Download semiconductor circuit templates

Diode Circuit Templates
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np junctions as diode template
0
p
current
meter
pn junction
anode
n
0
+
volt
meter
cathode
This diode is forward biased
np junctions as diode template
voltage regulator
+
R
1
v
voltage regulator
= 2 (0.6 volts)
out
R
1
+
+
+
-
+
0.6 volt
drop
v
out
(Regulated
output voltage
supply)
anode
(unregulated
input voltage )
-
v
out
(Regulated output
voltage supply)
+
+
cathode
2 Silicon diodes in series
2 Silicon diodes in series
np junctions as diode template
voltage regulator
1
+
(unregulated
input voltage )
+
+
v
out
(Regulated output
voltage supply)
(unregulated
input voltage )
+
+
+
-
-
2 Silicon diodes in series
+
R
1
-
R
voltage dropper
vout
1
v
= v - 0.6 volts
1
out
np junctions as diode template
voltage regulator
1
+
+
(unregulated
input voltage )
-
v
out
(Regulated output
voltage supply)
(unregulated
input voltage )
+
+
+
-
-
+
R
1
-
R
voltage dropper
vout
1
v
= v - 0.6 volts
1
out
2 Silicon diodes in series
These two silicon diode circuits will provide the same
regulated output voltage value if the unregulated input
supply voltage is the same for both.
np junctions as diode template
Finding the current that
flows through the diode
isn’t quite so simple.
One way to find this
current value is to create
a load line diagram.
R
(1000 ohms)
1
Value of Vout is
easy to find.
+
(5 volts)
v
out
For silicon diode
Vout is= 0.6 volts.
-
(1) Use a model to plot
the i vs v response for
the diode.
current
(2) Plot the conductance
value for R1 on the
same graph.
Voltage
np junctions as diode template
R
(1) Use a model to plot the i vs.
v response for the diode.
(1000 ohms)
1
Value of Vout is
easy to find.
+
(5 volts)
v
out
For silicon diode
Vout is= 0.6 volts.
-
Model for diode current flow.
600
i = i s + i se
500
current
(A)
(q)(v)
(k)(T)
-12
amps
Typical is value = 1.0 X 10
400
-23
k (Boltsman’s constant)= 1.38 X 10
J/K
-19
C
q (charge on an electron)= 1.60 X 10
300
200
1
2
3
4
Volt
5
6
np junctions as diode template
R
(1) Use a model to plot the i vs.
v response for the diode.
(+.5)
(-.5)
(+.3)
(+
(-2)
(q)(+.7)
(v)1)
(k) (T)
-13
8
-13
-2
-10
+2.2
+1.0
+5.0
10
+5.1 XXXi10
10 = i s + i s e
(1000 ohms)
1
Value of Vout is
easy to find.
+
(5 volts)
v
out
-
v
600
-2
500
current
(A)
-0.5
400
+0.3
300
+0.5
200
+0.7
1
2
3
4
Volt
5
For silicon diode
Vout is= 0.6 volts.
6
+1.0
i
-13
-10
-13
-10
-13
+1.0 X 10
-8
+2.2 X 10
-2
+5.0 X 10
2
+5.1 X 10
np junctions as diode template
(2) Plot the conductance value
for R1 on the same graph.
R
(v)
(+.7)
(+.5)
(-.5)
(q)(+.3)
(+1)
(-2)
(k) (T)
(5 volts)
diode model
for current
2
+5.1 X 10
i = is + is e
d
1
Value of Vout is
easy to find.
+
v
out
For silicon diode
Vout is= 0.6 volts.
-
The diode model for predicting current is a different
equation than the resistor model for predicting current.
600
The resistor model for predicting current is (of course)
Ohm’s law.
v
500
current
(A)
(1000 ohms)
ir =
400
R
+5.0
300
200
1
2
3
4
Volt
5
6
volts
ir =
3
1.0 X 10 Ohms
-3
i r = 5.0 X 10 Amperes
np junctions as diode template
(2) Plot the conductance value
for R1 on the same graph.
R
(v)
(+.7)
(+.5)
(-.5)
(q)(+.3)
(+1)
(-2)
(k) (T)
(5 volts)
diode model
for current
2
+5.1 X 10
i = is + is e
d
(1000 ohms)
1
Value of Vout is
easy to find.
+
v
out
For silicon diode
Vout is= 0.6 volts.
-
The load line is now defined by the following two points.
-3
( 0v , 5.0 X 10 A )
-3
( 5v , 0.0 X 10 A )
600
500
current
(A)
400
Current axis needs to
be expanded to plot
these two points.
300
200
1
2
3
4
Volts
5
6
ir =
v
R
+5.0
volts
ir =
3
1.0 X 10 Ohms
-3
i r = 5.0 X 10 Amperes
np junctions as diode template
(2) Plot the conductance value
for R1 on the same graph.
R
(v)
(+.7)
(+.5)
(-.5)
(q)(+.3)
(+1)
(-2)
(k) (T)
(5 volts)
diode model
for current
2
+5.1 X 10
i = is + is e
d
(1000 ohms)
1
Value of Vout is
easy to find.
+
v
out
For silicon diode
Vout is= 0.6 volts.
-
The load line is now defined by the following two points.
-3
( 0v , 5.0 X 10 A )
-3
( 5v , 0.0 X 10 A )
6
600
5
500
current
(A)
(mA)
4
400
3
300
2
200
1
2
3
4
Volts
5
6
The absolute value of the
slope of this line is the
conductance of the resistor.
np junctions as diode template
R
(2) Plot the conductance value
for R1 on the same graph.
(1000 ohms)
1
Value of Vout is
easy to find.
+
(5 volts)
v
out
For silicon diode
Vout is= 0.6 volts.
-
-3
-3
5.0 X 10 A
= 1.0 X 10 mhos
m =
5v
6
600
5
500
current
(A)
(mA)
The absolute value of the
slope of this line is the
conductance of the resistor.
4
400
3
300
2
200
(The inverse of the slope is the
resistance value for the resistor. )
1
2
3
4
Volts
5
6
np junctions as diode template
R
(2) Plot the conductance value
for R1 on the same graph.
(1000 ohms)
1
Value of Vout is
easy to find.
+
(5 volts)
v
out
For silicon diode
Vout is= 0.6 volts.
-
The operating point, (the amount of
current that flows through the circuit) is
at the intersection of these two plots.
6
5
current
(mA)
For this specific circuit, the current
flowing through R1 and the diode is
approximately 4.3 mA.
4
3
2
1
2
3
4
Volts
5
6