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Transistors

Discussion in 'General Electronics Discussion' started by mrmodify, Feb 1, 2011.

  1. mrmodify

    mrmodify

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    Feb 13, 2010
    With a transistor is the base always the controling lead, the collector the input and the emitter the output?
     
  2. shrtrnd

    shrtrnd

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    Jan 15, 2010
    Some tech perfectionist will correct my post, but I figured I better get on, before you zap your transistor.
    The voltage/current is from the base thru the collector.
    The emitter supplies the controlling voltage/current for the base/collector flow.
    Think of it like a water faucet.
    The base is the water pipe, the faucet is the collector, and the emitter is the valve controlling the water flow.
    Small emitter/base current flow, allows large base/collector current flow.
    The collector, would be the 'output', not the emitter.
     
  3. mrmodify

    mrmodify

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    Feb 13, 2010
    Thanks shrtrnd, Is your answer the norm? I am trying to understand a schematic and get route of the current flow and voltage through a transistor to maybe help me trouble shoot a power supply.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Have a look at common base, common collector, and common emitter amplifiers (use google). You'll find that the input and output can be in a variety of places.

    In all cases though, as shrtrnd has said, the effect of an input signal will be to alter the current through the base-emitter junction which will cause a larger change in the collector current.
     
  5. Resqueline

    Resqueline

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    Jul 31, 2009
    Yes, it seems that becomes my duty.. ;)

    The main voltage/current is from the collector thru the emitter.
    The base-emitter supplies the controlling voltage/current for the collector-emitter flow.
    Think of it like a water faucet.
    The collector is the water pipe, the emitter is the faucet, and the base is the valve controlling the water flow.
    A small base-emitter current flow allows a large collector-emitter current flow.
    The collector, or the emitter, would be the 'output', not the base.
     
  6. shrtrnd

    shrtrnd

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    Jan 15, 2010
    Nuts,
    Resqueline and *steve* are probably skilled engineers.
    I'm a lowly tech.
    In school, engineers are taught 'hole' flow for the direction of current flow.
    Techs are taught to think in terms of 'electron flow', as current flow.
    Using my water-pipe analogy: Think of a water pipe full of marbles. If you push one marble out of the pipe; engineers are taught to think of the absense of the marble at the other end of the pipe, (think a marble being an electron), so there's a positive 'potential' at the missing marble place, it wants another marble to replace the missing one.
    Techs are taught to think in terms of the marble that fell out of the pipe, if the marbles are electrons, the negative 'potential' of the electron movement, or 'electron flow'.
    I guess you should go with *steve* and Resqueline when they describe 'current flow', I think most of the world thinks of current flow that way.
    (Which is the 'hole flow' of current, not the electon flow).
    I think maybe some tech invented our understanding, to spite engineers.
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Actually our explanations are not specific to what model (conventional or electron flow) of current is being used. Since things are reversed between NPN and PNP transistors you can take your pick of those to use the model that you prefer :)

    Indeed, I steered away from specifying the direction of current flow for this precise reason (that it is opposite in NPN and PNP).

    The important point is that the BE junction is forward biased in normal operation and the CB is reverse biased. A small current flowing across the BE junction causes/allows a much larger current to flow between the collector and the emitter.

    From this relationship, it would appear that the base would always be the input, but if the OP looks at a common base amplifier it will be noticed that the base is not connected to the input signal. In this configuration the applied signal still varies the BE current albeit from the emitter rather than from the base. It's a bit like having an elevator where the car stands still and the building moves up and down around it :)

    Wikipedia provides a better explanation than I can do in a few minutes.

    I thank shrtrnd for his assumption, but I am more formally trained in the discipline that his ID appears to relate to.
     
  8. mrmodify

    mrmodify

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    Feb 13, 2010
    Hello. Am I correct in saying, if I understand that in the attached schematic C of H802 is supplying the voltage to turn on H801 and that C of H803 is controlling how much H802 regulates output of H801. I know schematic looks like B of H802 and C of H803 should be at 36.3 volts, I ask because at R801 one side has 47.9v and other side has 34v. with 12v across R801, R803 has 34v on one side and 21v on the other side. with 12v drop across it. The E of H801 and E of H802 follow the adjustment of R806. The most I can get out of E of H801 is 21 v. I have replaced all 3 transistors and C805. I measured voltage with black lead to chassis ground.
     

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  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    OK, this is kinda simpler than it looks :)

    H801/H802 are in a Darlington arrangement where they act pretty much like a single high gain transistor, Assume that the collectors are the collector of the transistor, the emitter of H801 is the emitter, and the base of H802 is the base.

    Power comes in through the fuse F601.

    The transistor H801/2 is turned ON via R803 and R801

    If just these components were in the circuit, the output (on the emitter of H801/2) would be essentially the full supply voltage (less losses in the transistor)

    The purpose of H803 and the components generally to the right of it are to rob base current from H801/2 in order to partially turn H801/2 off.

    Recall that an NPN transistor begins to turn on when current floes through the BE junction. This requires around 0.6V to overcome the the voltage drop through a diode. So, when the base of H803 gets more than 0.6V more positive than its emitter, (which is held at approx 7.4 volts by the zener) current will be robbed from the base of H801/2 and redirected to ground through H803.

    The source of the voltage for the base of H803 is a voltage divider formed by Rs 805/806/807. If you assume the 5k pot (R806) is centred, the 2 halves of the voltage divider measure 24.5k and 7.85k. This means the voltage on the base of H803 is approximately 1/4 of the output voltage. The output voltage is thus kept to approx the (zener voltage + 0.6) * 4, or 32 volts. This matches well with the expected 35 volts (and also means R806 will not be centred)

    All that's left are a couple of capacitors. In general they provide smoothing of the various voltage sources which (in the case of the first 2) will improve ripple rejection, and in the case of C807 will improve stability (I think)

    Yes the voltage on the base of H801/2 will be approx 36.3 V (2 diode drops higher than the emitter)

    The voltage drop across R801 and R803 should be the same. If they are different (the drop across R801 being higher) then I would suspect C805(?). It may be leaky. Try removing it and seeing if the voltages rise to their expected values. If they do, replace it with a new one.

    If C807 is leaky, the voltage will be regulated through the full range of R806, but at a lower voltage. If this is the case, removing C807 should restore the output to the correct voltage, and replacing it with a new one will fix things.

    A leaky C806 may cause the voltage to be adjustable by R806, but having it fall under load or simply not be adjustable beyond some position of R806. Again, removing it will restore the voltages, and if this occurs, replacing it will fix the problem.

    None of those capacitors look like they have a particularly hard life. If they are close to heatsinks then they may have been damaged over time by heat. Or they just might be old and leaky.

    It may simply be easier to replace the three capacitors and see if the problem is resolved, or you may wish to individually check each one. If any of them are at fault, I would be inclined to replace them all.
     
  10. mrmodify

    mrmodify

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    Feb 13, 2010
    Thanks everyone, Found problem yesterday. C807 had became leaky, I checked it a while back and I must have done something wrong, When I found C807 leaky I changed C805 & C806 also just as Steve suggested. Thanks for the last post from you Steve I have a lot better understanding. Unit was powered on last night and voltages are good except for voltage problem on pin 3, will check today. I still hve no relays coming in and no sound through amps only through right channel on headphones. My repairs to the AM & FM sections are still working.
     
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