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Semiconductor Electronics: Materials, Devices and

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3 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

In the circuit shown below, maximum zener diode current will be _________mA.

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Payal Gupta

Contributor-Level 10

$l = \frac{120 - 60}{4000} = 0.015 A$

Thus l₂ = l - L_L

= 0.015 – 0.006

= 0.009 = 9mA

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3 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

In a given circuit diagram, a 5 V zener diode along with a series resistance is connected across a 50 V power supply. The minimum value of the resistance required, if the maximum zener current is 90 mA be________ $Ω$ .

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Vishal Baghel

Contributor-Level 10

For 90 mA in zener diode, current in R should be greater than 90 mA.

$\frac{45}{R} = 90 * 1 0^{- 3} \Rightarrow R = \frac{45}{90 * 1 0^{- 3}} = 500 Ω$

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3 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

Consider a situation in which reverse biased current of a particular P – N junction increase when it is exposed to a light of wavelength $\leq$ 621 nm. During this process, enhancement in carrier concentration takes place due to generation of hole – electron pairs. The value of band gap is nearly.

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Vishal Baghel

Contributor-Level 10

$E = \frac{h c}{λ} = \frac{1242 e V - n m}{621 n m} = 2 e V$

minimum 2eV is required to emit.

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3 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

A zener diode having zener voltage 8V and power dissipation rating of 0.5W is connected across a potential divider arranged with maximum potential drop across zener diode is as shown in the diagram.

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Vishal Baghel

Contributor-Level 10

$P_{z} = V_{z} l_{z} \Rightarrow 0.5 = 8 l_{z} \Rightarrow l_{z} = l_{P} = \frac{1}{16} A m p$

When zener is connected across a potential divider arranged with maximum potential drop across zener diode, then

$V_{P} = V - V_{z} = 20 - 8 = 12$ volt Potential difference across protective resistance RP

$R_{P} = \frac{V_{P}}{I_{P}} = \frac{12}{\frac{1}{6}} = 192 A m p .$

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3 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

For the forward biased diode characteristics shown in the figure, the dynamic resistance at I_D = 3mA will be_________ $Ω .$

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Vishal Baghel

Contributor-Level 10

$R = \frac{Δ V}{Δ l} = \frac{0.75 - 0.70}{(5 - 3) * 1 0^{- 3}} = \frac{0.05 * 1000}{2} = 25 Ω$

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3 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

In the given figure, each diode has a forward bias resistance of 30 $Ω$ and infinite resistance in reverse bias. The current I₁ will be:

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alok kumar singh

Contributor-Level 10

In give diagram Diode 1 and 2 are in forward bias with R = $30 Ω$ and Diode 3 is reverse bias with R = infinite l₁ current is flowing through 20 $Ω$

So $\frac{l_{1}}{2} a n d \frac{l_{1}}{2}$ current will flow through Diode 1 and 2. As resistance is same applying KCl in ABCD Loop $- \frac{l_{1}}{2} * 130 - \frac{l_{1}}{2} * 130 - l_{1} * 20 + 200 = 0$

-100l₁ + 200 = 0

l₁ = 2

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3 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

Statement-I : To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output parallel to the load RL.

Statement-II : To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with RL.

In the light of the above statements, choose the most appropriate answer from the options given below:

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Vishal Baghel

Contributor-Level 10

Capacitor makes potential difference constant.

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3 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

If VA and VB are the input voltages (either 5V or 0V) and Vo is the output voltage then the two gates represented in the following circuit (A) and (B) are:-