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Diode Circuits
By
Professor Syed Idris Syed Hassan
Sch of Elect. & Electron Eng
Engineering Campus USM
Nibong Tebal 14300
SPS Penang
Application of diodes
•
•
•
•
•
•
Rectifier
Detector
Mixer
switching/switch
Phase shifter
Attenuator
Type of diodes
p-n junction Diode
Detector, mixer,
modulator(slow response)
Schottky barrier diode
Detector, mixer,
modulator (fast response)
PIN diode
Switching, attenuator
current controller
VARACTOR
Tuner, harmonic
generator, frequency
multiplier, VCO,
parametric amplifier.
Basic diode characteristic
V-I characteristic
Equivalent circuit
Package
components
+
V
-
I
Id
I
Lp
+
Vd
Is
V


I (V )  I s e V  1
-
Contact
resistance
Rs
Cp
Cj(V)
Junction
components
Rj(V)
where = q /nkT , q =charge, k=Boltzmann’s constant,
T = temperature, n = ideality factor and Is = saturation
current.
Continue
Let’s say diode voltage
V = Vo + u where Vo is a DC bias voltage and u is
a small AC signal voltage. We expand using Taylor series
Taylor series
1 "
f ( x)  f (a )  f (a )( x  a)  f (a)( x  a) 2 
2!
'
.... 
f
( n 1)
(a)( x  a)
(n  1)!
n 1
Reminder
 Rn
Taking f(x) = I(V), then x= Vo+u and a = Vo
By substituting, we have (x-a) = ( Vo+u -Vo)= u and the Taylor series for I(V) is
dI
1 2 d 2I
I (V )  I o  u
 u
 ....
2
dV V
2 dV
o
Vo
where


I (V )  I s e V  1
and


I o  I (Vo )  I s e Vo  1
By substituting
Continue
I (V )  I s e V  1 and I = I(V ) in the first derivative
o
dI
1
 Vo
  I se
  I o  I s   Gd 
dV V
Rj
o
o

where
Similarly in the second derivative, we have
d 2I
dGd

  2 I s e Vo   2 I o  I s    Gd  Gd'
dV V
dV 2 V
o
o
Then
I (V )  I o  u Gd 
u2
2

I o  I (Vo )  I s e Vo  1
Gd'  ...
(400)
Rectifier application
RF in
DC out
If the diode voltage consist of DC and small RF signal
V = Vo + uo cos wot
where Vo is a DC bias voltage and u cos wot is a small RF signal voltage. Then
by substituting into (400)
I (V )  I o  uoGd cos wot 
uo 2
 I o  uoGd cos wot 
2
uo 2
4
Gd' cos 2 wot  ....
Gd' 
uo 2
4
Gd' cos 2wot  ....
continue
Rearrange
I (V )  I o 
uo 2
4
Gd'  uoGd cos wot 
uo 2
4
Gd' cos 2wot  ....
AC harmonics current of frequency wo
and 2wo. This can be filtered off by
using lowpass filter
DC rectified current
Detector application
Modulated RF
Detected RF
Modulated signal representation
V  uo 1 m cos w mt  cos wot
where
m = modulation index
wm= modulation frequency
wo= RF carrier frequency
continue
I (V )  I o  uoGd 1  m cos w mt  cos w ot 
uo 2
2
Gd' 1  m cos w mt 2 cos 2 w ot  ....
m
m


