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Semiconductor Electronics PYQ | Ideal Diode Circuit | JEE Main 2025

Solve this JEE Main 2025 Physics PYQ on an ideal diode circuit using the ON-OFF diode analysis method. Learn how to identify forward and reverse bias

 

❓ Question

For the following circuit, assuming ideal diodes, determine the output voltage VoutV_{\text{out}}.

Semiconductor Electronics PYQ | Ideal Diode Circuit | JEE Main 2025

✍️ Solution

For an ideal diode:

  • Forward biased (ON): Acts as a short circuit (VD=0)(V_D=0).
  • Reverse biased (OFF): Acts as an open circuit (I=0)(I=0).

Check Diode D1D_1

The anode of D1D_1 is connected to

+5 V.+5\text{ V}.

If

Vout=+5 V,V_{\text{out}}=+5\text{ V},

then the cathode of D1D_1 is also at

+5 V.+5\text{ V}.

Hence,

Vanode=Vcathode.V_{\text{anode}}=V_{\text{cathode}}.

Therefore,

D1 is OFF.\boxed{D_1\ \text{is OFF}.}

Check Diode D2D_2

The anode of D2D_2 is connected to

0 V (Ground).0\text{ V (Ground)}.

The cathode is at

Vout=+5 V.V_{\text{out}}=+5\text{ V}.

Thus,

Vanode<Vcathode,V_{\text{anode}}<V_{\text{cathode}},

so D2D_2 is reverse biased.

Hence,

D2 is OFF.\boxed{D_2\ \text{is OFF}.}

Output Voltage

Since both diodes are OFF, no current flows through either diode.

The output node remains directly connected to the

+5 V+5\text{ V}

supply.

Therefore,

Vout=+5 V.V_{\text{out}}=+5\text{ V}.

Semiconductor Electronics PYQ | Ideal Diode Circuit | JEE Main 2025

✅ Final Answer

Vout=+5 V
\boxed{V_{\text{out}}=+5\ \text{V}}

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