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EE 462: Laboratory # 5
Dynamic Effects in P-N Junctions
by
Dr. A.V. Radun
Dr. K.D. Donohue (9/27/03)
Department of Electrical and Computer Engineering
University of Kentucky
Lexington, KY 40506
Laboratory # 5
I.



Pre-lab due at lab sessions October 7, 8, and 9
Lab due at lab sessions October 14, 15, and 16
Instructional Objectives
Perform an AC incremental analysis of a P-N junction over a range of reverse bias voltages.
Measure the reverse bias junction capacitance.
Measure the reverse recovery of a diode.
See Horenstein chapter 3.3
II.
Background
The I-V curves completely characterize diode behavior when the dynamic effects of its junction
are not considered. Behavior is constant, independent of frequency. However for high
frequency or for large scale switching behavior, the two charge storage mechanisms in real
diodes can create a significant dynamic behavior. The charge storage mechanisms occur at the
depletion region of the P-N junction and in the neutral regions adjacent to the depletion region.
These storage phenomena are modeled as capacitors as shown in Fig. 1. The capacitance at the
P-N junction depletion region in reverse bias mode is referred to as depletion capacitance or
junction capacitance, which is in parallel with the open circuit ideal diode. The effective circuit
in reverse bias mode is shown in Fig. 1b. Recall that capacitance relates an increase in voltage to
the charged stored (Q = C V). In the forward bias mode both charge storage mechanisms have
an effect and are modeled with capacitances shown in Fig 1a. However, the junction capacitance
is not significant when the diode is forward biased. One reason for its minimal effect is that the
voltage across the diode's junction capacitance is essentially constant (0.7V) and very little
current flows into the junction capacitance. In addition, the current through the junction
capacitance is much smaller than the forward current and typically can be neglected. On the
other hand, junction capacitance can have a significant effect when the diode is reversed biased
for two reasons. First, the reverse diode voltage is generally not constant. Second, the reverse
leakage current through a diode is very small and thus the current through the junction
capacitance can be much larger than the reverse leakage current.
Cj
Rs
rd
Rs
Cj
Cdif
(b)
(a)
Fig. 1. (a) Forward bias diode model (b) Reverse bias diode model with junction capacitance Cj,
diffusion capacitance Cdif, bulk material resistance Rs, and dynamic resistance rd.
The capacitance of a P-N junction is not a constant, but depends on the reverse bias voltage on
the junction, which changes the size of the depletion region. As a result, the growing or
shrinking depletion region changes the charge separations distance (unlike a conventional
capacitor with a constant distance between charges). Thus, the capacitance of a P-N junction is a
function of the P-N junction’s reverse voltage. The capacitance decreases as the reverse voltage
increases (distance between the charge increases C    A / d ). The junction capacitance of a
reversed biased junction is modeled by:
CJ VR  
CJO
m
 VR 
1  
 VJ 
(1)
where VR is the reverse bias voltage, VJ is the junction potential, m is the grading coefficient, and
CJO is the zero biased junction capacitance. Three constants (CJO, VJ, and m) must be specified
to determine the relationship between VR and CJ. The constant CJO has units of Farads, and is
referred to as the zero-bias (VR = 0V) junction capacitance. The junction potential VJ has the
units of Volts and is sometimes called the built-in potential. The junction potential’s value is
about 0.7V for silicon (Si) junctions. For Si diodes VJ changes slightly from 0.7V depending on
the doping levels on both sides of the junction, while VJ changes significantly from one
semiconductor material to another. The grading coefficient m is unitless and its value depends
on the nature of the P-N junction. If the P region changes abruptly (a step change) to the N
region at the junction, then m  0.5 and the junction is called an abrupt P-N junction. If the P
region changes linearly into the N region (graded junction), then m  3 .333 . High voltage power
diodes often go from P to intrinsic semiconductor to N semiconductor (PIN diode). This results
in m = 0, which implies a constant capacitance. The parameter m is found by plotting the log of
the capacitance versus the log of the voltage (m is the slope of this curve for voltages >>Vj).
The circuit in Fig. 2 can be used to measure the junction capacitance of a diode (P-N junction) as
a function of its reverse bias voltage. The diode is reverse biased by the DC source, and the
reverse bias voltage is varied by a relatively small sinusoidal AC signal. The DC voltage is
called the bias voltage while the AC voltage is called the incremental voltage. With a known AC
voltage amplitude and frequency, the diode’s reverse current is measured to determine the
junction capacitance at the DC value of reverse voltage.
A
IR
+
VAC
VR
-
VDC
Fig. 2. Elements to measure junction capacitance.
The second charge storage mechanism at a P-N junction is charge stored in the neutral regions
adjacent to the junction. The amount of charge stored is proportional to the forward current and
the proportionality constant is called the transit time and has the units of seconds.
 qVf