 I o  u oGd cos w ot  cosw o  w m t  cosw o  w m t  
2
2


2
2

m
m
Gd' 1 
 2m cos w mt 
cos 2w mt  cos 2w ot 
4
2
2

uo 2
m2
 m cos2w o  w m t  m cos2w o  w m  
cos 2w ot
2
2
2
m
m

cos 2w o  w m t 
cos 2w o  w m t 
4
4

Trigonometry relationship
1
sin x sin y  cosx  y   cos x  y 
2
1
cos x cos y  cosx  y   cosx  y 
2
1
sin x cos y  sin  x  y   sin  x  y 
2
1
cos x sin y  sin  x  y   sin x  y 
2
2 sin 2 x  1  cos2 x 
2 cos 2 x  1  cos2 x 
continue
From the eq. above we have several harmonics as shown with relative
amplitude.
Amplitude
1+m2/2
1+m2/2
k=uoGd/(uo2Gd’/4)
=4/(uo)
2m
k
m2/4
m
m2/4
wowm)
wowm)
wo
wowm)
wowm)
0
km/2 km/2
wo-wm
wo
wowm
wm
2wm
m2/2
m
w
As linear detector
As squared detector ~uo2Gd’/4
( ~uoGd)
Square-law region of diode detector
We are measuring power , thus square-law region is to be chosen since the
power measured is proportional to uo2. If we want to measure voltage , then
the linear region is the choice. For linear detector,we choose frequency at wo
and for square-detector at 2wo.. Using filter we can filter out the modulating
frequency wm.
log vout
1V
100mV
10mV
Square-law region
Saturation
Vout= uo2=Pin
1mV
100V
Noise level
10V
logPin
-30 -20 -10
0
10 20
30
(dBm)
Single-ended mixer
Downconverter
RF AMP MIxer
fRF
fRF
IF AMP
Lowpass
filter
fIF=f RF -f LO
RF input
fLO
Local
Oscillator
Bandpass
filter
MIxer
Upconverter
fIF
fIF=f RF +f LO
fRF
RF AMP
IF input
fLO
Local
Oscillator
The purpose of mixer is to convert either from one frequency to higher frequency
or vice versa. The advantages of conversion are (i) to reduce 1/f noise when
convert to lower frequency (ii) for easy tuning for a wide band with fixed IF and
(iii) frequency off-set between transmitter and receiver by using a single LO as in
Radar.
Simplest Single-ended mixer
DC bias
Combiner
RFC
Matching
network
vrcoswrt
LO
bandpass
filter
vicos(wr-wo)t
wr , wo ,
RFC w + w
r
o
vocoswot
•Uses nonlinearity of a diode property
•The output generated consist of frequencies spectrum dc component,
wr,wo,wr-wo, wr+wo.
•For IF, we filter out all frequencies except wr-wo.
•For upconverter, we filter out all lower frequencies and allow only wr+wo.
•Combiner either T-junction or directional coupler
•Matching network is to match the output of combiner to the diode circuitry.
analysis
Let’s
vRF  vr cos wr t and vLO  vo cos w ot
Then substituting into equation (400) and we have for the second term as
G 'd
vr cos w r t  vo cos w ot 2
2
G 'd 2

vr cos 2 w r t  2vr vo cos w r t cos w or t  vo 2 cos 2 w ot
2
G 'd 2 2

vr  vo  vr 2 cos 2w r t  vo 2 cos 2w ot  2vr vo cosw r  w o t
4
 2vr vo cosw r  w o t
DC
i



Figure of merit in mixer is its conversion loss, defined as
Lc  10 log
available RF input power
IF output power
dB 

Single Balanced Mixer Circuit
LPF
RF
diode
Zo
Zo
Zo / 2
/4
Zo
/4
Zo
LO
Diode
Zo
Zo / 2
Zo
LFP
* Note that, although it is not shown, the diodes required biasing and
matching network.
Advantages
•For either better input SWR or better RF/LO isolation
•Cancellation of AM noise from LO
analysis
Let’s
vRF  vr cos wr t and
Where vr<<vo and vn(t)<<vo
vLO  vo  vn (t )cos wot
Vn is a small random noise voltage
The voltages across the two diodes of 90o out of phase is given as
Diode 1
Diode 2



v1 (t )  vr cos w r t  90o  vo  vn  cos w ot  180o
 vr sin w r t  vo  vn  cos w ot




v2 (t )  vr cos w r t  180o  vo  vn  cos w ot  90o
 vr cos w r t  vo  vn sin w ot

Diode current
Assuming identical diodes so that diode currents can be represented as
i1  kv12 and


i2  kv22
(reverse polarity)
i1  k vr 2 sin 2 w r t  vo  vn 2 cos 2 w ot  2vr vo  vn sin w r t cos w ot
k 2
 vr 1  cos 2w r t   vo  vn 2 1  cos 2w ot 
2
 2vr vo  vn sin w r  w o t  sin w r  w o t


i2  k vr 2 cos 2 w r t  vo  vn 2 sin 2 w ot  2vr vo  vn  cos w r t sin w ot

k 2
  vr 1  cos 2w r t   vo  vn 2 1  cos 2w ot 
2
 2vr vo  vn sin w r  w o t  sin w r  w o t
Dc and lower frequency bands

