# IC Timer 555 Astable Multivibrator 50% Duty cycle

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

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?

• That circuit won't even oscillate. – EM Fields Nov 12 '16 at 5:16

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.

• 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. – Ramit Sawhney Nov 11 '16 at 21:20
• Yes, the ln term is wrong. Please edit your attempt into the question (add it at the end) and try to format it properly. – Spehro Pefhany Nov 11 '16 at 21:23
• I've formatted it by learning how much ever mathjax I could, please help me out. – Ramit Sawhney Nov 11 '16 at 21:56
• 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! – Ramit Sawhney Nov 11 '16 at 23:53