Calculating the Values of Passive Components in 1149.4 Environment - - PowerPoint PPT Presentation

Calculating the Values of Passive Components in 1149.4 Environment

3rd IEEE International Board Test Workshop

Teuvo Saikkonen, Juha-Veikko Voutilainen, and Markku Moilanen Department of Electrical and Information Engineering, University of Oulu, Oulu, Finland

Purpose

• Develop calculation methods alleviating the

problems encountered when using low cost instruments (external or BIST structures)

• Take also the loading effect into account in

calculations

Outline

• The measurement principle
• Calculations considering the loading effect
• Capacitors in a delta circuit
• Experimental results

Introduction

• The first general purpose 1149.4 IC (STA400)

is on the market now

• Results of passive component measurement

methods have been presented at BTW02 (Duzevik) and BTW03 (Saikkonen et al.)

• Goal: measure the component values with a

low cost instrumentation without phase measuring capability

• Some knowledge of the circuit needed a priori

Former measurement method

AT1 Rsense Function generator AT2 Voltmeter V AB2 AB1 To analog ground ABM1 ABM2 ABM3 ABM4

~

ABM1 Zx

• a default value for the

phase angle was needed

• the value was calculated

assuming that the components and their interconnections are OK

New measurement method

Funct. gen.

~

Rs1 Rs2 AT1 AT2 AB2 AB1 To analog ground ABM1 ABM2 Zx V AT2 STA400 as a multiplexer STA400 in the test mode

A01 A0 A1 A3 A2 A23

New measurement method

Z2 Z3 Z1a X RG R Rsw Rs1+ Rs2 Zx V1a V2a V3a V4a Z2 Z3 Z1b X RG R Rsw Rs2 Zx V1b V2b V3b V4b

New measurement method

If we denote, Ra = Rs1 + Rs2, Rb = Rs2 and R2 = RG + R +Rsw, we get

2 2 2 2 1 2 2 2 1 2 2 2 2 2 1

2 A V V Z Z R R R Z Z

a a a a a a

= = + = −

2 2 2 2 1 2 2 2 1 2 2 2 2 2 1

2 B V V Z Z R R R Z Z

b b b b b b

= = + = − and from which we get (with the assumption of constant impedances during the measurements)

1 2 1 2

2 2 2 2 2 2 2 2

− + = − + = B R R R A R R R Z

b b a a

New measurement method

We solve R2

( ) ( ) ( ) ( ) [ ]

1 1 2 1 1

2 2 2 2 2 2 2

− − − − − − = A R B R B R A R R

b a a b

and get the real and imaginary part of Zx

2 2 4 2 2 2 2 2 3 2 2

1 Z V V R V V Z R

b b b b

− + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − =

and

2 2 2 2

R Z X − =

If the circuit topology is known the component values can be calculated in simple cases.

Considering the loading effect

In exhaustive calculation we would end up in 4th order equations

( ) ( )

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

1 1 1 1 1 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + + ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + + + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + = X R R X X g X R R R R R g X R X X a X R R R a A

a in a a in in in

( ) ( )

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

1 1 1 1 1 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + + ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + + + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + = X R R X X g X R R R R R g X R X X b X R R R b B

b in b b in in in

a = Rgen + Rmux + Ra b = Rgen + Rmux + Rb g = Rgen + Rmux

Considering the loading effect

2 2 2 2 2 2 2 2 2 2 2

1 1 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + = X R X X R X R R R R A

in a in a 2 2 2 2 2 2 2 2 2 2 2

1 1 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + = X R X X R X R R R R B

in b in b

Rin is the input resistance of the voltmeter and Xin=1/ωCin, where Cin is the input capacitance of the voltmeter. X is the reactive part of Zx. The resistive part of Zx, Rx, can be calculated by subtracting the switch resistances from R2, if the the impedances are large enough so that the changes in the switch resistances can be neglected.

