BTEC NC - Electronic and Electrical Principles - DC Circuit Theory

Brendan Burr BTEC National Certificate in ElectronicsDC Circuit Theory

1


2


3


4


5


6


7

E+

-

R1 R2

R3

R5 R4

A B C

DEF

E= 300 VR1 = 20

ΩR2 = 30

ΩR3 = 40

ΩR4 = 10

ΩR5 = 50

Ω


Task 11.1a)

b) Total ResistanceRT = R1 +R2 + R3 + R4 + R5

RT = 20 + 30 + 40 + 10 + 50 RT = 150 Ω

c) Total CurrentIT = VT

RT

IT = 300 150 IT = 2A

d) VoltagesV = IRV = 2A x 20 ΩR1 = 40VV = 2A x 30 ΩR2 = 60VV = 2A x 40 ΩR3 = 80VV = 2A x 10 ΩR4 = 20VV = 2A x 50 Ω R5 = 100V

8

Therefore:40 + 60 + 80 + 20 + 100 = 300V


e) Potential DifferenceAD = R1 + R2 + R3

= 20 + 30 + 40= 90 Ω= 2A x 90 Ω= 180V

BE = R2 + R3 + R4

= 30 + 40 + 10= 80 Ω= 2A x 80 Ω= 160V

CE = R3 + R4

= 40 + 10= 50 Ω= 2A x 50 Ω= 100V

DF = R4 + R5

= 10 + 50= 60 Ω= 2A x 60 Ω= 120V

f) Voltage PotentialsB = 300V – (Voltage across R1)

= 300 – 40= 260V

D = 300V – (Voltage across (R1 +R2 + R3))= 300 – 180= 120V

Pd = 260 – 120= 140V

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Evaluation :

E = 300 V A = 300VA = 2A

R1 = 20 Ω B = 260VB = 2A

R2 = 30 Ω C = 200VC = 2A

R3 = 40 Ω D = 120VD = 2A


1.2a)

b) Total Resistance

RT (parallel) = R2 x R3

R2 + R3

= 20 x 3020 + 30

= 600 50

= 12 Ω

RT = R1+ RT (parallel) + R4

= 28 + 12 + 10= 50 Ω

c) Total Current

IT = VT

RT

IT = 1050

IT = 0.2A

10

E+

-

R1

R4

R2

R3

E= 10 VR1 = 28

ΩR2 = 20

ΩR3 = 30

ΩR4 = 10

Ω


d) Current in R2

Voltage across R1 = 0.2 x 28= 5.6V

Voltage across R4 = 0.2 x 10= 2V

Voltage across R1 + R4 = 5.6 + 2= 7.6V= 10 – 7.6= 2.4V

I1 = 2.420

= 0.12AI2 = 2.4

30= 0.08A

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Therefore the current acrossR2 = 0.12AR3 = 0.08A


Task 2

Colour BandsMeasured Value Ω

1 2 3 4Nominal Value Ω

% Tol

Max Valu

e

Min Valu

e Using DMMR1 Brown Grey Red

Gold 1 K 8 5% 1890 1790 1784

R2

Orange

White Red

Gold 3 K 9 5% 4095 3705 3860

R3 Brown

Black

Orange

Gold 10 K 0 5%

10500 9500 9820

R4 Red Red

Orange

Gold 22 K 0 5%

23100

20900 22400

R5 Blue Grey

Orange

Gold 68 K 0 5%

71400

64600 68400

Measured value using DMM

E (Volts

) 10VI

(mA) 9.410524731 x 10ˉRT

(Ohms) 106264VA 10 VVB 9.83 VVC 9.47 VVD 7.34 VVE 6.42 VVAB 0.16 VVBC 0.36 VVCD 2.1 VVDE 0.92 V

12


Task 3Diode Voltage = SIMULATION FORWARD BIASED

Measured Resistance = 1K ΩPower Supply

VoltageDiode

VoltageResistor Voltage Current

Vs (volts) Vf (V) Vr (V) If (mA)0 0 0 02 0.632 1.37 1.374 0.655 3.35 3.356 0.666 5.33 5.338 0.674 7.33 7.33

