ELEC1130 Lecture 110
2016/17
Quiz
School of Electronic
and Electrical Engineering
Is the diode on the right forward
biassed or reverse biassed?
Will current flow or not?
ELEC 1130
Lecture 110
–ve on left, +ve on
right
–
+
Diodes (2)
2016/17
ELEC 1130 Dr C. Trayner
Quiz
Summary
When a normal Si diode is
conducting, what is the
approximate voltage lost
across it?
1. 0 V
2. 0.7 V
3. 5 V
4. 230 V
Last time:
• Introduction to diodes
• Voltage and current sources
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Dr Chris Trayner
ELEC 1130 Dr C. Trayner
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Today:
• More details of diodes
• Understanding last week’s lab and preparing for
this week’s
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ELEC1130 Lecture 110
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Terminology - revision
Ways of thinking of the ideal diode
- revision
Names of electrodes
Forward biassed
current will flow
Anode
Cathode
+
+
–
Reverse biassed
current will not flow
Circuit symbol
Triangle may be filled in
or empty as you prefer
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Forward biassed
–
–
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I/V curve of the ideal diode
+
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=
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Open
circuit
=
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Open
switch
=
6
What did you find in the lab?
What sort of I/V curve did you get?
• Straight?
• Curving upwards?
• Curving downwards?
Forward bias: no
voltage drop
whatever the current
Reverse bias: no
current whatever the
voltage
∞Ω
The real diode
I
V
–
+
=
=
=
Reverse biassed
+
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–
Closed
switch
Short
circuit
0Ω
+
–
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ELEC1130 Lecture 110
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The real diode
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Do diodes obey Ohm’s Law?
Ohm’s law is not always true
But for most electronic
components
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The real diode – a surprisingly simple
approximation
Voltage
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Figure: Dr H. Eisele
~0.7 V
Tangent over
short section
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Current
• it is true
• or at least to a good enough
approximation
• or it is approximately true
under limited conditions
For diodes the third case is
occasionally useful
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Figure: Dr H. Eisele
Figure: Dr H. Eisele
2016/17
Resistors are simple to model
i.e. to calculate with
Diodes are more complicated
but we would still like a simple
model – even if an approximation
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How do we model this?
What if this approximation isn’t good
enough?
What equivalent
circuit can we
have for the rule
of thumb of 0.7 V?
An ideal diode in
series with a 0.7 V
ideal source
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Another fairly simple approximation
0.7 V
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Figure: Dr H. Eisele
Figure: Dr H. Eisele
(Rb is the currentlimiting resistor)
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How do we model this?
Diode
simulated by
a voltage
source and
an internal
resistance
Straight but sloping line
Behaves like a resistor but
for the offset
Figure: Dr H. Eisele
< 0.7 V
And an ideal
diode (here
shown
forward
biassed)
Figure: Dr H. Eisele
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So we have four models
Another fairly simple approximation
It looks scary
but try analysing it
Ideal diode
This is of the form
exp(kV)
k is a horizontal
scaling factor
Actually quite
large
0.7 V
<0.7 V & Rs
Exact equation
Figure: Dr H. Eisele
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ELEC 1130 Dr C. Trayner
Except for tiny or
–ve V, this is
smaller than
exp(kV) and can
be ignored
This is just a
vertical scaling
factor
Actually very small
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Exponentials – mysterious?
Exponentials – mysterious?
Consider ex (also written exp(x) )
e is just a number (about 2.718)
Compare with other examples of kx
2x: 1, 2, 4, 8, 16, …
10x: 1, 10, 100, 1000, 10000, ...
These grow fast, the curve gets steeper
These are the same sort of curve
The k in kx is just a horizontal scaling factor
They can be continued the other way for –ve x:
2x: 1/16, 1/8, 1/4, 1/2, 1, 2, 4, 8, 16, …
10x: 0.0001, 0.001, 0.01, 0.1, 1, 10, 100, 1000,
10000, ...
As x gets more and more –ve, kx tends
asymptotically towards zero
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exp(x)
exp(x) and exp(x)–1
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So we have four models
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So we have four models
When should you use which model?
Common sense should tell you
• Is the 0.7 V significant compared with the
voltages it will be handling?
• How much precision do you need?
Ideal diode
0.7 V
<0.7 V & Rs
Exact equation
Figure: Dr H. Eisele
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Dr Chris Trayner
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What can you do with diodes?
The half-wave rectifier
Power supplies: PSUs, battery chargers, …
Mains is
• nominally sine wave (waveform may be a bit rough),
• 50 Hz or 60 Hz in some parts of the world
• typically around 230 V RMS
This needs to be reduced to low voltages and turned from
AC to DC
AC
provided
by transformer
Load receives DC
(would normally be
smoothed)
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ELEC 1130 Dr C. Trayner
Figure: Dr H. Eisele
Transformers reduce the voltage
• more complicated with switched-mode PSUs
Diodes are part of the AC to DC conversion
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The full-wave rectifier bridge
Only uses the
+ve half
cycles
Four diodes
(the bridge)
Two conduct at
any time
• two for +ve
half cycle
• two for –ve
half cycle
50% of the time
is wasted
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Dr Chris Trayner
Figure: Dr H. Eisele
(actually a bit
more with
non-ideal
diodes)
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Figure: Dr H. Eisele
The half-wave rectifier
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ELEC1130 Lecture 110
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+ve half cycle
Follow the current
route
Convince yourself
that
• the red diodes
are forwardbiassed
• the black ones
are reversebiassed
• current flows
downwards
through the load
–ve half cycle
Follow the current
route
Convince yourself
that
• the red diodes
are forwardbiassed
• the black ones
are reversebiassed
• current flows
downwards
through the load
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Figure: Dr H. Eisele
The full-wave rectifier bridge
Figure: Dr H. Eisele
The full-wave rectifier bridge
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Half-wave vs full-wave
Rectifying and smoothing
Which is the more
economical (for a
given average
current)?
Discuss
Diodes only change from AC to DC
Leaves a very variable voltage
Needs smoothing
One capacitor will smooth it somewhat
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ELEC1130 Lecture 110
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Real power supplies
Traditionally you add a second capacitor (with a
resistor in between the two)
• This gave moderate smoothing with fairly large
capacitors
• but voltage dropped as loading current increased
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Figure: Dr H. Eisele
Still
somewhat
rough
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Modern approach is a regulator chip
• only needs one capacitor
• keeps the voltage fixed
• also sets a current limit
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Online homework:
google for a data
sheet for a 78XX IC,
e.g. 7805 for 5 V
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Summary
Diodes:
• one-way valves for current
• ideal and real diodes
• approximations to real diodes
• and the equation
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