IF frequency band
After low pass filtering, the remaining terms are dc and IF frequency terms, thus

k 2
i1  vr  vo  vn 2  2vr vo  vn sin w r  w o t 
2

k 2
i2   vr  vo  vn 2  2vr vo  vn sin w r  w o t 
2
Written the IF frequency wi = wr- wo then the IF current is
iIF  i1  i2  2kvr vo  vn sin wit  2kvr vo sin wit
where vn << vo . This show that the noise in the first order is cancelled by
the mixer while the desired IF signal combined in phase.
Anti parallel diode mixers
LO input
RF input
wr
Bandpass
filter for RF
wo
Lowpass
filter for LO
and IF
The LO is one-half of usual LO, I.e
wi
IF
output
Lowpass
filter for IF
1
w o  w r  w i 
2
The non-linearity of diode generates 2nd harmonic of LO to mix with RF(wr)
to produce desired IF. The anti parallel diode creates symmetrical V-I
characteristic that suppresses the fundamental product of RF and LO. It also
suppresses AM noise.
Double Balanced mixer

RF input

180o
hybrid
Zo
IF
output
LO input
Single -ended mixer produces output consisted of all harmonics. The
balanced mixer using hybrid suppresses all even harmonics of the LO.
Double balanced mixer suppresses all even harmonics both LO and RF.
Image rejection mixer
90o hybrid
Mixer A
3dB
power LO
divider
RF input
Zo
90o hybrid
Mixer B
LSB
IF out
USB
The RF with frequency wr= wo + wi will also produce the IF (wi) when
mixed with LO. The frequency produced will be USB(wr= wo + wi ) and
LSB(wr= wo - wi ) . The undesired frequency either USB or LSB is called
image frequency. The mixer can produce one single side band is used as
modulator.
Analysis
Let RF signal consist of both upper and lower sidebands
vr  vU coswo  wi t  vL coswo  wi t
Then input to mixer A and B
vU
vL
vr 
cosw o  wi t 
cosw o  wi t
2
2
v
v
vr B  U cos w o  wi t  90o  L cosw o  wi t  90o
2
2
A



After mixing with LO, wo , The IF’s produced by mixer are.
kvU
kvL
vi 
cos wi t 
cos wi t
2 2
2 2
kv
kv
vi B  U cos wi t  90o  L cos wi t  90o
2 2
2 2
A





Analysis
Both IF , then combined in the 90o hybrid produces LSB and USB.
vLSB



k
 vU cos w i t  vL cos w i t  vU cos w i t  180o  vL cos w i t
4
kv
 L cos w i t
2







k
vUSB  vU cos w i t  90o  vL cos w i t  90o
4
 vU cos w i t  90o  vL cos w i t  90o
kvU

sin w i t
2



Pin Diode Equivalent Circuit
PIN diode resistance
I
1
1000
10
100
100
10
1000
1
RF resistance (Ohm)
N+
P+
10000
Cp
0.0001
0.01
0.1
1
Forward bias (mA)
Lp
Rs
R1
C1
100
RF conductance
(mOhm)
symbol
Equivalent circuits for ON and OFF
states of PIN diodes
Li
OFF
state
Zr
Cj
Reverse bias will provide OFF state
Rr
Li
ON
state
Forward bias will provide ON state
Zf
Rf
Single-pole PIN diode Switches
R
SW
ON= RF out
OFF=No RF out
+V
RFC1
Series
RF in
C2
C1
RF out
Diode
RFC2
Note: C1 and C2 are dc block
R
SW
+V
Parallel
RFC
C1
RF in
Diode
C2
RF out
ON =No RF out
OFF= RF out
Simplified switching circuits
In general, the insertion loss
Series switch
Zo
2Vo
VL
Vo
Zd
+
VL
_
Zo
2Vo
Shunt switch
IL  20 log
IL  20 log
2Z o
2Z o  Z d
Zo
Zd
Zo
+
VL
_
IL  20 log
where
2Z d
2Z d  Z o

