Transcript The Ground Conundrum

The Ground Conundrum
Assignment: Find and research
papers on this subject, be
prepared to defend research
1
Grounding Thoughts
 Ground is only a convenience for a voltage
measurement reference
 The ground paradox
Ground is 0 volts
Ground is relative
 There is not an absolute ground
 Ground can be anywhere
 Any sources referenced to ground returns
power to ground
 All real sources have a reference to ground
The Ground Conundrum
2
Reduced Ground Definition
 The reason for the reduced ground concept is


related to the modeling of transmission lines in any
spice-like simulator.
We will develop the concept of reduced ground first
and subsequently illustrate why it is required.
The reduced ground:
Collapses return path circuits to a single reference node.
Signal measurement accuracy of a network is then
preserved by incorporating the return path effects as
elements added into the signal path.
Measurement of ground bounce is hidden.
 Return path may be power or ground.
 We will only concern ourselves with ground for now.
Reduce Power is a good research and development topic.
The Ground Conundrum
3
First Order View - Resistor Model
 This Simple model will be sufficient to
illustrate the ground reduction concept and
issues of creating multi-line reduced models
I2
-I2
I1
-I1
I1
-I1
Lets simplify a m-strip to simple resistors
Signal path
Return path
The Ground Conundrum
4
The lossless return path
5
 Calculate voltage at the load as a
reference
50  2V
50  50  2
2V
R_source=50
R_tline=2
The Ground Conundrum
 0.9804 V
R_load=50
The lossy return path
6
 Calculate the voltage at the load
 Notice this is less that the voltage on the previous
slide because the return path is considered here.
 Now lets use this voltage a reference
 The goal is to create a circuit that produces the same
load voltage and current but only has one ground
node.
50  2V
50  50  4
2V
R_source=50
R_tline=2
Rg_tline=2
The Ground Conundrum
 0.9615 V
R_load=50
The reduced ground circuit
7
 Thevinize the ground resistor into the
signal path.
 Both lossy ground circuit and the
reduced ground circuit produce 0.9615
volts at the loads.
2V
R_source=50
One ground
node
R_tline=2
Rg_tline=2
50  2V
50  50  4
The Ground Conundrum
R_load=50
 0.9615 V
Add another line!
8
 Lets keep with a resistor model and a reduced

ground path
Spice only allows a single node for return path for
multi conductor transmission line element.
50  2V
50  50  4
Tline model:
•2 lines plus return in
•2 lines plus return out
2V
2V
I2
I1
-I2
I4
-I4
R_source=50
R_tline=2
Rg_tline=2
R_load=50
R_source=50
R_tline=2
Rg_tline=2
R_load=50
This seems OK so far but consider –I1,
-I2, -I3, and -I4 are combined
I3
-I1
 0.9615 V
-I3
The Ground Conundrum
Take a closer look at where the current
are and the voltages are developed
It is possible to
collapse both
ground nodes into
one node but that
Z1
creates issues
9
There’s can be a
voltage drop between
these two nodes
Z12
Z2
Current in
plane
Z1g
Z12g
Z2g
The Ground Conundrum
9
Circuit Simulation Ground Rules –
Transmission Line Rules
1. Ground reference transmission lines
2. Include return currents in the
transmission path.
3. Do not use transmission line
reference node for return path
analysis
The Ground Conundrum
10
Connectors and Transmission Lines
 Cascading transmission lines is accurate
if reduced grounds are used.
 Cascading a connecter (or package) is a
different story.
Like the T line, start with a simple
resistor model for a connector.
Green is assigned for
ground pins
The Ground Conundrum
11
Matching up Connector and T-line signals
 The connector has 3 grounds
 The line model has 1 ground
 How do we connect ground?
?
?
?
?
?
?
The Ground Conundrum
12
Connector Model Usage
 Is this the model
usage for the
connector?
2V
Short all ground pins
together?
 Or is this the model
usage?
2V
Connect to ground
with a circuit.
The Ground Conundrum
13
Connector Interface on PWB
• Case 1 is when:
 Impedance between ground
pins and transmission line
pins are very small. Less than
0.1% of line impedance
across frequency range.
• Case 2 is when:
Transmission line
ends here. So
signal reference
is defined here.
 Impedance between ground
pins is significant. Greater
than 0.1% of line impedance
across frequency range
Connector ground
pin starts here.
Impedance
between ground
pins
The Ground Conundrum
14
Circuit Simulation Ground Rules –
Between T and Connector Rule
3. Short grounds at connector if
impedance between pins < 0.001*z0
for relevant frequencies.
4. Use circuit to model return path if
impedance between pins > 0.001*z0
for relevant frequencies.
1.
Or evaluate need
The Ground Conundrum
15
Example of Reduce Ground Connector
1.007  .03247  0.975
Ground reduced
coupled model
produces same
results
The Ground Conundrum
16
Reduced model connects to T line
 The connector’s 3 grounds have been
folded into the circuit.
 The T line model has 1 ground
 The following preserves crosstalk
 Most 2 D modelers can produce ground
reduced models for transmission lines
The Ground Conundrum
17
Circuit Simulation Ground Rules –
Connector and packages
5. Ground reduce connector and package
models
The Ground Conundrum
18
Reducing Ground: 3 Inductor Connector
I1
I2
-I1-I2
L11
L21
L12
L22
L13
L31
PIN 2
L32
L23
 Start with a 3 pin connector
 1nH self inductance
 0.2nH mutual between any leg
The Ground Conundrum
PIN 1
L33
GND
19
Create Current loop matrix
Create a 3x3 inductance matrix and a current matrix
 L1  1 L2  1 L3  1 
L  L1  2 L2  2 L3  2 


