# Best Topics for Hardware Design Engineer

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.
• Req = R1+R2+ …..+Rn
• Req = 100Ω+50 Ω = 150 Ω
• Applying Ohm’s Law,

I=   30V  =  30V = 0.24

Req     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

Req R1 R2          Rn

Req =[ 1   + 1  ]-1=60Ω

150Ω   100Ω

Ohm’s Law  I = 30V = 30V 0.5A

Req    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.

Req= 50Ω

I = 10V = 0.2A

50Ω

Vout = IR2 = (0.2A)(10Ω) = 2.0

• We can write a general  equation that describes the output of any two-resistor voltage divider.
• Req = R1 + R2
• I = Vs

Req

• Vout  = IR2
• Vout =[  Vs ] R2 = Vs R2

Req            R1+R2

•  Most generally, Vout  = Vs Rx

Req

• Attaching a load reduces the divider’s

Output voltage.

• Vout= Vs R2

R1+ R2

• Vout = Vs ( R2IIRL) Where R2II RL < R2

R1+ [R2II RL]

• If the load resistance is much greater the resistance of R2 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.
• RT= [1  +  1 ] -1

R1    R2

Iout = Is RT

Rx+RT

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

∑ Ij =0

j=1

• I1+ I2+I3+…..+In = 0
• Entering a node = positive (+I)
• Exiting a node = negative (-I)

Kirchhoff’s Current Law (KCL)-

• Find I1 and I2 using KCL.
• Ask Yourself: Do I need to find one current before I can find the other ?
• KCL at Node B

I2 + 12mA – 4mA =0

I2 = – 8ma

• KCL at Node A

-I1 – I2 + 2mA = 0

-I1 + 8mA + 2mA =0

I1 = 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)

I1 – I2 + Is = 0

( Vs – V1) –(V1 – V2) + Is = 0

R1               R2

• KCL at (2)
• I2 –I2 = 0

( V1 –V2 ) _ V2 = 0

R2             R3

Nodal Analysis with KCL-

• KCL at

I1 – I2 + Is = 0

( Vs – V1) –(V1 – V2) + Is = 0

R1               R2

V1  (- 1 – 1 ) + V2  (1 )  = -Ix – Vs

R1  R2            R3              R1

8 V1 = 4 + 12

7

V2 V1   R3 = 14 3

R2+R3    4+3

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

∑ Vj = 0

j=1

• V1 + V2 + V3 + …+ Vn = 0

Kirchhoff’s Voltage Law (KVL)-

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

Kirchhoff’s Voltage Law (KVL)-

• Find VAB 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

VAB = I (2Ω)

VAB = (- 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 (I1)
• -15+2K (I1 – I2) + 1K (I1) = 0
• -15+2K (I1 – I2) + 1K (I1) = 0

1K   1K                 1K         1K

-15mA + 2 (I1– I2 ) + (I1)=0

• KVL for (I2)

2K (I2) + 2K (I2 –I1 ) + 4K (I2 –I3 ) =0

2K (I2) + 2K (I2 – I1) + 4K (I2–I3 )= 0

2K          2K             2K            2K

I2 + (I2 – I1) + 2 (I2 + 5mA) =0

2 (3I1 – 2I2 = 15mA) → 6I1 – 4 I2 = 30mA

-I1 + 4I2 = -10mA → -I1 + 4I2 = -10mA

5I1         = 20 mA

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.

RTH   = RNO   = R

VTH = INO . R

INO = 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 VTh
• 1.) Detach the load resistor (if present).
• 2.) Find the open- circuit voltage Voc across the circuit’s output terminals.

Voc = VTh

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

Norton Equivalent Circuits-

• To find INO
• 1.) Detach the load resister (if present).
• 2.) Short circuit the output terminals
• Find the short-circuit current ISC through the shorted output.

ISC =INO

• 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.