# Micro Processor Questions

1. (TCO 4) What is the resolution for the HCS12 ATD in 10-bit mode with reference voltage of +5V? (Points : 5)

2. (TCO 4) A temperature sensor covers a range of 0 to 70 degrees C with an output sensitivity of 2 mV/degree C. The sensor is input to an ADC with voltage range of 0 to 4V. What is the value of the feedback resistor required for an inverting amplifier circuit required to interface the sensor? The resistor to the positive op-amp input is 1.0 kohm. (Points : 5)

3. (TCO 4) A position-sensitive potentiometer can measure 0 – 25 cm over a voltage range of 0 – 5V. If an ADC covering the same voltage range outputs 0111 1010, what is the distance measured? (Points : 5)

4. (TCO 4) A HCS12 ATD has been set-up to perform a right-justified, 10-bit, signed operation. What is the expected output for an input voltage of 3.09V? The voltage range is 0-5V.5V/4096 = 4.88 mV. (Points : 5)

5. (TCO 4) A HCS12 ATD is set to 8-bit unsigned mode. A temperature sensor measuring 0 to 100 degrees with an output of 25 mV/degree is attached to the input. The ATD low voltage reference is attached to ground and the high voltage reference between the two resistors is a voltage divider connected to +5V and ground. What value of resistors are required for the voltage divider to allow the ATD to output 1 step/degree change? (Points : 4)

1. $5V/2^10 =0.00488 V =4.88 mV$

2. maximum voltage of the temperature sensor is $V _{sensor}=70*0.002= 0.14 V$

$V_{out}/V_{sensor} = R1/R2$

$R1 = R2*V_{out}/V_{sensor} = 1000*4/0.14 = 28571 Ohms =28.57 kOhm$

3.   01111010 binary = 122 decimal
5 V……256

x V ……122

$x =122*5/256 =2.3828 V$

25 cm…..5V

Y………….2.3828 V

$Y = 2.3828*25/5 =11.91 cm$

4. $3.09 V/4.88 mV =632.8 decimal = 001001111000 binary$ right justified 12 bits signed

5. $25 mV/100 degree =0.25$ mv/Celsius degree

8 bit unsigned means 256 maximum decimal out

$5V/256 = 19.53 mV$

$R2/R1 = 19.53/0.25 = 78.12$

$R1 =1 Kohm$

$R2 =78.12 Kohm$ resistor from the reference to +5 V