How Electronics Started! And JLab Hits the Wall! In electronics, a - - PowerPoint PPT Presentation

How Electronics Started! And JLab Hits the Wall!

In electronics, a vacuum diode or tube is a device used to amplify, switch, other- wise modify, or create an electrical sig- nal by controlling the movement of elec- trons in a low-pressure space. Almost all depend on the thermal emission of elec-

• trons. The figure shows the components
• f the device. A demonstration of how

they work is here. The physics here also determines the limits on the amount of beam that can be produced in an accelerator.

Vacuum Diode – p.1/21

The Child-Langmuir Law

In a vacuum diode, electrons are boiled off a hot cathode at zero potential and accelerated across a gap to the anode at a positive potential V0. The cloud of moving electrons within the gap (called the space charge) builds up and reduces the field at the cathode surface to zero. From then on a steady current flows between the plates. Suppose two plates are large relative to the separation (A >> d2 in the figure) so that edge effects can be ignored and V , ρ, and the electron speed v are functions of only x.

d Electron Cathode (V=0) Anode(V=V ) A x

Vacuum Diode – p.2/21

The Child-Langmuir Law

• 1. What is Poisson’s equation for the region between the plates?
• 2. Assuming the electrons start from rest at the cathode, what is their

speed at point x?

• 3. In the steady state I, the current, is independent of x. How are ρ and

v related?

• 4. Now generate a differential equation for V by eliminating ρ and v and

solve this equation for V as a function of x, V0, and d. Make a plot to compare V (x) and the potential without the space charge.

• 5. What are ρ and v as functions of x?
• 6. Show that

I = KV 3/2 and find K. This the Child-Langmuir law.

d Electron Cathode (V=0) Anode(V=V ) A x

Vacuum Diode – p.3/21

Why Should You Care? The Physics 132 Picture

v

d

E v

d

positive negative

+

Area = A i d t

∆

I = JA = nqvdA = A

ρdV

I = 1

RV

V = IR

2007-10-16 11:35:02

Current (A)

0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18

Voltage (V)

1 2 3 4 5 6 7 8 9 10

Ohm’s Law

Ω 8 ± slope = 46 Ω = 46.5

meas

R

Vacuum Diode – p.4/21

Why Should You Care, Part Deaux?

January 11, 2006

Vacuum Diode – p.5/21

Coulomb’s Law and Superposition

Measuring the Electrostatic Force

• f Charges.

Use a torsion pendulum and charged spheres.

Vacuum Diode – p.6/21

Coulomb’s Law and Superposition

Measuring the Electrostatic Force

• f Charges.

Use a torsion pendulum and charged spheres.

Evidence for Coulomb’s Law.

Fix charge and vary distance.

n1.99 0.04 0.00 0.05 0.10 0.15 0.20 0.25 50 100 150 200 250 300 350 rm Θ deg Intermediate Lab Results

Vacuum Diode – p.6/21

Coulomb’s Law and Superposition

Measuring the Electrostatic Force

• f Charges.

Use a torsion pendulum and charged spheres.

Evidence for Coulomb’s Law.

Fix charge and vary distance.

n1.99 0.04 0.00 0.05 0.10 0.15 0.20 0.25 50 100 150 200 250 300 350 rm Θ deg Intermediate Lab Results

Evidence for Superposition.

Fix distance and vary charge on one sphere.

Q V Quadratic term: 0.5 0.3 1 2 3 4 5 6 7 20 40 60 80 100 V kV Θ deg Intermediate Lab Results

Vacuum Diode – p.6/21

Electric Field of a Ring

A ring of radius r as shown in the figure has a positive charge distribution per unit length with total charge q. Calculate the electric field E along the axis of the ring at a point lying a distance x from the center of the ring. Get your answer in terms of r, x, q. What happens as x → ∞?

r x q Q

Vacuum Diode – p.7/21

Electric Field of a Ring

A ring of radius r as shown in the figure has a positive charge distribution per unit length with total charge q. Calculate the electric field E along the axis of the ring at a point lying a distance x from the center of the ring. Get your answer in terms of r, x, q. What happens as x → ∞?

r x q Q

Vacuum Diode – p.7/21

Electric Field Lines

Point Charges Electric Dipole Positive Charges

Vacuum Diode – p.8/21

Properties of Electric Field Lines

Field lines start from positive charges (sources) and end at negative ones (sinks). Are symmetrical around point charges. Density of field lines is related to strength of field. Direction of field is tangent to the field line. Field lines never cross!

