- #1

MathsDude69

- 26

- 0

(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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- Thread starter MathsDude69
- Start date

- #1

MathsDude69

- 26

- 0

(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?

- #2

Phrak

- 4,265

- 2

[tex]P=fC \Delta V^2[/tex]

- #3

MathsDude69

- 26

- 0

Power = Frequency x Capacitance x Change in Voltage Squared

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

- #4

Bob S

- 4,662

- 6

LTSpice (and probably other simulation programs) have the NE555 in its library. Try it.

- #5

Phrak

- 4,265

- 2

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 disipated 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

[tex]v(t) = v_i \left( 1 - exp(-t/RC) \right)[/tex]

and

[tex]v(t) = v_f \left( exp(-t/RC) \right)[/tex]

where

[tex]i(t) = C \frac{dv}{dt} .[/tex]

You might be better off using the average,

[tex]\overline{I} = C \frac{\Delta V}{\Delta T} .[/tex]

Last edited:

- #6

Bob S

- 4,662

- 6

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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