Why DIODE is a non linear device???

• Ideal Diode:

➝ We know that an Ideal Diode is work same as an Ideal switch as discussed earlier.

➝Ideal Diode does not consume or Absorb any power from the source it means it has zero internal resistance and power loss.

Fig(a) Symbol of an Ideal Diode

• DEFINITION:-

➝"An ideal diode is one type of electronic switch (Ideal Switch), it allows to pass the current in only one direction ( ANODE TO CATHODE)".

Fig(a) working of an Ideal Diode

✱ DURING POSITIVE HALF CYCLE OF AC SUPPLY:-

⇨During the positive half cycle of ac supply P (anode) terminal of diode is connected to the +Vin and N (cathode) terminal of diode is connected to the-Vin of ac supply, so diode is in forward bias and start conducting.

⇨During forward biasing of diode current I is start to increasing and flows through the load.

✱DURING NEGATIVE HALF CYCLE OF AC SUPPLY:-

⇨During the Negative half cycle of ac supply P (anode) terminal of diode is connected to the-Vin and N (cathode) terminal of diode is connected to the +Vin of ac supply, so diode is in reverse biase.

⇨We know that diode doesn't allow to flow current in reverse biase, so that whole Negative half cycle of ac supply is BLOCKED across the Diode.

Fig (c) waveform of given circuit

• From the waveform we clearly understand that during positive half cycle of ac supply output is changed with respect to input but during the negative half cycle of ac supply, output remain ZERO during whole Negative half cycle.

➝So, from the above discussion and waveform we conclude that AN DIODE IT SELF SOWS NON LINEARITY, it means output does not follow input.

             ................Thank you 🙂................

How PRACTICAL DIODE works???

• To understand the working of practical diode, first we need to know the difference between an IDEAL DIODE and PRACTICAL DIODE. 

• Difference between an Ideal Diode and Practical Diode.

1. An Ideal Diode is behaves same as an Ideal switch. There is zero voltage drop across it.
2. But Practical Diode is replaced by its cut-in Voltage.
3. An Ideal Diode doesn't consume any power from the source during its operations.
4. Practical diode consume some amount of power from the source.
5. Ideal diode is replaced by short circuit when it is forward bias whereas an practical diode is replaced by an cut-in Voltage when it is forward bias. 

❋PRACTICAL DIODE:-

⇨Practical diode is conducted if an only the supply voltage is greater than its cut-in Voltage.

"Cut-in Voltage is the minimum potential required for satisfactory conduction of diode is known as a cut-in Voltage".

Fig.(a) Symbol of Practical Diode

• Cut-in Voltage is denoted by =Vλ
• If diode is Made up if GERMANIUM = Vλ = 0.3 Volts
• If diode is Made up of SILICON = Vλ = 0.7 Volts.

Fig(b) Use of Practical Diode in Circuit

Fig(c) Out put waveform

❇️ DURING POSITIVE HALF CYCLE OF AC SUPPLY:-

• At t=0 ; the value of Vin = 0 volts so diode D doesn't conduct this state of diode is called as a FORWARD BLOCKING MODE, when Vin> 0.7 Volt (for silicon) at that time diode is start to conducting, this state of diode is called as a forward conduction mode. During forward conduction mode diode is replaced by short circuit and current start to flow from P to N terminal of diode, Load and Vin and output voltage is obtained across the load RL.

❇️ DURING NEGATIVE HALF CYCLE OF AC SUPPLY:-

• During Negative half cycle of ac supply P terminal of diode is connected to the -Vin and N terminal of diode is connected to the +Vin of ac supply so diode is reverse biase and out put voltage across the load is zero during whole Negative half cycle as shown in waveform.

☺V-I characteristics of an practical diode:-

Fig(d) I-V characteristics of an practical diode

Where;  Vλ = Cut-in Voltage

                I0. = Reverse saturation current

• Reverse saturation current:-

" It is current of diode which is flows due to the minority carriers".

            ...................Thank you 🙂.................

How Ideal Diode is work??

• IDEAL DIODE:-

Fig(a) Symbol of Diode

⇒An ideal diode is behaves same as an Ideal switch.

• Behaviour of an Ideal Switch:

Concept of MUTUAL INDUCTUNCE and MAGNETIC COUPLING

• Definition of Mutual Induced emf:-

➝ " It is defined as the emf is induced in the coil due to the rate of change of flux is occurs in the nearest coil which is linked with it".