 L1  3 L2  3 L3  3 


 I1 
I  I2 
 
 I3 
 
Create current equations
 L1  1 L2  1 L3  1   I1 
  L1  1 I1  L2  1 I2  L3  1 I3  s 
L
   I   s   L  I  L  I  L  I  s 
L
L
1

2
2

2
3

2

  2
 1 2 1 2 2 2 3 2 3 
 L1  3 L2  3 L3  3   I3 
  L1  3 I1  L2  3 I2  L3  3 I3  s 

 


The Ground Conundrum
20
Use the return current definition
The current I3 is there return path for I1 and I2
I
3
I  I
1
2
Substitute
 L1  1 L2  1 L3  1   I1 
 L1  1 I1 s  L2  1 I2 s  L3  1  I1  I2  s 
L
 I
 s   L  I  s  L  I  s  L   I  I   s 
L
L
1

2
2

2
3

2
2
2 2 2
3 2
1
2 



 1 2 1
 L1  3 L2  3 L3  3   I1  I2 
 L1  3 I1 s  L2  3 I2 s  L3  3  I1  I2  s 





The Ground Conundrum
21
Equate to the voltage across the connector
Set up the new voltage relationships
 L1  1 I1  L2  1 I2  L3  1  I1  I2 
 L  I  L  I  L   I  I    s
 1 2 1 2 2 2 3 2 1 2 
 L1  3 I1  L2  3 I2  L3  3  I1  I2 


The Ground Conundrum
 V1 
 V2 
 
 V3 
22
The equivalent voltage at pin 1
23
 Remove s for now because its only common
factor
 Convert to columns with the matrix transpose
operation (T) and so we can use the column
function to extract V1 or column 0 minus V3
or column 2
 0
  L  I  L  I  L  I  I T 
  1  1 1 2  1 2 3  1  1 2  
  L  I  L  I  L  I  I  
  L  I  L  I  L   I  I   L  I  L  I  L   I  I  
  1  2 1 2  2 2 3  2  1 2   
1 3 1
2 3 2
3 3
1
2 
 1 1 1 2 1 2 3 1 1 2
  L  I  L  I  L   I  I   
  1 3 1 2 3 2 3 3 1 2  
 2
  L  I  L  I  L   I  I  T 
  1 1 1 2 1 2 3 1 1 2  


 0   L1  2 I1  L2  2 I2  L3  2  I1  I2 

 
  L  I  L  I  L   I  I   
  1 3 1 2 3 2 3 3 1 2  
The Ground Conundrum
 1 2  

 I Do
L the
 I  Lsame
  I for
I   V2-V3

  L  I  L  I  L   I 
2 1
2 2 2
3 2 1 2

 1 2 1 2 2 2 3 2 1

 I  L  I  L   I  I   
3 1
2 3 2
3 3 1 2  


 2




  L  I  L  I  L   I  I T 
  1  1 1 2  1 2  3 1  1 2  
  L  I  L  I  L   I  I  
  1  2 1 2  2 2 3  2   1 2  
  L  I  L  I  L   I  I   
  1 3 1 2 3 2 3 3 1 2  
1 1
2 1 2
3 1
24
 1
  L  I  L  I  L  I  I T 
  1 1 1 2 1 2 3 1 1 2  
  L  I  L  I  L  I  I  
  1 2 1 2 2 2 3 2 1 2  
  L  I  L  I  L  I  I  
  1 3 1 2 3 2 3 3 1 2  
  L  I  L  I  L  I  I
  1 1 1 2 1 2 3 1 1 2

 0   L1  2 I1  L2  2 I2  L3  2 I1  I2

  L  I  L  I  L  I  I
  1 3 1 2 3 2 3 3 1 2
 L


I  L
1 2 1
I  L
2 2 2
 I  I
3 2
1
2
L
I  L
1 3 1
I  L
2 3 2
 I  I
3 3
1
2

 2
T










Collect terms and put back in to matrix form
L1  1  L3  1  L1  3  L3  3 L2  1  L3  1  L2  3  L3  3


