each d type is initiated simultaneously by means of a common clock line
DESCRIPTION
Each D type is initiated simultaneously by means of a common clock line. 4 unused states, ie 001,011,101,111. S7 111. S6 101. S5 011. S4 001. S0 000. S3 110. S1 010. S2 100. B. A.C. B+C. - PowerPoint PPT PresentationTRANSCRIPT
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Each D type is initiated simultaneously by means of a common clock line.
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4 unused states, ie 001,011,101,111
S0 000
S1 010
S2 100
S3 110
S4 001
S5 011
S6 101
S7 111
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B
A.C
B+C
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Remove the two NOT gates and use the relevant Q output for A and B
0 1
0 1 1 0
1 0 1 1
1 1 0 1
S0 (00)
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goto captain
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PORTB,1
half
PORTB,1
half
btfss PORTA,0
goto loop
movf Wtemp,0
retfie
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In the event of an interrupt being initiated it immediately stores the contents of the working register in the temporary register Wtemp. This is because the working register may be used during the ISR and hence its contents changed.
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0.8V
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Y
X . Y .Z + X . Y . Z
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Use 7 comparators inputting into the priority- encoder, one comparator inputting into the overflow indicator, 8 resistor voltage divider and a 3 bit output.
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VLS= 2.3-0.7 = 1.6V
Input impedance =hfeRL=50 X 8= 400 ohms
Use a resistor divider network to provide positive DC bias
Use a push pull power amplifier which will reduce the effect of distortion but still produce crossover distortion
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V1=12-7.5=4.5V
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VL( approx) = VZ(1 + RF/R1), Max Value VL=7.5(1+1/3) = 9.98V ( approx 10V)
10V 7.5V
VS increases to 13V. Voltage across resistor increases but voltage across zener remains at 7.5V ie output voltage remains unchanged.
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The dummy gauge is included to compensate for any changes in ambient temperature conditions.
A B
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VA = ( 12/318.06 +340) X340 = 6.20V
VB = (12/680)X340 = 6V
V1= 6.00-6.20 = 0.20V
Gain = P/75K, P=75Kx50 =3750K
P=3750K 75K R= 3750K
Vout = +0.2 X 50 = =10v
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The thyristor must be Forward Biased
A positive gate pulse of sufficient voltage must be applied.
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24V 0V
0V 24V
0V 24V
S2
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VR= 24 -2 = 22V
R = VR/Ig = 22/60mA = 367ohms
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The de-coupling capacitors remove any unwanted DC from the input of the amplifier .
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√2500 = 50√2500 =50
BW =GBWP/50 = 1.6x106/50= 32KHz
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RF= 100K
R1 10K
R2 20K Vout
Gain on each channel is a max when the variable resistor are at minimum value ie = 0 ohms
Max gain on channel 1 =10, 100K/R1
Max gain on Channel 2 =100K/R2
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Fb=1/(2∏RC )=133Hz
Gain at frequencies above the break frequency =75/15 =5
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Fb