• EXPLANATION:-

Fig.(a) MUTUAL INDUCED VOLTAGE

Whre; 

             N1 = Numbers of turns of Coil A

             N2 = Numbers of turns of Coil B

             R.   = Variable Resistor

              G.  = Galvanometer

         Φ12. = Flux of coil A is linked with Coil B

         Φ21 = Flux of Coil B is linked with Coil A

➝ When supply voltage V is connected across Coil A then current I1 is flows through the coil A due to this current flux Φ1 is produced in the coil A, the direction of this flux Φ1 is found by RIGHT HAND CURL RULE as discussed in previous article.

➝ According to RIGHT HAND CURL RULE if the direction of current is in the direction of Curl then right hand thumb will represent the direction of magnetic field as shown in below figure.

➝ Some portion of this flux Φ1 is called as Φ12 is linked with Nearest Coil B, Now if we change the current I1 by changing the value of Variable Resistor R then flux Φ12 which is linked with Coil B is also change and due to this rate of change of Flux in Coil A, EMF is induced in nearest coil B and vice varsa. The direction of induced EMF in Coil B is found by RIGHT HAND CURL RULE.

• What is MUTUAL INDUCTUNCE?

⇨Mutual inductance is the property of the coil which Oppose the change of current in nearest coil.

                   M12 = N2 × Φ12/ I1 ...............(1)

                 M21 = N1 × Φ21 / I2 .............(2)

Now,. Suppose M12 = M21 = M

• Coefficient of Coupling (K):-

➝"Coefficient of Coupling K is describe the capability of flux linkages" OR it is defined as the fraction of the total flux that links with the Coil.

                 i.e.   K = Φ12/ Φ1 = Φ21/ Φ2 .......(3)

• Multiply equation number (1)&(2) with equation number (3) ..............

  M^2 = N1 × N2 × Φ12×Φ21/ I1×I2  × Φ1Φ2/ Φ1Φ2 .............(4)

• Rearrange term in above equation.......

    = N1Φ1/ I1 × N2Φ2/I2 × Φ12/ Φ1 × Φ21/Φ2 

     =  L1 × L2 × K × K 

• NOTE:- 

           L1 = N1Φ1/ I1

           L2 = N2Φ2/I2

             K = Φ12/ Φ1

             K = Φ21/Φ2

        ∴ M^2 = L1 × L2 × K^2

              M = K × √L1×L2 ...........(5)

        ∴ K = M/ √L1×L2 .............(6)

        ............Thank you 🙂.............

Concept of SELF INDUCTUNCE.

What is SELF INDUCTUNCE?

⇨ If an time varying current is passing through the Coil or Winding, it produces the time varying flux this flux is linked with it self and voltage is induced in the coil or Winding is called as a SELF INDUCED VOLTAGE.

                   V = L di/dt ................(1)

Fig.(a) SELF INDUCED VOLTAGE

• Why inductor oppose the rate of change of current?

➝we know that due to time varying current self induced voltage is induced in coil, this is shown by V. Magnetic field within it is try to maintain its value constant. But if at any intant supply current is disturb then this self induced voltage create a current which Oppose the rate of change of supply current (according to Lenz's law).

Now, the inductor L is expressed as,

                          L = NΦ/ I ................(2)

Where; N = Number of turns of Coil

               Φ = Flux linkage with coil

               I = Current passing through the

Coil

 

• Substitute equation number (2) in (1)......

                            I = NΦ/L ..................(3)

Now putting the value of I in equation number (1) we get.........

               V = L × d (NΦ/L) / dt

             ∴  V = L × 1/L × N  dΦ/dt

             ∴ V = N × dΦ/dt ..............(4)

• Comparing equation number (1)& (4)........

           L × di/dt = N × dΦ/dt

As seen "dt" is cancelled out ......

Finally we get value of INDUCTUNCE "L" is......

                       L = N × dΦ/ di 

        ............Thank you 🙂............

Concept of SUPER MESH.

"If an IDEAL CURRENT SOURCE (Dependent OR Independent) is common between two meshes then it is referred as a SUPER MESH by avoiding Ideal current source".

Q. Why Concept of SUPER MESH is developed?

➝ The concept of SUPER MESH is developed because it reduce the number of meshes in the network.

➝ In previous article ( If we doesn't have any idea about SUPER NODE then what we do???) We learn that if any ideal voltage source is present in the Network then we need to assume current of an Ideal Voltage Source similarly if any ideal current source is present in the network then we need to assume voltage of an Ideal current source. 

