Hello i'm new at this

Discussion in 'General Electronics Chat' started by ukiceman, Nov 28, 2009.

  1. ukiceman

    Thread Starter New Member

    Nov 22, 2009
    9
    0
    Hi Guys

    Firstly I’d like to say, what a great site :)
    As you can see this is my first posting so, I’ll jump right in with my question.

    Could somebody have a look at my graphic and tell me if I understand this properly? I’ve only been studying electronics for the last three months and, at 45 years old it’s slow going, I always get stuck with the maths.

    Regards
    Shaun
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Looks good for me. But remember that in saturation Ic=β*Ib don't hold any more.
    So RB_max=82K but in practical circuits we use RB smaller then 82K.
    3 to 10 times smallre.
    So R1=10K looks good
     
  3. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    As a general rule of thumb you need 1/10 the base current that you will get on a saturated collector current. It will work at smaller base currents, but if you want 100% use this rule.

    Something I picked up on this site.
     
  4. ukiceman

    Thread Starter New Member

    Nov 22, 2009
    9
    0
    I did ask myself why most examples I looked at used 10k on the base, yet applying the formulas in the examples I got 82k? Could you explain without getting deep why it doesn’t work? Now I’m even more confused :confused:

    Regards
    Shaun
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Hmm, Hfe is low when Vce is low,so the spec value for hfe will not work for saturation calculations.
    And that is why we must use the lower value for RB than we calculated.
    But if you select BJT with beta larger then hfe_min, your circuit will be work just fine with Rb=82K.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Shaun,
    You generally just use the hFE specification(s) when you are using the transistor as a linear amplifier.

    But in your application, you are using the transistor as a saturated switch. In the case of bjt's, (bipolar junction transistors; NPN, PNP) "saturation" means that an increase in base current (Ib) will have little or no effect on the collector current (Ic). This is because when the transistor is well saturated, Vce is very low; perhaps 200mV or less.

    When you are using a transistor as a saturated switch, power dissipation in the transistor is usually pretty low.

    2N3904 transistors have a maximum Ic rating of 200mA, but they have a practical limit of about 100mA. If you go much over that, they'll get mighty warm.

    Here is a plot from ONsemi's 2N3904 datasheet (page 6) for collector saturation:
    [​IMG]

    Vce is the Y axis, Ib is the X axis. Note that Vce is lowest when Ib=Ic/10. You can also get the idea that increasing Ib would not really help much in lowering Vce.

    2N2222 transistors have a max Ic of 800mA, but are best to keep at 1/2 that or less.
     
  7. ukiceman

    Thread Starter New Member

    Nov 22, 2009
    9
    0
    Thank-you for the replies everybody, I’ve replaced the offending 82k with a 10k and, nothing burned out so, that’s good. I’ll try digesting your posts tomorrow as I’ve been at it all day. Transistors used as logic gates is my next lesson :eek:

    Regards
    Shaun
     
Loading...