Physics help 2

Question

Question 3

3.1 Draw the symbols and characteristic curves of the following diodes. Also give an application of each diode:

3.1.1 Tunnel diode

3.1.2 Varactor diode

3.1.3 Zener diode

 

3.2 A power supply output voltage of 50 V DC when the load is disconnected. When the load is connected the voltage drops to 40 V DC.

Calculate the percentage voltage regulation.

 

 

 

 

 

Question 4

4.1 Choose a word from those in brackets. Write only the word next to the question number (4.1.1 – 4.1.5) in the answer book.

4.1.1 The field effect transistor is (less/more) sensitive to the temperature than the ordinary transistor.

4.1.2 The uni-junction transistor can turn (on/off) only when the emitter-to-base 1 is forward biased.

4.1.3 The input impedance of a field effect transistor amplifier is very (low/high).

4.1.4 The MOSFET is a metal-oxide, semiconductor field effect transistor, whose gate voltage control the (gate/drain) current.

4.1.5 The enhancement type MOSFET has no physical channel between drain and source (true/false).

4.2 Draw a neat, labelled characteristic surve of an N-channel, junction field effect transistor.

4.3 A common base amplifier produces an output current change of 4 mA with an input current of 2 µA. The output voltage changes with 6 V by means of 2 V input voltage.

Calculate the following:

4.3.1 hib

4.3.2 hrb

 

Question 5

5.1 Draw neat, labelled circuit diagrams of the following operational amplifiers:

5.1.1 Noninverter

5.1.2 Integrator

5.1.3 Inverter

5.1.4 Diffentiator

 

5.2 Calculate the input resistance of an integrator amplifier if C = 15 µF, Vin = 5 V and the rate of change of the output voltage is 20 v/sec.

 

 

Answer 1

(3.1.1)

Tunnel Diode :

A tunnel diode is a highly doped semiconductor and is mainly used for low voltage high frequency switching applications.

It works on the principle of Tunneling effect.

Circuit Symbol :

The tunnel diode is a two terminal device with p type semiconductor as anode and n-type semiconductor as cathode

The circuit symbol of tunnel diode is shown.

Characteristic Curve :

When the tunnel diode is forward voltage, the current increases and reaches the peak current.

If the voltage is applied to the tunnel diode beyond the peak voltage, the current starts decreasing due to tunneling effect. This is negative resistance region. Thereafter the current after the valley point starts increasing.

In reverse bias condition, tunnel diode conducts reverse bias current.

The V-I characteristics of the Tunnel diode is

Applications of Tunnel diode :

It is used as very high speed switches.

It is used as high frequency microwave oscillator.

It is used as an amplifiers and a mixer.

Answer 2

(3.1.2)

Varactor diode :

Varactor diodes or varicap diodes is a voltage controlled variable capacitance  type of semiconductor device.

Circuit Symbol :

A varactor diode uses a p-n junction in reverse and has a structure such that the capacitance of the diode varies with the reverse voltage.

Characteristic Curve :

When the varactor diode is forward biased, the diode acts as a p-n junction diode.

When the varactor diode is reverse biased, the depletion region size increases and the capacitance of the diode can be varied.

Applications of Varactor diode :

Varactor diodes is used in RF filters.

Varactor diode as used in voltage controlled oscillators.

Varactor diodes is used as RF phase shifters and resonant circuits.

Answer 3

(3.1.3)

Zener Diode :

Zener diode is a special type of the diode that allows current to flow even reverse bias condition.

Circuit Symbol :

Characteristic Curve :

When a zener diode is forward biased, it operates as a p-n junction diode.

When a zener diode is reverse biased, no current flows until the breakdown voltage is reached. Once the breakdown voltage is reached, there is sharp increase in the reverse current and acquires constant.

Applications of Zener diode :

It is used as Voltage stabilizers or regulators.

It is used as surge suppressors for device protection.

It is used as peak clippers, switching operations, reference elements.

Answer 4

(3.2)

Step 1:

The no-load voltage is 50 v.

The full load voltage is 40 v.

Find the percentage voltage regulation.

Percentage voltage regulation : ,

where is the no load voltage and is the full load voltage.

Percentage voltage regulation is .

Solution :

Percentage voltage regulation is .

Answer 5

(4.1.1)

The field effect transistor(FET) is less sensitive to the temperature than the ordinary transistor.

Explanation :

Increasing in temperature, tends to decrease the mobility of the charge carriers in the channel, reducing the current in FET.

Increasing the temperature also narrows the depletion width, in turn increasing the drain current.

(4.1.2)

The Uni - Junction transistor can turn ON only when the emitter-to-base 1 is forward biased.

Explanation :

The Uni - Junction Transistor is always operating with forward bias voltages.

For this purpose the Emitter Base 1 Junction is always kept forward bias with respect to Base 2.

(4.1.3)

The input impedance of a field effect transistor amplifier is very high.

Explanation :

 An amplifier must have very high input impedance because it is used to prevent loading effect at the input terminal.

(4.1.4)

The MOSFET is a metal-oxide, semiconductor field effect transistor, whose gate voltage control the drain current.

Explanation :

The operating principle of a MOSFET is a voltage controlled majority carrier device.

Movement of majority carriers in a MOSFET is controlled by the voltage applied on the control electrode (called gate).

The electric field produced by the gate voltage modulate the conductivity of the semiconductor material in the region between the main current carrying terminals called the Drain (D) and the Source (S).

(4.1.5)

The enhancement type MOSFET has no physical channel between drain and source.

The statement is true.

Explanation :

With no bias voltage applied to the gate terminal, there exists two back-to-back p-n junctions between the drain and the source.

No current flows from drain to source (the resistance will be on the order of 10¹² Ω).

Answer 6

(5.1.1)

Step 1:

Noninverter amplifier :

The input voltage V_in is applied directly to the non-inverting (+) input terminal, which means that the output gain of the amplifier becomes positive.

Therefore the output signal is in-phase with the input signal. So it acts as a non-inverting amplifier.

The voltage gain of the non-inverting amplifier is .

Answer 7

(5.1.2)

Step 1:

Integrator amplifier :

If 0 volts is applied to the input voltage, then there will be no current flows through the resistor, therefore no charging of the capacitor. Then the output voltage will not change. Hence the output voltage will be constant.

Apply positive voltage to the input, then the output of the op-amp will fall negative at a linear rate.

Now apply negative voltage to the input, then the output will rising at a linear rate.

The output voltage rate-of-change will be proportional to the value of the input voltage.

The ouput voltage of the integrator amplifier is .

Answer 8

(5.1.3)

Step 1:

Inverter amplifier :

The input voltage V_in is applied directly to the inverting (-) input terminal, which means that the output gain of the amplifier becomes negative.

Therefore the output signal is out-phase with the input signal. So it acts as a inverting amplifier.

The voltage gain of the inverting amplifier is .

Answer 9

(5.1.4)

Step 1:

Differentiator amplifier :

Apply linear positive voltage to the input, then the output of the op-amp will constant negative voltage.

Now apply linear negative voltage to the input, then the output of the op-amp will constant positive voltage.

The input voltage rate-of-change will be proportional to the value of the output voltage.

The ouput voltage of the differentiator amplifier is .

Answer 10

(5.2)

Step 1:

The capacitance of the integrator amplifier is .

Input voltage is .

If positive voltage is applied to the input, then the output of the op-amp will fall negative at a linear rate.

The rate of change of the outpuit voltage is .

The ouput voltage of the integrator amplifier is .

The input resistance of the integrator amplifier is .

Solution :

The input resistance of the integrator amplifier is .