If g << R2, X, Ra, Rb, Rin and Xin, the calculation simplifies, and we get a pair of 2nd order equations

Considering the loading effect

Equations can be expressed as

( )

2 1 2

2 2 2 2 2 2

= − − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − +

a in a in a a

R X X R R R R R X R c

( )

2 1 2

2 2 2 2 2 2

= − − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − +

b in b in b b

R X X R R R R R X R d where

2 2 2 2

1

in a in a

X R R R A c − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − =

2 2 2 2

1

in b in b

X R R R B d − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − =

Considering the loading effect

Solving X we get 2

2 in in

X eR X X − − =

where

2 2

1

b a b a in

R R c d R R c d R e − − + =

Considering the loading effect

4

2 2 2 2

= + +

in

cX hR fR

where

( )

2 2

1 e X c f

in

+ =

( )

⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − + =

in a a a in

R R R e R cX h 1 2 2

2 2

Substituting X into the first equation of the pair of the equations gives

Considering the loading effect

So, the equations for R2 and X are

⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − = 2 1

2

eR X X

in

f fcX h h R

in

2

2 2 2

− ± − =

Choosing between + and - in the equation above needs more analysis (normally the plus sign gives right values, but when measuring small capacitances we get right values with the minus sign!)

C1 C2 C3 A01 A23 A1 A2

V1(a) V2(a) V3(a) IA23 V4(a) V5(a) V6(a)

Capacitors in delta circuit

Current through A23

Capacitors in delta circuit

Current through A2 C1 C2 C3 A01 A23 A1 A2

V2(b) V1(b) V3(b) IA2 V5(b) V6(b) V4(b)

Capacitors in delta circuit

X Ctot ω 1 =

3 2 2 2 2 3 2 1 2 1 tot2 2 3 3 3 2 2 3 1 3 1 tot1

C C lC lC C C C C C C C C kC kC C C C C C C + + = + + = + + = + + =

(A23) (A2)

• The total reactance X can be obtained from the equations

presented

• The total capacitance of the network is obtained from the

equation

• Considering the network topology and the places of the

current source, Ctot is seen as

Capacitors in delta circuit

2 ) ( 6 2 ) ( 5 2 ) ( 6 2 ) ( 4 2 ) ( 6 2 ) ( 5 a a a a a a

V V V V V V k − − − − =

2 ) ( 6 2 ) ( 5 2 ) ( 6 2 ) ( 4 2 ) ( 6 2 ) ( 5 b b b b b b

V V V V V V l − − − − =

[ ][ ] [ ]

kl l k k C C k C l C

tot tot tot

− + + − + + = ) 1 )( 1 ( 1

2 1 1 2

• Determining voltage divisions gives us:
• Solving for C2 and C3, we get:

[ ][ ] [ ]

kl l k l C C l C k C

tot tot tot

− + + − + + = ) 1 )( 1 ( 1

1 2 2 3

Capacitors in delta circuit

• Finally C1 can be solved and calculated from

3 1

kC C =

• or from

2 1

lC C =

Impedance measurement results

• |Z| errors mainly < 2 %

– exceptions when the magnitude of the phase angle of one or more of the measured voltages approaches 0˚ or 90˚

• When measuring the component values the effect of

the magnitude of the phase angle can be much larger (the example on the next slides shows what happens with the 100 kΩ resistor and 6.8 nF capacitor in series when the measurement frequency is changed from 100 Hz to 1000 Hz)

Impedance measurement results

Cx Error as a Function of |argZ1a|

0,0 0,1 1,0 10,0 100,0 1000,0 2 4 6 8 10 12 |argZ1a| [degrees] Cx Error [%]

Impedance measurement results

1 2 3 4 5 6 2 4 6 8 10 12 |argZ1a| [degrees] Rx+ESR Error [%] |Zx| Error [%]

Delta circuit measurement results C1 ≈ C2 ≈ C3

• Quite good results were
• btained, when

capacitor values were larger or equal to approximately 1 nF (error > 5 % only when the reactance was too small compared with the sense resistors)

Delta circuit measurement results C1 ≈ C2 ≈ C3

• With smaller < 1 nF capacitors the loading

effect and STA400 pin and bus capacitances have a deteriorating effect to the results

• These can be taken into a account in

calculations → errors are decreasing considerably (in most cases!) as shown in the next tables

Delta circuit measurement results C1 ≈ C2 ≈ C3

Loading / pin C not considered Loading / pin C considered

Delta circuit measurement results C1 ≠ C2 ≠ C3

• As the ratio of the largest and the

smallest component value increases the accuracy degrades

• This problem is mainly due

selection of sense resistors Rs1 and Rs2 Small ratio Large ratio

Conclusions and further work

• A priori knowledge about the circuit still needed, but the

phase angle assumption is not necessary any more

• Optimizing the measurements needs more research
• Making the calculations a part of the automated

measurements will put the method towards real life applications

• Integrated analog BIST structures have to be simple to

be low-cost – calculations help to compensate for the deficiencies – calculation capacity probably no problem