Diode Circuits-Set-1

7 (B) Analog Circuits Topic-wise Previous Solved GATE Papers Diode Circuits-Set-1

[1992 : 2 M] The 6V Zener diode shown below has zero Zener resistance and a knee current of 5mA. The minimum value of R. So that the voltage across it does not fall below 6 V is

(a) 1.2 K $\Omega$

(b) 50 $\Omega$

(d) 0 $\Omega$

Ans. (c)

Explanation:- The current is calculated as,

I\;=\frac{10-6}{50}=\frac4{50}=80mA

I=I_z+I_L=I_{Zmin}+I_{Lmax}=I_{Zmax}+I_{Lmin}

I_Zmin=5mA

80=5=I_Lmax

I_Lmax=75 mA

I_{Lmax}=\frac{V_L}{R_{min}}

R_{min}=\frac{V_L}{I_{Lmax}}=\frac6{75\times10^{-3}}=80\Omega

The minimum value of R is 80 \Omega[/latex]

Therefore, option (c) is the correct answer.

[1993 : 1 M] The wave shape of V₀ in figure is

Ans.(a)

Explanation"-

Let us consider two cases,

Case 1 :During + ve half cycle

Diode D_A is forward bias, so D_A is short circuit. Diode D_B is reverse bias , so D_B is in conducting state when V _i > 4.1 V

Case 2: during -Ve half cycle

Diode D_B is forward bias, so D_Bis short circuit.

Diode D_A is reverse bias, so D_A is in conducting state when | V_i | > 4.1 V.

[1995 : 1 M] The Ebers Moll model is applicable to

(a) Bipolar junction transistors

(b) NOMS transistors

(c)Unipolar junction transistors

(d) Junction filed-effect

Ans. (a)

Explanation

Ebers Moll model is a composite model and is used to predict the operation of BJT all of its possible modes. It consists of two ideal diodes placed back to back with reverse saturation current I_EO and I _CO and two dependent current controlled sources shunting the diodes.

[1996 : 2 M] A Zener diode in the circuit shown in below figure has a knee current of 5 mA, and a maximum allowed power dissipation of 300 mW. What are the minimum and maximum load currents that can be drawn safely from the circuit, keeping the output voltage V₀ constant at 6 V?

(a) 0mA,180 mA

(b) 5mA, 110mA

(d) 60 mA,180 mA

Ans. (c)

explanation:- Given that,

Current through 50 $\Omega$ resistance =I

I=\frac{9-6}{50}=\frac3{50}=60mA

I_Zmin = 5 mA

P_Zmax = I_zmax V_z = 300mW

The Z_max is calculated as,

I_{Zmax}=\frac{300\times10^{-3}}{V_Z}=\frac{300\times10^{-3}}6=50mA

I=I_Zmin+I_Lmax=I_Zmax +I_Lmin

60=50+I _Lmin

I_Lmin=10mA

60=5=I_Lmax

I_Lmax= 55mA

The minimum and maximum load currents is 10 mA and 55 mA .

So, option (c) is the correct answer.

[1998 : 1 M] For small signal ac operation ,a practical forward biased diode can be modeled as

(a) a resistance and a capacitance

(b) an ideal diode and resistance in parallel

(d) a resistance

Ans. (d)

Explanation

A resistance : For small signal ac operation, a practical forward biased diode can be modeled as a resistance.

Option (d) is the correct answer.

[1998 : 1 M] For full wave rectification, a four diode bridge rectifier is claimed to have the following advantages over a two diode circuit.

1. Less expensive transformer

2. smaller size transformer and

3. suitability for higher voltage application of these

(a) only (1) and (2) are true

(b) only (1) and (3) are true

(d) (1),(2) ,as well as (3) are true

Ans. (d)

In general, A four diode bridge rectifier uses the smaller size of transformer, which is less expensive transformer and these rectifiers are suitable for higher voltage applications, because of low PIV rating required of each diode.

Option (d) is the correct answer. As all statements are true.

[1999 : 2 M] A dc power supply has a no- load voltage of 30 V, and a full-load of 25 V at a full -load current of 1 A. Its output resistance and load regulation, respectively, are

(a) 5 $\Omega$ and 20%

(b) 25 $\Omega$ and 20%

(d) 25 $\Omega$ and 16.7%

Ans. (b)

explanation:- The voltage regulation is given by,

Regulation = $\frac{V_{NL}-V_{FL}}{V_{FL}}=\frac{30-25}{25}=\frac15=20\%$

The output resistance is given as,

R_o=\frac{V_{DCNL-}V_{DCFL}}{I_{DC}}=5\Omega

Option (b) is the correct answer.

[2001 : 2 M] The transistor shunt regulator shown in the figure has a regulated output voltage of 10 V, when the input varies from 20V to 30 V. the relevant parameters for the zener diode and the transistor are :V_z =9.5, V_BE=0.5 v, $\beta$ =99 .Neglect power dissipated in the zener diode (P_Z) and the transistor (P_T )are

(a) P_z=75mW ,P_T=7.9 W

(b) P_z= 85mW , P_T =8.9 W

(d) P_z=115 mW ,P_T=11.9 W

Ans. (c)

Explanation:- The given circuit is,

I_1=\frac{30-10}{20}=1A\;\;(i.e.when\;I_z=0)

$I_E=I_c+I_Z$ .....(i)

I_B=I_z\;(as\;no\;current\;flows\;in\;R_B)

\beta=\frac{I_c}{I_B}=\frac{I_c}{I_z}

\Rightarrow I_c=\beta I_z

From (i) , $I_E\;=\beta I_z+I_z=\left(99+1\right)I_z$

I_E\;=100\;I_z

I_z\;=\frac{I_1}{100}=\frac1{100}=0.01A

The power dissipated in zener diode is given by,

P_z\;=V_zI_z=9.5\times0.01=95mW

I_c=99I_z=99\times0.01=0.99A\\

P_c=V_cI_c=10\times0.99=9.9W\\

Option (c) is the correct answer.

[2002 : 2 M] A zener diode regulator in the figure is to be designed to meet the specifications: I_L =10 mA,V₀ =10 V and V_in varies from 30 V to 50 V. The zener diode has V_z =10 v and I_zk (knee current)=1mA .For satisfactory operation

(a) $R\leq1800\Omega\\$

(b) $2000\Omega\leq\;R\;\leq2200\Omega\\$

(d) $\;R\;\leq4000\Omega\\$

Ans. (a)

Explanation: The given circuit is,

\frac{V_{in}-V_o}R\geq I_z+I_L(I_1)\;\;\;\;\left(I_z+I_L=I_1\right)

As it is given that V_in varies from 30 V to 50 V.

i) When V_in =30V

\frac{30-10}R\geq\left(10+1\right)mA

\frac{20}R\geq11mA

\Rightarrow R\leq1818\Omega

ii) when V_in =50V

\frac{40}R\leq11\times10^{-3}

R\leq3636

From (i) and (ii) $R\leq1818\Omega$

Option (a) is the correct answer.

[2003 : 1 M] The circuit shown in the figure is best described as a

(a) bridge rectifier

(b) ring modulator

(d) voltage doubler

Ans. (d)

Explanation :- A voltage doubler circuit outputs a DC voltage that is double the peak value of the AC input voltage, without using a transformer. It is voltage doubler circuit because here input voltage is AC and the output voltage is DC voltage with twice the peak value of the input AC voltage.

The given circuit is a voltage doubler circuit.

So, option (d) is the correct answer.

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