IC Timer 555 Astable Multivibrator 50% Duty cycle

Question

I'm given this configuration to solve for a IC-Timer 555 in astable mode to get a 50% duty cycle however I'm having some difficulties in the calculations.

The circuit is modified a bit as shown.

^{simulate this circuit – Schematic created using CircuitLab}

So during a HIGH output, Discharge is open and the capacitor charges in time 0.69R2*C2.

However, during LOW output, R1 is also grounded. In this case I'm having difficulty calculating the time period. If I could get some help in calculating the Tlow and thus proving that the duty cycle is 50%, that would be a big help.

$$What I'm getting is, Voltage across capacitor(t)=[Vcc/Ra*C + 2Vcc/3]e^-(1/C(R1||R2)

Now, a t=Tlow I should have Voltage across the capacitor as Vcc/3. On solving this I get Tlow = (R1||R2)*C *ln((3+2R1*C)/R1*C))

My attempt at the solution:

Applying KCL at the common node:

\$(Vcc-Vc)/R2=Vc/Rb + C\$*dVc/dt\$\$

Taking Laplace transform (Initial condition: \$ Vc=2Vcc/3\$ and simplifying:

\$ Vcc[1/R1 + 2C/3] = Vc *[1/R1 + 1/R2 + sC]\$

on Rearranging: \$ \frac{Vcc(\frac{1}{R1} + \frac{2C}{3})}{(C*[\frac{1}{R1*C}+\frac{1}{R2*C}+s])}\$ = Vc

Taking inverse laplace transform, putting Vc=Vcc/3 for t=Tlow, I get the above answer. Please help me out.

I'm getting the correct answer by solving only in time domain, while I'm satisfied with that, I'd like to know what possible mistake am I making in the method using laplace transform?

Answer 1

Calculate the Thévenin equivalent voltage. The equivalent resistance is R1||R2 when /DISCHARGE is low.

The duty cycle will only be 50% for a certain ratio of R1/R2.

You will then know the starting voltage, and the end voltage (2/3 and 1/3 Vcc respectively) and you will know the Thévenin equivalent voltage and resistance and you can calculate what ratio is required to get a 50% duty cycle.

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Okay, so the capacitor charges from \$V_I\$ to \$V_F\$ with time constant \$\tau = C2 \cdot R1||R2\$

If we have the /DISCHARGE pin go low at t = 0 we have:

\$v(t) = V_I + (V_F-V_I)\cdot (1-e^{t/\tau})\$

Substituting you get

\$t_D = \tau\ln(\frac{R1-2R2}{2R1-R2})\$

Which results in a nonlinear equation such that R1/R2 ~= 0.423 for 50% duty cycle.