RF Current Induced in an Ordnance Circuit

HIRSCH: CURRENT INDUCED IN AN ORDANCE CIRCUIT

RF CURRENT INDUCED IN AN ORDNANCE CIRCUIT

Stanley R. Hirsch

Martin-Marietta CorporationDenver, Colorado

INTRODUCTION

Summary

This paper describes and evaluates six meth-ods for predicting RF induced current in a typicalordnance circuit found in airplanes, missiles, or

space vehicles. The calculations here were

based on an incident RF field of 100 watts per

square meter at 2 megacycles.

Conclusions

Method 6, which is preferred, considers thesingle shielded portions of the circuit as a loopantenna then assumes that the effective apertureis the total area encompassed by the loop. Basedon Method 6, the typical ordnance circuit as de-picted in Fig. 1 is compatible with an electro-magnetic field up to 100 watts per square meterstrength. Method 6 yields 900 ma under theworst conditions, whereas the no fire currentis usually one ampere. However, our reliabil-ity people estimate the probability of all theassumptions occurring simultaneously is lessthan 0. 0000001, which is normally the prob-ability of failure of a missile structure notcompleting its mission.

If the mismatch between the bridgewire re-

sistance and the characteristic impedance of theattached twisted pair is considered, the voltagereflection coefficient

Pm 111

1 - IKI2 1 -(0.99)2 1 - 0. 9801

1

0.0199 0-2,where

P = Power delivered to the load

pm

Power that would be delivered where

the system is matched.

Current would be reduced by the square root ofthe power, or about 7. Therefore, instead of0. 900 there would be 129 ma through the bridge-wire (see results of Method 6).

BACKGROUND

Ambient RF fields (due to radars, microwavetransmitters, or other sources) may induce suf-ficient current in ordnance circuits to cause inad-vertent squib firing.

A typical ordnance circuit (Fig. 1) has beenanalyzed based on the assumptions in Chapter III.

Rl

z - z

P KK RR ZR +Z

0.25 - 50 (Ref 96)

50. 25

KR = ° 990

Z0= Characteristic impedance

R + jwL ohms (Ref 5, p 34)G + jwC

ZR Receiving end impedance (Loadimpedance or terminating imped-anc e)

Then, from Ref 4, p 569, mismatch loss = p

Bus

Fig. 1-Typical ordnance circuit.

1965 15

16

Fig. 1-Typical ordnance circuit (cont).

Blind use of the equations as found in the text-

books, papers, and other source literature can

lead to impossibly high or misleading low currentvalues.

ASSUMPT IONS

Each of the following assumptions would tendto decrease the actual current; therefore, thecalculated values are an unrealistically high max-

imum value of induced current:

1) The circuits are assumed to be hanging infree space and intercepting the maximumincident electromagnetic field;

2) Attenuation of the line is assumed to bezero;

3) All connection elements are assumed com-

plex conjugates (there are no losses due to

mismatch);

4) Access doors are open; therefore, enclo-sure skin attenuation is assumed to be zero;

5) An incident electromagnetic field of 100watts per square meter field strength (194volts per meter field intensity) is assumedto be continuously impinging upon the entireordnance circuit;

6) Resonance is assumed to exist at all fre-

MARCH

quencies;

7) Induced currents add algebraically ratherthan vectorally. Therefore, for Methods 1thru 5, the result is multiplied by twentybecause there are about twenty 2-in.lengths in Fig. 1 acting as theoretical an-

tennas.

METHODS

Method 1. The length that a shield may bestripped when making connections to either a ter-minal board or connector is limited to 2 in. bythe manufacturing processes. Method 1 assumes

that the unshielded length acts as a small dipoleantenna. Effective aperture and the resistance ofthe 2-in. length were calculated and substitutedinto an equation for induced current.

Method 2. This method substituted effectivelength instead of effective aperture. The maxi-mum of 2 in. was used as effective length. Ter-mination resistance of the remainder of thecircuit was calculated at 2 mc.

Method 3. Voltage induced in a 2-in. lengthwas computed. Characteristic (or surge) imped-lance of a twisted pair of #20 wire was calculated.The induced voltage was divided by the surge im-pedance to obtain a surge current.

Method 4. This method calculates the com-

plex impedance (mostly reactance) for each wire(at 2 mc) and divides this into the voltage inducedin the 2-in. length.

Method 5. The voltage induced in a 2-in.length is divided by the entire circuit impedance.

