Can I use 2N3906 Tranasistor for this?

Discussion in 'The Projects Forum' started by JDR04, Apr 11, 2013.

  1. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    I've attached the circuit in question.

    My objective is to turn on a fan when the temperature reaches a set limit.

    The voltage on pin 1 drops from 12V to 3.2V when the set limit is exceeded.

    Can I use the 2N3906 (PNP) transistor to activate a small fan (12V/240mA)?

    I would appreciate somebody telling me if my approach is correct or where I have gone wrong.

    Required current for fan is 240mA or 0.240A.
    Beta of transistor is 300. (not sure If I’m reading the datasheet correctly here)
    Voltage at pin one when limit is exceeded is 3.2V.

    Calculation for transistor turn on amperage is calculated as;

    0.240A divided by Beta 300 = 0.0008. So 1mA is required at base of transistor to switch fan on.

    3.2V - 0.7V = 2.5V.

    So, base current is calculated as; 2.5 is divided by 0.001 = 2500 or 2.5K
     
  2. takao21203

    Distinguished Member

    Apr 28, 2012
    3,577
    463
    They dont have a hFE of 300.
    You need more base current.

    It is not doing damage if you use higher base current.
     
  3. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Thanks, could you tell me what the hFE is then? Was my method for the base current calculation correct?

    Thanks JDR04
     
  4. tindel

    Active Member

    Sep 16, 2012
    568
    193
    Good work - you've got the right idea.

    If you are wanting to use the transistor as a switch (saturating the transistor) you must drive more base current. The datasheet probably says that the saturation voltage is 0.2V at a forced beta of 10... so use a beta of 10 instead. This will guarantee that the transistor is driven to saturation. The reason you use a beta of 10 is because the gain in transistors varies greatly even within the same lot. Assuming the beta is very low will ensure that you drive the transistor on.

    Also, the current will be (Vsource - Vpin1 - Vbe)/R... so your calculation was off just a hair.

    I calculate that you should use a base resistor of about 325ohms with a beta of 10.
     
  5. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Thanks tindel, that kind of explains it a bit more. So when I'm looking for the beta using the 2N3906 datasheet what am I looking for when it does not say beta?

    Could you also explain how you derived the beta 10 please.

    Thanks for your time. JDR04
     
  6. MrChips

    Moderator

    Oct 2, 2009
    12,425
    3,359
    While beta and hFE are not exactly the same thing, for all intents and purposes for general DC circuit analysis we use the same figures for beta and hFE.

    beta = 100 or even as high as 300 is not unusual.

    When the transistor reaches saturation mode beta is reduced.

    beta = 10 is not something one can derive. This is simply a good rule of thumb when you wish to drive the transistor into saturation mode.

    http://tymkrs.tumblr.com/post/13115454729/beta-hfe-gain-2n3904s-i-think-its-transistor
     
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  7. tindel

    Active Member

    Sep 16, 2012
    568
    193
    Sure -

    Looking at this datasheet: http://www.fairchildsemi.com/ds/2N/2N3906.pdf

    Page 2 has a Vce(sat) spec that is .4V when Ib=-5mA and Ic=-50mA. Ic/Ib=beta=10

    Note that the maximum Vc is 200mA (sorry I didn't catch this with my earlier post) so it won't drive 250mA to your motor without exploding, as I had assumed. You will need a different transistor. You could run two or three in parallel, just be aware that they won't share current evenly.

    I'd personally opt for a signal transistor that can handle a larger Ic...
    This would work: http://www.newark.com/on-semiconductor/mpsa92rlrmg/bipolar-transistor-pnp-300v-to/dp/09R9533 They are on sale for less than 5 cents each...
     
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  8. WBahn

    Moderator

    Mar 31, 2012
    17,716
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    Does your voltage drop smoothly from 12V to 3.2V (and below) as the temperature changes? If so, then the solution you have been discussing will gradually apply power to the fan and it will almost certainly reach full speed well before your set point is reached. This may be acceptable (and you can make adjustments to get the fan to speed up at a desireable point), but the two things to consider are that the fan may not like being teased this way and the transistor power may be excessive at some point because you've got both Ic and Vce being significant at the same time.

    An alternative (if you don't want to mess with opamps and the like) is to use a couple of transistors to make a simple comparator and driver circuit. You should be able to make a nice little circuit using a cheap transistor array IC , a discrete output transistor, and a few resistors.
     
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  9. crutschow

    Expert

    Mar 14, 2008
    12,993
    3,227
    Beta or HFE is used for bias and AC calculations with a BJT. For switching use, a value of 10 is commonly used to insure that the transistor is fully on, independent of whatever the actual value of beta is (since it varies widely from unit to unit).
     
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  10. WBahn

    Moderator

    Mar 31, 2012
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    And another reason for using it is that you often want to drive the transistor in very hard saturation, which results in low beta values, because you want to drive the Vce to very low values so as to limit the power dissipation in the transistor.
     
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  11. #12

    Expert

    Nov 30, 2010
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    The important point about gain at saturation is that as the voltage across the transistor gets very low, the gain becomes less. Years of experience tell us that calculating for a gain of 10 gives good results.
     
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