======================RF_ELECTRONICS======================
______________
 __ 
___ ___
___ 1 VCC 14 _________
   .22uF
___ ___ __
___ 2 13 ___ ___
  
___ ___ __
___ 3 12 ___ ///
 
___ ___
___ 4 11 ___
 
___ ___ Typical Digital
___ 5 10 ___ Supply Bypass
 
___ ___
___ 6 9 ___
 
___ ___
______ 7 GND 8 ___
  
__ ______________
///
For low frequencies the general rules are...
1) Supply bypass capacitors
2) Short leads
3) The greater the distance, the less the stray C
4) Watch for cross talk...etc
__\_
H  / 
____________________________
()_____________\/____________) R_cm
> I  V L_cm
_/_
\
SELF INDUCTANCE (uH) for ROUND WIRE (L_cm,R_cm)
L_uH=.002*L_cm*( ln(2*L_cm/R_cm) .75 )
When entering the RF world, you begin to see that having
short leads on your supply bypass capacitor is not telling you
the whole story. A good way to see this is to measure the impedance
of a 0.22uf bypass capacitor over frequency.
IMPEDANCE OF 0.1uF CAPACITOR
10 ..C.............................................
 C . . . L .
 C . . L .
 C . L . .
 C. . L . .
 C . L . .
1 ..............C...........L.....................
 . C .L . .
 . C L . .
 . C L. . .
 . R . . .
 . . . .
.1 ................................................
 . . .
 . . . .
 . . . .
 . . . .
 . . . .
______________________________________________ .
100KHz 1MHz 10MHz 100MHz 1GHz
above 10MHz, you really don't have
capacitor. self inductance takes over, and digital
supply current spikes really seeing a "bypass inductor"
The RF guys have long known this. they never use
things like 0.22uf capacitors any where critical like to
use those silvermica capacitors in picoFarads.
^ VCC
0.1uF /_\ Internal prop delays for a CS80CBI
_____________ inverter typ below 100pS.
 => I  _____
__ _< 
___ __   /\ Vout
 _  __ ___
 ___ __ =>I
 __  __ ______ ...
 /Vin\   __
 \___/ _ ___
  > ________ ...
______________ I<=
<= I __
///
current always flows in loops. Say you have an inverter
swinging from low to high. Current
coming out of inverter to charge the stray C. inverter
get this current from the supply bypass. current
will form the loop shown above.
_______\_____
I <  / 
________________ 
 \ 
_____ /\ \ \/ H Field
\  __\__ 
\  _____  Magnetic Radiation
\  \ 
\ > I \ 
\____________\ 
 
______/______
\
in terms of how much magnetic field
does current spike loop generate to understand size of
the "bypass inductor" . inductance of loop wire can show
the inductance is related to loop area.
_ _ Dia_inch
   
_   _
/ __ __ \
/ / ^ \ \ R_inch/Dia_inch > 2.5
/ / /\ \ \
    
  R_inch  
\ \ / /
\ \__ __/ /
\_  _/

SELF INDUCTANCE (uH) for a RING OF WIRE (R_cm, Dia_inch)
L_uH (R_inch/100)*( 7.353*log(16*R_inch/Dia_inch) 6.386)
area of the loop which defines the size of the
inductance. when you make the loop area small?
___\__ > I
_________/__________________
>()___________\/______________ 
_/____ _/___  
________\________\________ 
<()____________________________
< I ___\_
__ /
 __ _____
__   
*() ()  
\_/\_/\_/ L1 
Lb M _ _ _  Lb = L1 + L2 2*M
/*\/ \/ \ L2  if L1 = L2 =M
__  () ()  
 __ _____ then Lb=0
__
When return path flow right next to each other,
mutual inductance cancel out self inductance. Remember
the short lead rule? Suppose don't make bypass capacitor
leads short, rather just make return current flow next to
any signal current. ( make Loop area is small)
_____________________________
>()____________________________) < d_cm