L  L  L  L  L  L  L  L 
 1 2 3 2 1 3 3 3 2 2 3 2 2 3 3 3 
The Ground Conundrum
Apply values to the connector example
Example with values
K
L21
K13 K12 K23 .2
L1 L2
Find reduced inductance matrix
 L1  1 L2  1 L3  1 
 1nH .2nH .2nH 
 L L L    .2nH 1nH .2nH 
 1 2 2 2 3 2  

 L1  3 L2  3 L3  3   .2nH .2nH 1nH 


 L1  1  L3  1  L1  3  L3  3  L2  1  L3  1  L2  3  L3  3   1.6 0.8 


nH

 L1  2  L3  2  L1  3  L3  3  L2  2  L3  2  L2  3  L3  3   0.8 1.6 


Find new coupling factor for spice
Kp 
.8
2
Kp  0.5
1.6
The Ground Conundrum
25
Evaluate methods with spice
 Use testckt.sp as starting point and create
return_path_reduction.sp
 Insert the previous 3 pin connector example for the package
model
 Replace the single node vss with two node for vss in and out
The Ground Conundrum
26
Use library replacement
 Compare difference between received
voltage for the 3 pin model and the 2
pin return path reduced model.
 For the three pin case vss will only be
tied ground at the transmitter.
The Ground Conundrum
27
Use 400 ps UI to exaggerate effects
The Ground Conundrum
package
Printed Wiring
Board
package
Buffers
package
Data generator
package
Data generator
Buffers
Printed Wiring
Board
28
Main programs example
The Ground Conundrum
29
No measure voltage difference
The Ground Conundrum
30
31
Now look at vss and signal nodes
individually
These spikes can
cause simulator
instabilities. In
some circuits,
these spikes can
reach thousands
of volts.
The Ground Conundrum
31
Generalized Return Path Reductions
 Many 3D modelers have this operation as a
feature
 Start with s parameters
This can be acquired from a modeling tool or
measurements
Convert s to Z
Z

 

   1 
Zo  I  S
 I S
The Ground Conundrum
32
33
Look at voltage measurements relative
to row k
Determine reduction row k (ground)
 i
ZT 
 k
 T
 Z
Convert Z back to s



 
1  
1
S
I  Z Zo
 I  Z Zo
The Ground Conundrum
33
34
Lets look a elements required for a 3
pin resistor model
R1
R2
Rg
The Ground Conundrum
34
35
Create the return path matrix equation
 R1 0 0   I1 
 0 R1g 0    I2 



 0 0 R2   I2  I1 
The Ground Conundrum
 V1 
 V2 
 
 Vg 
35
Develop difference voltages
 0

T
   R1 0 0  
   0 R2 0  
 
 
   0 0 Rg  
 1

   R1 0 0 T 
   0 R2 0  
 
 
   0 0 Rg  
 2 
  R1 0 0 T    I1 









0 R2 0

I2
 R1 I1  Rg  ( I2  I1)
  0 0 Rg     I2  I1 

  

 2 
  R1 0 0 T    I1 
   0 R2 0      I2   R2 I2  Rg  ( I2  I1)
  0 0 Rg     I2  I1 

  

The Ground Conundrum
36
Implement resistor matrix spice
 R1  Rg Rg    I1 

 
 Rg R2  Rg   I2 
R1
R2
- 0V +
Rg*I2
- 0V +
Rg*I1
Rg
Rg
The Ground Conundrum
37
Part 2 Anatomy of 3D modeling
The Ground Conundrum
38
3-D Aspects of Ground
I2
I1
-I2
I3
-I1
-I3
Terminals
I4
• Current distribution in ground
plane is not at a point
-I4
• 3-D modeling accounts for
distribution.
• Defining a terminal port is
“point” assumption
 This point has the potential to
create circuit concatenation
issues
Voltage drop may exist across
reference nodes
The Ground Conundrum
39
For TEM, Cutting Up the Geometry is OK
An
Interesting
thought”
“Where is
ground?”
Ground
Reduction still
works OK
The Ground Conundrum
40
Multi-conductor T-Lines are Ground Reduced
A reasonably involved process that
Ansoft and other 2 D solvers can
do.
Ground
referenced
model
The Ground Conundrum
41
42
For Non TEM, Cascading Models Introduces Errors
Wave propagation
TEM
Wave propagation
Non TEM
E-H Field vectors
Cascading elements does not account for
the non transverse components correctly
The Ground Conundrum
Circuit Simulation Ground Rules Cascading Rule
6. Cut models on TEM or Quasi TEM
boundaries
Left and right half TEM need to match
The Ground Conundrum
43
44
Assignment: Use Ansoft to create a
ground reduced spice model
100 mils
Er=3.8
100 mils
100 mils
Mated Connector
50 mils
• Green is ground, purple is signal
• Board connect is on the bottom layer
• Components insert on top layer of PWB
The Ground Conundrum
Pins are 25 mil
diameter Cu
centered posts