Answer:
T1 = 1/12 x 48 MHz = 4 MHz
Timer’s clock period:
The time delay of one machine cycle is given below. We use this to generate the delay.
T2= 1/ 4000000= 0.25 µ sec
For delay of 10 ms:
1. Firstly divide the desired time delay value (10 ms) by the timer clock period.
N= 1 / 0.25µs
N= 10 ms / 0.25µsec
N=40000
2. Subtract the value of N from the maximum number of counts possible for 16 bit timer i.e. 2^16 = 65536.
M=65536-N
M=65536-40,000
M= 25536
Explanation:
Clock source:
