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help with understanding the mysteries of transistors

Discussion in 'General Electronics Discussion' started by mattsains, Jul 18, 2012.

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  1. mattsains

    mattsains

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    Jun 8, 2012
    Hi

    So, I'm sort-of-ok with using NPN BJTs as switches, so I decided to try out a Darlington pair to make a "human contact-ometer". When I hold an electrode in each hand, I want it to detect the tiny current running through me (I measure my resistance as 1.2MR between hands) and light up an LED.
    So I set up a Darlington like this:
    [​IMG]
    If I just touch the flylead, the LED switches on. I didn't expect this because I wanted it to switch on the LED only when I touch both leads (the second lead tied to 5V, not illustrated). I figured the Darlington was too sensitive so I added a resistor between the emitter of Q1 and the base of Q2. (I tried 1kR and 470kR)
    I also tried adding a resistor between the flylead and base, thinking the Darlington was picking up noise or something. (I was just winging it at this point)

    Then I figured this Darlington was just too crazy sensitive, so I replaced it with a single transistor like this:
    [​IMG]
    This worked a bit better, especially when I put a 1kR resistor in series with flylead2. It still made the LED glow slightly when touching only flylead2.

    Is the P2N2222 - or transistors in general - that sensitive? The P2N2222 has an HFE of around 35. Assuming I'm an ohmic resistor, I should only pass 4.2 microampere. In the darlington, as far as I know, shouldn't that give me x35x35 collector-emitter current? (ie., 5.1mA) Then as far as my dodgy maths is concerned, the darlington should give just enough current for the LED when I attach myself to both leads, not only one?

    What am I experiencing here? I hope I haven't shown my ignorance too much in this post
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    To some extent, a Darlington pair can be thought of as a single transistor with a very high gain. HOWEVER the way it is wired up means that the voltage across the collector and emitter will never fall much below 1.2V (for a normal transistor it can fal much lower than half of this).

    That can be a trap, but not a real issue in anything you've shown me so far :)

    Yep, sounds fair.

    You are picking up electrical noise which is strong enough to turn the transistor on.

    You might not think you have a circuit, but there is a very small capacitance between you and the rest of the circuit. And that's enough.

    Placement of that resistor is right. However it's not affecting the gain as much as it affects the input impedance. Your circuit now requires more current to turn on.

    Remember that you need about 1.4V on the base of the Darlington to turn it on. If you're just a resistor, then you need to be a small enough resistor that less than 7.6V of the 9V is dropped across you.

    Doing a bit of cheating math, if 1.4V is across 470k, then 7.6V must appear across you. (7.6/1.4)*470k = 2.5M. and for the 1k, (7.6/1.4)*1k = 5.4k

    Since you have measured your resistance as much greater than 5.4k and somewhat less than 2.5M, I would expect the 1K resistor totally failed to work (the LED stayed off) but the 470k resistor worked (or got pretty close)

    Not going to help with the noise, but the 1k resistor there means you can touch the probe to your battery without risking too much current flowing through the BE junction.

    What you could try is a capacitor between the base and ground. Something like 0.1uF might be sufficient.

    The 2N2222 almost certainly has a higher gain than that. you'll note the datasheet gives minimum gain figures

    No, I don't see much ignorance :)

    You are experiencing noise. If you replaced the LED with an earpiece (preferably a high impedance one) you might hear a jumble of radio stations.
     
  3. mattsains

    mattsains

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    Jun 8, 2012
    thanks for this blow-by blow dissection of my post! Very helpful.

    * I see now that I totally neglected that the HFE I was using was a minimum value and not a typical value. This is probably because I'm used to using them just as a digital switch to drive (comparatively) high current devices.
    * I also didn't account for the voltage drop across me nor the voltage required for the Darlington. My thinking now is this: the base-emitter resistor needs to be such that it is large enough to "steal" the voltage drop away from me and shift it inside the Darlington to cause it to switch.
    * Am I right in thinking that your proposed capacitor will be a filter by shorting AC to ground, but not affecting DC? (at least as far as I am concerned)
    * Would this circuit really work as a "two minute radio"? Would I get a half wave as an output?

    Thanks, this has been eye opening. I feel less scared to begin using transistors as proper amplifiers
     
  4. mattsains

    mattsains

    35
    0
    Jun 8, 2012
    Could you check my math/circuit theory? with a resistance of me and the protection resistor at 1'200'220R, supply of 5V, and darlington voltage of 1.5V, I calculate the base-emitter resistor to be around 514'380R.
    [​IMG]
    Circuit simulation in spice disagrees though, because I then get quite an impressive voltage across the darlington base-emitter of around 4.995V, and a voltage drop across me of around 0.052V. I expect it would be better if there was as much voltage drop across me as possible, with just enough on the darlington to set it off.

    I'm obviously doing something heinous with the theory

    EDIT: LTSpice attached
     

    Attached Files:

    Last edited: Jul 18, 2012
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Your calculations are probably correct, however they don't take into account the current drawn by the base of the transistor nor the characteristics of the transistor near cutoff (it doesn't just suddenly turn on at 0.6V, and off at 0.59999999999999V)

    What you are seeing is a real issue with digital logic. You might want the LED to be on or off, and some current into the base will turn it on, and some far lower amount will turn it off. However you're hovvering around in the middle and the transistor is neither fully on nor fully off. Any noise present will be amplified and seen at the output.

    Yes, the point of a capacitor is to shunt AC to ground. The larger the value you use, the less your circuit will be affected by high frequency noise, but also the slower it will be to respond to you touching the contacts.

    Yes, the AC amplified will be pretty much half wave. And the amplification is also likely to be extremely non-linear.

    Can it be a 2 minute radio? Well, yeah, maybe. But it will receive all stations as there is no way to tune it. You may hear a garble of speech and music. If you have a very strong signal for a single station, it may dominate.

    You've seen the problem. No amount of resistors and capacitors are going to completely solve it. The answer is the Schmitt trigger. You should google for a schematic. Here is one, but it doesn't give circuit values. You can use your existing circuit to drive this one. Simply replace the LED with a higher value resistor (say 1k to 10k) and take the output from the collector. I'm not sure that this circuit will be sensitive enough to just use your probes. You may also have to fiddle around with resistors to get a suitable hysteresis.

    You can also buy Schmitt triggers in ICs, but that's not as much fun
     
  6. mattsains

    mattsains

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    Jun 8, 2012
    thanks so much for your help! I learnt more in this thread than I did in all high school about electronics. (and I took science)
     
  7. mattsains

    mattsains

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    Jun 8, 2012
    also, could you recommend a project involving transistors that'd stretch my current levels of ineptitude? One where I could design the circuit myself, but would also be difficult enough for me to learn something
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Yeah, try making a simple multivibrator oscillator. Like this one.

    Nothing is critical. If you don't have the exact resistors or capacitors, the next value higher or lower won't really make any huge difference. Any NPN transistor you have on hand will work.

    Do you have a solderless breadboard?
     
  9. mattsains

    mattsains

    35
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    Jun 8, 2012
    That I do! Will check it out
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Great! Once you've built it and got it working, the next thing t do is to be able to describe how it works.

    To do that you'll have to go and read up on it, and understand what you're reading.

    Then try to explain it to me (explaining it to someone else in your own words is a great test of your understanding).

    Don't worry if you get stuff wrong, We can help with that!
     
  11. mattsains

    mattsains

    35
    0
    Jun 8, 2012
    want to do this during the course of the weekend.
    By the way, the circuit this thread was originally about seems to work great as a water sensor.
     
  12. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    Since you're using LT Spice I recommend playing with the 'DC Transfer Characteristics' function. It gives you a very clear picture of base current, voltage vs collector current and voltage.
     
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