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Inductance Measuring Method

IP.com Disclosure Number: IPCOM000075435D
Original Publication Date: 1971-Sep-01
Included in the Prior Art Database: 2005-Feb-24
Document File: 3 page(s) / 46K

Publishing Venue

IBM

Related People

Ruehli, AE: AUTHOR [+2]

Abstract

This method allows accurate measurement of small inductances, in the range of less than 100 nanohenries, with a conventional inductance bridge circuit by utilizing the concept of partial inductances and aids in the design of new equipment.

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Inductance Measuring Method

This method allows accurate measurement of small inductances, in the range of less than 100 nanohenries, with a conventional inductance bridge circuit by utilizing the concept of partial inductances and aids in the design of new equipment.

Fig. 1 shows an inductance measuring circuit where the external circuit 5, 6, 7, 8 and 9 represents, for example, the circuit pattern of a planar integrated circuit component of which the inductance is to be measured. A connection loop, represented by elements 2 and 3 is mounted perpendicular to the unknown loop to avoid complex coupling problems between the internal inductances of the measuring instrument and the external circuit. Element 1 represents the simplified internal inductance of the instrument. Also shown is a partial inductance 4. (The concept of partial inductances is discussed in IBM Technical Report TR 19.0181, "Inductance Calculations in a Complex Integrated Circuit Environment," by A. E. Ruehli.)

Fig. 2 represents the equivalent circuit for the connection loop to the instrument having an inductance (Lc) including that of partial inductance Lp44. 4 Lc = Sigma Lpii -2(Lp14 + Lp23). (1) i=1.

Fig. 3 represents the equivalent circuit of the unknown loop for which inductance (Lu) is to be measured. 9 Lu = Sigma Lpii -2(Lp57 + Lp68 + Lp79).
(2) i=5.

The total actual inductance of the two loops is: 9 L(u+c) = Sigma Lpii -2(Lp23 + Lp57 + Lp68 + Lp79 + Lp15 + (3) i=1 Lp17 + Lp19). i Not equal 4.

The sum of the separate inductances Lu and Lc is: 9 Lu + Lc = Sigma Lpii - 2(Lp14 + Lp23 + Lp57 + Lp68 + Lp79).(4) i=1.

Now, from equations (3) and (4) the unknown inductance, Lu, is: Lu = L(u +
c) - Lc - 2(Lp14 - Lpl5 - Lp17 - Lp19) + Lp44. (5).

If the coupling loop 1, 2, 3 and 4 is sufficiently long, that is if th...