D_cm Z_ohms sqrt(L/C)
 _____________________________
>()____________________________)
L_cm
C_pf/meter = 12.06/log(2*D_cm/d_cm)
L_uH/meter = 0.92*log(2*D_cm/d_cm)
Z_ohms = 276*log(2*D_cm/d_cm)
now a transmission line. current loop will still
generate a magnetic field. magnetic field is completely
confined in space between conductors. A transmission
line just like microwave wave guide. have current,
voltage, E fields and H fields. are happening all together
in a small area of space
Voltage dropping
current
\ > \
\____\ \______\
\ \ \ \
\ \ _
\ _ >E \ \ \
_ \ \ \/ _ /_ \
^ _ /_ v H _/ \ _V Voltage
\ _/ \ __/_ \ Pulse
\ __/_ _/ \ \
\ _/ \  _V
 voltage \
\  + \
\____\ \______\
\ \ \ \
\ > \ _
\ _  E \ \ \
_ \ \ \/ _ /_ \
^ _ /_ v H _/ \ _V
\ _/ \ __/_
\ __/_ >E _/ \
\ _/ \ \/ 
 v H \
\ \
\____\ \______\
\ \ \ \
current
<
Voltage increasing
Between conductors, capacitance. a fast
voltage pulse, transmission line form current
loops within itself. confine both the E field and
H field completely between conductors. So now signal
really both voltage and current and electromagnetic wave,
in same place. if signal current is differential
(twisted pair lines), transmission line will look like
no signal at short distance away. If only air bewteen
conductors, this wave will travel at speed of light.
( 1ns =about 1foot)
_____________________________
>()____________________________) < d_cm

D_cm Z_ohms sqrt(L/C)
 _____________________________
>()____________________________)
L_cm
C_pf/meter = 12.06/log(2*D_cm/d_cm)
L_uH/meter = 0.92*log(2*D_cm/d_cm)
Z_ohms = 276*log(2*D_cm/d_cm) Z_air =377 ohms
Rabbit ears 300 ohm Transmission line of
rabbit ears follow equation. Transmission lines have
resistance impedances. violate short lead supply bypass
capacitor rule and connect 5 ohm transmission line between
IC and supply bypass as shown below,now be bypassing
your supply with 5 ohm resistor all frequencies 10MHz
___\__ > I
_________/________________ 5 Ohm
>()___________\/_______________/\ __
_/____ _/___ \/ 
________\________\________ __
<()___________________________ ___ Bypass
< I ___\_ ______
0.22uF
In terms of a PC board layout is is common to have a
signal wire over ground plane with small enough dielectric
between. recommended spacings result in 50ohm trans line
_ I signal current
/
/
/ /
/ /_____/ / /
/ _____/ / / /
/_________________/ / /_ I return path
_________________/
simply laying out PC boardthat all signal current runs
next to return path, easy to have all high
frequency circuitry see nothing but transmission lines.
^ VCC
0.1uF /_\ Internal prop delaysCS80CBI
_____________ inverter typically below 100pS.
 => I  _____
__ _< 
___ __   /\ Vout
 _  __ ___ Signal Current Sees
 ___ __ Minimum Inductance
 __  __ 
 /Vin\    =>I ___________...
 \___/ _ _____
  > ________ Any Type of tran line
______________ I<= ___________...
<= I __
///
THIS IS WELL WORTH THE EFFORT.
1) High frequency energy completely confined
a) This greatly reduces crosstalk
Low RF radiation both in and out.
b) Signals travel faster (No Inductors) speed light
Efficient use of signal energy
2) termination ( Impedance at end =same as transmission Z)
a) Keeps the ringing down.
costto know where all signal current flows
and simply provide a path of least inductance
(Path of least indcutance = min loop area)
Supply bypass ____________
0.1uF  
_______ _________________
__ ___ 1 6 ___ __________ 50_Ohms
___ VCC  SOT6  CS 
_____________ _______\__________
 __ ___ 2 5 ___ /__________ 50_Ohms
__ ___ GND   DOUT 
\ / _________ _______\__________
V Local  ___ 3 4 ___ /__________ 50_Ohms
Gnd  VIN   CLK 
_  ____________ __
 _/\ ___ adcv0831 \ / Local
_ \/ V Gnd
use of this principle involved the building
of first SOT236 tester for an ADC as is shown.
In this case, the challenges were..
A) no separate analog and digital ground.ran out of pins
B) 3 foot cables connect the Eagle tester to the
SOT236 handler.
A standard cable size is 50 ohms change thinking for TD
using outer shield connector as current return paths,
In the RF world, if you can't transmit, you also
can't receive. tansmission line mentality introduces
Electromagnetic shielding as well as Electrostatic
shielding.
Travelling wave string on tension
tension s
mass_per_unit u
velocity sqrt(tension/mass_per_unit)

H ^
/\
 
______\ E
\ /
\ 
\ 
 \
 \
E /______\
\ \
  \
\/ \
V H \
_\
S
Electrical
c =sqrt(1/(e0*u0)) phi=u0*H eo*E
Z = E/H = sqrt(u0/e0) = 377ohms

