555 Timer RC Charging Capacitor Conundrum

AI Thread Summary
The discussion focuses on evaluating the power consumption of the external RC circuitry in a 555 timer during one cycle, particularly while the capacitor charges. The current decreases exponentially as the capacitor charges, and the voltage across the resistors is affected by this change. To find an average current reading between 1/3Vcc and 2/3Vcc, participants suggest using a piecewise function and calculating the average current based on the capacitor's voltage change over time. It is clarified that the power dissipated during charging and discharging is independent of the duty cycle. Additionally, the average current is influenced by the resistance ratio in the circuit, emphasizing the relationship between resistance, capacitance, and applied voltage.
MathsDude69
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Hello. I am currently trying to evaluate just how much power the external RC circuitry of a 555 timer uses over 1 cycle. Whilst the capacitor is charging the current falls exponentially. Naturally the voltage also falls through the resistors but this is couteracted by the rise in voltage of the capacitor as it charges. Basically what I am trying to find out is whether there is a way to get an average current reading between 1/3Vcc and 2/3Vcc using the equation
(4.5/101000 ⋅ e^(- x/0.4747)

(4.5 = Vcc)
(101000 is the total resistance in Ohms)
(0.4747 is the RC time constant)

Would the median value suffice?
 
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P=fC \Delta V^2
 
I don't mean to sound dense or anything but I've never tried this stuff before. The way I read your answer is:

Power = Frequency x Capacitance x Change in Voltage Squared

Is this right? And if so how does this tie in the the duty cycle?
 
LTSpice (and probably other simulation programs) have the NE555 in its library. Try it.
 
MathsDude69 said:
I don't mean to sound dense or anything but I've never tried this stuff before. The way I read your answer is:

Power = Frequency x Capacitance x Change in Voltage Squared

Is this right? And if so how does this tie in the the duty cycle?

Right. The power dissipated in charging and discharging the capacitor is independent of the duty cycle.

The current, at any time t, is obtained from constructing a piecewise continuous function from

v(t) = v_i \left( 1 - exp(-t/RC) \right)
and
v(t) = v_f \left( exp(-t/RC) \right)
where
i(t) = C \frac{dv}{dt} .

You might be better off using the average,

\overline{I} = C \frac{\Delta V}{\Delta T} .
 
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It is useful to review the astable NE555 circuit at http://www.daycounter.com/Calculators/NE555-Calculator.phtml
Scroll down to circuit model at bottom of page.The capacitor charges through the series resistance RA plus RB, but the charging time fraction (duty cycle) of charging time depends on the ratio (RA + RB)/(RA + 2 RB), so the average current depends on the ratio of RA/RB). Furthermore, although the frequency remains constant for given values of RA, RB, ad C, the average current increases linearly with the applied voltage VP.
 
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