New user, bipolar amplifier question

Discussion in 'General Electronics Chat' started by Michael S, Mar 25, 2013.

  1. Michael S

    Thread Starter New Member

    Mar 25, 2013
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    hello, I am an older user, trying to learn a new skill but am having difficulties because I have no teacher to ask. Hopefully I can ask this question coherently without looking too stupid.
    I am studying a common emitter amplifier. I am using a 2n4401 npn if I'm reading the datasheet correctly, it has a hFE of 40 at .001 amp. I have chosen that as my quiescent amperage. I'm using a Vcc of 10v my emitter resistor is 1k ohm which should give me a 1 volt drop at the emitter. the remaining 9 volts will drop through a 4.5k ohm collector resistor drawing 2mA when transistor is saturated. with an Ic of .001 and beta of 40, that makes my desired base current of .000025 amps at the quiescent collector current.
    I am biasing the base with a voltage divider with a desired voltage of 1.7 volts. My studies have told me that it is best to make the biasing divider network 10 times the base current. This is wherein my question lies. One source has stated that i should try to draw 11 times the base amperage between the base and Vcc while drawing 10 times the current between base and ground because one times is going through the base, but for the sake of simplicity (and to keep my head from exploding) I choose to keep it at 10 times. I chose 33.2k ohm between base and Vcc and 6.8k between base and ground (10 volts/ 40000 ohms = .00025 amps)
    Thevenin resistance is 5644 ohms at the base. Thevenin voltage is 1.7 volts. 1.7v/5644ohm=.0003 amps.
    Now, if I'm trying to keep the base current at .000025 amps, why has what I've tried to learn tell me to supply the base with .0003 amps? what am i missing here? Thank you for your time, Michael.
     
  2. crutschow

    Expert

    Mar 14, 2008
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    Here's a simple way to calculate the bias network values:

    For 1.7V at the base and .25mA divider bias current, the base to ground resistor value is 1.7V / .25mA = 6.8kΩ as you calculated.

    For the upper resistor current equal to .25m plus .025mA with a drop of 8.3V the resistor value is 8.3V / .275mA = 30.1kΩ.

    From that, the transistor will take the base current it requires to bias the emitter at ≈1V. You don't need to worry about the Thevenin resistance.

    The reason for having the divider current equal to at least ten times the base current is so that differences in hFE between transistors, which changes the amount of base current the transistor will take, will have only a small effect on the transistor bias current. (In effect you are trying to biasing the base at a voltage, not a current).

    If you want to design to the nominal hFE, typcially about 100, then the calculated upper resistor value would be slightly higher.
     
    Michael S likes this.
  3. patricktoday

    Member

    Feb 12, 2013
    157
    42
    Hello, most of your calculations sound quite correct. There are a couple things:

    * With your circuit as described the transistor won't be in saturation. Since you are targeting the emitter to be at 1V and therefore 1mA (.001A) is flowing across it, all of that current (minus the base current) will flow into your 4.5k resistor so you'll have ~1mA flowing into the 4.5k resistor as well which will drop ~4.5V (10-4.5)

    * The 5644Ω you're referring to would be the input resistance or impedance that an AC signal would see. For biasing the transistor, you are desiring 1.7 volts at the base, correct?, so say you start with 6.8k as you have; so 250uA will flow through that resistor (.00025). And you want 25uA (.000025) to flow into the base, correct? So you must have 25uA + 250uA or 275uA flowing down through your top bias resistor. But the way you set the two bias resistors will get you extremely close! :) (The higher the beta, the less the base current matters.)
     
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  4. Michael S

    Thread Starter New Member

    Mar 25, 2013
    10
    0
    Thank you both for your extremely helpful time.
    If i could follow up..
    Patrick, the saturation I spoke of was my attempt to set the Q-point at .001 amp. (midway between saturation and cut off Vcc-Ve / .002amp set the Rc at 4.5k ohms)
    I now understand, thanks to you both, the reason that the top resistor (R2) handles 11 times Ib (.000275 amps). Using ohms law on the bottom resistor (R1) is preferable to the voltage divider rule Vout=Vcc (R1/R1+R2) when biasing the base. I will change R2 from 33.2k ohm to 30.1k ohms. But wait- I was looking for 1.7 volts at the base, and the voltage divider rule tells me i'm now biasing it with 1.84 volts.
    And another thought, now this might reveal the holes in my self-education, but, in this circuit where the current (.000275 amps) flowing through R2 (30.1k) is split .00025 amp
    through R1 (6.8k) and
    .000025 through the base, to make 1.7 volts carry only .000025 amps, I need 68k ohms between base and ground. i'll cut .7 volts off at the b/e diode junction and another volt at the emitter resistor leaves me with zero volts and 25 ohm internal b/e resistance i don't know what to do with.
    I Assume the transistors internal Rbe of 25 ohms has something to do with it. the more i think the more my head hurts.
     
  5. patricktoday

    Member

    Feb 12, 2013
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    The voltage divider rule doesn't apply when you have a third pathway pulling current out of the connection point. So it really will be 1.7V if your base current really is 25uA (beta varies a lot and the datasheet specifies the _minimum_ beta). I like to start with where I _intend_ to set the bias points then use ohms law to determine how much current will be flowing through each path and, in this case, the current flowing down from the 30.1k resistor will branch into two paths, some going through the base and the rest flowing through the 6.8k resistor.
     
  6. crutschow

    Expert

    Mar 14, 2008
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    3,237
    The base impedance is approximately the emitter resistance times the transistor hFE, thus for a 1kΩ emitter resistor the minimum input impedance is ≈ 40kΩ for a transistor with minimum gain.

    And remember that the open-circuit voltage at the base from the bias network is reduced by the transistor base current flowing through the Thevenin equivalent bias network resistance of about 5.55kΩ.
     
  7. patricktoday

    Member

    Feb 12, 2013
    157
    42
    I think this is the correct formula to determine the base voltage point for arbitrary resistor values: if R1 is the top bias resistor, R2 is the bottom, Re is the emitter resistor and V is the resultant voltage at the base (~1.7V)

    {V\over R_2} + {{V-0.7\over R_{E}}\over B} = {{V_{cc} - V}\over R_1}

    The first term is the R2 current, the second is the base current and their sum is equal to the R1 current.

    (Yay! Got to try out the TEX tags :cool:)
     
  8. Michael S

    Thread Starter New Member

    Mar 25, 2013
    10
    0
    Thank you both.
    The impedance calculation helped me a great deal. I found this site in my attempt to get an answer to this question and find there is a wealth of information and helpful people. I think I will start again and read through this site to try to fill in some of the things I'm "fuzzy" on. It was fun to build this little pre-amp and it works fairly well. next up will be class ab.
     
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