Z r  Rr  j w Li  1 w C j
Zd  
Z f  R f  jw Li

Example
A single-pole switch is to be constructed using a PIN diode with the
following parameters: Cj= 0.1pF, Rr= 1W, Rf= 5 W , Li= 0.4nH. If the
operating frequency is 5 GHz and Zo= 50W, which circuit (series or shunt)
should be used to obtain the greatest ratio of off-to-on attenuation?
Ratio
10.05dB
state
ON
OFF

Solution

Z r  Rr  j w Li  1 w C j  1  j305.7W
Zd  
 0.5  j12.6W
Z f  R f  jw Li
Series switch
IL  20 log
2Z o
 0.11dB
2Z o  Z d
2Z o
IL  20 log
 10.16dB
2Z o  Z d
Ratio
7.04dB
Shunt switch
IL  20 log
2Z d
 0.03dB
2Z d  Z o
2Z d
IL  20 log
 7.07dB
2Z d  Z o
Other Single pole single throw PIN
Switches Configuration
D1
L- SPST
Switch
D2
D1
T- SPST
Switch
Single Pole
Single Throw
D3
D2
50W
Note:
Biasing are not shown,
just diodes configuration
50W
SPST Switches performance
Type
Isolation
Series

 X
10 log10 1   c

 2Zo

Shunt

Z 
20 log10 1  o 
2 Rs 

L

Z
10 log10 1  o
2 Rs


 X
  c
 2Z o
T



2






Insertion
2

Rs 
20 log10 1 
2 Z o 




2
2

Zo  
1 
 
R
s  


2

 Xc  
 
10 log10 1  
Z o  





Z
 10 log10 1  o
2 Rs


2

 X
   c

 2 Rs



2




 Z
10 log10 1   o

 2Xc




2

Rs
10 log10 1 
2Zo





2



 Z  Rs
  o
 2Xc



2

R
20 log10 1   s

 Zo




2







 Z  Rs
 10 log10 1   o

 2Xc




2



PIN diode switching operation
(Shunt diode)
Switch Configuration
Source
50W
Switch
Load
50W
By putting diodes in parallel will
reduce the total diode resistance
and thus improves isolation as
shown in graph.
AC
V
Isolation Vs Diode resistance
Equivalent circuit
Diode "ON"
-Switch "OFF"
Diode "OFF"
-Switch "ON"
Isolation (dB)
25
30
35
40
45
50
55
60
0
1
2
Diode resistance (ohm)
3
PIN diode switch (improving isolation)
Switch Configuration
Source
50W
Switch
Load
4
50W
AC
V
Isolation is maximum when the
transmission line is exactly 90o and
the effect of diodes similar to without
transmission line when its length equal
to 0o or 180o.
50W
Isolation vs line length
Isolation (dB)
Equivalent circuit
2550W
AC
1W
50W
35
40
45
50
55
60
65
70
75
80
85
Rd=1.5ohm
Rd=1ohm
Rd=0.5ohm
0
100
Line length(deg)
200
PIN diode switch(input impedance not 50W
Source
Rs
AC
V
Switch
Compensating
line
Load
50W
4
50W
Compensating line is a 90o transmission line to match the Rs
with 50ohm line.Rs is lower than 50 ohm.
All-shunt Diode Nonreflective
SPST Switch
B1
Input
Output

D3
W


D4
D1
D2
W
PIN diode switching operation
Switch Configuration
Source
50W
AC
V
(Serial diode)
By putting diodes in series will
reduce the total effective series
capacity, thus increase isolation.
Load
This is shown in graph below.
50W
Isolation vs Diode capacity
0
Isolation (dB)
5
Equivalent circuit
Diode "ON"
(Switch "ON")
Diode "OFF"
(Switch "OFF")
10
15
20
25
30
35
40
0
0.2
0.4
0.6
0.8
Diode capacity, Cd (pF)
1
PIN diode switching operation
In this case the optimum line
line is not 90o, but depend on
the diode capacity.
Switch Configuration
Source
50W
Switch
Load
Isolation vs Line length
50W
0
4
V
50W
Isolation (dB)
AC
10
20
30
40
50
60
70
0
50
100
Line length(deg)
150
Single pole double throw PIN diode switches
Series
Output 1
Ouput 2
Input
/4
Shunt
Output 1
/4
Output2
Input
Operation
•One diode is biased in low
impedance state with another
diode in the high impedance state,
so that input signal can be
switched to one output to the
other by reversing the diodes state
or biasing.
•The quarter-wave lines limit of
the shunt circuit limit the
bandwidth.
PIN diode application
(TR switch)
dc supply
Antenna

