Question

In space charged limited region, plate current in a diode is \(10\,mA\) for plate voltage \(150\,V\). If the plate voltage is increased to \(600\,V\), then the plate current will be

• A. 10 mA
• B. 40 mA
• C. 80 mA
• D. 160 mA

Hint:

In space charge limited region, Child’s law applies: \(I \propto V^{3/2}\). Increase voltage by factor \(k\), current increases by \(k^{3/2}\).

Answer 1

The Correct Option is C

Step 1: Use Child’s law for space charge limited current.
For diode in space charge limited region:
\[ I \propto V^{3/2} \] Step 2: Apply ratio form.
\[ \frac{I_2}{I_1} = \left(\frac{V_2}{V_1}\right)^{3/2} \] Step 3: Substitute values.
\[ I_1 = 10\,mA,\quad V_1 = 150\,V,\quad V_2 = 600\,V \] \[ \frac{I_2}{10} = \left(\frac{600}{150}\right)^{3/2} = (4)^{3/2} \] Step 4: Simplify.
\[ 4^{3/2} = (\sqrt{4})^3 = 2^3 = 8 \] So,
\[ I_2 = 10 \times 8 = 80\,mA \] Final Answer: \[ \boxed{80\ mA} \]