In this tutorial, we are going to learn about Best Topics for Hardware Design Engineer

**Below are the best topics for Electronics Engineer**

- Linear Circuit Elements
- Ohm’s Law
- Series & Parallel Circuits
- Voltage & current Dividers
- Kirchhoff’s Current Law (KCL)
- Nodal Analysis
- Kirchhoff’s Voltage Law (KVL)
- Loop Analysis
- Source Transformation
- Thevenin & Norton Equivalence
- Superposition Theorem

**Linear Circuit Elements- **

- Resistor Ohm Ω
- Capacitor Farad F
- Inductor Henry H
- Voltage Source Volt V
- Current Source Ampere A

Below is the image for symbol of above components which are basic use in circuit.

**Nodes, Branches, and Loops-**

- A node is a junction of connecting wires. Every point on a node is at the same potential (same voltage).
- A branch just another name for any circuit element between two nodes.
- A loop is a closed path that begins and ends at the node.

**Ohm’s Law-**

- V = Voltage
- I = current
- R = Resistance
- V = IR or I = v or R=
- Ohm’s Law makes it possible to solve for an unknown current, voltage or resistance.
- In this case, As per circuit below circuit we will calculate the resistance in this circuit by Ohm’s Law.

R= 75Ω, V = 30 Now I = 30/75 = 0.44

**Series Circuits-**

- There is only one path for current so the current is the same everywhere in the circuit.
- R
_{eq}= R_{1}+R_{2}+ …..+R_{n} - R
_{eq}= 100Ω+50 Ω = 150 Ω - Applying Ohm’s Law,

I_{= } 30V _{= } 30V _{= }0.24

R_{eq } 150Ω

**Parallel Circuits-**

- All resistors share a common voltage, but the current through each depend on the value of each

depend on the value of each resistor.

- 1
_{=}1_{+}1_{+}…._{+}1

R_{eq} R_{1} R_{2} R_{n}

R_{eq }=[ 1 + 1 ]^{-1}=60Ω

150Ω 100Ω

Ohm’s Law I _{=} 30V _{=} 30V 0.5A

R_{eq} 60Ω

I _{=} 30V _{=} 0.2A

150Ω

I _{=} 30V _{=} 0.3A

100Ω

**Voltage Dividers-**

- A voltage divider creates a voltage which is some fraction of its voltage source
- It is a series circuit where the output voltage is (usually) taken across the second resistor.

R_{eq}= 50Ω

I = 10V = 0.2A

50Ω

V_{out} = IR_{2} = (0.2A)(10Ω) = 2.0

- We can write a general equation that describes the output of any two-resistor voltage divider.
- R
_{eq }= R_{1}+ R_{2} - I = Vs

R_{eq}

- V
_{out }= IR_{2} - V
_{out}=[ Vs ] R_{2}= Vs R2

R_{eq} R_{1}+R_{2}

- Most generally, V
_{out }= V_{s}Rx

R_{eq }

- Attaching a load reduces the divider’s

Output voltage.

- V
_{out}= V_{s}R2

R_{1}+ R_{2}

- V
_{out}= V_{s}( R2IIRL) Where R_{2}II R_{L}< R_{2}

R_{1}+ [R_{2}II R_{L}]

- If the load resistance is much greater the resistance of R
_{2}then the output voltage only drops a bit. - i.e. The load current should be much less than the current
**.**

**Current Dividers-**

- A current divider creates a current which is some fraction of its current source.
- It is a parallel circuit where the output current is observed in one of the branches.
- R
_{T}= [1 + 1 ]^{-1}

R_{1} R_{2}

I_{out }= I_{s} RT

R_{x}+R_{T}

**Kirchhoff’s Current Law (KCL)-**

- Kirchhoff’s current Law states that all current entering anode equals all currents exiting a node.

- The sum of all current in a node =0.

n

∑ I_{j} =0

j=1

- I
_{1}+ I_{2}+I_{3}+…..+I_{n}= 0 - Entering a node = positive (+I)
- Exiting a node = negative (-I)

**Kirchhoff’s Current Law (KCL)-**

- Find I
_{1}and I_{2}using KCL. - Ask Yourself: Do I need to find one current before I can find the other ?
- KCL at Node B

I_{2} + 12mA – 4mA =0

I_{2} = – 8ma

- KCL at Node A

-I_{1} – I_{2} + 2mA = 0

-I_{1} + 8mA + 2mA =0

I_{1} = 10mA

**Nodal Analysis with KCL-**

- Nodal analysis is a process that uses KCL to determine node voltages.
- A reference node (ground) is used to make life easier.
- A current through a resistor is described by Ohm’s Law

I = Vx –Vy

R

- To find node voltage, write a KCL equation for each node, except the reference node. (N-1)

**Nodal Analysis with KCL-**

- KCL at (1)