Maxwell's Equations
__
\/ dot J = delta_rho/dt
__
\/ cross E = delta_B/dt V=delta_Phi/dt
__
\/ cross H = J+ delta_D/dt H =I*N
__
\/ dot D = rho
__
\/ dot B = 0
SUPPLY_BY_PASS
______________
 __ 
___ ___
___ 1 VCC 14 _________
   .1uF
___ ___ __
___ 2 13 ___ ___
  
___ ___ __
___ 3 12 ___ ///
 
___ ___
___ 4 11 ___
 
___ ___
___ 5 10 ___
 
___ ___
___ 6 9 ___
 
___ ___
______ 7 GND 8 ___
  
__ ______________
///
world of supply supply bypass decoupling has been changing
due to the ever higher and higher speeds at which digital
circuit can now operate. A standard off_the_shelf digital
circuit with its supply by_pass is shown above. At
speeds which digital circuits using CS80CBI process
at, this 0.1uF capacitor really is not a capacitor.
SELF INDUCTANCE (uH) for ROUND WIRE (L_cm,R_cm)
__\_
H  / 
____________________________
()_____________\/____________) R_cm
> I  V L_cm
_/_
\
L_uH=.002*L_cm*( ln(2*L_cm/R_cm) .75 )
Inductances in the nanoHenry range are not hard to come
by given the package lead sizes. equation for single wire
inductance is give above.
IMPEDANCE OF 0.1uF CAPACITOR
10 ..C.............................................
 C . . . L .
 C . . L .
 C . L . .
 C. . L . .
 C . L . .
1 ..............C...........L.....................
 . C .L . .
 . C L . .
 . C L. . .
 . R . . .
 . . . .
.1 ................................................
 . . .
 . . . .
 . . . .
 . . . .
 . . . .
______________________________________________ .
100KHz 1MHz 10MHz 100MHz 1GHz
The 0.1uF supply bypass capacitance may have the impedance
as is shown above. Above a certain frequency, the size
of the capacitor no longer matters. Rather, it is how the
capacitor is connected that defines the impedances.
_______\_____
I <  / 
________________ 
 \ 
_____ /\ \ \/ H Field
\  __\__ 
\  _____  Magnetic Radiation
\  \ 
\ > I \ 
\____________\ 
 
______/______
\
Current really always flows in loops. Wher current flows,
a magnetic field. However it is the Area which current
loop encircles determins how much magnetic field gets
radiated. To have inductor, you need space for magnetic
field. Little space, little magnetic field and inductance.
SELF INDUCTANCE (uH) for a RING OF WIRE (R_cm, Dia_inch)
_ _ Dia_inch
   
