Transcript Slide 1

Lecture 6
o Aim of the lecture
 Appreciation of
 Current Density
 Drift Velocity
 Kirchhoff’s Laws
 Voltage Loop Law
 Current Node Law
 RC Circuits
 Response to step Voltages
 Charge and discharge
o Main learning outcomes
 familiarity with
 Kirchhoff’s Laws and application to circuits
 Typical current densities and drift velocities
 Calculation of RC time constant
 Charging and discharging
Reminder:
o Electrons are made to drift in an electric field caused by an external voltage.
 They loose energy in collisions with the fixed atoms
 They therefore do not accelerate
 They drift at constant speed
Consider a wire with a voltage across it.
Remember:
atoms fixed in place
electrons move
oCurrent density is
 the current per unit cross-sectional area of the wire
 if this is too great the wire can melt
 as the current density goes up, the wire will get hot
 this makes its resistance higher.
 bigger currents need bigger wires
Current = i
Area = A
Current density, r = i/A
Current density, r = i/A
For example,
up to 5A currents in household wires
use 1mm2 copper wires
r = 5/0.0012 = 5 x 106 A/m2
This is not the maximum the wire could take, but it is
a safe limit for use in houses.
r = 5/0.0012 = 5 x 106 A/m2
At 5A in a 1mm2 wire,
there is 5 x 106 A per m2 in the wire
This sounds large, so how fast
are the electrons moving?
5 x 106 A = 5 x 106
electrons/second
1.6 x 10-19
= 3.1 x 1025 per second
That’s a lot, 3 x 1025 peas would cover the earth to a depth of ~1km
To work out electron speed, need
density of electrons in the wire.
Copper has density = 8.9 g/cm3
Copper atom has molecular weight = 63.546 g/mol
Avogadros number is 6.02 x 1026 atoms/mol
So there are 6.02 x 1023 x 146085.5 = 8.5 x 1028 atoms/m3
And the drift speed = current/(density x area x charge/electron)
Which makes this guy look fast!
= 5/(8.5 x 1028 x 1 x 10-6 x 1.6 x 10-19) = 0.4 mm/sec
which is about 1.3 x 10-3 km/hr
This is actually
a VERY thin wire
wound in a spiral
The electrons in a wire don’t move far normally
o In an incandescent light bulb (one with a wire)
 the wire is very thin
 the electrons are drifting fast
 about walking pace (!)
 which is why the wire gets very hot
 once the electrons get through the bulb
 they move slowly again
This is one place where the water in a pipe analogy is
a little weak – the water is hardly moving at all to be a good picture
Kirchhoff’s Laws
o These are effectively
 energy conservation
 charge conservation
o Applied to circuits
Current Law
Charge cannot be destroyed, so
the sum of currents flowing into a node
is equal to the sum of currents flowing out
( hence it is vital to understand that capacitors do NOT store charge)
Analogy:
If water flows into a junction
Then
the volume of water flowing in
equals the total flowing out
So
ID+IC+IB = IA
Note that IIN = Ia + Ib
capacitors do NOT store charge.
Ia
IIN
Ib
This is the reason we have been so ‘determined’ that
capacitors should not be thought of as ‘storing’ charge
- if they could then IIN would not necessarily be Ia+Ib.
Voltage Law
The sum of the voltage drops round a closed loop is zero
Recall that voltage is a measure of potential
And remember that gravitational potential behaves in a similar way
If a mass is moved round a closed path in a gravity field
the sum of Dmgh round its path must be zero.
Just says that if you start at one height and end up at that same
height then the sum of all the changes in height must be zero
Electric potential is the same, if you move round a closed loop
then the sum of the changes in voltage must be zero.
Example
This is the old fashioned
symbol for a resistor,
it is still used a lot
Be careful defining the sign.
You MUST measure the
voltages in the same direction
On all voltages round the loop
Svi = 0 round loop
SIi = 0 into node
Prof. Kirchhoff
Note that the loop laws are true for all the loops in a circuit
Altogether there are 7 loops in the circuit above (3 shown)
Find the others.
Kirchhoff’s Laws are used in working out what the
currents and voltages are in a network of components.
Each loop and each node yields an equation
These then form a set of simultaneous equations which can
be solved to find the currents and voltages.
Not necessarily an easy way to do it,
but formulaic and usually gives the right answer.
With this definition of
directions, all the currents
will be positive.
Sometimes called Kirchhoff’s
First and Second Laws
Note that in this diagram,
voltage ‘a’ will have the opposite
sign to all the others
RC Circuits
Switch
Vr=0
VC=0
Q=CV
I = dQ/dt = CdV/dt
Svi = 0 round loop
SIi = 0 into node
RC Circuits
A ‘long’ time after the switch is closed.
Vr=0
Vc=emf
Q=CV
I = dQ/dt = CdV/dt
Svi = 0 round loop
SIi = 0 into node
RC Circuits
Now open switch again
Vr=0
Vc=emf
Q=CV
I = dQ/dt = CdV/dt
Svi = 0 round loop
SIi = 0 into node
RC Circuits
Then close the other way
Vr=0
=emf
But now there
is a circuit with
a resistor across
an energised capacitor
(‘charged’ capacitor)
Q=CV
I = dQ/dt = CdV/dt
Vc=emf
Svi = 0 round loop
SIi = 0 into node
RC Circuits
Vr
Svi = 0 round loop
So Vr = -Vc
Vc
The current flowing
round the circuit
is the same everywhere
So, using
I = dQ/dt = CdV/dt
then
The solution is
-t/RC
I = Vr/R = -CdVV
r/dt
r = Ae
Where A is a constant and is the voltage
0, say V0
so Vr/Rat+time
CdV=
r/dt = 0
(in this case V0 is what was called ‘emf’ earlier)
RC Circuits
Vr
Vc
A ‘discharging’ capacitor obeys the equation
Vr = V0e-t/RC
RC is called the time constant for the circuit
The voltage drops from V0 to V0/e in a time RC
Having made the point that capacitors do not store charge,
we will now adopt the usual convention of talking about
charging and discharging.
0
Kirchhoff’s voltage law now states Vbattery + Vr + Vc = 0
and again the current is the same round the loop
students will be able to show:
V = Vbattery (1 – e-t/RC)
The bottom axis is shown here in terms of multiples of RC, so it is
a ‘universal’ plot.
Note that the current and voltage both have exponential forms,
as the voltage increases, the current decreases (or vice versa)
o Finally, these are differential equations.
o To find particular solution requires boundary conditions
o For step voltages (switches for example)
determined from the conditions at t=0
Need to evaluate voltages and currents just after switch moved
Important:
 The voltage across a capacitor cannot change instantaneously
 Because E=CV2/2 so if the voltage change instant,
implies infinite power
o The voltage across a capacitor CANNOT change instantly
o The current through a capacitor CAN change instantly
o The voltage and the currents for a resistor can BOTH change instantly
Recipe for analysing RC circuits
o Develop differential equation using
 Kirchhoff’s Laws
 V=IR for resistors
 V=Q/C for capacitors
o Establish boundary conditions by
 Working out conditions just before switch moved
 Evaluating what changes will occur just after using
 V across capacitors unchanged
 I through capacitor can change
 V and I can both change in a resistor
This may be easy, or it may not depending on circuit.
It takes practice.
Practice!
Analysis of RC networks is not an academic exercise, nearly all
electronics will contain them.
BUT they are mostly used with
•They are used as
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 noise suppressors
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