Q   T  If   T  Is  e nkT  1


(2)
where T is the transit time, which typically varies from 10s to about 10ns depending on
junction processing, and If is the forward bias current. In the second part of Eq. 2, If is
substituted out with the Shockley Equation, where Is is the saturation current (on the order of
10-14 A for small signal diodes at 300K), q is the charge on an electron (1.6x10-19C), T is the
junction temperature in kelvins, k is Boltzmann's constant (1.38x10-23 J/K), n is a the emission
coefficient (typically between 1 and 2), and Vf is the voltage drop over the diode. This charge
storage mechanism is very nonlinear leading to a very nonlinear capacitance. The stored charge
is significant for forward bias and nearly zero for reverse bias. This charge storage affects the
diode's turn-off properties, delaying its turn-off time. The turn-off delay is called reverse
recovery and the delay time is called the reverse recovery time.
The reverse recovery time is close in value to the transit time, but not exactly equal to it. The
reverse recovery time depends on the transit time and the circuitry connecting the diode. The
basic diode rectifier circuit shown in Fig. 3 can be used to study diode reverse recovery. The
resistor Rsense performs 2 functions; it is the load for the circuit in forward bias mode and is a
resistor used to sense the current in the diode.
+
Vin
Vsense
Rsense
-
Fig. 3. Test circuit to measure reverse recovery time.
III.
Pre-Laboratory Exercise
1. Using Q = C Vc, apply simple definitions and differential calculus to show the current in the
capacitor is given by i C  C
2.
3.
4.
5.
6.
7.
8.
9.
dVC
dt
if C is a constant. Determine the capacitor current if the
capacitance is not a constant but rather depends on the capacitor voltage (C = C(VC)).
Explain how to compute the diode’s junction capacitance from the measured AC current, the
AC voltage, and the frequency. Explain why the DC voltage does not enter into the
calculation.
Build a SPICE model of the circuit in Fig. 4 for a diode with parameters IS = 9x10-14A, n = 1,
CJO = 23pF, VJ = 0.7V, m = 0.336, and T =5s.
The source consists of a DC part and an AC part, which in SPICE can be created with a
single transient source. Make the AC part of the source 1V P-P at 100kHz. For the
simulation set up, determine a good step size for this simulation, and explain what bad things
happen if the step size is too small or too large. When you set up your simulation (set the
time step and duration) make sure you do not check “use initial conditions” so SPICE will
compute them. This way your simulation immediately reaches steady state.
When you observe the diode voltage, it will have both an AC and a DC part. Only the AC
part is of interest. The DC part can be removed using a coupling capacitor and a resistor as
shown in Fig. 4, so that the AC voltage across R is close to the AC voltage over the diode.
Assume the resistance is 1M and explain how to compute the Ccouple value. Verify that a
Ccouple of 200pF results in the AC measurement over the diode being approximately equal to
the voltage over R.
Simulate and measure the capacitance of the diode for VDC = 1V, 2V, and 4V. Thus make the
DC part of the Spice transient source equal to 1V, 2V, and 4V. Run your simulation for three
cycles. Plot VAC and IR for 2V DC reverse bias voltage. You may normalize one of them so
they both fit on the same plot (on SPICE in edit plots, select "Add New Plot to Graph" and it
will allow you to perform mathematical expressions on the circuit quantities it computed to
create a new quantity for plotting).
Compute the diode’s junction capacitance for DC reverse bias voltages equal to 1V, 2V, and
4V using Eq. 1. Compare your SPICE results to those obtained with Eq. 1. Try to explain any
discrepancies you observe.
A student hooks up their signal generator and lab power supply as shown in Fig. 5 to create
an AC plus DC source. Explain what problems this circuit will have and how they can be
fixed.
Make a SPICE model of the circuit in Fig. 3 using Rsense = 10k, and Vin equal to a 20Vp-p
(goes from -10V to +10V) square wave at 10kHz. Use the SPICE pulse input with the rise
and fall times equal to 1s. Compute and plot the voltage across the sense resistor with the
diode parameters of pre-lab problem 4, first making T =0s and then making T = 5s. What
should your time step be? What do you observe?
Ccouple
A
Single Source in
Spice
IR
R
+
VAC
VR
-
VDC
Fig. 4. Test circuit model for performing a junction
capacitance measurement.
Signal
Generator
VAC
VAC
red
DC lab
power
supply
black
VDC
VDC
green
Fig. 5. Example for Pre-lab Problem 9 (Do not try
to implement this in the lab!!!!!! Never put the
signal generator in series with the lab power
supplies, you can destroy the signal generator)
IV.
Laboratory Exercise
1. Fabricate the circuit in Fig. 6 using one of the variable lab power supplies for the DC source
and the signal (function) generator for the AC source. Make the AC source a 1Vp-p 100kHz
sine wave.
2. The 10k resistor is being used as a current sense resistor. (Discussion: What is the
constraint on the voltage across this resistor and on its value such that it serves this
purpose adequately?)
3. Set the DC source to 1V, 2V, 4V, 8V, 16V, and 32V keeping the AC voltage at 1Vp-p.
Measure the voltage from the diode cathode to ground. (Discussion: Does this voltage have
a DC part, AC part, or both?) Record the voltage waveform for a DC voltage equal to 1V,
4V and 32V.
4. Measure the voltage across the 10k resistor for each DC voltage in Exercise 3.
(Discussion: Does this voltage have a DC part, AC part, or both? Explain.) Record the
voltage waveform for a DC voltage equal to 1V, 4V and 32V.
5. Compute the current through the diode’s junction capacitance at each DC voltage and the
corresponding diode’s junction capacitance. Plot this capacitance versus reverse bias DC
voltage.
6. Fabricate the circuit in Fig. 3 with Rsense = 10k.
7. Make the input voltage a 20Vp-p square wave. Measure and record the current sensing
voltage at 100Hz, 1kHz, 10kHz, and 100kHz. (Discussion: Describe how the peak reverserecovery currents and the reverse-recovery times change with frequency?)
0.05F
100k
1N4001
VDC
Rsense =
10k
+
VAC
VR
-
Fig. 6. Test Circuit for Measuring Junction Capacitance.