_   _
/ __ __ \
/ / ^ \ \ R_inch/Dia_inch > 2.5
/ / /\ \ \
    
  R_inch  
\ \ / /
\ \__ __/ /
\_  _/

L_uH (R_inch/100)*( 7.353*log(16*R_inch/Dia_inch) 6.386)
equations of inductance for a ring of wire is given
above. If current loop is made to runthat currents
flowing in opposite directions are next to each other,
you get a transmission line.
___\__ > I _______
_________/________________ 
>()___________\/_____________ 
_/____ _/___ 50Ohms 
________\________\________ 
<()__________________________ 
< I ___\_ _______
/
From outside world, transmission lines have magnetic
fields cancelling out. In terms loop current,
impedances look like a resistor. Depending on spacing of
the wires,it is common to set this resistance value to be
typically 50 ohms.
Without inductance, the moment you apply signal voltage,
you instantly get signal current. Things happen close to
speed of light.
Voltage dropping
current
\ > \
\____\ \______\
\ \ \ \
\ \ _
\ _ >E \ \ \
_ \ \ \/ _ /_ \
^ _ /_ v H _/ \ _V Voltage
\ _/ \ __/_ \ Pulse
\ __/_ _/ \ \
\ _/ \  _V
 voltage \
\  + \
\____\ \______\
\ \ \ \
\ > \ _
\ _  E \ \ \
_ \ \ \/ _ /_ \
^ _ /_ v H _/ \ _V
\ _/ \ __/_
\ __/_ >E _/ \
\ _/ \ \/ 
 v H \
\ \
\____\ \______\
\ \ \ \
current
<
Voltage increasing
w0=sqrt(L*C) = sqrt(dL*dC*X^2) =X*sqrt(dL*dC)
velocity=(2*pi()/sqrt(dL*dC)
transmission linethought of as a "wave_guide".
E and H field travel down the such the E field is
set by signal voltage and H field is set by signal
current.Given transmission line typically 50 ohms, power,
voltage, current, etc are all defined.
Using a transmission line really win/win situation.
you remove inductance from equation which greatly speeds
things up. also prevent electromagnetic radiation leaving
or entering circuit. you confine the signal
energy to flow where you want, everything gets better.
^ VCC
0.1uF /_\ Internal prop delays for a CS80CBI
_____________ inverter typically typically 100pS
 => I  _____
__ _< 
___ __   /\ Vout
 _  __ ___ Signal Current Sees
 ___ __ Minimum Inductance
 __  __ 
 /Vin\    =>I ___________...
 \___/ _ _____
  > ________ Any Type of trans line
______________ I<= ___________...
<= I __
///
Inside chip, things can now happen very fast. prop delays
for inverter CS80CBI typically under 100 picoseconds!
10GHz bandwidth. If digital circuit is connected to
outside world shown above, current loop can be confined in
space consisting of transmission line and supply bypass
capacitor. to make this whole path look real in impedances
by having currents flowing in opposite directions next to
each other so that their magnetic fields cancel out.
Another way
say it to make lead lengths of bypass capacitor as
short as possible.
Supply bypass ____________
0.1uF  
_______ _________________
__ ___ 1 6 ___ __________ 50_Ohms
___ VCC  SOT6  CS 
_____________ _______\__________
 __ ___ 2 5 ___ /__________ 50_Ohms
__ ___ GND   DOUT 
\ / _________ _______\__________
V Local  ___ 3 4 ___ /__________ 50_Ohms
Gnd  VIN   CLK 
_  ____________ __
 _/\ ___ adcv0831 \ / Local
_ \/ V Gnd
This relationship ofsupply bypass to signal path
ofADC seen in the lab using the test circuit shown
Because of ESD requirements, all pad are going to have
around 5pF of capacitance. every digital input or output
pin going to cause some supply current to flow through
0.1uF capacitor and ADCV0831's ground lead. ,
sample and hold only has access to internal ground node
to which references its Analog input voltage. When change
lead length of supply bypass capacitor, can modulate
analog input voltage which is what graph shows below.
First Code
Transition
^
/_\

2.2_lsb _ * *
 *
2.1_lsb _ *
 * *
2.0_lsb _
 * *
1.9_lsb _
 *
1.8_lsb _ * *

______________________________________\
       /
2mm 3mm 4mm 5mm 6mm
Supply bypass Lead length
This type input modulation consistant. should not effect
linearity of the ADC. present test system may have
linearity modulating error. major code transition on test
is coming close to having missing code while the lab
set up does not correlate This error only possible if
the Data output signal from DUT is getting back into
the analog signal line.
____ Sample and Hold 10nsec after this edge
__
\ /
V
____ CHIP SELEC ___
_______________________________________________
_ __ __ __ __ __ __ __ __ __ __
                    
  V  V  V  V  V  V  V  V  V  CLOCK
_ _ _ _ _ _ _ _ _ _ __
________________________________________
        
............Bit7Bit6Bit5Bit4Bit3Bit2Bit1Bit0 ...
TRI STATE _________________________________
TRI STATE
Data Output signal comes well after external analog
voltages capture, reducting feedback from output to input
should not be that hard.
input sample and hold circuit need an external digital
signal to tell when to capture analog signal. Care
taken in design to have few of things happening possible
inside die untilsample and hold process is done. still
youhave a digital input signal and how much time to wait
until you sample?
Supply bypass ____________
0.1uF  
_______ _________________
__ ___ 1 6 ___ __________ 50_Ohms
___ VCC  SOT6  CS 
_____________ _______\__________
 __ ___ 2 5 ___ /__________ 50_Ohms
__ ___ GND   DOUT 
\ / _________ _______\__________
V Local  ___ 3 4 ___ /__________ 50_Ohms
Gnd  VIN   CLK 
_  ____________ __
 _/\ ___ adcv0831 \ / Local
_ \/ V Gnd
offical end point offset will probably be defined as what
part measures in a common lab set up. On SOT production
tester, things more challenging in that nothing
can come up close to DUT. all cable are at least
2 feet long. have to generate a local ground close
to DUT such that really apply an accurate Analog input
voltage. capacitor form local ground to input appears to
common mode out most of trouble input pin draws no current
and some resistors can be added if they help.
Hopefullycan geterrors small or at least consistant.
best to spec end point offset numbers separately
on the spec so they can be handled independently.