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. enter image description here

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?

  • \$\begingroup\$ That circuit won't even oscillate. \$\endgroup\$
    – EM Fields
    Nov 12, 2016 at 5:16

1 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.

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.

  • \$\begingroup\$ What I'm getting is, Voltage across capacitor(t)= [Vcc/RaC + 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)) Is this correct? I feel the log term is wrong. \$\endgroup\$ Nov 11, 2016 at 21:20
  • \$\begingroup\$ Yes, the ln term is wrong. Please edit your attempt into the question (add it at the end) and try to format it properly. \$\endgroup\$ Nov 11, 2016 at 21:23
  • \$\begingroup\$ I've formatted it by learning how much ever mathjax I could, please help me out. \$\endgroup\$ Nov 11, 2016 at 21:56
  • \$\begingroup\$ Thankyou! I solved it completely in the time domain and got the same as well. However, if I take the laplace transform and then it's inverse, I'm getting something vague. Is the change \$ CdVc/dt\$ <=> \$ C*(sVc-\frac{Vcc}{3}) \$ incorrect? I'm just trying to figure out what I did wrong because maybe I'm unclear with a concept! \$\endgroup\$ Nov 11, 2016 at 23:53

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