element14 Community
element14 Community
    Register Log In
  • Site
  • Search
  • Log In Register
  • About Us
  • Community Hub
    Community Hub
    • What's New on element14
    • Feedback and Support
    • Benefits of Membership
    • Personal Blogs
    • Members Area
    • Achievement Levels
  • Learn
    Learn
    • Ask an Expert
    • eBooks
    • element14 presents
    • Learning Center
    • Tech Spotlight
    • STEM Academy
    • Webinars, Training and Events
    • Learning Groups
  • Technologies
    Technologies
    • 3D Printing
    • FPGA
    • Industrial Automation
    • Internet of Things
    • Power & Energy
    • Sensors
    • Technology Groups
  • Challenges & Projects
    Challenges & Projects
    • Design Challenges
    • element14 presents Projects
    • Project14
    • Arduino Projects
    • Raspberry Pi Projects
    • Project Groups
  • Products
    Products
    • Arduino
    • Avnet Boards Community
    • Dev Tools
    • Manufacturers
    • Multicomp Pro
    • Product Groups
    • Raspberry Pi
    • RoadTests & Reviews
  • Store
    Store
    • Visit Your Store
    • Choose another store...
      • Europe
      •  Austria (German)
      •  Belgium (Dutch, French)
      •  Bulgaria (Bulgarian)
      •  Czech Republic (Czech)
      •  Denmark (Danish)
      •  Estonia (Estonian)
      •  Finland (Finnish)
      •  France (French)
      •  Germany (German)
      •  Hungary (Hungarian)
      •  Ireland
      •  Israel
      •  Italy (Italian)
      •  Latvia (Latvian)
      •  
      •  Lithuania (Lithuanian)
      •  Netherlands (Dutch)
      •  Norway (Norwegian)
      •  Poland (Polish)
      •  Portugal (Portuguese)
      •  Romania (Romanian)
      •  Russia (Russian)
      •  Slovakia (Slovak)
      •  Slovenia (Slovenian)
      •  Spain (Spanish)
      •  Sweden (Swedish)
      •  Switzerland(German, French)
      •  Turkey (Turkish)
      •  United Kingdom
      • Asia Pacific
      •  Australia
      •  China
      •  Hong Kong
      •  India
      •  Korea (Korean)
      •  Malaysia
      •  New Zealand
      •  Philippines
      •  Singapore
      •  Taiwan
      •  Thailand (Thai)
      • Americas
      •  Brazil (Portuguese)
      •  Canada
      •  Mexico (Spanish)
      •  United States
      Can't find the country/region you're looking for? Visit our export site or find a local distributor.
  • Translate
  • Profile
  • Settings
555 Timers
  • Technologies
  • More
555 Timers
Forum Understanding 555 timer in astable mode but stuck on capacitors
  • Blog
  • Forum
  • Documents
  • Polls
  • Files
  • Members
  • Mentions
  • Sub-Groups
  • Tags
  • More
  • Cancel
  • New
Join 555 Timers to participate - click to join for free!
Actions
  • Share
  • More
  • Cancel
Forum Thread Details
  • State Verified Answer
  • Replies 14 replies
  • Answers 7 answers
  • Subscribers 53 subscribers
  • Views 8338 views
  • Users 0 members are here
  • capacitor
  • 555 timer
Related

Understanding 555 timer in astable mode but stuck on capacitors

opalko
opalko over 4 years ago

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)

 

image

 

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":

 

image

 

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.

 

image

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

 

Cheers

 

Message was edited by: Robert Opalko same day

  • Sign in to reply
  • Cancel

Top Replies

  • jw0752
    jw0752 over 4 years ago +6 verified
    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…
  • jw0752
    jw0752 over 4 years ago in reply to opalko +6 suggested
    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…
  • jw0752
    jw0752 over 4 years ago +5
    PS you are doing an excellent job of presenting your questions. I would bet that several times you have actually answered them yourself in the process of preparing the question. John
  • jw0752
    0 jw0752 over 4 years ago

    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.