➝so from above discussion we conclude that concept of SUPER MESH is developed to avoid assumption of Voltage of an Ideal current source.

• Important points regarding SUPER MESH:-

1. A SUPER MESH has no current of it's own.
2. A current source in SUPER MESH provide equation which is necessary for solving of Mesh Current.

☺ Consider the Network as shown in below figure, Find the Value of current "I" ______in Amp.

Fig.(a) Find the value of I

• SOLUTION:-

If any ideal current source is COMMON between two meshes then super mesh is form by avoiding Ideal current source present in the network.

Fig(b) Represent Super Mesh 

In this figure ;

                I1 =  current flow through Mesh(loop) 1

                I2 =  Current flow through Mesh (loop) 2

Blue Dotted Line Represent SUPER MESH please observed in Figure.

• KVL OF SUPER MESH:-

      -10 + I1×1 + I2×3 + 5 = 0 

   ∴ I1 + 3I2 = 5 .................(1)

• KCL OF SUPER MESH:-

          I2 - I1 = 2 .................(2)

NOTE; in figure I2 is in the Direction of 2 AMP current source so it's direction is taken as POSITIVE and I1 Oppose the 2AMP current source so its direction is taken as a NEGATIVE please seen in figure (b).

• Adding equation number (1)&(2) we get.....

           I1 + 3I2 =5

           -I1 + I2 = 2                       

         ______________

                    4I2 = 7 

So,.           I2 = 7/4 AMP.

AS SHOWN IN FIGURE ;

                          I2 = I 

                 

                      

                  I = 7/4 AMP.   

                 

Can we do supernode questions by using only nodal analysis?

• Answer is YES.

➝ If we don't know about super node then we used simple node equation to solve any network contain super node.

• Before understanding this concept first we need to understand basics of Practical voltage source and Ideal Voltage Source:-

✱Practical Voltage Source:

➝ Voltage source in series with Internal Impedance (Zin) is referred as practical voltage Source.

➝Value of this internal Impedance is as low as possible,  Because if the value of Zin is high then Voltage drop across Zin is also high so voltage at load point or Terminal is Low. So it is required to choose the value of Zin is as low as possible.

Fig (a) Practical voltage source

✱ Ideal Voltage Source:

➝ Internal Impedance Zin of any practical voltage source is tends to ZERO is referred as Ideal Voltage Source.

➝It is important to know that "CURRENT OF ANY IDEAL VOLTAGE SOURCE IS NOT ZERO" , Because resistance of any ideal voltage source is tends to Zero not exactly Zero so that CURRENT  through the source is tends to Infinite, It means  Ideal Voltage Source is capable to deliver unlimited power.

Fib(b) Ideal Voltage Source

➡️ Consider the Network as shown in below figure, Find the Value of Vp_____& Vq_______in Volts Without using concept of SUPER NODE.

In this above network 5V source is Ideal Voltage Source so we assume current of an Ideal voltage source is "I". Direction of this current is assuming Negative terminal to Positive Terminal ( because current is always flow from Higher potential to lower potential).

• Applying Simple Node equation at Node Vp..............

       Vp-10/1 + Vp/2 - I =0 .............(1)

• Applying Simple Node equation at Node Vq.....,.......

       Vq/3 + Vq-10/4 + I =0 ..............(2)

➝Adding equation number (1)&(2).............

At that time "I" will be canceled out Please observed in equation.........

After adding equation number (1)&(2).........

   Vp-10/1 + Vp/2 + Vq/3 + Vq-10/4 = 0 ........(3)

Please observed equation number (3) is Same as Super Node equation......

      Vp- Vq = 5 .............(4)

Now find out Vp and Vq from Equation Number (3) & (4)............

Multiply by 12 in Equation Number (3)........

12Vp - 120 + 6Vp + 4Vq + 3Vq - 30 =0

Simplified........

18Vp + 7Vq =150 .............(5)

Now multiply 7 in Equation Number (4).........

7Vp - 7Vq = 35 ............(6)

Adding equation number (5)&(6)...........

After adding we get 

        25Vp = 185

∴ Vp = 7.4 Volt

Put value of Vp in equation Number (4.).....

            5 = Vp - Vq

        ∴ Vq = 7.4 - 5

           Vq = 2.4 Volt

So without using concept of SUPER NODE we can easily solving Any questions.

          ...............Thank you 🙂...............