Here's another more complicated schematic. What is the purpose of each diode in this circuit ?
View attachment 189657

Without D1 and D2, we charge C1 with R1+R2 (none can be zero, since then we would have a short at charge or at discharge) and discharge C1 through only R2. Adding D1 allows us to "bypass" R2 when we charge C1, so only R1 will govern the RC constant for C1. But we decrease the allowable voltage by V_D1. D2 may be then required for some ratio of t1/t2 (if I remember correctly).

 

Without D1 and D2, we charge C1 with R1+R2 (none can be zero, since then we would have a short at charge or at discharge) and discharge C1 through only R2. Adding D1 allows us to "bypass" R2 when we charge C1, so only R1 will govern the RC constant for C1. But we decrease the allowable voltage by V_D1. D2 may be then required for some ratio of t1/t2 (if I remember correctly).

This question was asked in post b#6, and answered in posts 8 and 9

 

I used my circuit simulation software - Circuit Wizard to see what was happening in the second 555 circuit. I didn't know that current can and does flow into the discharge pin from the capacitor. I thought that current only flowed from the discharge pin to ground. Thankfully, I monitored the circuit in said software and It clarified a few things and I've learned some things as well.

Thanks guys.

 

(...) Thankfully, I monitored the circuit in said software and It clarified a few things and I've learned some things as well.

It may be obvious for many (in particular, for those who were already knowing the answer), but as for WHY to use D1 (instead of answering about HOW the circuit works) is that without D1, the charging time will be proportional to R1+R2 and the discharging time to R2. And as R1 cannot be negative, that would make impossible to have a charging time lesser than the discharging time and so, impossible to have the (almost) full range from 0 to 100% for the duty cycle for the output, as 50% will become one of the limit ( using R2 as close as 0 as possible). The diode D1 make possible to get almost the full range from 0 to 100%.
See Wikipedia, among other places, to get the theoretical timing when using the diode D1.

 

the charging time will be proportional to R1+R2

No, the current only flows through R1, because D2 will stop the current flows through R2, please check the circuit in post #6 again.

Edit : What you said was that the D1 should in series with a resistor, and the present R1 is the same, but R2 should be labels as R3 and the resistor in series with D1 should be labels as R2, so the Charging time is R1+R2+C1, that is a normal circuit, but in this thread was no R2.

 

No, the current only flows through R1, because D2 will stop the current flows through R2, please check the circuit in post #6 again.

Edit : What you said was that the D1 should in series with a resistor, and the present R1 is the same, but R2 should be labels as R3 and the resistor in series with D1 should be labels as R2, so the Charging time is R1+R2+C1, that is a normal circuit, but in this thread was no R2.

I thought that I clearly stated in my explanation about the WHY someone would use D1, in post #6, started by a case with NO D1 and NO D2 present. That seems to be a source of confusion. So, again, WITHOUT D1 (and D2), the charging time is proportional to R1 + R2 and the discharging time, proportional to R2, making impossible to have a duty cycle in the full range of 0% to 100%. Thus, adding D1 give a possible duty cycle for (almost) the said full range.

Note that the charging time is proportional to (R1+R2) * C1, so IT IS proportional to R1+R2 since C1 is also a constant. Again, for a more "precise" theoretical formula, without and with the diode, you can take a look at https://en.wikipedia.org/wiki/555_timer_IC

I am really confuse about your sentence telling that there is no R2 ... in post #6. It seems that there are two different discussions here. I always considered the case of post #6 in my interventions, here.

 

@vanderghast
About the circuit without D1 and D2, that is a standard 555 oscillator circuit, I didn't mention that, because I was assuming that the TS already knew it, so when you mentioned that then I can't say that you were wrong, but to me it is talking too much.

I thought that you may also using the way as mine, but probably you didn't, the circuit shown as Fig-01 is the same with the TS, and that was also shown in the TI's datasheet, and I using that very less, you can see the R1, R2 and R3 in the circuit Fig-02 comparing to Fig-01 and it will explain to what your confusion, the R1 in the Fig-01 can be set 1K~1M, so I set it as 1K in the Fig-02, Fig-03, Fig-04, that is to avoid R1 less then 1K.