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

V1  (- 1 – 1 ) + V2  (1 )  = -Ix – Vs        R1  R2           R1           R1
  • KCL at (2)
  • I2 –I2 = 0

( V1 –V2 ) _ V2 = 0

   R2             R3

V1(1 ) + V2 ( -1  –  1  )=0     R2                         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

V1 = 14 V

                                                                          V2 V1   R3 = 14 3

                                                                                  R2+R3    4+3

V2 = 6V =Vo

 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

3/1– 2/2 = 15mA
  • 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

4I2 –I1 = – 10mA

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

                                                                               -I1 + 4I2 = -10mA → -I1 + 4I2 = -10mA

                                                                                                                          5I1         = 20 mA

I1   = 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.

      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.

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