Fastest way to get rid of accessive bits (chop off bits)

Hello,

So far this is (I think) my fastest way to chop of accessive bits from a 
byte (in Delphi).

// Mission:
// Value contains a binary value/code.
// BitCount specifies which bits of the value starting from bit position 
0/lsb should remain.
// so BitCount 5 means: Bits 0 to 4 must remain. Bits 5 to 7 must be 
cleared.
// The accessive bits beyond bit count are to be zero-ed/cleared.

var
    Bitcount : byte;
    Value : byte;
    Mask : byte;
begin
    Value := 255; // could be anything
    BitCount := 5; // could be zero to 8

    // 5 instructions, no memory usage, best one so far.
    Mask := not (255 shl BitCount);
    Value := Value and Mask;

    writeln( Value ); // displays 31.
    readln;
end.

// generated assembler for the method:

Project1.dpr.18: Mask := not (255 shl BitCount);
00409151 8BC8             mov ecx,eax
00409153 B0FF             mov al,$ff
00409155 D2E0             shl al,cl
00409157 F6D0             not al
Project1.dpr.19: Value := Value and Mask;
00409159 22D8             and bl,al

I am pretty sure there is a faster way in assembler.

I am mostly looking for high level code Delphi code.

But it could still be interesting to see any faster low level assembler
code 
in case I want to write an assembler routine in the future.

Who can optimize it ? :):):)

I tried different high level techniques, I already tried one high level 
technique using a larger mask and ****fting the low bits into the high bits

and then use the high byte to immediatly mask.. reducing the not 
instruction... but delphi compiler generated/used memory access which is 
probably worse.

But when coding directly in assembler it's probably possible to prevent
the 
memory access ;)

I haven't tried yet in assembler because I leave that fun to you guys the 
assembler-fan-boys :)

Can you write something better than the delphi compiler ? ;)

Bye,
  Skybuck.