In this lab, we will determine and demonstrate the operation of the Darlington pair by using the basic electronic laws. We will build a circuit to operate a motor running at a certain a voltage and current. We will determine the values to operate that motor by using the regulations of two transistors and three different power sources. We will learn the basics of the function generator and the oscilloscope.

1) Compute all voltages and currents through all elements at Vin = 0 V,

IR28 = 0 mA, IR18 = 0 mA, IB = 0 mA;

because by Ohm’s Law ( V = IR ), if there is no input current, there is no voltage drop through any element in the base branch of the Tr5 transistor.

VBE = 0 V, no voltage here because there is no input voltage to provide it with.

Collector:

IR21 = 0 mA, VR21 = 0 V;

because there is no base voltage or current going through the base to emitter junction of the transistor, there can not be any voltage drop or current going through R21.

VCE = 5 V;

because of no voltage drop at R21, the voltage supplied from the power supply continues through that resistor and through the transistor’s collector to emitter junction.

VCB = 5 V;

because VCB = VCE - VBE = 5 V - 0 V = 5 V.

Tr5 Emitter/Tr6 Base: (base here is respect to Tr6)

VB = +5 V;

because of voltage from the CE junction of Tr5.

IB = 0 mA;

because no current is flowing through the transistor.

VBE = 0 V;

because needs at least 0.7 V to operate from the input voltage

Collector/Motor:

Imotor = 0 mA, Vmotor = 0 V;

because there is no input voltage to operate the motor nor any input voltage to give

voltage drop or current flowing through any elements in the circuit.

VCE = +3 V;

because of voltage from the battery power supply

VCB = +3 V;

because VCB = VCE - VBE = 3 V - 0 V = 3 V.

IE = 0 mA;

because no current flow in the circuit.

VE = +3 V;

because the voltage continues from the battery power supply since there is no voltage drop in the motor and no voltage through VBE.

Tr5 and Tr6 are both in cut-off regions. This is because of any range of positive input voltage, the current through both transistors will be zero.

2) compute all voltages and currents through all elements at Vin = +5 V,

VBE1 = 0.7 V, VBE2 = 0.7 V; diode-like drops in BE junction of transistors.

V1 = VBE1 + VBE2 = 0.7 V + 0.7 V = 1.4 V

VR28 = Vin - V1 = 5 V - 1.4 V = 3.6 V

IR28 = VR28 / R28 = 3.6 V / 4.7 k ohm = 0.766 mA

VR18 = Vin - VR28 = 5 V - 3.6 V = 1.4 V

IR18 = V1 / R18 = 1.4 V / 4.7 k ohm = 0.298 mA

IB = IR28 - IR18 = 0.766 mA - 0.298 mA = 0.468 mA

power supply to collector: +5V

Tr5 is in saturation, because by ß = IC / IB = 41 mA / 0.77 mA = 89 < 200

Beta is way smaller than the beta of the transistor, so Tr5 is in saturation in order to operate the transistor.

VCE1 = 0.2 V, because the transistor is in saturation

VBE2 = 0.7 V, diode-like voltage drop in

V2 = VCE1 + VBE2 = 0.2 V + 0.7 V = 0.9 V

VR21 = VCC - V2 = +5 V - 0.9 V = 4.1 V

IC = IR21 = VR21 / R21 = 4.1 V / 100 ohm = 41 mA

VCB = VCE - VBE = 0.2 V - 0.7 V = -0.5 V

IE = IB + IC = 0.468 mA + 41mA = 41.468 mA

Tr6 is in saturation, because by ß = IC / IB = 38 mA / 0.77 mA = 82 < 200

Beta is way smaller than the beta of the transistor, so Tr6 is in saturation in order to operate the transistor.

IB = 41.468 mA; IB of the Tr6 transistor is the same as the IE of the Tr5 transistor.

VBE2 = 0.7 V; diode-like drops in BE junction of transistors.

battery power supply: +3 V

Vmotor = +2.8 V

Rmotor = varies

Imotor = since Rmotor varies, the current varies inversely proportion to the resistor through the motor

VCE = 0.2 V

VCB = VCE - VBE = 0.2 V - 0.7 V = - 0.5 V

IE > 41.468 mA, since we don’t know the exact value for IC, we can only say that the current flowing in the emitter of Tr6 is more than that of the base current.

3) compute smallest input voltage that make motor run,

Vin = R28I1 - V1

V1 = I2 R18 ,

I1 = IB + IC ,

Vin = R28I2 + R28/R18 + V1 ,

Vin = 2.8 V1 + R28IB ,

Vin => 2.8V

4) The function of the Tr5 transistor is to build up a base current from the input current and to act a primary switch in the circuit. The Tr6 transistor is to supply current to the motor from the input voltage and the collector current. It also acts as a secondary switch. The R21 resistor functions to keep the Tr5 transistor in saturation instead of active-linear. Because you want saturation, the R21 need a quite high resistor to control current from the +5 V input. The R28 resistor is to control and regulate the amount of base current and voltage from the input voltage because too much current can hurt the Tr5 transistor. Because you want a very low current, a very high resistor is added. The R18 resistor is to make the motor run at a higher voltage than the normal 1.4 V. This is because we want the resistor to run higher than the 2.5 V offset, so by adding a large resistor, we can make the motor run a little above the offset voltage set by the industry.

5) The resistance of the motor had a variable measurement. It ranged from 1 ohm to 10 ohm.

