Understanding 555 timer in astable mode but stuck on capacitors

Question

So I am still plowing my way through Platt's Make: Electronics and into chapter 4 working with the 555 timer IC. The astable mode of the timer is giving me trouble as I don't understand how

When C1 acquires 2/3 of the positive voltage powering the circuit, the 555 reverses its output on Pin 3 from positive to negative and forces C1 to discharge itself through R2.

(This last sentence is actually from Make Magazine issue 10 where Platt is explaining the same thing as in chapter 4 of his book - see image below)

What I am confused about is the discharging through R2. I am obviously stuck on what is going on with C1. As I understand it, it first gains a positive charge on what I will call the top plate in the picture above, and a negative charge on the bottom plate (next to where the label C1 is). Maybe I have read too many analogies of water with electricity but I think of the capacitor as a balloon (in fact there are many analogies in Platt's book like this). So how is the balloon releasing the pressure in what seems like a reverse direction back through R2? Here is a similar circuit I found on circuitlab albeit the pins arranged differently and it shows that reversing of "flow":

Is there not +5V into R2 from R1 at the same time C1 is discharging??

Finally here is Platt's "internal" view of the 555 timer in astable mode. I assume pin 7 gets grounded to pin 3 but I am still confused about the seemingly reversal of flow through R2.

I've read through this many times and still stuck. Any additional help is appreciated!

Cheers

Message was edited by: Robert Opalko same day

jw0752 · Accepted Answer

Hi Robert, During the discharge part of the cycle the Pin 7 is attached to ground. (see the blue - in the last picture). Once pin 7 is connected to Ground the electricity in C1 starts to move through R2 and into the Ground. The grounding of Pin 7 does also connect R-1 to Ground but this has nothing to do with R2 and C1 at this time since there is a ground between them. The last picture also shows that Pin 3 is connected to ground at this time but here again it has nothing to do with R2 and C1 since there is a ground between them. See if this helps? Keep in mind that the ground that I drew outside of Pin 7 is actually inside the chip and it is temporary. As soon as the charge on C1 hits 1/3 of Vcc it once again goes open and allows the current coming from R1 to recharge C1 through R2. John

jw0752 · Answer

Hi Robert, The polarized capacitors must be used in the correct polarity as they will fail if more than a token voltage is applied in the reverse direction. The polarity on the capacitor is not like that of a diode. In the case of the polarized capacitor if you hook it up in the proper polarity it will work and if you hook it up in the incorrect polarity it will conduct current, get hot and potentially explode provided there is enough energy available in the circuit. Generally speaking the electrolytic capacitors have higher values of capacitance for their size than do the non-polarized types. There are other differences too but probably not important to this application. The number and variety of components that can be used to make any particular circuit application is what makes electronics an art. The 555 that you are working with is a great example. There are literally hundreds of different applications that it can be applied to. Check out: http://www.talkingelectronics.com/projects/50%20-%20555%20Circuits/50%20-%20555%20Circuits.html If there is ever any conflict between what I am telling you and what jc2048 is telling you your best bet is to listen to him. He is a real expert while i am just an enthusiast. John

fmilburn · Answer

This is one of those times when animated graphics might help. The 555 timer was not intuitive to me at first either from just the schematics and block diagrams. Karen on the Learning Circuit might be helpful. Also this video

jc2048 · Answer

When you charge a capacitor, the increasing imbalance of charge causes the voltage between the plates to increase and

the resulting electric field between the plates then stores energy. When you discharge the capacitor, that

energy is then powering the discharge circuit.

Personally, I don't much like bucket [water] and balloon [air] analogies. A capacitor isn't simply a

container that you to pour something in to. When you add charge to one plate, you always remove the same

charge from the other. It's the resulting imbalance between the two that stores energy and it can be either

way round (leading to a voltage in either direction across the capacitor). Negative amounts of water and

negative pressures are a bit weird, when you think about it. [Yes, the maths works with 'negative water'

too, but what's the point? You're then struggling to explain an analogy that's supposed to make things

easier to understand.]

When we charge or discharge a capacitor, that flow of charge constitutes a current. Current paths are loops

'driven' by voltage sources.

Here is the current path [red] when the capacitor charges

and here is what happens when the 555 connects pin 7 to ground [black] and the capacitor discharges

There are now two paths. The current through R1 [red] now flows back to the supply through the 555. At the

same time, the capacitor voltage drives a current through R2 [green] that returns to the other side of the

capacitor through the 555. That current discharges the capacitor. [As it discharges, the voltage across the

capacitor reduces until it reaches the lower threshold, at one third the supply voltage, at which point the

555 removes the connection from pin 7 to gound and the capacitor starts charging again.]

colporteur · Answer

Great response JW. I wanted to acknowledge your contributions. I have indicated in my past posts, the responses from members to question posted have value. Someone took the time to respond. This post is a perfect example demonstrating that. To individuals that have an understanding the response may seem trivial. To the learner, stuck in spot in theory that seems incongruent, small clarification such as what you have provided is immense in furthering their progress. I doff my chapeau to you, in acknowledgement!

jc2048 · Answer

I've just found a bag with some 555s in, so I thought I'd show you the voltage across C1 in the

circuit of 'Figure 2'.

I chose a supply voltage of 6V so that the lower and upper thresholds would be 2V and 4V (the

thresholds, unless you modify them, are 1/3 and 2/3 of the supply voltage), as that makes it

easy to consider on the oscilloscope trace.

You can see that the voltage across the capacitor always remains positive [the '1' channel

marker on the left of the grid shows where 'ground' [zero volts] is, and the vertical scale is 1V per

division], so it would be possible to use a polarised capacitor here.

When the capacitor is charging, the voltage goes up, and as it discharges, it goes down.

Jan Cumps · Answer

'negative water'

I'm not a big fan of the analogy either, but it helps to get the discussion started.

I try to explain negative water with a "very empty" bucket. Again not a good example, just a hook to get the discussion going without having the attendee's eyes glaze over or have them walk away after a few seconds.

jc2048 · Answer

so it would not matter in this example if the cacpacitor was polarized or non-polarized?

For this circuit, the voltage at the top plate of the capacitor always remains between the supply voltages, so it's always more positive that the end connected to ground and there's no problem using a polarised capacitor.

why do we use polarized caps instead if it doesn't matter?

If you want long timing intervals [slow oscillations], you need high values of capacitance, and electrolytics are the cheap way to do that.

Understanding 555 timer in astable mode but stuck on capacitors

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