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Why does my transistor desaturate so slowly

  1. Feb 14, 2012 #1
    Hey guys I am playing with my new oscilloscope i got myself with income tax, and ran into a puzzling question.
    I am trying to generate an inverted pulse signal, and while my transistor is saturating quickly (driving the output to ground), when i turn the input signal off, it takes its sweet time before it cuts off the ground, and the voltage is allowed to build up from the pull up resistor.

    Someone on a different board suggested "check your ground", but I don't know what else to check. it is a 22g wire from the ground pin the IC is using, and plugged into the breadboard next to the transistor.

    The closest thing I can find searching on here is that the solderless breadboards mess with higher frequencies, but i don't know if that is what is happening here.

    What do you guys think? I can take a picture of the breadboard if that would help but it is as simple as it gets. :)

  2. jcsd
  3. Feb 14, 2012 #2
    Try to add a resistor 10K between and GND and diode 1N4148 parallel to 1K resistor, diode anode toward the base.
    Last edited: Feb 14, 2012
  4. Feb 14, 2012 #3
    This is called saturation recovery time. When the BJT collect/base diode turn on, it takes time to remove the charge to recover. This is the reason they came out with schottky TTL in the 70s. If you connect the cathode of the schottky diode to the base, anode to the collector. When the collector try to swing over 0.3V below the base, the schottky turn on and drag the base along so the collect/base diode never turn on. With this the recovery time is a lot faster. Try put a schottky diode like what I describe and you'll see it recover a lot faster.
  5. Feb 14, 2012 #4
    I also do some measurements of this circuit


    And I get almost the same recovery time.
    And the only think that helps was to add anti-saturation diode.

    And speed-up capacitor help as well.

    And the full diagram

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    Last edited: Feb 14, 2012
  6. Feb 14, 2012 #5
    But the key is still the schottky. You put a 1N4148, that is a normal diode, it help a little, but nothing like the schottky. The cap is only secondary. You should get delay down to few nS, not close to 100nS.
  7. Feb 18, 2012 #6
    Learning the causes and work-abouts of transistor behavior can be a fun past time.

    When it comes to speed, you can really slow it down by either driving it with too much current, or by turning the transistor completely off and starving it for current when it attempts to come on.

    The most common issue is the one you're seeing. Just like Yungman said, you have extra charge carriers loafing around in the base region. Using you general transistor, you can help with the turn off in these ways:

    1 - Turn down the turn-on drive so you don't have as many extra carriers.
    2 - Turn your generic transistor into a schottcky transistor by adding a high speed schottky diode.
    3 - Pull the extra charge out during turn off with a speed-up cap.

    For case 1, take note that there is a comprimise between too little base current and too much. Granted the transistor is saturated when your driving with the collector current. However, unless it's got a really low Hfe, it's probably well saturated at 1/20th the collector current or even less. Some people stick to the rule 10x the minimum you'd need over the range of Hfe. Unless you're preparing it to be neutron hard, I think 5x is good.

    For case 2, the schottky diode between collector and base is good, but remember that it needs to be a schottky signal diode. The ones used for power supplies have a HUGE capacitance, especially near zero volts.

    Case 3 is my favorite. In this one, you place a series RC in parallel with the driving resistor such that during the instant of turn on, there is excessive current momentarily. Then, the RC allows the drive circuit to yank charge carriers away - thus speeding the turn-off operation.

    There is a drawback in that you can actually make the base emitter voltage go negative during the turn off. If it should go too far negative, say below about 4 volts, than the base-emitter junction will begin to behave as a zener. Continual use of the base emitter junction as a zener is okay for use as a zener, but it's detrimental to the long term Hfe of the transistor.

    The answer is to place a resistor between the base and emitter that prevents the turn off pulse from going below about -2 volts. This resistor also helps diverting drive current to ground and pulling charge out of the base-emitter during turn off.
  8. Feb 18, 2012 #7
    Well after working with it and suggestions from here and elsewhere. This is where I ended up.

    I have the frequency back up to 750 kHz which is where I wanted it to begin with (I had slowed it down by 4x earlier)

    The key that I found to decreasing the fall time was from the 2N2222A datasheet, one of the graphs buried in the datasheet talked about collector current vs. turn off time. So I put more current through the transistor when it was on, and when I turned it off it was a MUCH faster fall time. It went from 300ns fall time to 55ns fall time. So I’m happy with that so far. I will try the shottky diodes when they get here (I just hope I ordered the right kind now, I ordered "1N5819-TPCT-ND" from digikey)

    Here is the print and waveform I ended up with (this waveform is at the slower frequency)

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