# 555 timer - demonstrating formulas

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
Hello..

I have an assignment with the 555 IC in the astable mode.

I'm trying to demonstrate the tHIGH and tLOW formulas.

tHIGH =ln (2)*(Ra+Rb)*C

tLOW = ln (2)*Rb*C

Now, for the tLOW I'm using the discharge capacitor equation:
τ = Rb*C

vC = V0*e^(-t/τ)
⇔(1/3)*Vcc = (2/3)*Vcc*e^(-t/τ)
⇔1/2 = e^(-t/τ)
⇔ln (1/2) = -t/τ
⇔t = ln (2)*Rb*C

Now for the tHIGH I'm using the charge equation formula: vC = V0*(1-e^(-t/τ)), with τ = Ra + Rb

vC = V0*(1-e^(-t/τ))
⇔(2/3)*Vcc = (1/3)*Vcc*(1-e^(-t/τ))
⇔2 = 1 - e^(-t/τ)
⇔ln (1) = -(-t/τ)

but this is not looking correct... ln (1) = 0...

Where am I going wrong?

#### RBR1317

Joined Nov 13, 2010
550
For the discharge, the capacitor discharges from an initial value of 2/3 down to zero. Find t when the capacitor reaches 1/3.

For the charge, the capacitor charges from an initial value of 1/3 up to 1. Find t when the capacitor reaches 2/3.

Your equation seems to be charging the capacitor from zero up to 1/3, and finding when the capacitor reaches 2/3.

#### Jony130

Joined Feb 17, 2009
5,176
For charging phase, the capacitor charges from an initial value of 1/3 up to 2/3 .
So to find a time of a charging phase, first you need to find "t" when the capacitor reaches 2/3 minus "t" when capacitor reaches 1/3.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
For the discharge, the capacitor discharges from an initial value of 2/3 down to zero. Find t when the capacitor reaches 1/3.

For the charge, the capacitor charges from an initial value of 1/3 up to 1. Find t when the capacitor reaches 2/3.

Your equation seems to be charging the capacitor from zero up to 1/3, and finding when the capacitor reaches 2/3.
The maximum capacitor voltage is 2/3 of Vcc and the minimum voltage is 1/3 of Vcc. I didn't understood why you say it discharges from 2/3 down to 0 V.

Also I'm not understanding why you say the capacitor charges from 1/3 to 1. It charges from 1/3 of Vcc up to 2/ of Vcc, if I'm not mistaken.

For charging phase, the capacitor charges from an initial value of 1/3 up to 2/3 .
So to find a time of a charging phase, first you need to find "t" when the capacitor reaches 2/3 minus "t" when capacitor reaches 1/3.
Is the formula I'm using not the general case? Because I'm having the idea that it is not the general formula for the charging phase!

I have already solved this problem because I found a different formula in my teacher's notes that says:

vC = Vcc - (Vcc - V0)*e^(-t/(τ))

I could solve the problem with this formula, but I couldn't quite understand what is the general formula for the capacitor charge phase and why this formula allows me to solve the problem!

#### Jony130

Joined Feb 17, 2009
5,176
The most common formula for charging phase in RC circuit looks like this:
Vc = Vcc*(1 - e^-t/RC) And this formula assumes that, the capacitor is discharged (Vc = 0V) at the beginning.
So to find time needed to charge the capacitor from 1/3Vcc to 2/3Vcc we are force to use the difference in time.
First we need to find T for 2/3Vcc minus the T for 1/3Vcc

Vc = Vcc*(1 - e^-t/RC) ---> Solve for T = RC•ln(Vcc/(Vcc - Vc))

T = RC•ln(Vcc/(Vcc - 2/3Vcc)) - RC•ln(Vcc/(Vcc - 1/3Vcc)) =

T = RC•ln[ (Vcc - 1/3Vcc) / (Vcc - 2/3Vcc) ] = RC•ln[ (1 - 1/3) / (1 - 2/3) ] =

= RC•ln[ (2/3) / (1/3) ] = RC•ln (2)

The general formula for capacitor charging phase looks like this
Vc = Vcc + (Vo - Vcc)*e^(-t/(τ)) and in this formula Vo includes the initial capacitor voltage.

Last edited:
• PsySc0rpi0n

#### MrAl

Joined Jun 17, 2014
7,656
Hi,

The maximum capacitor voltage is 2/3 of Vcc and the minimum voltage is 1/3 of Vcc. I didn't understood why you say it discharges from 2/3 down to 0 V.

Also I'm not understanding why you say the capacitor charges from 1/3 to 1. It charges from 1/3 of Vcc up to 2/ of Vcc, if I'm not mistaken.

