Rail voltage responses to current pulses are analyzed using recursive differential equations and two different current pulse models on the worst case assumption that passive decoupling effects of bypass capacitors are all that we have working in our favor..

It is shown in this particular case that rail voltage excursions in response to current pulses of 3 nSec rise and fall times do not exceed 40 mV per Ampere of peak pulse current.

The recursive differential equations are as follows:

The current pulses are modeled using two different analytic expressions. Each is normalized to a peak current of one Ampere and 3 nSec pulse rise and fall times.

Resistor Rx is added to the paralleled decoupling capacitors to simplify the differential equation derivation. ( See e = ix * Rx above. ) Rx is chosen as 1000 Ohms in the presented calculations, a value much larger than the capacitor impedances over the frequencies of interest.

A sine tip model of the current pulses is demonstrated below:

The following code provides the verification of the above that the sine tip model of " i " has the required rise and fall times:

10 CLS:SCREEN 9:COLOR 15,1:YSTART=170:XSTART=250:RAD=150

20 PRINT "save "+CHR$(34)+"trigfns1.bas"+CHR$(34):PRINT:PI=3.14159265#

30 PRINT "save "+CHR$(34)+"a:trigfns1.bas"+CHR$(34):PRINT:PRINT

40 C$="##.## Emax at ###.##ø ":D$="##.##### MHz"

50 E$="##.## nSec at ###.##ø "

60 REM THETA is the arcsin of A/C.

70 A1=.1:A2=.9:C=1:THETA1=ATN(1/SQR((C/A1)^2-1)):THETA2=ATN(1/SQR((C/A2)^2-1))

80 THETA3=PI-THETA2:THETA4=PI-THETA1

90 TNSEC=3:TRISE=TNSEC/1E+09:F=(THETA2-THETA1)/(2*PI*TRISE)

100 PRINT USING C$;SIN(THETA1),THETA1*180/PI,SIN(THETA2),THETA2*180/PI,

110 PRINT USING D$;F/1000000!:PRINT:GOTO 190

120 X=50*TNSEC/10:Y=E*50

130 CC=XSTART+X:DD=(320-Y-YSTART):IF KK<>0 THEN LINE (AA,BB)-(CC,DD)

140 AA=CC:BB=DD:KK=1:RETURN

150 LOCATE 8,1:PRINT USING E$;TNSEC,THETA*180/PI:RETURN

160 LOCATE 9,1:PRINT USING E$;TNSEC,THETA*180/PI:RETURN

170 LOCATE 10,1:PRINT USING E$;TNSEC,THETA*180/PI:RETURN

180 LOCATE 11,1:PRINT USING E$;TNSEC,THETA*180/PI:RETURN

190 FOR TNSEC=0 TO 15 STEP .001:T=(TNSEC-3)/1E+09:THETA=2*PI*F*T:E=SIN(THETA)

200 IF E<0 THEN E=0

210 IF ABS(THETA-THETA1)<.0002 THEN GOSUB 150

220 IF ABS(THETA-THETA2)<.0002 THEN GOSUB 160

230 IF ABS(THETA-THETA3)<.0002 THEN GOSUB 170

240 IF ABS(THETA-THETA4)<.0002 THEN GOSUB 180

250 GOSUB 120:NEXT TNSEC

A similar analysis ( not shown here ) yields an offset cosine model of the current pulse, again normalized to a peak current of one Ampere and 3 nSec pulse rise and fall times.

The two expressions for the current pulses are as follows.

For the sine tip:

i = Sin ( 2 * pi * F * t ) where F = 5409157 Hz = 54.01957 MHz and where i<0 then i = 0.

For the offset cosine:

i = 0.5 * ( 1 - cos ( 2 * pi * F * t ) where F = 9838909 Hz = 98.38909 MHz and where when t<0 then i=0 and when F * t >1 then i=0.

Rail voltage responses to both of these current pulse models are examined.

In terms of the actual electrical components, we have for this example, the following:

The results for sine tip and offset cosine current pulses are similar and are compared as follows:

The rail voltage departures from +5V are as many mV as shown per ampere of pulse current. Rounded upward, the rail voltage departures may be taken as 40 mV per Ampere of pulse current.