Method 6. The effective aperture is consid-ered as the area encompassed by the singleshielded wire in the current limiting resistor(Rl, Fig. 1) portion of the circuit. The resist-ance of the entire circuit at 2 mc is considered as

the terminating resistance.

RESULTS

Current Induced in theMethod Circuit at 2 mc (amp)

1 126, 600 (for 20 two-inch lengths)

2 13. 72 (for 20 two-inch lengths)




6 0.9

IEEE TRANSACTIONS ON ELECTROMAGNETIC COMPATIBILITY

Description of Wires

Wire LengthNumber (in.) Type Notes

A 65 #12 TPSJ Paired with F

B 107 #20 TPSJ Paired with C

C 108 #20 TPSJ Paired with B

D 12 #16 Single Shield

E 14 #16 Single Shield

F 65 #12 TPSJ Paired with A

G 93 #12 TPSJ Paired with J

H 33 #14 Single Shield

I 32 #14 Single Shield

J 90 #12 TPSJ Paired with G

Note: 1. TPSJ is twisted pair, shielded, jacketed.2. Shields are terminated on one end.


EVALUATION OF METHODS

Method 6 is the most logical approach. Con-sidering the effective aperture to be actual looparea is sufficiently pessimistic, particularly ifall the assumptions above are true.

Method 1 can be ignored because the effectiveaperture of a 2-in. dipole has no meaning. Fur-thermore, the equation used to derive effectiveaperture would yield an impossibly high area.

Method 2 is similar to Method 5 except that thereal part of the terminating impedance is consid-ered as the load seen by each 2-in. length.

Methods 3 and 4 are similar and should beexpected to yield similar results. Results sup-

ported this.

Method 5 ignores the loop antenna effect of thesingle shielded portion of the circuit (particularlythe current limiting resistor and its leads).

Induced Current-Method 1

Derivation of Equation for Induced Current.The power (W) in watts in the terminating imped-ance (Z ) of an antenna is:

W =I Rt (Ref 2, p 43)t

[11

where I is the current produced through Zt and Rtis the real part of Zt

I R 377A = t

e 2E[51

Solving for the induced current:

2AI =5/ e

377 Rt

[61

I is in amperes, E in volts/meter, A in square

meters, and R is in ohms.

High Frequency Resistance of Copper Wire.Reference 3 contains information for calculatingthe high frequency resistance of copper wire:

R 2 828+ 250

[7]

R is the resistance in ohms at frequency (f)

f is the frequency in cps

R is the dc resistance in ohms0

x 10 da

d diameter of the wire in centimeters

a = 0. 01071.../T

The resistance of each wire at 2 mc is shownin Table 1.

Effective aperture A is related to the power

by:

A = W/P (Ref 2, p 43)e

(2]

where P is the Poynting Vector (power density) in

watts/square meter.

Substituting Eq [ 1] in Eq [2]:

A IR IPe

[3]

However,

P = E2/Z (Ref 2, p 44) [4]

where E is the field intensity in volts/meter andZ is the intrinsic impedance of the medium. Forfree space, Z = 120r 377 ohms.

Table 1 Wire Number vs Resistance at 2 mc

Substituting Eq [4] in Eq [3]:

1965 17

dc Resistance R at 2 mc LengthWire No. (ohm) Fac tor (ohm) (in.)

0.0119 11.25 0.1 34 90

A 0.00864 11.Z5 0.0970 6 5

B 0. 0891 4. 6 0.41 0 107

C 0.0899 4.6 0.413 108

D 0.00396 7.18 0.0 Z84 1 2

E 0.00462 7.18 0.0332 14

Ez 0.00865 11.25 0.972 65

G 0. 0 1 Z 3 7 11. Z5 0. 1 39 9 3

H 0.00439 8.95 0. 392 33

I 0.006656 8.95 0.0596 32

Resistor RI 1.6 *5 8.0

Br idg e w i r e 0.2 5 *4 1. 0

2.09 10.8044 619

*Estimated


Calculation of Factor from Table 1.

Rf = R (x + 0. 25)

Factor.,c X + ohmns

Gauge dia (cm) 10 da 2.828 2.828 (per 1000 ft)

#20 0. 08118 12.3 4.35 4.6 10.15

#16 0. 1291 19.55 6.93 7.18 4.016

#14 0.1628 24.6 8.7 8.95 2.525

#12 0.2053 31 11 11. 25 1. 588

x = 10 da

d = diameter in cm

a = 0.01071VT

a 0.01071 Vx10 = 15.144 at 2 mc

IOa - 151. 44.

