# 555 Timer ASAP need to do my presentation

Discussion in 'Homework Help' started by Aikawa Aikun, Sep 21, 2016.

1. ### Aikawa Aikun Thread Starter New Member

Sep 21, 2016
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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.[/url][/IMG]
http://imgur.com/a/uGVq4

Jun 4, 2014
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3. ### Alec_t AAC Fanatic!

Sep 17, 2013
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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!

Jul 11, 2016
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5. ### Aikawa Aikun Thread Starter New Member

Sep 21, 2016
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-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

6. ### wayneh Expert

Sep 9, 2010
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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.

Aikawa Aikun likes this.
7. ### Alec_t AAC Fanatic!

Sep 17, 2013
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Nearly right. The diode is not a true short-circuit.

8. ### Aikawa Aikun Thread Starter New Member

Sep 21, 2016
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what can i say about the diode?
can i say like this?
"the charging will bypass R2 and charge through R1 ?"

9. ### wayneh Expert

Sep 9, 2010
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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.

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10. ### Aikawa Aikun Thread Starter New Member

Sep 21, 2016
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i get it.. thank

11. ### Aikawa Aikun Thread Starter New Member

Sep 21, 2016
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thank all for helping me