10 0.68 9.32 9.3212 0.686 11.3 11.314 0.69 13.3 13.316 0.693 15.3 15.318 0.696 17.3 17.320 0.699 19.3 19.322 0.702 21.3 21.324 0.705 23.3 23.326 0.707 25.3 25.328 0.708 27.3 27.330 0.71 29.3 29.3

Diode Voltage = SIMULATION REVERSE BIASEDMeasured Resistance = 1K Ω

Power Supply Voltage

Diode Voltage

Resistor Voltage Current

Vs (volts) Vf (V) Vr (V) If (mA)0 0 0 0-2 -2 -0.1 0-4 -4 -0.2 0-6 -6 -0.3 0-8 -8 -0.4 0

-10 -10 -0.5 0-12 -12 -0.6 0-14 -14 -0.7 0-16 -16 -0.8 0-18 -18 -0.9 0-20 -20 -1 0-22 -22 -1.1 0-24 -24 -1.2 0-26 -26 -1.3 0-28 -28 -1.4 0-30 -30 -1.5 0

13

1N4001 Diode Simulation

0

5

10

15

20

25

30

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Diode Voltage (Vf)

Cu

rren

t (I

f)


The following chart simulates the characteristics of a 1N4001 Silicon Diode. As you can see no current is passed through the diode when I reverse biased, this can protect components from possible damage and can also enable four diodes to act as a rectifier, to change ac to dc.The graph shows that the curve begins to get less and appears to be straightening, however in theory the line will never be straight. The voltage across the diode will keep having lesser changes, but there will always be change.

14


Diode Voltage = 1N4001 FORWARD BIASEDMeasured Resistance = 986 Ω


Diode Voltage


Vs (volts) Vf (V) Vr (V) If (mA)0 0.08 0 02 0.59 1.44 1.454 0.63 3.34 3.356 0.65 5.33 5.368 0.67 7.25 7.29

10 0.68 9.22 9.2812 0.69 11.19 11.2714 0.7 13.15 13.2816 0.7 15.13 15.318 0.71 17.13 17.3620 0.71 19.08 19.422 0.72 21.2 21.624 0.72 23.2 23.726 0.72 25.2 25.828 0.73 27.2 27.930 0.73 29.2 30.1

Diode Voltage = 1N4001 REVERSE BIASEDMeasured Resistance = 986 Ω


Diode Voltage


Vs (volts) Vf (V) Vr (V) If (mA)0 0 0 0-2 -2 0 0-4 -4 0 0-6 -6.01 -0.001 0-8 -8 -0.001 0

-10 -10 -0.001 -0.001-12 -12 -0.001 -0.001-14 -14.01 -0.001 -0.001-16 -16 -0.001 -0.001-18 -18.01 -0.001 -0.001-20 -20 -0.002 -0.001-22 -21.99 -0.002 -0.002-24 -24 -0.002 -0.002-26 -26 -0.002 -0.002-28 -27.99 -0.002 -0.002-30 -29.99 -0.003 -0.003

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1N4001 Diode Practical

-2

2

6

10

14

18

22

26

30

-30 -25 -20 -15 -10 -5 0

Diode Voltage (Vf)

Cu

rren

t (I

f)


After repeating the above simulation for real, I noticed that a small amount of current actually leaks through at -10V whilst in the reverse biased. Although it is too small to see on the graph there is a slight degradation in the diode. There is a difference in the simulation and real life, as the simulation would act on perfect conditions, such as the voltage being exactly 10V. However in real life there are always tolerances, due to the manufacturing methods.

