In a vacuum diode, electrons are "boiled" off a hot cathode, at potential zero, and accelerated across a gap to the anode, which is held at positive potential ${\mathit{V}}_{\mathbf{0}}$. The cloud of moving electrons within the gap (called space charge) quickly builds up to the point where it reduces the field at the surface of the cathode to zero. From then on, a steady current flows between the plates.

Suppose the plates are large relative to the separation ($\mathit{A}\mathbf{>}\mathbf{>}{\mathit{d}}^{\mathbf{2}}$in Fig. 2.55), so

that edge effects can be neglected. Thenlocalid="1657521889714" $\mathit{V}\mathbf{}\mathbf{,}\mathit{\rho }$and $\mathit{v}$(the speed of the electrons) are all functions of x alone.

(a) Write Poisson's equation for the region between the plates.

(b) Assuming the electrons start from rest at the cathode, what is their speed at point x, where the potential is$\mathit{V}\left(x\right)$

(c) In the steady state,localid="1657522496305" $\mathit{I}$is independent of $\mathit{x}$. What, then, is the relation between p and v?

(d) Use these three results to obtain a differential equation for$\mathit{V}$, by eliminating$\mathit{\rho }$and$\mathit{v}$.

(e) Solve this equation for $\mathit{V}$as a function of $\mathit{x}$,${\mathit{V}}_{\mathbf{0}}$and $\mathit{d}$. Plot $\mathit{V}\left(x\right)$, and compare it to the potential without space-charge. Also, find$\mathit{\rho }$and$\mathit{v}$as functions of .

(f) Show that

$\mathit{I}\mathbf{=}\mathit{k}{\mathit{V}}_{\mathbf{0}}^{\mathbf{3}\mathbf{/}\mathbf{2}}$

and find the constant$\mathit{K}$. (Equation 2.56 is called the Child-Langmuir law. It holds for other geometries as well, whenever space-charge limits the current. Notice that the space-charge limited diode is nonlinear-it does not obey Ohm's law.)

Here,$I$ is the steady current flow between the plates, $A$is the area between the plates,$d$ is the distance between the plates, $v$is the speed of the electrons, $\rho$is the volume charge density.

$V,\rho ,v$are all the functions of x plane alone.

a)

The expression for Poisson’s equation for the region between the plates is,

$\frac{{\partial }^{2}V}{\partial x^2}=\frac{-{\rho }_x}{{\epsilon }_0}$ …… (1)

b)

Write the formula for speed of the electrons using law of conservation of energy.

$\frac{1}{2}m{v}_x^2=V_{x}e$

$v_x=\sqrt{\frac{2e{V}_x}{m}}$ …… (2)

Here,$v_{x}e$ is the Potential energy of the electrons,$m$ is the mass.

Therefore, the speed at cathode from rest at point x is $v_x=\sqrt{\frac{2e{V}_x}{m}}$.

c)

Write the expression for volume charge density,

$$\begin{gathered} \rho =\frac{dq}{V} \\ dq=\rho V \end{gathered}$$

Here, the value of volume $V$can be taken as,$area\times smallportionoflength=\left(Adx\right)$

Therefore,

$dq=\rho \left(Adx\right)$

Write the formula for rate of flow of charge.

$I=\frac{dq}{dt}$ …… (3)

Differentiating the above equation,

$$\begin{gathered} I=A\rho \frac{dx}{dt} \\ =A\rho V \end{gathered}$$

Here, current is independent of x .

At steady state $I$is remains constant.

Thus, as can be seen there is inverse relation between $\rho$and v .

d)

Using the equation (1), (2) and (3)

$$\begin{gathered} \frac{{\partial }^2{V}_x}{\partial x^2}=\frac{-{\rho }_x}{{\epsilon }_0} \\ =\frac{-I_0}{{\epsilon }_{0}A{v}_x} \\ =\frac{-I_0}{{\epsilon }_{0}A}\sqrt{\frac{m}{2e{V}_x}} \\ =\frac{K}{\sqrt{V_x}} \end{gathered}$$

Here, $K=\frac{-I_0}{{\epsilon }_{0}A}\sqrt{\frac{m}{2e}}$is constant.

Thus, role="math" localid="1657524366758" $\frac{{\partial }^{2}V}{\partial x^2}=K{V}_x^{\frac{-1}{2}}$ …… (4)

Hence,role="math" localid="1657524374548" $\frac{{\partial }^{2}V}{\partial x^2}=K{V}_x^{\frac{-1}{2}}$ is the required differential equation.

e)

Let’s consider$V_x=a{x}^b$ …… (5)

Differentiate equation (5) twice,

$$\begin{gathered} \frac{d{V}_x}{dx}=ab{x}^{b-1} \\ \frac{d^2{V}_x}{dx}=ab\left(b-1\right)x^{b-1} \end{gathered}$$ …… (6)

From the equation (4) and (5)

$$\begin{gathered} K{V}_x^{-\frac{1}{2}}=ab\left(b-1\right)x^{b-2} \\ \frac{K}{\sqrt{a}}{x}^{\frac{-b}{2}}=ab\left(b-1\right)x^{b-2} \end{gathered}$$

Equating the power of x,

$$\begin{gathered} b-2=\frac{-b}{2} \\ b=\frac{4}{3} \end{gathered}$$

After substituting the value of$b$, equation (5) is written as

$V_x=a{x}^{\frac{4}{3}}$

At x=d ,

$V=V_0$

$$\begin{gathered} V_0=a{d}^{\frac{4}{3}} \\ a=\frac{V_0}{d^{{\displaystyle \frac{4}{3}}}} \end{gathered}$$ …… (7)

Substitute equation (7) in equation (1) and (2).

${\rho }_x=-{\epsilon }_0\frac{V_0}{d^{{\displaystyle \frac{4}{3}}}}\left(\frac{4}{9}\right)\frac{1}{x^{{\displaystyle \frac{2}{3}}}}$ …… (8)

$v_x=\sqrt{\frac{2e{V}_0}{m}}\frac{x^{{\displaystyle \frac{2}{3}}}}{d^{{\displaystyle \frac{4}{3}}}}$ …… (9)

The above figure shows graph of variation of potential with distance. The dotted line represents as without space charge is linear.

f)

Now combine equation (8) and (9) and solve as further,

$$\begin{gathered} I=A{\rho }_x{v}_x \\ =A{\epsilon }_0=\frac{V_0}{d^{{\displaystyle \frac{4}{3}}}}\left(\frac{4}{9}\right)\frac{1}{x^{{\displaystyle \frac{2}{3}}}}\sqrt{\frac{2e{V}_0}{m}}\frac{x^{{\displaystyle \frac{2}{3}}}}{d^{{\displaystyle \frac{2}{3}}}} \\ =\frac{4A{\epsilon }_0}{d^2}\frac{V_0^{{\displaystyle \frac{2}{3}}}}{9x^{{\displaystyle \frac{4}{3}}}}\sqrt{\frac{2e}{m}} \\ =K{V}_0^{\frac{3}{2}} \end{gathered}$$

Here, $K$is constant and the value is $K=\frac{-4A}{9}\sqrt{\frac{2e}{m}}\frac{{\epsilon }_0}{d^2}$.