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I am trying to analyse a 555 based timer circuit:

enter image description here

My current understanding of the circuit is that the output is changed by the voltage at pins 2/6 (let's call it V). If V drops below 1/3 Vcc, output is set high and if V gets greater than 2/3 Vcc, the output is set low and the discharge pin is open to ground.

The timing is provided by charging/discharging C3. The charging is done through the R1/RV2 pair and discharging through R2/RV1 pair.

Now, my questions related to this:

  • What is V at the time the circuit is connected to supply? To me, it seems like it's floating without a reference voltage, since C3 is empty (i.e. basically open circuit) and D2 does not conduct since OUT is low. (I am pretty sure I am wrong here, but I don't know why/where).

  • How to determine the period and the duty time? (I am looking more for a formula and some pointers to understanding its derivation)

  • The examples of 555-based timers I have seen around usually charge C3 from Vcc. In this case, it is charged from the output pin. Are there any advantages/disadvantages of this approach?

  • What is the role of D1?

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  • \$\begingroup\$ That's an odd little 555 circuit. Can you share with us it's purpose or the function it performs? If this is coursework, they could be using the 555 in an unconventional way to force a deeper understanding of it's function. It's an extremely versatile chip. \$\endgroup\$ – K H Oct 6 '18 at 0:09
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    \$\begingroup\$ I can tell you at a glance that D1 is there to prevent inductive kickback from the relay coil from damaging the rest of the circuit. \$\endgroup\$ – Hearth Oct 6 '18 at 0:13
  • \$\begingroup\$ @KH This is a circuit from a small electronics kit, velleman.eu/products/view/?id=339209. I'm just trying to understand how it works/the design decisions behind it and trying to come up with a way of setting more precise timings rather than empirically turning the potentiometer (which is 1M, not very easy to reliably set to some value). \$\endgroup\$ – Paul92 Oct 6 '18 at 0:18
  • \$\begingroup\$ By using the output pin instead of the discharge pin you get a 50% duty cycle, however this is a mixed topology to make you think about each cycle. \$\endgroup\$ – Sparky256 Oct 6 '18 at 0:20
  • \$\begingroup\$ The TLC555 is not a precision timer like a XTAL device is. Yes, you can trim it to 1.000000 KHZ with NPO grade capacitors and 0.1% resistors, but as the hours pass it will drift more than a XTAL will. \$\endgroup\$ – Sparky256 Oct 6 '18 at 0:24
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  • What is V at the time the circuit is connected to supply? To me, it seems like it's floating without a reference voltage, since C3 is empty (i.e. basically open circuit) and D2 does not conduct since OUT is low. (I am pretty sure I am wrong here, but I don't know why/where).

V is 0 volts. And no, it isn't open-circuit. A capacitor will "hold" 0V just as well as it holds any other voltage.

  • How to determine the period and the duty time? (I am looking more for a formula and some pointers to understanding its derivation)

In astable mode, V is cycling between 1/3 Vcc and 2/3 Vcc. It does this at a rate determined by both the capacitance and resistance that is applied during each phase of operation.

It can be shown that the exponential curve moves from 1/3 to 2/3 (or vice-versa) in a time that is equal to \$-\ln (0.5) \cdot R \cdot C\$, or 0.693×R×C, which is the formula used in the datasheet.

EDIT: Specifically, the voltage across the capacitor (during discharge) is $$v(t) = V_0 e^{-\frac{t}{RC}}$$ We want to know how long it takes to fall from 2/3 Vcc (\$V_0\$) to 1/3 Vcc. Therefore, we set the expression equal to 1/3 Vcc and solve for t: $$\frac{1}{3}V_{CC} = \frac{2}{3}V_{CC} \cdot e^{-\frac{t}{RC}}$$ Isolate the exponential: $$\frac{1}{2} = e^{-\frac{t}{RC}}$$ Take the logarithm of both sides: $$\ln \frac{1}{2} = -\frac{t}{RC}$$ And finally, isolate t: $$-\ln \frac{1}{2}\cdot R \cdot C = t$$ A similar formula applies when charging from 1/3 Vcc to 2/3 Vcc using a supply voltage of Vcc.

  • The examples of 555-based timers I have seen around usually charge C3 from Vcc. In this case, it is charged from the output pin. Are there any advantages/disadvantages of this approach?

This circuit topology allows the timer to operate with a duty cycle that is less than 50% — in this case, it can be MUCH less than 50%. A disadvantage is that the charging voltage is not exactly Vcc (and this gets worse if there's a significant load on the output), which throws off the timing calculation for the "on" time.

  • What is the role of D1?

The relay is an inductive load, which means that there's a significant voltage spike when you turn it off that could damage the output of the IC. D1 "shorts out" this spike to protect the rest of the circuit.

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  • \$\begingroup\$ Regarding the formula, do you have any pointers/resources understanding its derivation? When calculating the "on time", R is R1+RV2 and for the "off time", R is R2+RV1, is that correct? \$\endgroup\$ – Paul92 Oct 6 '18 at 1:35
  • \$\begingroup\$ The formula is derived from simple exponential decay. I'll edit into my answer. Yes, you have the correct resistor combinations. \$\endgroup\$ – Dave Tweed Oct 6 '18 at 1:42

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