is there any simple solution around?

“Simple” as in “totally trivial”? Afraid not. Some clutter must be expected.

You did not show us what you “… tried myself with - long - expressions …”, nor did you indicate what you consider a " - long - " expression, so this may be of no use to you.

Nevertheless …

This one works for up to 24 fraction bits, yielding precision of approximately 7 decimals. Redirect A1 to whatever cell your number is contained in.

```
=DEC2BIN(A1) & "," & DEC2BIN( MOD(A1 ; 2^0) * 2^8 ; 8 ) & DEC2BIN(MOD( A1 ; 2^(-8) ) * 2^16 ; 8 ) & DEC2BIN( MOD( A1 ; 2^(-16) ) * 2^24 ; 8 )
```

This utilizes the `DEC2BIN()`

function in a stepwise manner. That function is limited to 10 bit signed binary. Here a choice is made to proceed in byte size steps, so only 8 bits are used for each step.

Utilizing the full unsigned capacity of the function (9 bits) requires that ~~you liberate yourself~~ *I liberate myself* from the byte-at-a-time mindset, which yields this formula:

```
=DEC2BIN(A1) & "," & DEC2BIN( MOD( A1 ; 2^0 ) * 2^9 ; 9 ) & DEC2BIN( MOD( A1 ; 2^(-9) ) * 2^18 ; 9 ) & DEC2BIN( MOD( A1 ; 2^(-18) ) * 2^27 ; 9 )
```

I have not considered two’s complement conversion or large integer part, so the formula needs more work to support negative numbers or values exceeding 256.

Note also the use of comma as fraction separator (immediately after the first `DEC2BIN()`

). This corresponds to the decimal separator used in the question, but depending on setting you may want to use a different character there.