576
21 Diodes
21.5 Applications and Special Diode Devices
In the following, various electronic applications of diodes are discussed. The most important special
diode types are introduced. Optoelectronic applications (involving absorption and emission of photons)
are treated below (Sect. 22). The applications of Schottky diodes are discussed in [1791].
21.5.1 Rectification
In a rectifier, the diode has to supply a high resistivity for one polarity of the bias and a low one to the
other polarity. In Fig. 21.63a, a single-path rectification method is shown. Only the positive half-wave
can pass the load resistor R L (Fig. 21.63b). In Fig. 21.63c, the characteristic of a Si diode is shown. Of
course, the voltage drop across the diode can only range in the 1-V regime. In order to make the setup
work, the load resistor has to be considered. The total current is given by I = I s [exp(eU d /nkT ) − 1].
The total voltage U is split between the voltage drop across the diode U d and that over the load
resistance U L = R L I . The current is therefore given by
I =
U − U d
R L
.
(21.169)
For sizeable currents the voltage drop across the diode U d is between 0.7 and 1 V. The characteristic
is linear between about 1 and 220 V (Fig. 21.63d). Typically, the voltage U L is low-pass filtered with
a capacitor parallel to the load resistor. The effective voltage is the peak voltage divided by 2.
The drawback of the single diode rectifier is that only the positive half-wave contributes to a dc
signal. The setup in Fig. 21.63e (bridge rectifier) allows both half-waves to contribute to the dc signal.
The effective voltage in this case is the peak voltage divided by
√
2.
The forward resistance in the static (R f ) and dynamic (r f ) case are (for βV f > 3)
R f =
V f
I f
∼ =
V f
I s
exp
−
e V f
n kT
(21.170a)
r f =
∂ V f
∂ I f
=
n kT
e I s
exp
e V f
n kT
∼ =
n kT
e I f
.
(21.170b)
For reverse bias we have (β|V r | > 3)
R r =
V r
I r
∼ =
V r
I s
(21.171a)
r r =
∂ V r
∂ I r
=
nkT
eI s
exp
e |V r |
n kT
.
(21.171b)
Thus, the dc and ac rectifications ratios are given by
R r
R f
= exp
e V f
n kT
(21.172a)
r r
r f
=
I f
I s exp
e |V r |
n kT
.
(21.172b)
21 Diodes
21.5 Applications and Special Diode Devices
In the following, various electronic applications of diodes are discussed. The most important special
diode types are introduced. Optoelectronic applications (involving absorption and emission of photons)
are treated below (Sect. 22). The applications of Schottky diodes are discussed in [1791].
21.5.1 Rectification
In a rectifier, the diode has to supply a high resistivity for one polarity of the bias and a low one to the
other polarity. In Fig. 21.63a, a single-path rectification method is shown. Only the positive half-wave
can pass the load resistor R L (Fig. 21.63b). In Fig. 21.63c, the characteristic of a Si diode is shown. Of
course, the voltage drop across the diode can only range in the 1-V regime. In order to make the setup
work, the load resistor has to be considered. The total current is given by I = I s [exp(eU d /nkT ) − 1].
The total voltage U is split between the voltage drop across the diode U d and that over the load
resistance U L = R L I . The current is therefore given by
I =
U − U d
R L
.
(21.169)
For sizeable currents the voltage drop across the diode U d is between 0.7 and 1 V. The characteristic
is linear between about 1 and 220 V (Fig. 21.63d). Typically, the voltage U L is low-pass filtered with
a capacitor parallel to the load resistor. The effective voltage is the peak voltage divided by 2.
The drawback of the single diode rectifier is that only the positive half-wave contributes to a dc
signal. The setup in Fig. 21.63e (bridge rectifier) allows both half-waves to contribute to the dc signal.
The effective voltage in this case is the peak voltage divided by
√
2.
The forward resistance in the static (R f ) and dynamic (r f ) case are (for βV f > 3)
R f =
V f
I f
∼ =
V f
I s
exp
−
e V f
n kT
(21.170a)
r f =
∂ V f
∂ I f
=
n kT
e I s
exp
e V f
n kT
∼ =
n kT
e I f
.
(21.170b)
For reverse bias we have (β|V r | > 3)
R r =
V r
I r
∼ =
V r
I s
(21.171a)
r r =
∂ V r
∂ I r
=
nkT
eI s
exp
e |V r |
n kT
.
(21.171b)
Thus, the dc and ac rectifications ratios are given by
R r
R f
= exp
e V f
n kT
(21.172a)
r r
r f
=
I f
I s exp
e |V r |
n kT
.
(21.172b)