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

Question

In space charge limited region, the plate current is given by Child's law

i_{p} = K V_{p}^{3/2}

Thus,

\frac{i _{p_{2}}}{i _{p_{1}}} = (\frac{V _{p_{2}}}{V _{p_{1}}})^{3/2} = (\frac{600}{150})^{3/2}

= (4)^{3/2} = 8

or

i_{p_{2}} = i_{p_{1}} \times 8 = 10 \times 8 m A = 80 m A

Tardigrade Admin · Accepted Answer

In space charge limited region, the plate current is given by Child's law ip=K Vp3 / 2 Thus, (ip2/ip1)=((Vp2/Vp1))3 / 2=((600/150))3 / 2 =(4)3 / 2=8 or ip2=ip1 × 8=10 × 8 mA =80 mA

Q. In space charge limited region, the plate current in a diode is $10 m A$ for plate voltage $150 V$ . If the plate voltagte is increased to $600 V$ , then the plate current will be

10 mA

40 mA

80 mA

160 mA

In space charge limited region, the plate current is given by Child's law $i_{p} = K V_{p}^{3/2}$
Thus, $\frac{i _{p_{2}}}{i _{p_{1}}} = (\frac{V _{p_{2}}}{V _{p_{1}}})^{3/2} = (\frac{600}{150})^{3/2}$
$= (4)^{3/2} = 8$
or $i_{p_{2}} = i_{p_{1}} \times 8 = 10 \times 8 m A = 80 m A$

Q. In space charge limited region, the plate current in a diode is 10mA for plate voltage 150V. If the plate voltagte is increased to 600V, then the plate current will be