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I've seen below 555 circuit which produces a square wave of 50% duty cycle. I am quite new to electronics and I saw Ben eater's 555 timer tutorial where he used discharge pin for capacitor discharging. One Resistor was very very low compared to Second resistor and created 50-ish% duty cycle. I've seen below circuit from Tutorial and In below circuit Charging and discharging is being done on output pin, when it's low capacitor is sinking into it and vice versa. It's okay but can anybody explain the working of R1 in below image? Article says that

Resistor R1 is used to ensure that the capacitor charges up fully to the same value as the supply voltage

enter image description here

But when capacitor voltage becomes > 2/3VCC, it will trigger capacitor discharging so capacitor will never reach full VCC. And also what if we do not use R1 at all? will is suppose to work?

Excuse my lack of knowledge on subject as I am new and also excuse my bad english.

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2 Answers 2

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Firstly (and probably not that important) is that your hyper-link is not going to where you probably intended so, if there's something really important here that folk need to know then, fix that link.

But when capacitor voltage becomes > 2/3VCC, it will trigger capacitor discharging so capacitor will never reach full VCC

Quite correct but, I think what the rather lazy statement is all about is that the capacitor charging voltage follows a trajectory that would eventually make the final charge voltage equal to the supply voltage: -

enter image description here

Image from here.

It's okay but can anybody explain the working of R1 in below image?

This is the main route by which charge enters the capacitor and this route is normal for most 555 circuits I've ever seen.

And also what if we do not use R1 at all?

It will still work according to this diagram (similar ones in many other places): -

enter image description here

Image taken from here.

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  • \$\begingroup\$ I've fixed the link. Thank you for the answer and very well explained. \$\endgroup\$
    – usmandroid
    Commented Dec 17, 2021 at 15:51
  • \$\begingroup\$ "This is the main route by which charge enters the capacitor and this route is normal for most 555 circuits I've ever seen." Disagree in this case. See my answer. \$\endgroup\$
    – AnalogKid
    Commented Dec 17, 2021 at 19:35
  • \$\begingroup\$ @AnalogKid it's just an observation on my part and not a hard fact on every 555 circuit ever produced. Be reasonable. \$\endgroup\$
    – Andy aka
    Commented Dec 17, 2021 at 19:41
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    \$\begingroup\$ Please forgive me. \$\endgroup\$
    – AnalogKid
    Commented Dec 17, 2021 at 23:16
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The rate at which a capacitor charges and discharges through a resistor depends on the voltage the other end of the resistor is connected to. The problem is this: while the two internal trip points are set to exactly 1/3 Vcc and 2/3 Vcc (+/- resistor tolerances), the 555 output stage does not swing to exactly GND and Vcc. Worse (and the real problem) is that how close it gets to each of the two rails is different. For a bipolar 555, the output can get down to within 1/2 volt of GND, but the high output might be over 1 volt away from Vcc. This introduces a small asymmetry into the output waveform.

There are two common solutions. First is to change to a CMOS 555, because its output stage is way more symmetrical in its output impedance and headroom. Second is to pee in a small charge-up current to make up for the lower charging voltage. That is what R1 is doing.

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