I. I. + _. Reverse bias. Reverse breakdown. Forward bias. V. V. V F. V ZK. Last time…. Focus. and its (complicated) IV relationship. we introduced the diode. Today we will…. focus on the relevant area of the diode IV graph
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I
+
_
Reverse
bias
Reverse
breakdown
Forward
bias
V
V
VF
VZK
Last time…
Focus
and its (complicated)
IV relationship.
we introduced the
diode
Today we will…
+
_
V
REALISTIC DIODE MODELI
V
+
_
V
IDEAL DIODE MODELI
I
Forward bias
+
_
V
Reverse bias
V
+
_
V
LARGESIGNAL DIODE MODELI
I
+
+

Forward bias
VF
V
Reverse bias
V
VF

+
_
V
SMALLSIGNAL DIODE MODELI
I
+
slope = 1/RD
+

VF
Forward bias
Reverse bias
V
V
RD
VF

IL
INL
+
VL

+
VNL

Nonlinear
element
Linear circuit
SOLVING CIRCUITS WITH NONLINEAR ELEMENTSLook at circuits with a nonlinear element like this:
A nonlinear element with its own IV relationship, attached to a linear circuit with its own IV relationship.
To find the solution graphically,
INL
graph the nonlinear IV relationship,
fL(VNL)
graph the linear IV relationship in terms of INL and VNL (reflect over yaxis),
x
fNL(VNL)
VNL
and find the intersection: the voltage across and current through the nonlinear element.
1 kW
Find VNL.
Assume realistic diode model with
I0 = 1015 A.
INL
+
_
+
_
IL
+

2 V
VNL
VL
Either substitute into 3. and solve
or determine graphically that VNL = 0.725 V
1 kW
Find VNL.
Assume smallsignal diode model with
VF = 0.7 V and RD = 20 W.
INL
+
_
+
_
IL
+

2 V
VNL
VL
Either substitute into 3. and solve
(VNL – 0.7) / 20 = (VNL 2) / 1000
or determine graphically that VNL = 0.725 V
1 kW
Find VNL.
Assume smallsignal diode model with
VF = 0.7 V and RD = 20 W.
INL
+
_
+
_
IL
+

2 V
VNL
VL
Either substitute into 3. and solve
0 = (VNL  2) / 1000
or determine graphically that VNL = 2 V