In this
article, we will learn how to analyze circuits
that utilize forward conducting diodes. We will also talk about the exponential
model, constant voltage drop model, as well as the ideal diode model.
Having read the previous article in this series, you should have an understanding of the
diode terminal characteristics. From this previous knowledge, we will now delve
into analyzing circuits that utilize
forward conducting diodes. The figure below, Figure 1.1, illustrates a circuit
that consists of a DC source, a resistor, as well as a diode. We will analyze this circuit and determine the diode’s current
ID and voltage VD. For a better understanding, we will
represent the diode with a model.
A diode operating in the
forward region can be most accurately described by the exponential model.
However, due to it being extremely nonlinear, it is very difficult to use.
Anyways, we will illustrate its accurate yet severe nonlinear behavior by analyzing the
circuit shown in Figure 1.1.
Recall from the previous article that IS is a constant for a
given diode at a given temperature. There is a formula that is used to
calculate the value for IS, the reverse saturation current, in terms of the diode's
temperature and parameters that we will not explain in this article.
When VDD is high enough to cause the
diode’s forward current (ID) to be much larger than its saturation current, we can
approximate the current–voltage relationship with the following exponential
equation:
ID=ISeVD/nVTID=ISeVD/nVT
where VT is approximately 25 mV at room
temperature and n is a constant influenced by the diode’s
physical structure and fabrication process. For simplicity, in this article we
will assume that n = 1.
This equation indicates that
the forward current increases exponentially with respect to the forward voltage.
There is another equation
that regulates how the circuit operates and is obtained by writing out a
Kirchhoff loop, which results in
ID=VDD−VDRID=VDD−VDR
We will use Equations 1.1 and
1.2 by plotting their relationships and characteristics on a current vs.
voltage plane. The intersection point of the two graphs plotted on the same
axes is the solution, or the operating point Q. The two equations
are illustrated in Figure 1.2.
The curve that is plotted in
Figure 1.2 represents the exponential diode equation, or Equation 1.1, and the
straight line that slopes downwards represents Equation 1.2. This straight line
is referred to what is known as the load line. This load line represents the
constraint of other parts of the circuit place on a non-linear device (such as
a diode or even a transistor). The exponential curve is the diode's
characteristic curve. The operating point of the circuit has coordinates that
give the values of ID and VD.
By analyzing the
exponential diode model graphically, we can gain a better understanding
visually how the circuit will work. However, to actually perform such an
analysis is considered to be very tedious and requires much effort, especially
for complex circuits.
The solutions to Eqs 1.1 & 1.2 can be found through a rather
straightforward iterative process, which is illustrated in the following
example.
Analyze the circuit in Figure 1.1
with R = 2 kΩ and VDD = 5 V to determine the current ID as well as the diode's voltage VD. We can assume that the diode's current
of 2 mA at a voltage of 0.7 V.
With VD = 0.7 V and Equation 1.2, the
current can be found as follows:
ID=VDD−VDRID=VDD−VDR
=5V−0.7V2kΩ=2.15mA=5V−0.7V2kΩ=2.15mA
By applying the equation
illustrated in the previous article, we can obtain a more accurate value
for VD.
V2−V1=2.3VTlogI2I1V2−V1=2.3VTlogI2I1
If we assume that 2.3VT = 60 mV, we have
V2=V1+0.06V⋅logI2I1V2=V1+0.06V⋅logI2I1
Using the diode's voltage and
current as V1 =
0.7 V and I1 =
2 mA, as well as I2 =
2.15 mA provides a value of V2 = 0.70188 V. The values from the first iteration are ID = 2.15 mA and VD = 0.70188 V. Iterating a second
time provides the following values:
ID=5V−0.70188V2kΩ=2.149mAID=5V−0.70188V2kΩ=2.149mA
V2=0.70188V+0.06log[2.149mA2.15mA]V2=0.70188V+0.06log[2.149mA2.15mA]
=0.70187V=0.70187V
This second iteration
provides values of ID = 2.149 mA and VD = 0.70187 V. Since the second
iteration provided values that are extremely close to those of the first
iteration, the iteration process is complete. Thus the solution is ID = 2.149 mA and VD = 0.70187 V.
Of all the models, the
simplest and most commonly used in modeling the
diode is the constant-voltage-drop (CVD) model. The CVD model uses the fact
that any forward conducting diode has a voltage drop that fluctuates in a
rather narrow range (~ 0.6 to 0.8 V). In this model, we assume that the voltage
is at a constant value of 0.7 V. This assumption is better explained in Figure
1.3.
The CVD model is perhaps one
of the most often utilized in the beginning stages of analysis and design of
any circuitry. Also, at any point, if there is little to no information
detailing the diode’s characteristics, this model is used quite frequently as
well.
Lastly, if we apply the CVD
model to solve our previous example, we are provided with
VD = 0.7 V
thus,
ID=VDD−0.7VRID=VDD−0.7VR
=5V−0.7V2kΩ=2.15mA=5V−0.7V2kΩ=2.15mA
which isn't too far off from the values
previously obtained with a more extensive, exponential model.
In many instances,
applications that involve voltage values that are much greater than that used
in the CVD model (0.6 - 0.8 V), neglect the diode's voltage drop completely
whilst computing the diode current. This is what is known as the ideal diode
model. For the circuit that was used in the previous example with its values,
we see that applying this model provides us with
VD=0VVD=0V
ID=5V−0V2kΩ=2.5mAID=5V−0V2kΩ=2.5mA
For an analysis this quick,
such an answer would not be an entirely bad estimate. However, the 0.7 V CVD
model provides a much more accurate result almost no more work needed to be
done. This model is great for determining which diodes are on or are off in a
circuit containing multiple diodes.
In this article, we discussed
the characteristics of forward conducting diodes, the exponential model, the
CVD model, as well as the ideal diode model. As of now, you should have an
understanding of each model, what its applications are, as well as be able to
solve problems using each model.
In the next article, we will
build off of this article and talk about the small signal model and how the
diode forward drop is used in voltage regulation.
Thank you for reading. If you
have any questions or comments, please leave them below!