Each D type is initiated simultaneously by means of a common clock line

Preview:

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 Presentation

Citation preview

Each D type is initiated simultaneously by means of a common clock line.

4 unused states, ie 001,011,101,111

S0 000

S1 010

S2 100

S3 110

S4 001

S5 011

S6 101

S7 111

B

A.C

B+C

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)

goto captain

PORTB,1

half

PORTB,1

half

btfss PORTA,0

goto loop

movf Wtemp,0

retfie

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.

0.8V

Y

X . Y .Z + X . Y . Z

Use 7 comparators inputting into the priority- encoder, one comparator inputting into the overflow indicator, 8 resistor voltage divider and a 3 bit output.

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

V1=12-7.5=4.5V

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.

The dummy gauge is included to compensate for any changes in ambient temperature conditions.

A B

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

The thyristor must be Forward Biased

A positive gate pulse of sufficient voltage must be applied.

24V 0V

0V 24V

0V 24V

S2

VR= 24 -2 = 22V

R = VR/Ig = 22/60mA = 367ohms

The de-coupling capacitors remove any unwanted DC from the input of the amplifier .

√2500 = 50√2500 =50

BW =GBWP/50 = 1.6x106/50= 32KHz

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

Fb=1/(2∏RC )=133Hz

Gain at frequencies above the break frequency =75/15 =5

Fb

Recommended