     

    image

     

    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

    • Cancel
    • Vote Up +6 Vote Down
    • Sign in to reply
    • Verify Answer
    • Reject Answer
    • Cancel
  • jw0752
    0 jw0752 over 4 years ago

    PS you are doing an excellent job of presenting your questions. I would bet that several times you have actually answered them yourself in the process of preparing the question.

     

    John

    • Cancel
    • Vote Up +5 Vote Down
    • Sign in to reply
    • Verify Answer
    • Cancel
  • fmilburn
    0 fmilburn over 4 years ago

    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 

    • Cancel
    • Vote Up +5 Vote Down
    • Sign in to reply
    • Verify Answer
    • Reject Answer
    • Cancel
  • jc2048
    0 jc2048 over 4 years ago

    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

     

    image

     

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

     

    image

     

    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.]

    • Cancel
    • Vote Up +5 Vote Down
    • Sign in to reply
    • Verify Answer
    • Reject Answer
    • Cancel
  • opalko
    0 opalko over 4 years ago in reply to jc2048

    Thank you for this, so it would not matter in this example if the cacpacitor was polarized or non-polarized?  I guess that is the part I am struggling with.  I think I was thinking of a capacitor like a diode with the charge only flowing in one direction but this doesn't appear to be the case.  And also, why do we use polarized caps instead if it doesn't matter? 
    Thank you for the reply!
    Cheers

    • Cancel
    • Vote Up +3 Vote Down
    • Sign in to reply
    • Verify Answer
    • Cancel
  • opalko
    0 opalko over 4 years ago in reply to jw0752

    Thank you that makes it more obvious now with your diagram.  I should have realized R1 would flow the same way into ground.  See my question below to Jon about polarized vs non-polarized caps..would that matter? 

     

    I wish I could mark 2 answers as correct here, both are such good explanations.


    Cheers

    • Cancel
    • Vote Up +2 Vote Down
    • Sign in to reply
    • Verify Answer
    • Cancel
  • jw0752
    0 jw0752 over 4 years ago in reply to opalko

    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

    • Cancel
    • Vote Up +6 Vote Down
    • Sign in to reply
    • Verify Answer
    • Reject Answer
    • Cancel
  • Jan Cumps
    0 Jan Cumps over 4 years ago in reply to jc2048
    '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.

    • Cancel
    • Vote Up +3 Vote Down
    • Sign in to reply
    • Verify Answer
    • Reject Answer
    • Cancel
  • colporteur
    0 colporteur over 4 years ago in reply to jw0752

    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!

    • Cancel
    • Vote Up +4 Vote Down
    • Sign in to reply
    • Verify Answer
    • Reject Answer
    • Cancel
  • jc2048
    0 jc2048 over 4 years ago in reply to opalko

    I wish I could mark 2 answers as correct here, both are such good explanations.

    Give it to John: he got to the finishing post first. I was just adding a little more detail to the answer he'd already given.

    • Cancel
    • Vote Up +2 Vote Down
    • Sign in to reply
    • Verify Answer
    • Cancel
>
element14 Community

element14 is the first online community specifically for engineers. Connect with your peers and get expert answers to your questions.

  • Members
  • Learn
  • Technologies
  • Challenges & Projects
  • Products
  • Store
  • About Us
  • Feedback & Support
  • FAQs
  • Terms of Use
  • Privacy Policy
  • Legal and Copyright Notices
  • Sitemap
  • Cookies

An Avnet Company © 2025 Premier Farnell Limited. All Rights Reserved.

Premier Farnell Ltd, registered in England and Wales (no 00876412), registered office: Farnell House, Forge Lane, Leeds LS12 2NE.

ICP 备案号 10220084.

Follow element14

  • X
  • Facebook
  • linkedin
  • YouTube