BradtheRod this not seem to work in all frequencies...... Isnt it??
Adding a battery is a method that could work at low frequencies. However at high frequencies it is likely to distort the waveform.
Screenshot of the simulation (theoretical):
Notice the output goes below 0V.
Also notice the battery receives a charging current of a few mA when the input is -4V. Some types of batteries cannot tolerate this.
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I modified the above by putting the diode in series.
This schematic could do the job:
The output will not reach 5V if:
a) the diode has too high a forward volt spec,
or
b) the following stage draws too much current.
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Yes, the resistors create a virtual ground.
If the load draws too much current, the output level will drop. This will require raising the supply V at the 555 IC.
There will be a question about whether the output is referenced correctly to the next stage's ground rail.