16


Diode Voltage = SIMULATION 5.1 Zener Diode FORWARD BIASED

Measured Resistance = 1K ΩPower Supply

VoltageDiode

VoltageResistor Voltage Current

Vs (volts) Vf (V) Vr (V) If (mA)0 0 0 02 0.632 1.37 1.374 0.655 3.35 3.356 0.666 5.33 5.338 0.674 7.33 7.33

10 0.68 9.32 9.3212 0.686 11.3 11.314 0.69 13.3 13.316 0.693 15.3 15.318 0.696 17.3 17.320 0.699 19.3 19.322 0.702 21.3 21.324 0.705 23.3 23.326 0.707 25.3 25.328 0.708 27.3 27.330 0.71 29.3 29.3

Diode Voltage = SIMULATION 5.1 Zener REVERSE BIASEDMeasured Resistance = 1K Ω


Diode Voltage


Vs (volts) Vf (V) Vr (V) If (mA)0 0 0 0-2 -2 -0.2 0-4 -4 -0.4 0-6 -5.02 -0.978 978-8 -5.05 -2.95 -2.95

-10 -5.06 -4.94 -4.94-12 -5.07 -6.93 -6.93-14 -5.08 -8.92 -8.92-16 -5.08 -10.9 -10.9-18 -5.09 -12.9 -12.9-20 -5.09 -14.9 -14.9-22 -5.1 -16.9 -16.9-24 -5.1 -18.9 -18.9-26 -5.1 -20.9 -20.9-28 -5.1 -22.9 -22.9-30 -5.11 -24.9 -24.9

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5.1 Zener Simulation

-25

-20

-15

-10

-5

0

5

10

15

20

25

30

-5.2 -4.2 -3.2 -2.2 -1.2 -0.2

Diode Voltage (Vf)

Cu

rren

t (I

f)


The Zener diode is mainly used as a stabilising component. This is so large amounts of current can get through, however small amounts (that below the stated value on the Zener diode) will not. In the simulation I used a 5.1 zener diode, as this cannot be altered, this means that at 5.1 volts the diode will begin to let the current through and allow the rest of the circuit to be powered.This is shown on the graph above through the x axis. At 5.1 volts, along the diode voltage (Vf) line, it clearly shows that the zener diode begins to break down and begins to let more and more current through.Whilst in the forward biased the zener diode acts in a very similar fashion to the normal silicon diode, 1N4001.

18


Diode Voltage = 2.7 Zener FORWARD BIASEDMeasured Resistance = 999 Ω


Diode Voltage


Vs (volts) Vf (V) Vr (V) If (mA)0 0.00 0.00 0.002 0.71 1.30 1.304 0.73 3.22 3.226 0.75 5.19 5.198 0.76 7.17 7.17

10 0.77 9.15 9.1612 0.77 11.13 11.1614 0.78 13.09 13.1416 0.78 15.08 15.1618 0.78 17.06 17.1820 0.78 19.12 19.3022 0.79 21.20 21.5024 0.79 23.20 23.5026 0.79 25.20 25.7028 0.79 27.30 27.8030 0.79 29.20 29.90

Diode Voltage = 2.7 Zener REVERSE BIASEDMeasured Resistance = 999 Ω


Diode Voltage


Vs (volts) Vf (V) Vr (V) If (mA)0 0.00 0.00 0.00-2 -1.76 -0.24 -0.20-4 -2.33 -1.69 -1.70-6 -2.57 -3.44 -3.40-8 -2.73 -5.33 -5.30

-10 -2.84 -7.17 -7.20-12 -2.93 -9.11 -9.10-14 -3.01 -11.04 -10.97-16 -3.07 -12.87 -12.93-18 -3.13 -14.82 -14.92-20 -3.18 -16.72 -16.86-22 -3.22 -18.61 -18.54-24 -3.26 -20.70 -21.00-26 -3.30 -22.80 -23.10-28 -3.34 -24.70 -25.20-30 -3.36 -26.60 -27.20

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2.7 Zener Practical

-30

-26

-22

-18

-14

-10

-6

-2

2

6

10

14

18

22

26

30

-3.40 -2.90 -2.40 -1.90 -1.40 -0.90 -0.40 0.10 0.60

Diode Voltage (Vf)

Cu

rren

t (I

f)


When putting the zener diode to practice, I used a 2.7 zener diode, so the voltage will be lower when it breaks down, however there should be similar results. Current was leaking through at 2 volts, even though it shouldn’t until 2.7 volts. This proves further the slight inconsistency of the mass produced components.The 2.7 zener diode, seems to break down much more slowly than the 5.1 zener diode.