dc block
Transmitter
dc block
Receiver
D1
D2
Dc supply “ON” for transmitting. D1 and D2 will conduct,
allowing the signal from transmitter to go to antenna and
any signal go to receiver will be shorted. When dc supply
“OFF”, D1 and D2 will not conduct, thus allowing only signal
from antenna flow to the receiver.
PIN diode application
(Reflective phase shifter)
RF input
A
B

RF output
D
C
90o Hybrid

2  r 


Switched line phase shifter
2
in
out
1
•Using two single pole and double throw switches to route the signal
between one of two transmission lines of difference length.
•The phase difference is    1   2  . This circuit is a broadband &
reciprocal phase shifter so that it can be used as receiver as well as
transmitter.
•Disadvantages-resonance when the length is multiple of /2 and frequency
is shifted due to diode capacitance.
PIN diode application
(8-steps phase shifter)
D2
D2

D1
RFC
A
RFC

D2
D1
RFC RFC
B
D1

RFC RFC
C
PIN diode application
(8-steps phase shifter)
• When A, B or C is set “1” then D1 and D2 will
conducted allowing the RF go straight.
• When A,B or C is set “0” then D1 and D2 will not
conducted and the RF signal will experienced
phase shift according to the length of U -line.
 2 is 90o phase shift, /4 is 45o phase shift and
8 is 22.5o phase shift.
Switching equivalent phase shift
A
B
C
Phase shift
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
157.5o
o
135
112.5o
90o
67.5o
o
45
22.5o
0o
PIN diode application Bridged T attenuator
Attenuation factor
is defined as
D2 (Rs2)
Zo
Zo
RF output
RF input
D1 (Rs1)
Vout
a
Vin
Attenuation is small when D2 is forward biased (low impedance) and D1 is
reverse biased (high impedance). Conversely , attenuation is large when D2 is
reverse biased (high impedance) and D1 is forward biased (low impedance).

Zo 

A  20 log 1 
 Rs1 
where
Zo2  Rs1  Rs 2
PIN diode application  attenuator
D3 (Rs3)
RF Input
RF Output
D1 (Rs1)
D2 (Rs2)
Attenuation is small when D3 is forward biased (low impedance) and D1and D2
are reverse biased (high impedance). Conversely , attenuation is large when D3 is
reverse biased (high impedance) and D1 and D2 are forward biased (low
impedance).
 Rs3 

A  20 log 1 
 Rs 2 
PIN diode application
(intermodulation attenuator)
+20V
IN 50/75 W
OUT 50/75 W
R4
R5
R1
R3
D1
R6
D2
Vin=0-20V
R1, R2 2.2k
R3, R4 1k
R5, R6 75ohm
D1, D2 UM9301unitrode
All capacitors are 470pF ceramic
R2
At high input voltage and
low attenuation, D1 tends to
conduct signal.R1 and R2
set the current and isolate
the dc. D2 tends to be off.
At low input voltage and
high attenuation, D1 tends to
be off. D2 tends to bypass
the signal to ground. R3 and
R4 set the current and isolate
the dc. R5 and R6 maintain
the characteristic impedance
PIN diode application
(intermodulation attenuator)
Attenuation of signal after applying Vin for frequency 100MHz
to 400MHz
Attenuation (dB)
Input Voltage Vs Attenuation (dB)
0
100MHz
10
200MHz
400MHz
20
0
5
10
15
Input Voltage (V)
20
PIN diode application
(intermodulation attenuator)
Return loss is less than 10 dB. It seem the impedance
characteristic is maintain.
Return loss (dB)
Input Voltage Vs Return loss (dB)
10
100MHz
15
200MHz
20
400MHz
25
0
5
10
15
Input Voltage (V)
20
Attenuator
Attenuator
(transmission mode)
Input
Coupler
Coupler
Zo
Diode
Bias
Diode ON-attenuated
Diode OFF- transmitted
Input
Attenuator
(Reflection mode)
Output
Diode
Zo
Coupler
Diode
Output
Diode
Bias