# Derivation of voltage for vacuum diode space charge region

• Nuahaun
In summary, the differential equation relating voltage and position for a vacuum diode in the space charge region can be solved by a simple separation of variables, but for the life of me I can't tell where it is.

#### Nuahaun

I have found the 1D differential equation relating voltage and position for a vacuum diode in the space charge region, which is

$$\frac{d^2V}{dx^2} = constant * V^{-1/2}$$

and I know the solution to be

$$V(x) = V_0 \left(\frac{x}{d}\right)^{4/3}$$

which is found by multiplying both sides by $V' = \frac{dV}{dx}$ and then integrating the following expression with homogeneous boundary conditions:

$$\int V' dV' = constant*\int V^{-1/2} dV$$

What I don't understand is why this trick is even necessary. As far as I can tell the differential equation can be solved by a simple separation of variables, which gives an answer of

$$V(x) = V_0 \left(\frac{x}{d}\right)^{4/5}$$

The two answers are different so obviously it's a mistake to separate variables, but for the life of me I can't tell where it is. Could anyone enlighten me? Thanks.

Nuahaun said:
What I don't understand is why this trick is even necessary. As far as I can tell the differential equation can be solved by a simple separation of variables, which gives an answer of

I'm not exactly sure what you mean by "separation of variables" in this case. To apply separation of variables, $V$ would have to be a function of more than one variable; but in this case, it is a function of only $x$. SOV is typically used to turn PDEs into ODEs, not to solve an ODE.

I can only guess that what you meant to say was that you tried treating it as a "separable differential equation"; but you have a second order DE...how exactly does one integrate $$V^{1/2} d^2 V$$ and $$dx^2$$?

yeah wasn't totally sure what was meant, however a trick to reduce it to 2 first order DEs as follows:

First write
P = V' = $$\frac{dV}{dx}$$
then
P' = V" = $$\frac{dp}{dx}$$ = $$\frac{dp}{dV}$$.$$\frac{dV}{dx}$$ = p.$$\frac{dp}{dV}$$

otherwise just assume solutions V~xy diff & equate coeffs

Last edited:
Thanks for your replies. My mistake was that I though I could solve $V^{1/2} d^2 V = dx^2$, which stemmed from an incomplete understanding of the method of separation of variables (I took the fact that you could solve something like $d^2x/dt^2= F/m$ using double integration, and incorrectly applied that to the differential equation above). I reviewed my notes, and now see why that isn't possible.

## 1. What is the purpose of studying the derivation of voltage for vacuum diode space charge region?

The derivation of voltage for vacuum diode space charge region is important because it helps us understand the behavior of a vacuum diode and how it functions as a component in electronic devices.

## 2. How is the voltage for vacuum diode space charge region calculated?

The voltage for vacuum diode space charge region is calculated using the Child-Langmuir Law, which takes into account the physical properties of the diode such as its dimensions and the charge density in the space charge region.

## 3. What factors affect the voltage for vacuum diode space charge region?

The voltage for vacuum diode space charge region is affected by the physical properties of the diode, such as the distance between the cathode and anode, the temperature, and the work function of the cathode material.

## 4. How does the voltage for vacuum diode space charge region affect the performance of the diode?

The voltage for vacuum diode space charge region directly affects the current flow in the diode. As the voltage increases, the space charge region becomes smaller and the current flow increases. This can also lead to changes in the diode's electrical properties such as its capacitance and frequency response.

## 5. Can the derivation of voltage for vacuum diode space charge region be applied to other electronic devices?

While the Child-Langmuir Law is specific to vacuum diodes, the concept of space charge region can be applied to other electronic devices such as transistors and capacitors. However, the calculation of the voltage may differ based on the device's specific characteristics and operating conditions.

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