Equations for Calculating Effective Aperture.Reference 1 gives these equations for finding the

effective aperture (Ae):

For a half-wave dipole, A - 64% for [8e 4 -t

unshielded lead lengths < 0. 5 %. (Ref 4, p 703)

where % is the wave length in meters;

3 x 10 meters/secfreq in cycles/sec

G%2A = (Ref 4, p 676) [9

e 4 c

R of a 2-in. length of #20 wire = 60 6

ohms/ft - 0. 00159 ohms for a 2-in. length.

R at 2 mc = 0. 00159 x 4.66 0. 00732 ohms.

3 x 0Wave length at 2 mc (Eq [4]), - 6= 150

2 x 10

meters. X2 = 150 = 22, 500 square meters.

Effective aperture (Eql8]),

1. 642 1.64 x 22, 500A = =- =e 4it 12.56

- 2940 square meters.

E AInduced current (Eq[6]), I 37

t

37, 700 x 2940 294, 000377 x 0. 00732 0. 00732

40 x 10 = 6330 amp for one two-inch-length.

Evaluation of Method 1. The result is too highto be acceptable. Analysis of the method yieldstwo fallacies:

1) Use of the resistance of the 2-in. length isincorrect. Resistance of the rest of thecircuit should have been used;

2) The equation for effective aperture cannotbe used for these conditions. The veryconcept of effective area for a 2-in. lengthof wire is meaningless.

Induced Current-Method 2where

G =5 (L is unshielded lead length) [10]

Substituting Eq (101 in Eq [91

A 5LEe 16r [11]

Calculation of Induced Current in a 2-in.Dipole Antenna. For a field of power density P =

2100 watts/square meter, using Eq [4] E = PZ

Still using Eq[6], but substituting effectivelength (chosen at 2 inches to give a maximumvalue) and the resistance of the remainder of thecircuit at 2 mc for Rt, the current induced in the

circuit is:

I- 7, 700 x 0. 0508 meters377 x 10. 8

- 0. 686 amp for one two-inch length.


E = \/100 x 377 = V/37700 Z 194 volts/meter.

From Eq [3] the factor for #20 is 4. 6.

The equation V = E9 (Ref 2, p 50) is used, [12]

where E is the effective field intensity at a shortdipole ( \/ 10 or shorter)

18 MARCH

195HIRSCH: CURRENT INDUCED IN AN ORDANCE CIRCUIT

e is the length of the dipole

V is the voltage induced in the dipole.

III E - 9.86 volts 0. 197 amp0

for one two-inch length.

Calculation of Induced Voltage in Short Dipole. Induced Current-Method 4

E 194 volts/meter

0. 0508 meters

V EQ = 194 x 0. 0508 = 9.86 volts.

Calculation of Characteristic Impedance (ZO).For a lossless line (Chapter III. 2), the charac-teristic impedance (Z )is

Z=4 -[13]o C

where

L is in henrys/unit length

C is in farads/unit length

Z is the impedance seen by the wave as it

travels down the line.

The capacitance (C) for the twisted pair of #20

wire used is 77. 8 x 10 farads/ft.

The inductance is given by

[14]

Method 4 uses Eq [12] to calculate the voltageinduced in a 2-in. length of unshielded wire, thendivides this voltage by the impedance, whichconsists mainly of reactance at the frequenciesselected. The twisted pair of #20 wires, B andC, are connected to the bridgewire (Fig. 1).These wires are considered as a transmissionline fed by a generator of 9. 86 volts (Fig. 2).

Conditions. G and R (Fig. 2) are considerednegligible. If G is not negligible, it will shortsome of the current flowing, causing still lesscurrent to flow to the next part of the line. If Ris not negligible, it will reduce the current flow-ing in the circuit, and will also reduce the cur-rent transmitted to the next section of line.

calculations (at 2 mc). The pair of wiresfrom the bridgewire are 108 in. long. Twoinches of shield on each end are stripped back,giving a length of 104 in. as the terminating cir-cuit. The capacitance and inductance of #20TPSJ (twisted pair, shielded, jacketed) are 77. 8

picofarads/ft and 16.4 x 10 henrys/in.77.8 -12

For 104 in., C =7 x 104 x 1012

674 picofarads.

