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

3rd IEEE International Board Test Workshop Calculating the Values of Passive Components in 1149.4 Environment 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 ABM1 ABM1 ABM2 Function ~ generator R sense Z x ABM4 ABM3 - a default value for the AT1 phase angle was needed To analog AT2 ground - the value was calculated Voltmeter V AB1 AB2 assuming that the components and their interconnections are OK

New measurement method A0 A01 R s1 ABM1 Funct. A1 ~ gen. Z x R s2 ABM2 A3 AT2 AT1 A2 To analog AT2 ground V A23 AB1 AB2 STA400 as a multiplexer STA400 in the test mode

New measurement method R G V 4a R R sw R s1 + R s2 Z x Z 2 Z 1a X Z 3 V 3a V 2a V 1a R G V 4b R R sw R s2 Z x Z 2 Z 1b X Z 3 V 2b V 3b V 1b

New measurement method If we denote, R a = R s1 + R s2 , R b = R s2 and R 2 = R G + R +R sw , we get − = + 2 2 2 − = + 2 2 2 Z Z 2 R R R Z Z 2 R R R 1 b 2 b 2 b 1 a 2 a 2 a and 2 2 2 2 Z V Z V = = = = 2 1 b 1 b 2 B 1 a 1 a A 2 2 2 2 Z V Z V 2 2 b 2 2 a from which we get (with the assumption of constant impedances during the measurements) + + 2 2 2 R R R 2 R R R = = 2 a 2 a b 2 b Z 2 − − 2 2 A 1 B 1

New measurement method We solve R 2 ( ) ( ) − − − 2 2 2 2 R A 1 R B 1 = b a [ ] R ( ) ( ) 2 − − − 2 2 2 R B 1 R A 1 a b and get the real and imaginary part of Z x ⎛ ⎞ 2 V V ⎜ ⎟ = − 2 2 = − + − 2 2 X Z R 3 b 4 b R Z 1 R Z and ⎜ ⎟ 2 2 2 2 2 2 ⎝ ⎠ V V 2 b 2 b 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 4 th order equations 2 ⎡ ⎤ 2 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ 1 R 1 X ⎢ ⎥ + + + + 2 2 1 a a ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ + + 2 2 2 2 R R X X R X ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ in 2 in 2 = 2 A 2 ⎡ ⎤ 2 ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ 1 R R 1 X ⎢ ⎥ + + + + 2 2 a 1 g g ⎜ ⎟ ⎜ ⎟ ( ) ( ) ⎢ ⎥ + + + + 2 2 2 2 R R R X X R R X ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ in 2 a in 2 a 2 ⎡ ⎤ 2 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ 1 R 1 X ⎢ ⎥ + + + + 2 2 1 b b ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ + + 2 2 2 2 R R X X R X ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ in 2 in 2 = 2 B 2 ⎡ ⎤ 2 ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ 1 R R 1 X ⎢ ⎥ + + + + 2 2 b 1 g g ⎜ ⎟ ⎜ ⎟ ( ) ( ) ⎢ ⎥ + + + + 2 2 2 2 R R R X X R R X ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ in 2 b in 2 b a = R gen + R mux + R a b = R gen + R mux + R b g = R gen + R mux

Considering the loading effect If g << R 2 , X, R a , R b , R in and X in , the calculation simplifies, and we get a pair of 2 nd order equations 2 ⎡ ⎤ 2 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ 1 R 1 X ⎢ ⎥ = + + + + 2 2 2 A 1 R R ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ a a + + 2 2 2 2 R R X X R X ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ 2 2 in in 2 ⎡ ⎤ 2 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ 1 R 1 X ⎢ ⎥ = + + + + 2 2 2 B 1 R R ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ b b + + 2 2 2 2 R R X X R X ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ 2 2 in in R in is the input resistance of the voltmeter and X in =1/ ω C in , where C in is the input capacitance of the voltmeter. X is the reactive part of Z x . The resistive part of Z x , R x , can be calculated by subtracting the switch resistances from R 2 , if the the impedances are large enough so that the changes in the switch resistances can be neglected.

