together.

8. LED turns on at 0.90 V, it appears that the voltage we got in part 3 was only a value in range of voltages that allow the LED to turn on. The 0.9 V can be used as a minimum amount of input voltage, but certainly any reasonable value above it will be able to light the LED up.

9.

11. Some application include the regulation of voltage across a electric device. The device may not get a high doseage of voltage. If it does, the LED will turn on, signaling the user to lessen the amount of collecter voltage entering the circuit. The electric device may be another computer or maybe a refrigator or alarm clock or etc. Using the LED, the light can be used to signal users of a potential problem that too much voltage is being applied.

In this lab, we learned that the transitor is a much better way to regulate voltage and thus current through a bigger circuit. We learned that the transitor may not apply with KVL due to a saturation and cut-off regions. We learned that the function of the elements in our circuit helps regulate the base and collector currents before the regulation of the transitor. Regulation of a circuit is needed to operate a certain electronic device. We also learned that the higher the base current, the high the collecter current, thus a lower voltage and operating point. So in this case, a better functioning transitor, you must need a low value for the base current.