Algorithm for calculating prescaler and period of a timer

Question

I want to set frequency of a timer in a microcontroller. Frequency is determined by Prescaler and Period (ARR). Both of these are 16bit integer numbers.

$$f_{tim}=\frac{f_{clock}}{prescaler\cdot period}$$

Can you suggest an efficient way (not just trying all possible values) to find values that result in frequency as close as possible to the desired one?

• user16324: Prescaler values are usually quite restricted, e.g. to powers of 2. So the first step is to discover what the restrictions are, from the MCU's datasheets and programming guides.
• Jiří Maier: As I said: Prescaler is any 16bit unsigned integer. (STM32L4)
• user16324: then you are looking for factorisation algorithms.
• old_timer: obviously you dont need to try all possible values. the smaller the prescaler the more accurate it is going to be, so you dont need to try all possible values, but can quickly narrow in on the solution without too many iterations. couple of easy ways to do that.
• the busybee: Since prescaler and period are both integers, their product is also an integer. As @user_1818839 said, it's a factorisation problem; apply mathematics. If you can, define the clock frequency as a integer multiple of the timer frequency. Would you mind to edit your question providing some more context?

Answer 1

See if you like this version better.

A set of two integer factor (X & Y), that produce closest integer to a real number (N).

In case, factors of N do not exist, or factors are not usable:
Consider a division, N / X = Y + R, where X and Y are integer, and R is a remainder.
Since, R < X, the smaller divisor (X) is, the smaller remainder (R) results.
Meantime, from X * Y ~ N, the larger Y gets, the smaller X is.
Thus, Find the largest Y, then calculate X.

A random example:

Clock speed: 16Mhz
Prescaler: 16bit
Divider: 8bit
Desired output frequency: 115Hz

Operation:

Scale = 16Mhz / 115Hz = 139130.4348 ---(1)
The largest from the 16bit prescaler (Y) = 65535 ---(2)
The smallest divisor, X = N / Y = 139130.4348 / 65535 = 2.122994 ---(3)
Since Y cannot be larger, X = 3 ---(4)
Thus, recalculating, N / X = 139130.4348 / 3 = 46376.81159, Y = 46377 ---(5)

Output clock = 16Mhz / (46377 * 3) = 16Mhz / 139131 = 114.999532814 ---(6)

• jay: Now, I see @old_timer gave the answer already. Sorry about that.
• Jiří Maier: Tkank you for answer. However I am still not convinced that this gives the best possible values. It Is clear that R < X, so small X is guaranteed to have small R, but R could be even smaller "by chance" for any large X.
• Jiří Maier: Correction: but R*X could be even smaller "by chance" for any large X there may even be a value of X for which R is zero.
• jay: @JiříMaier Yes, R could be zero, if appropriate factors exist. You do not need to hand factorize. Search for "factor calculator".