Analog Electronics - 1 - Electrical Engineering (EE) MCQ

Voltage at secondary winding of the coil is V_s

Voltage across diode

V_p → V_{s max}

V_n → 12 V

For conduction:

V_p > V_n


Therefore the diode is conducting for the period.

For full wave rectifier using inductor filter at output is

When,

D₁ → ON; D₂ → OFF: R_AB = 10 kΩ

D₂ → ON; D₁ → OFF: R_AB = 10 kΩ

∴ The impedance by the circuit across the terminals A and B  of 10 kΩ

Clamping network “Clamp” a signal to different DC level.

During the Internal O → T/1 → Diode in the ON state during this internal output voltage is directly across the short circuit and V_o = 0 volts

When input switch to -V  state → diode OFF

Applying Kirchhoff’s voltage law around loop we get

-V -V -V_o = 0

V_o = -2V

So output wave form will be,

The above circuit can be redrawn as:

Assume diode is off:

Voltage across 20 k resistor connected to left of diode is 

This will turn on the diode 

the equivalent circuit is now:

Now the two 20k resistors are in parallel whose equivalent is 10k

this 10 k is in series with other 10k resistor 

9V divides equally among them

V₀ = 4.5V 

Cuurent I = 4.5/20 mA

= 0.225 mA

A) In graph ‘A’ the output follows input, but above certain positive input, output stays constant. The circuit is called positive clipper.

B) Similarly the graph ‘B’ is of negative clipper.

C) In this path positive and negative side are clipped, it is a double-ended clipper.

D) In this input output follows relation V_out = |V_in|. This is a full-wave rectifier circuit.

Due to filter capacitor the output voltage is given by waveform

Output DC voltage  V_avg = V_m = √2 × 230
= 325.3 V

The load current is given by 

This current discharges the capacitor to the load resistance during the interval (t₂ to t₃)

 The time period of AC voltage. 

= 20 ms (for 50 Hz)

Thus the charge supplied by the capacitor to the load resistance during this interval will be

∆Q = I_LT = 6.51 × 20 × 10^-3

= 0.1302 coulombs

Peak to peak ripple voltage

Given circuit can be written with ideal diode

Is input with


During positive values of V_i < s – V_k i.e. V_i < 4.3 V the diode is in on-state, the diode can be taken as short circuit. The output voltage is simply be voltage across 5V voltage source.


 

⇒ V_o = s – 0.7 V

= 4.3 V

For V_i > 4.3 V the diode is in off-state the circuit become open circuit hence V_o = V_i
 

The circuit is in on-state for entire negative cycle. Hence V_o = 4.3 V

The wave form of output voltage is
 

Time period = 1/f = 1/100 = 10msec

 

1) During +Ve ½ cycle

V_in = + Ve diode is not forward bias
 

2) During – Ve cycle

V_in = - Ve diode is Reverse Bias

Average value  = 0

Opening the current source we get the AC small-signal circuit (η = 1)


I = 157 uA