6) measure all voltages and currents through all elements at Vin = 0 V,

R28 = 4.68 K ohm

R18 = 4.68 K ohm

R21 = 98.2 ohm

IR28 = 0 mA

VR28 = 0 V

IR18 = 0 mA

VR18 = 0 V

IB = 0 mA

VBE = 0 V

IR21 = 0 mA

VR21 = 0 V

VCE = 4.54 V

VCB = 4.51 V

Tr5 Emitter/Tr6 Base: (base here is respect to Tr6)

VB = 4.54 V

IB = 0 mA

VBE = 0 V

Collector/Motor:

Imotor = 0 mA

Vmotor = 0 V

VCE = 3.06 V

VCB = 3.04 V;

IE = 0 mA;

VE = 3.02 V

The CE junctions of both transistors were surprising off by a little, around a 15% error. This may be caused by the tolerance of the resistors and motor and that the voltmeters can’t read well under low current and voltages. All and all, the values seem near each other. Which at the end is zero for most elements in currents and voltages because since there is no input voltage, no current flows.

7) compute all voltages and currents through all elements at Vin = +5 V,

VBE1 = 0.773 V

VBE2 = 0.848 V

VR28 = 3.33 V

IR28 = 0.712 mA

VR18 = 1.674 V

IR18 = 0.358 mA

IB = 0.370 mA

power supply to collector: +5V

VCE1 = 0.340 V

VR21 = 3.78 V

IC = 37.8 mA

VCB = - 0.13 V

IE = 35.5 mA

IB = 35.5 mA

VBE2 = 0.848 V

battery power supply: +3 V

Vmotor = 2.69V

Rmotor = 1 to 10 ohms

Imotor = varying, good approximate is 164 mA

VCE = 0.2 V

VCB = - 0.15 V

IE = varying, good approximation is 166 mA, but also gives 191.2 mA without motor

We find that our calculations and measurements are pretty close together. The motor however causing a varying resistance, causing the current in the collector and emitter branches in the Tr6 transistor to have vary ranges. And also the voltmeters and ammeters causing different values through our elements in the collector and emitter regions of Tr6 as well. The multimeters caused our CE junction values for both resistors to have more different values than expected with higher percent error. This is because the

voltage is so low, the mulitmeters have trouble reading it.

8) The motor starts running at Vin = 2.72 V when increase Vin from 0 V.

The value we got here is a little bit lower than predicted, but that is factored by the tolerance in the resistors and the multimeters differences. So we got a near match.

9) The motor stops running at Vin = 2.62 V when decrease Vin from 5 V.

The motor stops at a lower voltage than the value we got from #8 because there might be some voltage and current remaining in the circuit while we kept turning the input voltage knob.

10) When we remove R18 resistor, the motor runs when the input voltage is 1.37 V. The resistor takes some current and voltage away from the original base branch so the motor will run at a much higher voltage. This is done to keep the motor on a voltage higher than the offset voltage of 2.5 V used by industry.

11)

The plot does indeed follow Ohm’s Law once the motor is running. So the variation of the resistance in the motor still follow Ohm’s Law, but in a sinusoidal way.

1)

R2 = 100 ohms

R1 = 100 ohms

f = 1 kHz

Vin = 5 V

Vin = Vo sin (wt + ø)

V1 = 2.5 V

V2 = 2.5 V

t = 1 / f = 1 / 1000 Hz = 1 millisecond

2)

R2 = 103.0 ohms

R1 = 101.6 ohms

f = 1 kHz

Vin = 5.00 V

V1 = 2.50 V

V2 = 2.50 V

t = 4.3 boxes of 250 µs = 1.080 milliseconds

KVL holds because the circuit in series: Vin = V1 + V2 = 2.50 V + 2.50 V = 5.00 V

Also look at graph in next pages.

3) When we increase the frequency, V1 and V2 both have decreased periods. When we decrease the frequency, V1 and V2 both have increased periods. When we increase the amplitude, V1 and V2 both have decreased periods. When we decrease the amplitude, V1 and V2 both have increased periods. The reasons for this occurrence is that the frequency is the inverse of the period, and vice versa, f = 1 / t, t = 1/ t. The more frequency, the period most decrease to fit the frequency (and vice versa).

1) R1 = 100 ohms

f = 2 kHz

Vin = 8 V

Vin = Vo sin (wt + ø)

t = 1 / f = 1 / 2000 Hz = 0.500 milliseconds

VD = 3.30 V

VR = 4.70 V

2) R1 = 101.6 ohms

f = 2.000 kHz

Vin = 8.00 V

VD = 3.30 V

VR = 4.70 V

t = 0.500 milliseconds

KVL does hold by virtue of Vin = VD + VR = 3.30 V + 4.70 V = 8.00 V

Graphs on next pages.

3) When we increase the frequency, VD and VR both have decreased periods. When we decrease the frequency, VD and VR both have increased periods. When we increase the amplitude, VD and VR both have decreased periods. When we decrease the amplitude, VD and VR both have increased periods. The reasons for this occurrence is that the frequency is the inverse of the period, and vice versa, f = 1 / t, t = 1/ t. The more frequency, the period most decrease to fit the frequency (and vice versa).

4) Graphs on different page.

The graphs of the DC at 2 V appear to be the same when the DC is at 0 V.

In this lab, we learned the function of the Darlington pair. It is used to operate motors and devices in electronics once a certain amount of input voltage is given into a circuit. We learned the circuits of transistors have saturation and active-linear regions and saturation region does not necessarily mean it doesn’t operate. We learned that industry set an offset voltage of 2.5 V for operation of motors and devices. We’ve demonstrated the period and the frequency are inversely proportional to each other. We also learned the basics of the function generator and the oscilloscope.