Code used for the evaluation of the rail voltage responses is shown as follows:

10 CLS:SCREEN 9:COLOR 15,1:PI=3.14159265#:XSTART=140:YSTART=40

20 PRINT "save "+CHR$(34)+"pul_caps.bas"+CHR$(34):PRINT:YS=142:REM YS=160

30 PRINT "save "+CHR$(34)+"a:pul_caps.bas"+CHR$(34):PRINT:PRINT:GOTO 80

40 Y=-E*1000:X=300*T/1.5E-08:GOTO 60

50 Y=I*100+60:X=300*T/1.5E-08:GOTO 60

60 CC=XSTART+X:DD=(320-Y-YSTART):IF KK<>0 THEN LINE (AA,BB)-(CC,DD)

70 AA=CC:BB=DD:KK=1:RETURN

80 READ CAP1,N1,R1,SRF1, CAP2,N2,R2,SRF2, CAP3,N3,R3,SRF3, CAP4,N4,R4,SRF4

90 REM uF N Ohms MHz uF N Ohms MHz uF N Ohms MHz uF N Ohms MHz

100 DATA 100,1,.125,.318, 10,1,2.5, 1 , 0.1,10,.08, 11, 0.01,12, .2 ,40

110 CAP1=CAP1*N1:CAP2=CAP2*N2:CAP3=CAP3*N3:CAP4=CAP4*N4

120 CAP5=3000:R5=.001:SRF5=290

130 R1=R1/N1:R2=R2/N2:R3=R3/N3:R4=R4/N4:RX=1000

140 C1=CAP1/1000000!:C2=CAP2/1000000!:C3=CAP3/1000000!

150 C4=CAP4/1000000!:C5=CAP5/1E+12

160 SRF1=SRF1*1000000!:SRF2=SRF2*1000000!:SRF3=SRF3*1000000!

170 SRF4=SRF4*1000000!:SRF5=SRF5*1000000!

180 L1=1/4/PI^2/SRF1^2/C1:L2=1/4/PI^2/SRF2^2/C2:L3=1/4/PI^2/SRF3^2/C3

190 L4=1/4/PI^2/SRF4^2/C4:L5=1/4/PI^2/SRF5^2/C5

200 DT=1E-14:FMHZ=54.09157:REM Sine tip frequency.

210 REM FMHZ=98.38909:REM Offset cosine frequency.

220 F=FMHZ*1000000!:FOR T=0 TO 1.5E-08 STEP DT

230 I=SIN(2*PI*F*(T-3E-09)):REM Sine tip model.

240 REM I=.5-.5*COS(2*PI*F*T):IF F*T>1 THEN I=0:REM Offset cosine model.

250 IF I<0 THEN I=0

260 IX=I-I1-I2-I3-I4-I5:E=IX*RX:EL1=E-EC1-I1*R1:EL2=E-EC2-I2*R2

270 EL3=E-EC3-I3*R3:EL4=E-EC4-I4*R4:EL5=E-EC5-I5*R5:DI1=EL1*DT/L1

280 DI2=EL2*DT/L2:DI3=EL3*DT/L3:DI4=EL4*DT/L4:DI5=EL5*DT/L5:DEC1=I1*DT/C1

290 DEC2=I2*DT/C2:DEC3=I3*DT/C3:DEC4=I4*DT/C4:DEC5=I5*DT/C5

300 EC1=EC1+DEC1:EC2=EC2+DEC2:EC3=EC3+DEC3:EC4=EC4+DEC4:EC5=EC5+DEC5

310 I1=I1+DI1:I2=I2+DI2:I3=I3+DI3:I4=I4+DI4:I5=I5+DI5

320 IF EMAX<E THEN EMAX=E

330 IF IMAX<I THEN IMAX=I

340 KK=0:GOSUB 40:GOSUB 40:KK=0:GOSUB 50:GOSUB 50:NEXT T

350 PRINT IMAX,EMAX

The lines that execute the recursive differential equations are 260 thru 310.

As shown, this code is set for the sine tip model. It may be changed to the offset cosine model by removing "REM" from line 210 and line 240.

One caution when running this code: It executes very slowly. Be patient, go have a cup of coffee and come back later.

I have never seen AC current response to capacitors explained so simply and intelligently. Thank you for enlightening the rest of us. Have you ever considered trying to forecast the existence of; and proving mathematically the electrical possibiiity of, a Net zero energy "power consumptionless" electric motor effect? I have worked for 25 years to develop one. I believe I have finally done just that. I would love to have the numerical proof of the existence of the effect, that I believe I have discovered. I regret; that I probably could not afford, your quality of analysis. But thank you for your input. Steve Hardison

Posted by: Steve Hardison | October 28, 2011 at 08:43 AM