Considering the loading effect Equations can be expressed as ⎛ ⎞ ( ) 2 R 2 R ⎜ ⎟ + − + − − = 2 2 2 a a c R X 2 R 1 R X R 0 ⎜ ⎟ 2 a 2 a ⎝ ⎠ R X in in ⎛ ⎞ ( ) 2 R 2 R ⎜ ⎟ + − + − − = 2 2 2 b b 2 1 0 d R X R R X R ⎜ ⎟ 2 b 2 b ⎝ ⎠ R X in in 2 ⎛ + ⎞ 2 R R ⎜ ⎟ = − − 2 a a where c A 1 ⎜ ⎟ 2 ⎝ ⎠ R X in in 2 ⎛ + ⎞ 2 R R ⎜ ⎟ = − − 2 b b d B 1 ⎜ ⎟ 2 ⎝ ⎠ R X in in

Considering the loading effect Solving X we get X = − − in X X eR 2 in 2 d − R R 1 a b c = + e where d R − 2 2 R R in a b c

Considering the loading effect Substituting X into the first equation of the pair of the equations gives 2 cX + + = 2 in 0 fR hR 2 2 4 ( ) = + 2 2 f c 1 X e where in ⎛ + ⎞ ( ) R ⎜ ⎟ = + − 2 2 a h cX 2 R e 2 R 1 ⎜ ⎟ in a a ⎝ ⎠ R in

Considering the loading effect So, the equations for R 2 and X are − ± − 2 2 h h fcX = in R 2 2 f ⎛ ⎞ 1 = − ⎜ + ⎟ X X ⎜ eR ⎟ in 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!)

Capacitors in delta circuit Current through A23 V1(a) V2(a) C1 V5(a) V4(a) IA23 A23 A2 C2 C3 A01 A1 V3(a) V6(a)

Capacitors in delta circuit Current through A2 V2(b) V1(b) C1 V5(b) V4(b) IA2 A23 A2 C2 C3 A01 A1 V3(b) V6(b)

Capacitors in delta circuit • The total reactance X can be obtained from the equations presented • The total capacitance of the network is obtained from the equation 1 = C tot ω X • Considering the network topology and the places of the current source, C tot is seen as 2 C C kC = + = 3 + 1 3 C C C (A23) tot1 2 2 + + C C kC C 1 3 3 3 2 C C lC = + = + 2 1 2 (A2) C C C tot2 3 3 + + C C lC C 1 2 2 2

Capacitors in delta circuit • Determining voltage divisions gives us: − 2 2 V V 5 ( a ) 6 ( a ) = k − − − 2 2 2 2 V V V V 4 ( a ) 6 ( a ) 5 ( a ) 6 ( a ) − 2 2 V V 5 ( b ) 6 ( b ) = l − − − 2 2 2 2 V V V V 4 ( b ) 6 ( b ) 5 ( b ) 6 ( b ) • Solving for C 2 and C 3 , we get: [ ][ ] + + − l 1 C k C C k = tot 1 tot 1 tot 2 C [ ] 2 + + − ( k 1 )( l 1 ) kl [ ][ ] + + − k 1 C l C C l = tot 2 tot 2 tot 1 C [ ] 3 + + − ( k 1 )( l 1 ) kl

Capacitors in delta circuit • Finally C 1 can be solved and calculated from C = kC 1 3 • or from C = lC 1 2

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

Impedance measurement results 6 5 4 Rx+ESR Error [%] 3 |Zx| Error [%] 2 1 0 0 2 4 6 8 10 12 |argZ1a| [degrees]

Delta circuit measurement results C1 ≈ C2 ≈ C3 • Quite good results were obtained, 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 R s1 and R s2 Small ratio Large ratio