Why would you expect it to work the same with or without diodes?
The diodes isolate the charge resistance from the discharge resistance so that the two can independently control their respective times.
Without the diodes, the two resistances interact such that the full duty-cycle range cannot be achieved.

I expect it to work the same because charge path and discharge path stay the same. Left side of the potentiometer for charging and right side for discharging. When charge resistance grows the discharge resistance decreases. I don't understand why you would need diodes and also that bridge between them.

 

Therein lies your problem. The basic 555-timer astable circuit works for 50-100% duty cycle.
For lower than 50% you need the diodes.

See:
https://electronicsclub.info/555astable.htm

https://www.electronics-tutorials.ws/waveforms/555-circuits-part-1.html

I need to check that out.

 

Left side of the potentiometer for charging and right side for discharging.

But the two sides are not isolated.
When discharging from the right side, current is still being supplied from the left side.

 

But the two sides are not isolated.
When discharging from the right side, current is still being supplied from the left side.

And? Same thing happens with the diodes.

 

And? Same thing happens with the diodes.

Not the way the diodes are configured.
Look at the circuit with the diodes.

 

Not the way the diodes are configured.
Look at the circuit with the diodes.

And what does that current supply have to do with discharging?When discharging happens Pin 7 is grounded so all the current goes to pin 7 anyway.

 

And what does that current supply have to do with discharging?When discharging happens Pin 7 is grounded so all the current goes to pin 7 anyway.

When charging, the capacitor sees the left part of the pot resistance.
When discharging, the capacitor seed the parallel value of the left and right pot resistances.
So the two resistance are not isolated from the capacitor during charge and discharge as they are with the diodes, thus reducing the duty-cycle range that can be achieved.

 

When charging, the capacitor sees the left part of the pot resistance.
When discharging, the capacitor seed the parallel value of the left and right pot resistances.
So the two resistance are not isolated from the capacitor during charge and discharge as they are with the diodes, thus reducing the duty-cycle range that can be achieved.

I think It is starting to make sense. If my picture is correct.

 

I think It is starting to make sense. If my picture is correct.

Yes.
The only charge path is through D1 and the only discharge path is through D2.

 

Yes.
The only charge path is through D1 and the only discharge path is through D2.

Yeah I knew that. I thought there is current being supplied from the left side of the potentiometer even when discharging. Just like what happens without the diodes. Then I realised that the current doesn't go there and instead takes the "zero" resistance route to the ground.

Thanks. That was quite confusing.