L = 16.4 x 10 x 104 = 170.6 x 10 henrys

X = 2ftf = 2 x 3.14 x 2 x 106 = IZ.56 x 10where

L is the inductance in henrys/meter

B is spacing between the wires

r is the radius of a wire

- 4 ln 4 = 1.386 4 In 4 = 5.444 4 In 4 + 1r

6.444

-7L = 10 (6. 444) henrys/meter x 0.3048 ft/meter

= 1.96(o henrys /ft

1 9-

x 10 57.96x10 vrZ~5Z S0 ohms.

0 77. 8 x 10 12

6 -12 = 8W~C 12. 56 x10 x 674x10 8460 x106

- =118

Induced Current.Fig. 2-Representation of a short section of wire as a

transmission line.

191965

io- 7ln

B+I = L

r

20

-L= 12.56 x 10

Neglecting R, Z

IEEE TRANSACTIONS ON ELECTROMAGNETIC COMPATIBILITY b

6 170.6 x 10 7 =214 Capacity vs Gauge.

1 Twisted Pair Shielded Wire=X= wL-

Gauge Capacity (picofarads/ft)

Z 214 - 118 = 96 ohms

V 986 v= 0. 105 amp in one wire

Z 96 ohmsdue to an antenna of 2 in. The current inthe bridgewire circuit is then 2 x 0. 1050. 210 amp for one two-inch length.


The impedance is calculated for the entire loopas a series circuit. The induced voltage in a2-in. length is divided by the impedance. Tables1 and 2 contain the values used for calculations.

Table 2 Wire and Component Characteristics

Impedance of Current Limiting Resistor (Rl)

R 1.6 ohms

L = 40 x 10 henrys

!ARCH

#12 120.3

#14 110

#16 98 (96 used to simplifycalculations)

#20 77.75

Sample Calculations for Table 3.

#14.

C 100 x 1012 farads/ft

-12~108 x 10 farads/ft

9 x 10 1 farads/in.

2 x 105 cps,cw = 12.56 x 10

5 -12wC 12.56 x lOx 9 xlO 113 x 10

1 1 x 10= 8. 85 x 10 ohms/in.

WDC 113

#16.

C = 98 x 10 farads/ft -8 x 10 farads/in.

5xIxo cps 12. 56 x 10

Frequency

52 x 10 12.5 x 10

2 x 10 12.5 x 10

2 x 10 12.5 x 10

2 x 108 12.56 x 108

2 x 109 12.56 x 109

F = (L (ohms)(R is negligible)

50. 24

5 -12 -7

C 12.56 x 10 x 8 x 10 = 100 x 10

0 = 1 x 10 ohms / in.'D)C I

2

502.4

5, 024

50, 240

502, 400

2 x 1010 12.56 x 1010

#20.

C = 77. 75 x 10 farads/ft . 6. 5 x 10 12

fa rads / in.

5 - 12 -7C 12.5x 10 x6.S5xlIO = 81.S5xlO0

I = 5=x 10 os51. 25 x 10 ohms/in.

WC 81. 5

#20 = 10. 1Q/1000 ft = 0. 01Q/ft = 0.000833 0 /in.

#16 = 4.0 Q/1000 ft = 0. 000330/in.

#14 = 2.5Q/1000 ft = 0.0025Q/ft 0.0002080/in

#12 = 1.6Q/1000 ft = 0.0016Q /ft= 0. 000133Q /in

Vendor Wire to Resistor = #16 Squib = 0. 25 7

Vendor Wire to Contactor = #14 RI = 1.6Q

1.

5, 024, 000


Table 3 Inductive and Capacitive Reactance in Ohms/Inch

#12.

C = 120.3 picofarads/ft = 10 x 10 farads/in.

5 -12wC = 12.56 x 10 x 10 x 10

1 1000 x 10 8 x 104ohms/in.w C 125. 6

Wire D - #16.

Length = 12 - 4 = 8 in.

X (ohms)Frequency (cps) L

0.2x105 0.170

2 x

2 x

2 x

2 x

Impedance of Each Wire

Wire C - #20.

Length 108 - 4 (stripped back) = 104 in.

