# What is the diode for in the astable IC 555

I am trying to get my head into electronics, so my teacher challenged me to find out what this diode does. I have done a ton of experiments and i just cant put my finger on it... Everything from complete analysis with and without it.

Here is a picture of it i threw together quickly. I am kinda in a hurry so i had no time to open up EWB or even PS for that matter so sorry for bad drawing:

http://prntscr.com/iikdt9

• Welcome to EE.SE! This appears to be a homework question. As such, you need to show us your work so far, and explain which part of the question you're having trouble with. For future reference: Homework questions on EE.SE enjoy/suffer a special treatment. We don't provide complete answers, we only provide hints or Socratic questions, and only when you have demonstrated sufficient effort of your own. Otherwise, we would be doing you a disservice, and getting swamped by homework questions at the same time. See also here. – Dave Tweed Feb 23 '18 at 2:48
• If you edit your question, there is a circuit editor built into this site. Looks like a pencil and some components. Or Google it. How hard is that? 1st hit – StainlessSteelRat Feb 23 '18 at 3:01
• It is not a homework question. It is like a bonus question my teacher told me to do because he thinks im more advanced then the others (because my brother is a good electronic eng.). So this is basically out of the material we are suppose to learn in school. He told me to look up the IC 555 because i can make some cool stuff with it and practice. That is all. – KeyBasher Feb 23 '18 at 3:21
• School + teacher + question = homework. He challenges you. You come here. Ask question. Expect answer. How do you learn something by taking shortcuts? – StainlessSteelRat Feb 23 '18 at 3:27
• After a few days i came back at it with a clear head. SO i came to the conclusion that when the capacitor is being charged, the time to charge or T0 is smaller then the time it is discharging or TP. Which means that when it is charging, only R1 is active and R2 is not because the diode is pollarized directly. But when it is discharging into the circuit, we have voltage higher on the chatode and thus that part of the cct is not active and both R1 and R2 are in action causing the TP to go higher: s1: t = R1*C; s2: t = (R1 + R2)*C; – KeyBasher Feb 26 '18 at 22:48