Transcript No Slide Title

BASIC SEMICONDUCTOR ELECTRONIC
CIRCUITS
Introduction of two basic electronic elements: diode and transistor
LEARNING GOALS
Diodes
structure and four modeling techniques
Transistors
MOSFETs and BJTs.
MOSFETs in switching and amplification
PASSIVE DEVICES IN INTEGRATED CIRCUITS
Inductor
Resistor
Capacitor
DIODES
Structure
Symbol
I-V convention
Ideal diode
Forward bias
Ideal diode
Reverse bias
Ideal diode
I - V curve
Actual diode I - V characteristic
I S  reverse saturation current
q  charge of electron
k  Boltzmann' s constant
T  Temperatu re in degrees Kelvin
 qVD 
I D  I S e nkT  1


n  'ideality' factor
Typically: I S  1014 A, n  1
I - V curve of an actual diode
For high power diodes: I S  1010 A, n  2
q
 39 at room temperatur e
kT
Next we develop two
approximations to the
actual I - V curve
Constant voltage model
Circuit
equivalent
I - V curve
Piecewise linear model
Comparison of models
Circuit
equivalent
I - V curve
LEARNING EXAMPLE Using the ideal model, find the voltage V1
V1  12[V ]
Diode is forward biased by the source
Forward bias
V1  4[V ]
Diode is reverse biased
V1  10[V ]
Reverse bias
V1  1[V ]
LEARNING EXAMPLE
Analyze the protection circuit using diode models
Simplifying
assumption
vin  v s
When diodes are
reverse biased
Thevenin
equivalent
of driving
network
Ideal diodemodel
0  vin  Vdc
Under-voltage with piecewise linear
model
Using source superposition
vinu 
rd
Rs
vs 
VF
rd  Rs
Rs  rd
Using a similar way for over-voltage
 Von  vin  Vdc  Von
Under-voltage protection model
using diode constant voltage model
vino 
rd
RS
vS 
(Vdc  VF )
rd  Rs
rd  Rs
The figure in the next slide compares
the three models
Vdc  3V
Von  0.7V (constant voltage model)
VF  0.6V 
 (piecewise linear model)
rd  10 
RS  40
Analysis using non linear diode model
Simplifying
assumption
vin  v s
When diodes are
reverse biased
Under-voltage scenario. Top diode open
v S  v in
 iD  0
RS




i D  I S e 39v D  1  I S e 39vin  1 
v D  v in
vin  v S
RS
1  vin  v S 
vin  ln 
 1
39  Rs I s

Implicit function. Must be evaluated
numerically
Is=1e-14;% MATLAB SCRIPT TO ANALYZE
n=1;
% UNDER-VOLTAGE
Rs=40;
vs=[-2:.2:2]';
vin=[];
for k=1:length(vs)
x0=[vs(k);vs(k);Rs;Is];
kcl=inline('((x(2)x(1))/x(3)+x(4)*(exp(-39*x(1))-1))^2');
x=fminsearch(kcl,x0);
vin=[vin;x(1)];
end
plot(vs,vin,'bo',vs,vin,'r'),
title('UNDER-VOLTAGE WITH NONLINEAR
DIODE')
xlabel('Source Voltage(v_S)[V]'),
ylabel('Input Voltage(v_{in})[V]'),grid
TRANSISTORS
STRUCTURE OF MOSFETs
The oxide is an insulator allowing zero
current into the gate
The gate voltage controls resistance
from source to drain (in linear range)
Cutoff state
Bulk internally
connected to source
threshold
conducting
N - channel MOSFET in the linear range
Complete characteristics of n-channel
MOSFET
VDS  Ron I D
Linear characteristic
Threshold voltage VT  1V
A cutoff region
B linear region
C saturation region
STRUCTURE OF BJTs (Bipolar Junction Tansistors)
NPN transistor
VBE  constant
Active mode : IC   I E , IC   I B   