I_{1} – I_{2} + I_{s} = 0

( Vs – V1) –(V1 – V2) + I_{s} = 0

R_{1} R_{2}

V_{1} (- 1 – 1 ) + V_{2} (1 ) = -I_{x} – Vs R_{1} R2 R1 R1 |

- KCL at (2)
- I
_{2}–I_{2}= 0

( V1 –V2 ) _ V2 = 0

R_{2} R_{3}

V1(1 ) + V_{2} ( -1 – 1 )=0 R_{2 }R_{2} R_{3} |

Nodal Analysis with KCL-

- KCL at

I_{1} – I_{2} + I_{s} = 0

( Vs – V1) –(V1 – V2) + I_{s} = 0

R_{1} R_{2}

_{ } V_{1} (- 1 – 1 ) + V_{2} (1 ) = -I_{x} – Vs

R_{1} R2 R3 R1

8 V_{1} = 4 + 12

7

V_{1} = 14 V |

V_{2} V_{1} R3 = 14 3

R_{2}+R_{3} 4+3

V_{2} = 6V =V_{o} |

**Kirchhoff’s Voltage Law (KVL)-**

- Kirchhoff’s Voltage Law states that all voltage” drops’ must equal all voltage “rise” in a closed loop.
- That is , the sum of all voltages in a closed loop = 0

n

∑ V_{j} = 0

j=1

- V
_{1}+ V_{2}+ V_{3}+ …+ V_{n}= 0

**Kirchhoff’s Voltage Law (KVL)-**

- Voltage Rise (-V)
- Voltage Drop (+V)
- Voltage Drop (+V)

**Kirchhoff’s Voltage Law (KVL)-**

- Find V
_{AB}using KVL. - First, we need to find the current around the loop.
- KVL starting at B

15+3/+1/-6+3/ + 2/ = 0

9 = – 9I

I = – 1A

V_{AB} = I (2Ω)

V_{AB} = (- 1 A) (2Ω) = -2V

**Loop Analysis with KVL-**

- Loop analysis is a process that uses KVL to determine loop currents.
- Loop current is assigned to all independent loops.
- To solve for all loop currents. Write a KVL equation for all independent loops and then solve.

**Loop Analysis with KVL-**

- KVL for (I
_{1}) - -15+2K (I
_{1}– I_{2}) + 1K (I_{1}) = 0 - -15+2K (I
_{1}– I_{2}) + 1K (I_{1}) = 0

1K 1K 1K 1K

-15mA + 2 (I_{1}– I_{2 }) + (I_{1})=0

3/_{1}– 2/_{2 }= 15mA |

- KVL for (I
_{2})

2K (I_{2}) + 2K (I_{2} –I_{1} ) + 4K (I_{2} –I_{3} ) =0

2K (I_{2}) + 2K (I_{2} – I_{1}) + 4K (I_{2}–I_{3} )= 0

2K 2K 2K 2K

I_{2} + (I_{2} – I_{1}) + 2 (I_{2} + 5mA) =0

4I_{2} –I_{1} = – 10mA |

2 (3I_{1} – 2I_{2} = 15mA) → 6I_{1} – 4 I_{2} = 30mA

-I1 + 4I2 = -10mA → -I1 + 4I2 = -10mA

5I_{1} = 20 mA

I_{1} = 4mA |

**Source Transformation-**

- Thevenin circuit : Voltage source in series with a resistor
- Norton circuit : Current source in parallel with a resistor
- Source transformation allows simple conversion between these circuits.

R_{TH} = R_{NO} = R

V_{TH} = I_{NO} . R

I_{NO} _{=} VTH

R

**Thevenin’s & Norton’s Theorems-**

- Thevenin Equivalent Circuit
- Norton Equivalent Circuit

**Thevenin’s & Norton’s Theorems-**

- To find R :
- 1.) Detach the load resistor (if present).
- 2.) Set all sources equal to 0.

Voltage source → Short- circuit

Current source → Open –circuit

- Find the equivalent resistance “looking in’’ from the two output terminals.

**Thevenin Equivalent Circuits-**

- To find V
_{Th} - 1.) Detach the load resistor (if present).
- 2.) Find the open- circuit voltage V
_{oc}across the circuit’s output terminals.

V_{oc} = V_{Th}

Use any form of analysis (nodal, loop, voltage divider source transformation, etc.)

**Norton Equivalent Circuits-**

- To find I
_{NO} - 1.) Detach the load resister (if present).
- 2.) Short circuit the output terminals
- Find the short-circuit current I
_{SC}through the shorted output.

I_{SC} =I_{NO}

- Find VAB using Norton’s theorem.

** Superposition Theorem-**

- Superposition says that a circuit with multiple sources can be solved by this process:
- 1.) Set all sources = 0, except one.
- 2.) Solve necessary currents and voltages, using only that source.
- 3.) Repeat Step 2 for each source.
- 4.) “Superimpose” the solutions onto each other.
- Use the symbol prime (‘) to differentiate variables with the same name.