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Unformatted text preview: ECE 3040B Dr. Alan Doolittle
Microelectronic Circuits
Homework #4 Solutions
Due 2/28/01 1.) Pierret 5.3
This problem is a practical example of a “HighLow” (high doping to low doping)
junction. This type of junction is often used with a pn junction to improve the
performance of the metal contacts at the ends of the diode. (b) Start with either Jp=0 or Jn=0 (similar to equation 5.6) and solve for E (eq. 5.7). 2.) Pierret 5.5 3.) a.) Given,
−1 2
D n n i2 D p ni
+
where I o = qA
Ln N A L p N D kT I o is the " reverse saturation current" I = Io e qV A We have all parameters necessary except the diffusion coefficients and diffusion lengths.
But,
1000(0.0259) = 25.9 cm 2 / S for electrons
kT
=
q
250(0.0259 ) = 6.475 cm 2 / S for holes Ln, p = Dn, pτ n , p = 25.9(10e − 6 ) = 0.0161 cm for electrons
6.475(1e − 6 ) = 0.00254 cm for holes 2
2
2
Dn ni2 D p ni
25.9(1e10)
6.475(1e10)
I o = qA
+
= 1.6e − 19(0.00456 )
+
Ln N A L p N D
0.0161(1e18) 0.00254(1e14 ) I o = 1.86e − 12 A or 1.86 pA b.) If a positive voltage is applied to the “pside” the diode is forward biased. +
VA
 Dn, p = µ n , p c.)Consider the law of the junction on either side of the depletion region,
At x = − x p At x = xn p = p o + ∆p n
p= but
p = NA
np = ni2 e but ,
n = ND so,
qV A ni2
n 2 qVA
+ i e kT − 1
ND ND np = ni2 e kT ni2
n 2 qVA
+ i e kT − 1
NA NA n= ni2 qVA kT
e
∆p n ( x = x n ) =
−1
ND −1 n = no + ∆n p kT n
e
NA qVA ∆n p ( x = − x p ) = 2
i so,
qV A kT ( + 0.35 np = (1e10) 2 e 0.0259 = 7.4e25 cm −3 which is much greater than in equilibrium no po = ni2 = 1e 20 cm −3
Note that I could have well asked you the what the excess carrier concentration was at the
depletion region edges. In that case the answer would not be identical on opposite sides
of the depletion region due to differences in doping on the two sides.
d.) If a positive voltage is applied to the ntype side of the junction, the diode is reverse
biased.
e.) As follows from part c,
np = n i2 e qV A kT np = (1e10) 2 e − 0.35 0.0259 ( = 1.4e14 cm − 3 which is much less than in equilibrium n o p o = ni2 = 1e20 cm −3 f.) Since in equilibrium (no bias), no excess minority carriers are injected from the ptype side, the hole concentration is merely, po=ni2/ND = 1e6 cm3 .
For the forward bias case in part b, excess minority carriers are injected across the
junction and recombine in the ntype quasineutral region as they move away from the
junction. As derived in class the expression for the excess hole concentration is,
qVA
ni2
∆p n ( x' ) =
e kT − 1 e (− x '/ LP )
ND for x' ≥ 0 or
e 0.35 1e14
∆p n ( x' = 1um) = 7.3e11 cm −3 0.0259 − 1 e ( −[1e− 4cm / 0.016cm ]) ∆p n (1e10)2
( x' = 1um) = For the reverse bias case in part c, minority carriers are stripped away from the depletion
region edges by the electric field. Thus, ) ) qVA
ni2
∆p n ( x' ) =
e kT − 1 e ( − x '/ LP )
ND for x' ≥ 0 or
e −0.35 0.0259 1e14
∆p n ( x' = 1um) = −9.9e5 cm −3 − 1 e ( −[1e− 4 cm / 0.016cm ]) ∆p n (1e10)2
( x' = 1um) = You will notice that this is almost all the minority carriers we have in this region (~po). It
should make since that you can never take away more carriers than you had in
equilibrium. However, in the previous forward bias case (see above) you can add more
minority carriers.
4.)Consider the following plots to scale of the diode current, I=1e13(eVa/0.02591)
2 Current [A] 2.5 8.0E07 Current [A] 1.0E06 6.0E07
4.0E07
2.0E07 1.5
1
0.5
0 0.0E+00 2 1.6 1.2 2.0E07
0.4
0 0.8 0.4 0.8 2 1.6 1.2 0.8 0.5
0.4
0 0.4 0.8 Voltage [V] Voltage [V]
1.0E+02
1.0E+00 Current [A] 1.0E02
1.0E04
1.0E06
1.0E08
1.0E10
1.0E12 2 1.5 1 1.0E14
0.5
0 0.5 1 Voltage [V] a.) From the curve with a maximum current scale (yaxis maximum) or 1e6 A, the turn
on voltage appears to be ~0.4V
b.) From the curve with a maximum current scale (yaxis maximum) or 2.5 A, the turn
on voltage appears to be ~0.8V
c.) When the data is plotted in a semilog plot, it is clear that the diode current is
continuously increasing with increasing forward bias. However, changes in current
from say, 1e10 A to 1e8A are not noticeable on the linear scale. Thus, for the linear
scales, when the currents approach the range of the scale maximum, the current
rapidly increases with increasing voltage, thus, appearing to “turn on” the diode. 5.) Consider the following expressions for the junction capacitance and the built in
voltage.
NN
kT
ln A 2 D
q
ni or, qK S ε o N A N D
1
2 ( N A + N D ) kT
NN
ln A 2 D − V A
q
ni C junction = A Vbi = qK S ε o N A N D
1
2 ( N A + N D ) (V bi − V A ) C junction = A using numeric values making sure to use the correct units for the dielectric constant,
1
1e18 N D 0.0259 ln (1e10) 2 1.6e − 19(11.8)8.85e − 14 1e18 N D
2
(1e18 + N D ) − (−3) or
N D = 5.5e15 cm − 3 b.) What is really being asked here is to take the large signal solution (bias point of the
diode) and determine the small signal capacitance and resistance at that bias
condition. From part a, Cdiode=50pF. Making use of the fact that the diode current
in reverse bias is essentially the negative of the saturation current, the resistance of
the diode is, RDiode = 1 / g d = VT
0.0259
≈
=∞
ID + IS − IS + IS c.) The output voltage, Vout, for the time varying input signal, VAC, can be found by
noticing that Vout is part of a reactive voltage divider (that is a voltage divider that
uses complex impedances). Note: due to a typing error, L1=L3. Thus,
1
jϖ C Diode
1
1
Vout = V AC
=
=
2
1
1 − ϖ L1C Diode 1 − (2.5e − 14)ϖ
+ jϖ L1
jϖ C Diode
This type of circuit would have “infinite” voltage gain at a specific frequency, when
w2=1/L1Cdiode. In practice the inductors and capacitors have additional resistive
components (are not perfect reactive parts) and would thus result in a large but finite
voltage gain. 2 C junction = 50e − 12 = 0.00456 d.) Plugging Va=9 V into the equation in part a, Cdiode=31.2pF and Rdiode still is
infinity. Notice that this would change the resonance frequency of the circuit. Using
this approach, one could select a given frequency to amplify will not amplifying all
other frequencies. Thus, a radio, cell phone or similar receiver circuit might employ
such a diode variable tuning element.
e.)
1
w2 =
L1C Diode 2πf = 1
L1C Diode
1
2π 1
= 1 MHz for part b
1
1
5e − 4(50e − 12)
f=
=
2π L1C Diode
1
1
= 1.27 MHz for part d
2π 5e − 4(31.2e − 12)
Thus a substantial amount of tuning can be achieved by changing the diode bias voltage f.) At 1 MHz, neglecting the effect of Cdiode, the voltage divider becomes,
Vout = V 1 RDiode
thus ,
RDiode + jϖL1 RDiode RDiode
RDiode
= 0.001 =
RDiode + j 3141
+ j 2π (1e6)(5e − 4) RDiode = 3.14159 Ω
but RDiode is the small signal resis tan ce of the diode,
RDiode = rD = VT
V
≈ T in forward bias (I D >> I S )
ID + IS ID Thus ,
I D = 0.0259 3.14159 = 8.2 mA This would be the current bias needed to attenuate the signal by the requested amount.
Thus, the same circuit that gave a large voltage gain in reverse bias could be used to
“blank” or “zero” the output voltage when the diode is now placed into forward bias. 6.) Jaeger 3.42 7.) Jaeger 3.46 ...
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This note was uploaded on 02/17/2010 for the course ECE 3040 taught by Professor Hamblen during the Spring '07 term at Georgia Institute of Technology.
 Spring '07
 HAMBLEN

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