Frequency (cps) XL (ohms) Xc (ohms) z1 (ohms)

2 x 105 22 1200 1197.8

2 x 106 22 120 98

2 x 107 220 12 208

2 x 108 2,200 1 2 2198.8

2 x 109 22, 000 0. 12 22, 000

2 x 10 220, 000 0.012 220, 000

X -21. 2 x 10 x 104 in. = 2.2 ohms at 2 x 10 cpsL

Xc 1. 25 x I0/104 =l, 200 ohns at 2xl10 cps

2 x

XL

xC

106

1 08

109

,10

XC (ohms)

1.25 x 104

1. 7 1. 25 x 103

17 125

17 0 12. 5

1,700 1.25

10 - 17,000

-3=21.2 x10 x8 =0.

0. 125

170 ohms at 2 x

Z (ohms)

1,25 x 104

1. 25 x 103

1 08

157.5

1, 7 00

17, 000

10 cps

1x10= 1.25 x 10 ohms

izi = /R + X

If R<< X,

IZi ViS

X ( L --

L C

21

x ( C) (ohms/in.)

X~~~~~~L

Frequency c (cwL) (ohms/in.) #16 #14 #12 #20

2 x 105 12.56 x 105 Z1.2 x 10-3 51x 105 S.85 x 104 S x 104 1. 25 x 1052x106 12.56x106 21.2xx1x04 8.S5x103 8x10 l.25x104

2 x 107 12.56 x 107 21. 2 x 10'I

I x 103 S. 85 x 102 S00 1. 25 x 103

2 x 108 12.56 x 108 21.2 1 x 100 88.5 80 125

2 x 109 12. 56 x 109 212 1 x 10 8.85 8 12.5

2 x 10 12. 56 x 1010 2120 1 0.885 0.8 1.25

-7 B -7L/a = henrys/meter = 10 (4 In -+ 1) = 10 x 6.444 henrys/meter B/r = 4. ln 4 = 1. 386,r7r8 41n 4 = 5. 444,

1 meter = 39.37 in. 6.444x 10 /39.37 = 1.64x 10 henrys/in. 4 In 4 + 1 = 6.444

To obtain XL for any length, multiply XL given above by length in inches.

To obtain X for any length, divide X given above by length in inches.

1965

MARCH22 IEEE TRANSACTIONS ON ELECTROMAGNETIC COMPATIBILITY

Wire E - #16. Wire G - #12.

Length = 14 - 4 = 10 in. Length = 93 - 4 = 89 in.

Frequency (cps) L

2 x 105 0.212

2 x 106 2.12

2x 107 21.2

2 x 108 212

2 x 109 2,120

2 x1010 21,200

X 21.2 x x1x0 = 0.21L

X =- lxlO ohmsC 10

Xc (ohms)

1 x 104

1 x 103

1 00

10

0. 1

L2 ohms

Wire F - #12.

Length = 65 - 4 = 61 in.

X (ohmns) X (ohms)Frequency (cps) L C

2 x 105 1.38 1. 23 x 103

2 x 106 13.8 123

2x107 138 12.3

82 x 10° 1,380 1.23

2 x 109 13, 800 0. 123

2 x 10 138, 000 0. 0123

X 21.2 x 10 x 65 = 1.38 ohmsL

8 x104 3X - 65 1. 23 x10 ohms

Z (ohms)

1 x 104

1, 000

78.8

202

2, 119

21, 200

Frequency (cps) XL (ohmCs)X (ohms)

52 x 10 1.89 900

62 x 10 18.9 90

2 x 107 189 9

82x10 1,890 0.9

92 x 10 18,900 0.09

2 x 101 189, 000 0.009

X 21.2 x 10 x 89 = 1.89 ohmsL

8 x 1O4X =

8900 ohms

Wire H - #14.

Length = 33 - 4 = 29 in.

Z (ohms)

1, 23 0

109.2

125.7

1, 3 79

13, 800

138, 000

Frequency (cps) XL (ohms) XC (ohms)

2 x 105 0.615 3050

2x106 6.15 305

2 x 107 61.5 30.5

2 x 108 615 3.05

2 x 10 6, 150 0.305

2 x 10 61, 500 0.0305

X =21.2 x 10 x 29 = 0.615 ohmsL

x 885X' - 3050 ohmsC 29

Z (ohms)

888

81

180

1, 889

18, 900

189, 000

Z (ohm s)

3, 050

299

31

612

6, 150

61, 500

231965 HIRSCH: CURRENT INDUCED IN AN ORDANCE CIRCUIT

Wire B - #20. Wire J - #12.

Length = 107 - 4 = 103 in. Length = 90 - 4 = 86 in.

