Connect with us

How do I figure out these resistors?

Discussion in 'General Electronics Discussion' started by xgabrielx, Oct 19, 2015.

Scroll to continue with content
  1. xgabrielx

    xgabrielx

    6
    0
    Jan 1, 2010
    I was looking at 555 timer circuits. I can see what this does...it is supposed to turn the transistor on and off at about 10khz. But I haven't really looked into transistors yet. I'm kinda learning stuff either as I need it, or, like now, when I see something that raises a question. I guess the one closer to the IC is like the grid resistor on a vacuum tube? They haven't labelled them...what does the other one do? The one between the base and emitter? How would I calculate the values of both using this circuit as an example? Oh, and how critical is it? Surely there's a wide range of what works? I don't know how this will react as the battery voltage drops from the 9.8v or whatever of a fresh battery, but I'm just interested in the actual maths, not their specific circuit. I guess if they just use a value somewhere between a fully charged battery and a "dead" battery, everything will be within acceptable tolerances?
    I haven't looked into transistors much, but I thought to use them as a switch, you just needed a current flowing from base to emitter, and that as long as it was enough current for that tranmsistor, then collecter to emitter would simply function like an open switch. (or closed..can never remember which way round, because it's a bad choice of word) Did I miss some things? I guess I did, bceause I don't see a point in those resistors based on what I've read so far. Is there some relationship between the base to emitter current and the collector to emmitter current that must be kept within a certain range? Is that the point of the resistor on the left?
    Then, what's the one from base to emitter for? Is it some non-essential thing, or does it have to be a specific value?[​IMG]
     
  2. KJ6EAD

    KJ6EAD

    1,114
    157
    Aug 13, 2011
    I count 10 question marks. Would you care to focus on one?
     
  3. xgabrielx

    xgabrielx

    6
    0
    Jan 1, 2010
    Well, I guess it can be narrowed down to 2 questions. What do they do, and how do you calculate what values are required?
     
  4. Martaine2005

    Martaine2005

    2,775
    745
    May 12, 2015
    Hello,
    They are both voltage dividers of sorts.
    The 8k2 and 68k charge the capacitor. When charged the 68k discharges it.
    The component values used determine the frequency, ie flash rate. on/off rate.
    The datasheet explains the math involved Here.

    Martin
     
    Last edited: Oct 19, 2015
  5. xgabrielx

    xgabrielx

    6
    0
    Jan 1, 2010
    Oh, I know ;D I meant the two that don't have values next to them. Sorry, I should have been more clear about that
     
  6. Martaine2005

    Martaine2005

    2,775
    745
    May 12, 2015
    No, you were quite clear.:)
    But I cannot help with the math. I have a friend that helps me with that!!!
    The resistor from the base to 0v is a pull down resistor. It is to do with leakage or noise from collector to emitter. It basically keeps the transistor turned off .
    Would you believe I have edited this about 10 times!! . Because I don't wont to start a war!!
    Only because I can't win it.:D



    Martin
     
    Last edited: Oct 19, 2015
    Arouse1973 likes this.
  7. davenn

    davenn Moderator

    13,589
    1,872
    Sep 5, 2009
    LOL yup

    PLEASE desist from any argument/discussion on that topic in this thread. ... anything further like that and I will delete those posts

    cheers
    Dave
     
  8. Martaine2005

    Martaine2005

    2,775
    745
    May 12, 2015
    It never fails......Just different players...Lol.:)

    Martin
     
  9. hevans1944

    hevans1944 Hop - AC8NS

    4,475
    2,085
    Jun 21, 2012
    The 555 has a "totem-pole" output. See attached PDF. The output on pin 3 is either pulled up toward Vcc or pulled down toward ground. You can use just about any value resistor in the range of 1 kΩ to 10 kΩ to provide base current for the transistor that is switching the inductor on and off. The resistor from the base to the emitter is optional. You can safely omit it. If you want to leave it in, choose any value greater than 1 kΩ.
     

    Attached Files:

  10. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    heavans,
    How would a beginner learn if you try to "chew " it for him...?

    xgabrielx,
    If you want to learn, come back to this thread .
    there is much more on this issue...
     
  11. davenn

    davenn Moderator

    13,589
    1,872
    Sep 5, 2009
    OK a move of several posts to a new thread and a deletion of a bunch of others

    KEEP THIS THREAD ON TOPIC
    please just discuss the purpose and values of the 2 specified resistors

    Dave
     
  12. GPG

    GPG

    452
    66
    Sep 18, 2015
    The 555 will pull down much better than up So the b-e resistor is immaterial. The resistor from pin3 should saturate the transistor. If you calculate it based on a low beta this will do. The idea is called forced beta
    http://whites.sdsmt.edu/classes/ee320/notes/320Lecture24.pdf
    In your case the relay is 38Ω and the supply 9V so relay current = ~240 mA
    Forcing beta to say 20 = Ib~ 20 mA
    Under worst case the 555 Vout will be ~ V+ -1,5 and Vbe = .6v so the voltage available to drive 20mA is~7V
    So 7V/ 20mA =~350Ω Nearest E12, 380Ω And that is being conservative.
     
    Last edited: Oct 25, 2015
  13. eetech00

    eetech00

    95
    8
    Nov 17, 2014
    Basically, you need to know:
    Base input voltage
    Relay drive current
    Gain of transistor

    Hi..

    Example:
    Vb = 9v <--base input voltage (same as (Vcc) supply voltage)
    Ic = 236mA <--Collector current (current to drive relay = 9v/38 = 236mA (estimate))
    hfe = 100 <--Transistor Gain = 100
    Vbe = 0.7v <-- Voltage drop across Transistor BE junction

    How much base current to pass 236mA through collector?:
    Ib = Ic/hfe = 0.236/100 = 0.00236
    But leave headroom, so instead, use 300mA for Ic
    Ib = Ic/hfe = 0.300/100 = 0.003

    Base Resistor:
    Rb = (Vb-(0.7v))/Ib = 9v-(0.7v)/0.003 = 2767
    Use nearest value = 2.7k

    Estimate Resistor from base to ground
    10 * Rb = 27670 use 27k

    Make sure Tip41c is on heatsink.:cool:
     
  14. GPG

    GPG

    452
    66
    Sep 18, 2015
    .Correcting error:
    Forcing beta to say 20 = Ib~12 mA
    Under worst case the 555 Vout will be ~ V+ -1,5 and Vbe = .6v so the voltage available to drive .12mA is~7V
    So 7V/ 12mA =~583Ω Nearest E12, 560Ω.
    https://www.fairchildsemi.com/datasheets/TI/TIP41C.pdf
    beta may be as low as 30
    Using a logic level fet makes things easier.
     
  15. dorke

    dorke

    2,342
    665
    Jun 20, 2015

    Taking that circuit as a theoretical drill only.

    Correction:
    For the relay to be on, we need the Tr. in saturation.
    (deep saturation that is ,not just on the verge of it)

    The Beta we need is the minimum one (take into account temp,IC current, etc.)
    for TIP41C,Beta would be 15.
    BTW, this Tr. is an overkill for driving that relay.

    VBE sat=0.8V
    VCEsat=0.2V

    IC=8.8/38=231ma
    IB=231/15=15ma

    So design for:
    IB=at least 20ma.
    VBEsat=0.8V

    and that TR. doesn't need an heat sink at all!
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-