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Transistors

Discussion in 'General Electronics Discussion' started by TechNotes, Sep 5, 2011.

  1. TechNotes

    TechNotes

    5
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    Sep 5, 2011
    I am trying to find a good resistance value for the base and collector of a npn transistor. I have no idea how to do this so please help. I have a MPS650GOS transistor. The page I ordered from, the datasheet, and the schematic is below.

    The schematic I am working from: http://www.instructables.com/files/orig/FJ4/7RE1/GICYB6CI/FJ47RE1GICYB6CI.png

    Datasheets:
    http://parts.digikey.com/1/parts/1007871-trans-npn-ss-2a-40v-to92-mps650g.html

    http://www.onsemi.com/pub_link/Collateral/MPS650-D.PDF

    Thanks!
     
  2. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010
    What is your required current? Divide this by the HFE (transistor DC gain) and you will know your base current. If you know the base voltage, you should just be able to use Ohms law to figure the resistor needed to provide the current. Don't forget, you will need to make sure the transistor is on, so take into account the base-emitter junction voltage drop.

    edit: Formula is (Vb - Vbe)/Ib
     
    Last edited: Sep 5, 2011
  3. TechNotes

    TechNotes

    5
    0
    Sep 5, 2011
  4. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010
    Base-emitter junction is typically 0.7V for a silicon based transistor. Transistor DC gain is 'HFE' found in the datasheet(I looked, it is anywhere from 40 to 75 based on Ic, it is always best to use the low number which gives you some extra breathing room). The base voltage is what your circuit will be providing to the base of the transistor. Since all your transistors are being powered by the MEGA32, it would probably be listed in the MEGA32 datasheet.

    Vb - Voltage applied to the Base of the transistor.
    Vbe - base-emitter junction voltage drop, explained above
    Ib - Base current, you will need to figure this out on your own, based upon the current you need divided by the HFE of the transistor.
     
    Last edited: Sep 5, 2011
  5. TechNotes

    TechNotes

    5
    0
    Sep 5, 2011
    Thank you so much! This really helped
     
  6. TechNotes

    TechNotes

    5
    0
    Sep 5, 2011
    I have another question. How do you find the value of the resistor on the collector using the same transistor and schematic as stated above?
     
  7. rogerk8

    rogerk8

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    Jul 28, 2011
    This depends on what you want to do. Normally you want the transistor to be able to swing equally posively and negatively so you leave half of the rest of the voltage (the voltage over the emitter-resistance in a high gain CE-configuration is normally low and may be disregarded) over the transistor and the rest over the collector-resistance. Using Ohm's law it is then easy to calculate the resistance needed. Hope this will help. KR/Roger
     
  8. TechNotes

    TechNotes

    5
    0
    Sep 5, 2011
    Sorry, I really am a newbie. What is high gain CE-configuration? And what do you mean by I want the transistor to be able to swing equally positively and negatively? I also don't see how using Ohm's Law will help. Please explain.

    Thanks!
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,178
    2,690
    Jan 21, 2010
    By CE, Roger means "Common Emitter". This is the fairly common connection you see for a transistor where (for an NPN transistor) the emitter is grounded (often via a relatively small resistance), the input is to the base of the transistor (often via a resistor) and the load or a load resistor is placed between the collector and the positive supply rail.

    Here is a very simple example:

    [​IMG][​IMG]

    In practice, unless the transistor is simply used to switch something on and off, some form of biasing is required. Biasing turns the transistor part way on. This means that an audio dignal input to the base of the transistor can turn the transistor on more (with positive half cycles) and off more (negative half cycles).

    Without the biasing, only half of the waveform would be amplified. For audio signals this would produce huge amounts of distortion.

    A very simple form of biasing is to attach a resistor between the base and some positive bias voltage.

    [​IMG]

    The disadvantage of this circuit is that the output voltage is highly dependant on the transistor characteristics and the temperature. In addition any DC voltage on the input will affect the bias, and the DC voltage on the output will affect the bias of any following stage.

    It is clear that there is no positive and negative output. The output MUST always be positive with respect to ground. The output is more properly described as AC with a DC bias, or perhaps DC with ripple :).

    In practice, you might use a voltage divider to set the bias point as this makes the amplifier more stable. An emitter resistor provides negative feedback (which can be bypassed with a capacitor to provide higher gain at higher -- audio -- frequencies). Capacitors are used between stages to isolate the DC bias between each stage. It is the capacitor that allows you to have a truly AC (absent of DC bias) input or output.

    So eventually you get a design that looks like this:

    [​IMG]

    Not all single transistor amplifier stages look like this, there are many variations. In addition there are other ways to eliminate DC bias that do not require capacitors.

    However, understanding a basic common emitter amplifier will help you in a huge number of cases where you need to use a transistor,

    Oh, and you are right in questioning the statement about ohms law. In determining what value resistors to use there is more to it than just ohms law.

    For more information, you might like to check this out. Page 2 is where it starts talking about the common emitter amplifier.
     
    Last edited: Sep 8, 2011
  10. rogerk8

    rogerk8

    176
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    Jul 28, 2011
    Better late than never...I like your explanation of my kind of stupid way of trying to help. It even learned me some. And you're right, there's lots more to properly bias a transistor than Ohm's Law. It is kind of delicate actually. But mostly the gain is close to the Rc/Re-ratio. It is however not exactly this because it depends on input driving impedance (Rs), hie, Re and hfe. I have studied this closely like the amateur I am and have come to this conclusion. Yet my Germanium pre-amp using a single transistor works just fine (Rc/Re=10 but actual gain ended at 7). Take care! BR/Roger
     
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