Frequency (cps) XL (ohms) Xc (ohms) Z (ohms)

2 x 105 2.18 1210 1,208

2 x 106 21.8 121 99

2 x 107 218 12.1 206

2 x 108 2, 180 1.21 2,179

2 x 109 21,800 0.121 21,800

2 x 1010 218, 000 0.0121 218, 000

X 21.2 x 10 x 103 = 2.18 ohmsL

125 x 103C 103

Wire A - #12.

Length = 65 - 4 = 61 in.

Frequency (cps) XL (ohms) X (ohms) (ohms)

2 x 105 1.29 1310 1, 309

2 x 106 12.9 131 118

2 x 107 129 13. 1 116

2 x 108 1, 290 1.31 1, 289

2 x 109 12, 900 0. 131 12, 900

2 x 1010 129, 000 0. 0131 129. 000

-3X = 21.2 x 10 x 61 = 1.29 ohmsL

80 x610 13 10 ohms

Frequency (cps) XL (ohms) Xc (ohms) Z (ohms)

2 x 10 1.82 930 928

2 x 106 18.2 93 75

2 x 107 182 9.3 173

2x 10 1,820 0.93 1,819

2 x 109 18, 200 0.093 1s, 200

2 x 10 182, 000 0. 0093 182, 000

-3X = 21. 2 x 10 x 86 = 1. 82 ohmsL

380 x 1

C 86 930 ohms

Wire I - #14.

Length = 32 - 4 = 28 in.

Frequency (cps) XL (ohms) XC(ohms)

2 x 105 0. 594 3160

2 x 106 5.94 316

2x 107 59.4 31.6

2x 10 594 3.16

2 x 109 5, 940 0.316

2 x 10 59, 400 0. 0316

X 21.2 x 10 x 28 = 0.594 ohmsL

Z (ohms)

3, 160

310

27.8

591

5, 940

59, 400

X .5 x 10 3160 ohms28


Circuit Impedance vs Frequency and CurrentInduced.

(60 sq in. ) of the current limiting resistor loopexposed to the incident rf field, then Eq [61yields:

Frequency Z Resistor Z Circuit I per 2-in.(cps) Z Wire (ohms) (ohms) (ohms) length I Total

2 x 10 35, 471 C: 50.24 L 35,421 0.279 ma 5.55 ma

2 x 106 3439.2 C* 502. 4

2 x 10 5530. 3 L 5, 024

2 x 10 12, 315.3 L* 50, 240

2936. 8 3.36 ma 67.2 ma

5, 904 1.67ma 33.4 ma

62, 555 0. 1575 ma 3. 15 ma

2 ~2x

",

I = EAe 19 1550377R 377 x 10.8

t= '0. 359 =0. 6 amp

637, 700 x 155

4071.6

[161"I 601550 converts square inches into square

meters.

2 x 109 123, 509 L* 502, 400 625, 909 15.75 .a

2 x 10 1, 235, 100 L* 5, 024, 000 6, 259, 100 1. 575,ua

Induced Current.

V = Ef for short dipole (Ref 2, p 50)

If E = 194 volts/meter g = 0. 0508 meters for a2-in. line

V induced = 194 x 0. 0508 9. 86 volts per 2-in.line.

There are about twenty 2-in. lengths (Fig. 1),

At 100 mc %. = 3 meters. Therefore, a 2-in. an-tenna (0. 0508 meters) can be considered a shortdipole up to 100 mc.

Method 6

If it is assumed that the effective aperturecannot be greater than the actual physical area

An additional 300 ma can be added for the 30 sqin. loop containing the motorized switch. Thetotal current induced, then, is 0. 900 amp.

REFERENCES

1. STL Interoffice Correspondence 6110-6307-DU-000 GM 6416.6-247 to A. Hoffman from M.Rosenthal entitled YLR87-AJ-3 and YLR91-AJ-3 Ordnance Initiator Ignition Leads;requirement for, 24 January 1962.

2. Krauss, John, Antennas, McGraw Hill, 1950.

3. Handbook of Chemistry and Physics 39thEdition, Chemical Rubber Publishing Company,p 3022.

4. Reference Data for Radio Engineers 4thEdition, International Telegraph and TelephoneCompany.

5. Transmission Lines and Networks by W. C.Johnson, McGraw Hill, 1950.

'i:C indicates the reactive impedance is capacitive,L indicates the reactive impedance is inductive.

24 MARCH

Documents

RF Current Induced in an Ordnance Circuit