BJT switch basics again

Discussion in 'General Electronics Chat' started by dlatch, Feb 8, 2017.

  1. dlatch

    Thread Starter Member

    May 15, 2016
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    Do I need a pull up resister at point A? How do I determine the Vbe?

    The pnp is the common 3906 The Hall switch is non latching. The circuit works I just want to make sure it is highly robust and reliable.

    How high can I go with the Base resistor? Are the currents the only consideration or do things get twitchy if the resistors are at maximums and currents at minimums?

    Please deconstruct this simple circuit.

    BJt switch basics.gif


    Moderator's Note:
    A jpg or gif file will be better than a pdf file when the circuit is so easy, and it will be more easy to see, the pdf already deleted and modified to gif file.
     
  2. MrAl

    AAC Fanatic!

    Jun 17, 2014
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    Hi,

    You dont need to use a pdf for such a simple document like this.

    You probably need a pullup where you say because of the low normal operating current in the collector.
    The pullup is to reduce the effect of the transistor leakage current collector to base, which tends to turn on the transistor. With the pullup, the leakage current has somewhere to go besides into the base thus keeping the transistor more in the 'off' state.
    For this particular circuit you could put a pullup on the collector of the first transistor instead if you wanted to as long as the 22k combined with the pullup was still low enough to circumvent the effect of the leakage current, but because the 22k is so large it may not work as well there. A calculation would tell you for sure.

    The calculation of the value of either of these resistors is based on the maximum leakage current and the DC current gain as found on the data sheet and a target base emitter voltage of less than 0.4 volts.

    Here's a small jpg for the benefit of other readers...
     
    Last edited: Feb 8, 2017
  3. #12

    Expert

    Nov 30, 2010
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    Pretty much, no. If this circuit is going to get hot you can look on the datasheet for high temperature leakage. Icbo = Amps collector to base with the emitter open. Not on the Fairchild datasheet. The only leakage current on there is 50 nanoamps. So you need to dump 50 nanoamps at a voltage below the base emitter useful region. Let's say 0.2 volts. R=.2/50e-9 = 4 megs. Anything below 4 megs should shut down that transistor under any combination of voltage and temperature.
    Look on the datasheet. 0.66 volts at 1 ma collector current.
    You want 1 ma through the collector so you want 100 ua through the base.
    Guessing you're going to lose 0.66 volts in the 2N3906 and another 0.66 volts in the Hall Effect device:
    (12V- 1.32V)/1e-4 = 106.8K
    That seems pretty high, but the math says it's good. If the Hall Sensor wants to work in a different current range, you could lower the base resistor to 1 ma which would be 10.68K.
    Personally, I like to stay in the 1 ma range to keep impedances low enough to avoid picking up stray noise.
    So, there's the math and there's an opinion.
     
    cmartinez likes this.
  4. hp1729

    Well-Known Member

    Nov 23, 2015
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    What Hall Effect device are you using? What is the output like? You may not need the NPN at all.
     
  5. dlatch

    Thread Starter Member

    May 15, 2016
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    The NPN is the Hall switch open collector output.

    I must learn to post the image straight to the body rather than as an attachment. I used pdf because my scanner goes there by default.

    Great answers...just what I was looking for. Thanks expert....I knew on the numbers I could go much higher on the base resistors but I was wondering if that makes things twitchy. I'll take your math and your opinion and leave them where they are.
     
  6. MrAl

    AAC Fanatic!

    Jun 17, 2014
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    Hi,

    Well this is interesting, one guy says 'yes' and one guy says 'no', so which is it?
    But wait a minute, the guy who said 'no' then later says you need a 4 Meg resistor :)
    So i guess it's not 'no then?

    For this transistor we are lucky the leakage itself isnt too bad and so we might get away with no pullup resistor. Not what i like to see in a new design though, and we also have to consider the switching speed. What we dont know yet is how fast this has to switch.
    Hall effect devices are not particularly fast, so we probably luck out there too :)
     
  7. hp1729

    Well-Known Member

    Nov 23, 2015
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    Two answers? so try it without the resistor and see if it works okay. Why waste time asking opinions?
     
  8. #12

    Expert

    Nov 30, 2010
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    For room temperature operation, I wouldn't use a pull-up resistor. I used to design with big, hot transistors, so I had to know what to do with high temperature leakage. The fact that I know how doesn't mean you need to treat your circuit like you can cook breakfast on it.
     
  9. MrAl

    AAC Fanatic!

    Jun 17, 2014
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    Hi,

    Well sorry to say but that doesnt work all the time because the transistor characteristics can change.
    It may work in this case though because i dont think he's that worried about it as temperature increases.
     
  10. MrAl

    AAC Fanatic!

    Jun 17, 2014
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    Hi again,

    Yes i have to agree in part. I'm more used to working with switching circuits that have to be able to turn off fast too. In that case no base emitter resistor would be just crazy. In fact, no reverse bias would be crazy too :)
     
  11. dlatch

    Thread Starter Member

    May 15, 2016
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    The circuit works. I want to understand best practice. Explain what a base emitter resistor does. And the way to select them.

    Most tutorials on this topic make no mention of them.
     
  12. MrAl

    AAC Fanatic!

    Jun 17, 2014
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    Hi,

    A base emitter resistor would be used to shunt some of the leakage current from the reversed biased collector base diode. Without some base emitter resistor the transistor might turn on at least partially. How you get this resistance can be in several ways.

    The other function is to sweep charge carriers out of the base region in order to turn the transistor off fast. Because you are using a Hall device though you probably dont have to worry about this but it does depend on how fast you want that transistor to turn off. I am guessing that it is not that critical for your application here. For maximum speed you would need an actual reverse bias supply to create a reverse bias for the base emitter diode so you would need not just a base emitter resistor but also another supply voltage, or be able to bias the emitter down from +Vcc in order to obtain that reverse bias.
    I was very concerned with this back in the mid 1980's and found a good book in the Rutgers University library on this topic. I might have a copy of the pages where that topic was discussed in great detail. I'll have to look for that. There may be something online these days though too.
    Of course there is also the possibility of keeping the transistor out of saturation in the first place which reduces switching time also. I would bet you dont need that either here though. If you do, then i guess we would have to talk about that as well.
     
    Last edited: Feb 10, 2017
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