If I use an 8 MHz oscillator with accuracy of 25 ppm and the PLL multiplies it to 32 MHz, then the accuracy is also multiplied (x4 = 100 ppm) or is it not? Thanx.
4
-
\$\begingroup\$ Fundamentally, it is not. However this depends on correct design of the PLL loop filter. \$\endgroup\$– user16324Commented Mar 9, 2023 at 20:30
-
\$\begingroup\$ As @user_1818839 states, theoretical accuracy stays the same, because both the frequency of the oscillator and the frequency error are multiplied by the same amount. If an egg costs US$1, 12 eggs cost US$12, same (exorbitant) ratio. \$\endgroup\$– DrMoishe PippikCommented Mar 9, 2023 at 20:39
-
2\$\begingroup\$ Properly speaking (and in spite of quite a bit of current usage), if there's a PLL and frequency multiplication is happening, the whole device (PLL and divider) is a frequency sythesizer. But yes, assuming it's working right if the input is off by 25PPM, then so will the output be. \$\endgroup\$– TimWescottCommented Mar 9, 2023 at 20:54
-
1\$\begingroup\$ ppm stands for parts per million and it doesn't scale; it's a ratio. \$\endgroup\$– Andy akaCommented Mar 9, 2023 at 21:02
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
The accuracy stays the same.
If you feed in exactly 8 MHz, and PLL multiplies by exactly 4, you get exactly 32 MHz out.
If you feed in 8.1 MHz, you get 32.4 MHz out. They are both 1.25% (12500 ppm) larger than the original frequencies.