1
  common emitter current gain ( 1)
  common base current gain
B
C

 IB
V BE

E
Linear model for BJT in active mode
MODELING MOSFET SWITCHING APPLICATIONS
Ron depends on gate voltage.
Cutoff
Basic switching circuit
Linear
Region
Typical values range from a few hundred Ohms
for low-current applications to milli-ohms for
currents in the tens of Amperes
Cutoff
(switch open)
In the next slide we quantify
this non-ideal behavior
‘Switch closed’
VS
R
2
Ron
Ideally,Vo  0
Vo 
The ratio,
Ron
, is more important than the actual value of Ron
R
LEARNING EXAMPLE
The MOSFET is used as switch to energize and de-energize an
inductor. Determine the output voltage with the “switch closed”
R1  RL  1,VS  10V , Ron  0.25
Vo  0.1V  Ron  0.01R  10m
Vo
 0.17  Vo  1.7V
VS
Circuit after all transients have
subsided (R=1 Ohm)
Von 
VS
R
2
Ron
Ron
 0.25
R
LEARNING EXAMPLE The MOSFET is used to switch 10A of current. The voltage drop
across the FET in the on state must be less than 4% of the supply
voltage. Find the maximum value of R_on
Von
Von  10 A  Ron  0.04  40
Ron  0.16
MODELING AMPLIFIER APPLICATIONS
MOSFET in saturation
Linear model : id  gm v gs
VGS  (by v gs )  I D  (by id e)
id applied to a resistor, R, will generate
vo  ( Rgm )v gs
Rgm  1  amplification
Due to non-zero slope of I - V curves in saturation
Linear (small signals) model for MOSFET
Common Source Amplifier
MOSFET in saturation range
Define drain-source current.
Together with bias voltage they define operating point
Bias voltage.
Helps to define location on I - V curve.
The operating point determines the
values of the parameters gm , rds
in the linear model
Linear model relating changes in
gate voltage to changes in output
voltage
RD also helps to determine gain
LEARNING EXAMPLE
Given: gm  50mA / V , rds  2k.
Determine RD to produce a gain of - 50
vo  (rds || RD ) gm v gs  A  (rds || RD ) gm
 0.05[V / V ]
2k  RD
 50  RD  2k
2k  RD
LEARNING EXAMPLE
If rds  2k  25%, what are the expected minimum and maximum gains
rds min  2k (1  0.25)  1.5k  Amin  0.05(1.5k || 2k )  42.9
rds max  2k (1  0.25)  2.5k  Amax  0.05(2.5k || 2k )  55.5
PROBLEM
If one must assure a gain of at least 50 in absolute value what should
be the value for Rd?
LEARNING EXAMPLE
Find the value of R_d such that maximum average power is
transferred to the load at 30krad/s    3 104
 ZTH  R  jL  1  j 3 k
LOAD
Because there are dependent sources we
must determine open-circuit voltage
and short-circuit current
ZTH 
VOC
I SC
Open circuit
gm  1.5mS , rds  110k
R  rds || RD
KCL :
Small signals model
For maximum power transfer the Thevenin
impedance of the amplifier must be the
complex conjugate of the load
VOC  Vgs
VOC
 gmVgs 
0
1
R
jC1
VOC  Vgs
( jC1  gm ) R
jC1 R  1
EXAMPLE (continued)
Determination of short-circuit current
I SC   gmVgs  jC1Vgs  ( jC1  gm )Vgs
VOC  Vgs
( jC1  gm ) R
jC1 R  1
ZTH 
ZTH
VOC
R
1  jC1R


I SC jC1 R  1 1  jC1R
R
C1R2  1000  j 3000 

j
2
1  (C1 )
1  (C1 )2
 C1R  3  R 
3
 10k
4
9
3 10 10
10k  rds || RD  110k || RD
RD  11k
CASCADING COMMON SOURCE AMPLIFIERS
v gs 2  (rds1 || R1 ) gm1v gs1   A1v gs1
vo  (rds2 || R2 ) gm 2v gs 2   A2v gs 2
vo  A1 A2v gs1
ANALYSIS OF A DC-DC CONVERTER
When S2 is open and S1 closed, the inductor
stores energy. A large capacitor helps to keep
output voltage constant
When S2 is closed and S1 open, the inductor
transfers energy to capacitor
Idealized circuit
1 2
Vin ton D  duty cycle
WL  LI p ; I p 
2
L Ts  switching period
1
Practical implementation of booster
Vin2 D 2Ts2
f

t on  DTs  W L 
s
Ts
2L
Turns FET on or off. When FET is off
diode becomes direct biased
1
WC  C (Vo2  Vo2 )  energy transferre d to
2
Assuming W L  WC  Wload and solving for Vo
capacitor
Vo  voltage at end of cycle
Vo   voltage at beginning of cycle
Energy transferre d to load
Vo2
Wload 
Ts
Rload
2W L
2Vo2
Vo 

 Vo2
C
RloadCf s
USING EXCEL TO COMPUTE OUTPUT OF DC-DC CONVERTER
Vin[V]
L[H]
C[F]
D
fs
Rload
5
1.00E-05
1.00E-06
0.6
1.00E+05
200
n
Time(ms)
0.00E+00
1.00E-02
2.00E-02
3.00E-02
4.00E-02
5.00E-02
6.00E-02
7.00E-02
8.00E-02
9.00E-02
1.00E-01
1.10E-01
1.20E-01
1.30E-01
1.40E-01
1.50E-01
1.60E-01
1.70E-01
1.80E-01
1.90E-01
2.00E-01
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Excel formula to compute W_L
=$B$1^2*($B$4)^2/(2*($B$2)*($B$5)^2)
Excel formula to compute Vo (cellD10)
Wl
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
4.50E-05
Vo
0
9.486833
13.0767
15.6173
17.5929
19.19789
20.53541
21.66871
22.64022
23.48024
24.21135
24.85097
25.41286
25.90815
26.34595
26.73384
27.07819
27.3844
27.65709
27.90024
28.11727
=SQRT(2*C10/$B$3-2*D9^2/($B$6*$B$3*$B$5)+D9^2)
Plot of DC-DC converter output voltage
MATLAB SCRIPT TO COMPUTE AND PLOT THE BOOSTER OUTPUT
%booster.m
%Solves the booster circuit in BECA 7th Ed, page648
%%%%%%%%% define circuit parameters
Vin=5;
L=1e-5;
C=1e-6;
D=0.6;
fs=1e5;
Rload=200;
%%%%%%%%% define iteration array
n=[0:30];
tms=1e3*n/fs; %time in msec
%%%%%%%%% initialize vectors and do one-time computations
Wl=Vin^2*D^2/(2*L*fs^2);
t1=2*Wl/C;
kt2=1-2/(Rload*C*fs);
vo=zeros(size(n)); %initialize output array
%%%%%%%%% do the iterations
for k=2:length(n)
vo(k)=sqrt(t1+kt2*(vo(k-1))^2);
end
%%%%%%%%% display results
plot(tms,vo,'bo',tms,vo,'r'),title('DC-DC CONVERTER OUPUT')
xlabel('Time(ms)'),ylabel('V_o(V)'), grid
DIODES
MOSFETs