The normal silicon diode, 1N4001, graph is partly similar to the zener diode, in the forward biased. The 1N4001 degrades very slowly, holding the current when in the reverse biased. The zener diode is designed to degrade, at the specified voltage. Both of the diodes conduct at around 0.7V.

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Task 44.1a)

b)ABEFE1 –V1 +V2 = 0E1 = V2 – V1

BCDEE2 – V3 – V2 = 0E2 = V3 + V2

Therefore:ABEF = E1 = (R2 x I2) – (R1 x I1)E1 = (6 x I2) – (35 x I1)

BCDEE2 = (R2 x I2) + (R3 x I3)E2 = (6 x I2) + (4 x I3)

ABCDEFE1 + E2 = V1 + V3

17 + 34 = (35 x I1) + (4 x I3)51 = (35 x1) + (4 x I3)

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E2

I1 I3

I2

I2

I1 I3

E1 V2

V1

A

V3

R1

R3

R2

B C

DEF

E1 = 17 VE2 = 34 VR1 = 35 ΩR2 = 6 ΩR3 = 4 Ω

A

B

C


Kirchhoff’s Current Law

I1 = I2 - I3I2 = I3 - I1I3 = I1 + I2

Substitute D into B

E2 = (6 x I2) + (4 (I1 + I2))E2 = (6 x I2) + (4 x I2) + (4 x I1)E2 = (10 x I2) + (4 x I1)

Simultaneous Equation

A 17 = ( 6 x I2) – (35 x I1) (x 10)E 34 = (10 x I2) + ( 4 x I1) (x 6 )

A 170 = (60 x I2) – (350 x I1)E 204 = (60 x I2) + (24 x I1)

374 = 0 + (374 x I1)

374 = 374 x I1

I1 = 374 374

I1 = 1 A

c) Currents

C 51 = (35 x I1) + (4 x I3)51 = (35 x 1 ) + (4 x I3)51 = 35 + (4 x I3)

51 – 35 = 35 + (4 x I3) – 35 (minus 35 from both sides)

16 = 4 x I 3 (divide both sides by 4) 4 4

I3 = 16 4

I3 = 4 A

D I2 = I3 – I1

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E

D


I2 = 4 – 1I2 = 3 A

d) Voltages

V1 = I1 x R1

V1 = 1 x 35V1 = 35 V

V2 = I2 x R2

V2 = 3 x 6V2 = 18 V

V3 = I3 x R3

V3 = 4 x 4V3 = 16 V

Therefore:

e) Power Dissipated in R2:

P = I2 x V2

P = 3 x 18P = 54 Watts

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E2

I1 I3

I2

I2

I1 I3

E1 V2

V1

A

V3

R1

R3

R2

B C

DEF

E1 = 17 VE2 = 34 VV1 = 35 VV2 = 18 VV3 = 16 VR1 = 35 ΩR2 = 6 ΩR3 = 4 ΩI1 = 1 AI2 = 3 AI3 =4 A


4.2

A1 = 180 A A7 = 120 AA2 = 20 A A8 = 40 AA3 = 160 A A9 = 40 AA4 = 160 A A10 = 120 AA5 = 180 A A11 = 180 AA6 = 300 A A12 = 220 A

A12 = A1 + A9A A1 = A2 + A3

A1 = 20 + 160A1 = 180

B A5 = A2 + A4

A5 = 20 + 160A5 =180

C A6 = A5 + A7

A6 = 180 + 120A6 = 300

D A7 = A3 - A8

A7 = 160 - 40A7 = 120

E A4 = A9 + A10

A4 = 40 + 120A4 = 160

F A6 = A10 +A11

A6 = 120 + 180A6 = 300

G A12 = A8 + A11

A12 = 40 + 180A12 = 220

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G

E

FA C

B

D

A1

A3

A2

A4

A8

A5

A7

A6

A10

A11

A9

A12

BTEC NC - Electronic and Electrical Principles - DC Circuit Theory

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Transcript of BTEC NC - Electronic and Electrical Principles - DC Circuit Theory