Vacuum Diode – p.9/21

Electric Flux

ΦE = EA ΦE = EA cos θ = E · A ΦE = E · dA

Vacuum Diode – p.10/21

An Example of Electric Flux

A nonuniform magnetic field is described by

• E = (3.0 N/C − m)xˆ

x + (4.0 N/C)ˆ y

pierces the Gaussian cube shown in the figure. What is the flux through the cube? Note the orientation of the coordinate system.

Vacuum Diode – p.11/21

Gauss’s Law

What is the flux from a point charge q at the origin through a sphere centered at the origin of radius r?

Vacuum Diode – p.12/21

Differential Surface and Volume Elements

Cartesian Coordinates Cylindrical Coordinates Spherical Coordinates

y z x x y z (x,y,z)

θ φ θ φ

dA = dxdy dA = ρdφdz dA = r2d cos θdφ dτ = dxdydz dτ = ρdφdzdρ dτ = r2d cos θdφdr

Vacuum Diode – p.13/21

Applying Gauss’s Law: Nuclear E

The nucleus of a gold atom has a radius R = 6.2 × 10−15 m and a positive charge q = Ze where Z = 79 is the atomic

• number. Assume the gold nucleus is spherical and the

charge is uniformly distributed in the volume. What is the electric field for any r?

Vacuum Diode – p.14/21

Applying Gauss’s Law: Nuclear E

The nucleus of a gold atom has a radius R = 6.2 × 10−15 m and a positive charge q = Ze where Z = 79 is the atomic

• number. Assume the gold nucleus is spherical and the

charge is uniformly distributed in the volume. What is the electric field for any r?

5 10 15 20 25 0.0 0.5 1.0 1.5 2.0 2.5 3.0 rfm E1021 NC Electric field of a gold nucleus

Vacuum Diode – p.14/21

An Example of Applying ∇ × E

A nonuniform magnetic field is described by

• E = (3.0 N/C − m)xˆ

x + (4.0 N/C)ˆ y

pierces the Gaussian cube shown in the figure. Is this a conservative field?

Vacuum Diode – p.15/21

Applying V

• 1. What is the potential energy of a point charge?
• 2. What is the potential inside and outside a uniformly

charged sphere of total charge q and radius R?

• 3. What is the electric field for the previous question?

Vacuum Diode – p.16/21

The Child-Langmuir Law

In a vacuum diode, electrons are boiled off a hot cathode at potential zero and accelerated across a gap to the anode at a positive potential V0. The cloud of moving electrons within the gap (called the space charge) builds up and reduces the field at the cathode surface to zero. From then on a steady current flows between the plates. Suppose two plates are large relative to the separation (A >> d2 in the figure) so that edge effects can be ignored and V , ρ, and the electron speed v are functions of only x.

d Electron Cathode (V=0) Anode(V=V ) A x

Vacuum Diode – p.17/21

The Child-Langmuir Law

• 1. What is Poisson’s equation for the region between the plates?
• 2. Assuming the electrons start from rest at the cathode, what is their

speed at point x?

• 3. In the steady state I, the current, is independent of x. How are ρ and

v related?

• 4. Now generate a differential equation for V by eliminating ρ and v and

solve this equation for V as a function of x, V0, and d. Make a plot to compare V (x) and the potential without the space charge.

• 5. What are ρ and v as functions of x?
• 6. Show that

I = KV 3/2 and find K. This the Child-Langmuir law.

d Electron Cathode (V=0) Anode(V=V ) A x

Vacuum Diode – p.18/21

Using Mathematica To Solve the DE

Vacuum Diode – p.19/21

The Child-Langmuir Law: Results

Comparing the Child-Langmuir Law with the no-space-charge solution.

Blue ’132’ picture Red ChildLangmuir 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 V units of Vmax Current units of Imax

Vacuum Diode – p.20/21

More Child-Langmuir Law Results

Comparing the Child-Langmuir Law in vacuum with a copper wire.

Red copper wire Blue vacuum 0.0 0.2 0.4 0.6 0.8 1.0 100 105 108 1011 1014 V megavolts Current Density Am2 Wires vs. Vacuum

JCU = 1 ρ V d ρ = 1.7×10−8 Ω−m JCL = 4ǫ0 9d2

• 2e

me V 3/2 d = 0.1 m

Vacuum Diode – p.21/21