Is the formula I'm using not the general case? Because I'm having the idea that it is not the general formula for the charging phase!

I have already solved this problem because I found a different formula in my teacher's notes that says:

vC = Vcc - (Vcc - V0)*e^(-t/(τ))

I could solve the problem with this formula, but I couldn't quite understand what is the general formula for the capacitor charge phase and why this formula allows me to solve the problem!

The difference is when there is an initial capacitor voltage you have to change the charge 'formula' to include that voltage because that changes the voltage across the cap at t=0. The discharge formula does not change because that already includes the initial voltage, which is the starting voltage.

The charge formula when there is no initial voltage (Vc=0 at t=0) looks like this:
Vc=Vs*(1-e^(-t/RC))

but as noted, this does not work when the cap has initial voltage at t=0.

To handle the initial cap voltage we have to compensate for that initial voltage. We can either use a shift in voltage level or a shift in time to accomplish this. There are other more direct methods but i assume you did not learn them yet.

To shift the voltage level we can imagine momentarily that the initial cap voltage is zero and ground is at the negative of the initial cap voltage. So we subtract it from the source voltage:
Vc=(Vs-Vc0)*(1-e^(-t/RC))

and now that we have that shifted formula, we must shift ground again to get back to the real circuit, so we add the cap voltage back to that:
Vc=(Vs-Vc0)*(1-e^(-t/RC))+Vc0

and that is the basic voltage shifted formula, which can be simplified if desired.

A second method is to shift time instead. We know that we need to solve for the charge time between 1/3 of Vs and 2/3 of Vs, so we know what the cap voltage is at t=0, it is 1/3 of Vs. Since we know what this cap voltage is and we know the more basic charge formula:
Vc=Vs*(1-e^(-t/RC))

we can solve that for t and thus we know how long it would take to charge to 1/3 of Vs. We will call this time T1 for short.

Knowing T1 now, to get the time to charge from 1/3 of Vs to 2/3 of Vs using the basic charge formula we shift the time by T1 so that the time 't' now only represents the time between those two levels, and not from 0 to 2/3 of Vs anymore. So first shift the time by T1:
T=t+T1

and then use the basic equation again:
Vc=Vs*(1-e^(-T/RC))

and solve this for t (not for T). The solution should again yield t=ln(2)*RC.

You can use the first method to solve, then use the second method to test your answer from using the first method. You should get the same result for both methods or you did something wrong If you like, you can also prove that a slow change in Vs does not alter the charge and discharge times.

Last edited:
• PsySc0rpi0n

#### RBR1317

Joined Nov 13, 2010
550
The maximum capacitor voltage is 2/3 of Vcc and the minimum voltage is 1/3 of Vcc. I didn't understood why you say it discharges from 2/3 down to 0 V. Also I'm not understanding why you say the capacitor charges from 1/3 to 1. It charges from 1/3 of Vcc up to 2/ of Vcc, if I'm not mistaken.
When looking at a capacitor charge (or discharge) equation there are two defining points: the initial value at t=0, and the final value at t=∞. The charge/discharge equations for the 555 timer don't know that at some time along the charge/discharge curve a switching point will be reached and the circuit will change. When the circuit changes, a new charge/discharge equation will take effect with the switching point being the new t=0.

Note that the value of the exponential term, e^(-t/τ), equals 1 at t=0 for the initial value, and equals 0 at t=∞ for the final value. In this 555 circuit the capacitor charges to Vcc and discharges to zero; however, it never reaches the final value because the circuit changes at the switching points. But the charge/discharge equations are written as though the capacitor will never reach a switching point in the time between t=0 and t=∞.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
Ok, I think I got it...

I'll try the time shifting method later to confirm my results that I got using the other formula.

Now I have another problem that I need help. I'm going to start a new thread!

#### WBahn

Joined Mar 31, 2012
25,899
The maximum capacitor voltage is 2/3 of Vcc and the minimum voltage is 1/3 of Vcc. I didn't understood why you say it discharges from 2/3 down to 0 V.

Also I'm not understanding why you say the capacitor charges from 1/3 to 1. It charges from 1/3 of Vcc up to 2/ of Vcc, if I'm not mistaken.
The RC charging/discharging circuit is not omniscient -- it only knows the final values it is moving toward based on the circuit configuration at that time. It has no idea that something is monitoring the voltage and will change the circuit at some point in the future. Hence, the capacitor is charging TOWARDS a final voltage of Vcc, but is stopped when it reaches 2/3 Vcc. Similarly, the capacitor is discharging TOWARDS a final value of 0 V, but is stopped when it reaches 1/3 Vcc.