With the diode in place, the output waveform's duty cycle may be adjusted to less than 50% if desired.
Explain why the diode is necessary for that capability.
Show the current flow for charging and discharging of capacitor, C1.

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http://imgur.com/a/uGVq4

 

Does this help:http://www.doctronics.co.uk/555.htm#more_astables

 

Welcome to AAC!
Since this is your homework you are expected to show your best effort so far. We can then point you in the right direction. We don't do your homework for you!

 

best effort so far

 

my answer
-the capactor will charge through R1 only because the R2 is shorted out by Diode. because of that, less resistance when charging than discharging.
-current flow for charging is from R1 to diode. current flow for discharging is through R2

is my answer true?
sorry for ask the answer before trying and sorry because my english is bad.
it my bad.

 

That's correct. Just be aware that the diode drops ~0.7V across itself when conducting. This fact does not affect the qualitative answer, but would be important if you try to calculate the timing.

 

-the capactor will charge through R1 only because the R2 is shorted out by Diode. because of that, less resistance when charging than discharging.

Nearly right. The diode is not a true short-circuit.

 

Nearly right. The diode is not a true short-circuit.

what can i say about the diode?
can i say like this?
"the charging will bypass R2 and charge through R1 ?"

 

The diode will conduct when/if the voltage drop across R2 exceeds the diode drop I mentioned earlier. This means some current is still passing through R2, because there is a voltage across it.

 

The diode will conduct when/if the voltage drop across R2 exceeds the diode drop I mentioned earlier. This means some current is still passing through R2, because there is a voltage across it.

i get it.. thank

 

thank all for helping me