## Superposition Theorem with Solved Example

Hi friends, here are going to learn a superposition theorem. We will also solve on simple example using superposition theorem.

### Statement:

Superposition theorem states that, the responce in any element of LTI linear, bilateral network containing more than one sources is the sum of the responses produced by the sources each acting independently.

The superposition theorem is not applicable for the power, as power is directly proportional to the square of the current which is not a linear function.

### Steps:

1)      Select any one source and short all other voltage sources  and open all current sources if internal impedance is not known. If known replace them by their impedance.

2)      Find out the current or voltage across the required element, due to the source under consideration.

3)      Repeat the above steps for all other sources.

4)      Add all the individual effects produced by individual sources to obtain the total current in or across the voltage element.

Example: Using the superposition theorem, determine the voltage drop and current across the resistor 3.3K as shown in figure below.

Solution:

Step 1: Remove the 8V power supply from the original circuit, such that the new circuit becomes as the following and then measure voltage across resistor.

Here 3.3K and 2K are in parallel, therefore resultant resistance will be 1.245K.

Using voltage divider rule voltage across 1.245K will be

V1= [1.245/(1.245+4.7)]*5 = 1.047V

Step 2: Remove the 5V power supply from the original circuit such that the new circuit becomes as the following and then measure voltage across resistor.

Here 3.3K and 4.7K are in parallel, therefore resultant resistance will be 1.938K.

Using voltage divider rule voltage across 1.938K will be

V2= [1.938/(1.938+2)]*8 = 3.9377V

Therefore voltage drop across 3.3K resistor is V1+V2 = 1.047+3.9377=4.9847

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## Full wave bridge rectifier:

This post provides an information about full wave bridge rectifier. We will also see its working principle and advantages and disadvantages.

In full wave Bridge rectifier a transformer and four diodes are used. During the positive half cycle of secondary voltage, the diodes D2 and D4 are forward-biased, but diodes D1 and D3 do no conduct. The current is through D2, R, D4 and secondary winding.

During the negative half cycle, the diodes D1 and D3 are forward-biased, but diodes D2 and D4 do not conduct. The current is through D1, secondary winding, D3 and R.

The load current is in the same direction in both half-cycles. Therefore a unidirectional (d.c.) voltage is obtained across load resistor.

Average voltage = Vdc = 0.636 Vp = 2Vp/π

Where Vp = peak value of secondary voltage.

Since each diode conducts for only half cycle, the current rating (Io) of the diodes must be at least – half of the dc load current. i.e. 0.5 Idc.

Each diode must withstand a peak inverse voltage equal to the peak secondary voltage. PIV = Vp.

Therefore the PIV rating of the diodes must be greater than Vp. As the output is a full-wave signal, the output frequency is double the input frequency.

The maximum efficiency of bridge rectifier is 81.2%

Advantages of full wave bridge rectifier:

• Centre tap on the secondary of the transformer is not necessary.
• Small transformer can be used.
• For a bridge rectifier circuit PIV per diode is one-half of the value for each diode in a full-wave rectifier.

• In this type two extra diodes are used.
• The voltage regulation is poor.

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## Half Wave Rectifier:

This post provides an information about half wave recifier and its working principle.

In this type only one diode is used. Generally a step-down transformer is used to provide the required secondary voltage. The transformer isolates the load from the line. This reduces the possibility of electric shock.

In the positive half cycle of secondary voltage, the diode is forward biased for voltages greater than the offset voltage. The offset voltage is 0.7 V for silicon diodes and 0.3 V for germanium diodes. This produces a half sine wave of voltage across the load resistor.

In the negative half cycle, the diode is reverse biased. The load current drops to zero.

The load current is always in the same direction. This provides rectification.

Average Voltage = Vdc = Vp/π = 0.318 Vp

Where, Vp = peak value of voltage across secondary. The value of direct current, the diode can handle is called as ‘Current rating of diode’ (Io).

Peak Inverse Voltage – In the negative half – cycle, the diode is reverse-biased. All secondary voltage appears across the diode. The maximum negative (reverse) voltage appearing across the diode is called the ‘Peak Inverse Voltage’ (PIV). To avoid break down, the PIV must be less than PIV rating of the diode. For half wave rectifier PIV = Vp.

Maximum efficiency of half wave rectifier is 40.6 %.

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