Base to Ground resistor in BJT application

Discussion in 'General Electronics Chat' started by EngIntoHW, Aug 19, 2010.

  1. EngIntoHW

    Thread Starter Member

    Apr 24, 2010
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    Hi,

    I'd like to ask a couple of questions please regarding the below circuit.

    1. What is the role of Rbe? What is it needed for?

    2. Is Rbe connected to Ground (on its bottom side) because the Emitter is grounded?
    I mean, should it generally go from Base to Emitter (in case the Emitter is not grounded).

    Thanks.

    [​IMG]
     
    Last edited: Aug 20, 2010
  2. marshallf3

    Well-Known Member

    Jul 26, 2010
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    Generally used to form a voltage divider with the input reistor to limit Vbe on higher gain transistors that require very little Ibe.

    At least that's what I use them for, the Spice people probably have a more technical answer.
     
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  3. Wendy

    Moderator

    Mar 24, 2008
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    Vbe isn't really necessary, but it does guarentee that the transistor actually turns completely off, and doesn't false trigger to noise. In some circumstances it might be needed, such as when the input of the transistor through Rin is open.
     
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  4. EngIntoHW

    Thread Starter Member

    Apr 24, 2010
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    How does it ensure that the transistor turns off?
    Is it by discharging Cbe?
    Any other means?
     
  5. tom66

    Senior Member

    May 9, 2009
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    I thought it only affected MOSFETs but if the gate is not connected the output will float or oscillate because of the Miller effect or even if you move your hand near them due to electromagnetic interference. Or something like that. I'm not sure if it applies to BJTs because they are current controlled.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Rbe helps a great deal to decrease turn-off time, particularly if when Vin (the source to the left of Rin, which should actually be called Rbase) is either present (say, 5v) or "floating" (no current path to Vcc or Ve.)

    Without Rbe, the transistor may stay on for quite a bit longer than you'd expect. At low frequencies, you may not even notice it. However, if you're operating at speeds in the 10s of kHz, the turn-off delay becomes much more significant.

    Usually, you'll want Rbe to be roughly 3 to 5 times the value of Rbase.

    Calculating Rbase for using an NPN transistor as a saturated switch:

    Rbase ~= (Vin-Vbe)/(Ic/10)
    where:
    Vin = the voltage to the left of Rbase, referenced to the transistors' emitter, when the transistor is to be turned on.
    Vbe = 0.7v if the desired collector current is <= 1/3 the maximum transistor collector current as specified in the datasheet. If you're getting near 1/2 the maximum specified collector current, use 0.8v. I do not recommend exceeding 1/2 the maximum specified collector current.
    Ic = the desired collector current.

    So, if you have an LED that needs 20mA current, and Vin is 5v, you'd then calculate:
    Rbase = (5v-0.7v)/(20mA/10) = 4.3/0.002 = 2150 Ohms. 2.2k and 2.1k are the closest standard E24 values. If you're going to use Rbe, then select the 2.1k resistor.

    Rbe can then be anywhere from about 6.2k to 10.5k, and work pretty well. If you're going to be switching the transistor at higher frequencies, use the lower Rbe.
     
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  7. EngIntoHW

    Thread Starter Member

    Apr 24, 2010
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    Thank you very much.
    It helps me to understand it much much better.

    I'd like to discuss another point please.
    What if Vin doesnt remain floating when it's not HIGH, but neither goes to ground level.
    Say it goes to 0.5V when it's LOW.

    How does Rbe help in this case?
     
  8. marshallf3

    Well-Known Member

    Jul 26, 2010
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    Exactly as I mentioned in my post. A 5K input resistor with a 10K to ground will drop Vbe by 1/3 putting it closer into the "sure it's turned off" range.

    I don't deal with logic as much as i do analog and I'm often needing to switch the base of a small signal transistor with a 12V control input voltage. It's just easier to divide it down providing you can still get sufficient Ibe flowing. It also limits a lot of the guesswork involved when a transistor is rated to have an Hfe of 100 - 300
     
  9. EngIntoHW

    Thread Starter Member

    Apr 24, 2010
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    I see, so having Rbe equals 10 times Rin isn't a good idea.

    About your second point, how does Rbe relate to having HFE ranged from 100 to 300?

    Thank you for your kind help.
     
  10. marshallf3

    Well-Known Member

    Jul 26, 2010
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    10x dosn't enter into it, you just want to make sure you're not putting too much current through the BE junction.

    Most BJTs are hooked up such that you have sort of a combined input circuit. The BE junction is going to sit around 0.6 or so volts regardless, and to eliminate wasted power from heating up the BE junction while at the same time keeping the drive current from the previous stage to a minimum you'd only want to put a bit more current through it than you'd need out depending on the gain of the transistor using commmon formulas. If the transistor in question has a minimum gan of 100 that means you need to forward bias the BE junction with at least 1/100th the current as you expect the collector-emitter junction to be able to pass.

    Transistors vary widely in exact gains, it wouldn't be uncommon to find a pile of the exact same part # transistor yet when you meaured them you could find gain variations between any two of 100% or more. In properly designed bias circuits this isn't as much of an effect on operation as you might think, youv'e just got to accept the fact that a 1 mA input bias might allow a 300 mA output change or it might only allow a 100 mA change capability.

    Rbe is just part of the bias circuit. If I was driving the base with 12V through a resistor I'd probably want some of the divider current to flow through Rbe to ensure that I didn't have to totally rely on voltage drop across the input resistor to maintain sifficient base drive current as well as not chancing going over or under the operating window.

    Were I a bit more awake I could probably explain it better, for now let's just say tht's the way I've usually done it in low power circuits as have most other designers.
     
  11. campeck

    Active Member

    Sep 5, 2009
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    No one has said this but Isn't that known as a pulldown resistor?
     
  12. EngIntoHW

    Thread Starter Member

    Apr 24, 2010
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    Hey Marshal, thanks again!

    As the BJT is saturated, and not in its linear region, there's no significance to HFE, is there?

    So I'm still not sure how HFE relates to this application.
     
  13. hgmjr

    Moderator

    Jan 28, 2005
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    In my experience, this resistor has been referred to as a base-return resistor.

    Actually, labeling the resistor rbe could be a bit confusing since that is the designation for the intrinsic base-emitter resistance.

    hgmjr
     
  14. EngIntoHW

    Thread Starter Member

    Apr 24, 2010
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    Hey hgmjr.

    So you're talking on a situation where Vin is left float when not HIGH.
    However, when VIN goes to 0.5V when LOW, then Rbe just forms a voltage divider to decrease VB from ~0.5V to say 0.3V, right?

    BTW, I thought that in any case, the BE junction couldn't get turned on from 0.5V, could it?
     
  15. hgmjr

    Moderator

    Jan 28, 2005
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    Actually, the best way to calculate the effect of the base-return resistor is to calculate the Thevenin equivalent voltage and resistance and then use that to determine the effective base current being delivered to the transistor.

    hgmjr
     
  16. marshallf3

    Well-Known Member

    Jul 26, 2010
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    Partially true, however how you hard saturate the transistor has a lot to do with the efficiency at which the circuit operates. Not too terribly important when running from a power supply and if heat isn't a consideration, however in battery powered apps &/or where the least amount of waste heat is desired it pays to take the extra time (and 2 cents) to somewhat optimize the circuit.
     
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  17. EngIntoHW

    Thread Starter Member

    Apr 24, 2010
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    Hey,

    Thanks a lot!

    Still about the 0.5V when VIN goes LOW.
    How could 0.5V turn on a diode? (BE junction).
     
  18. Wendy

    Moderator

    Mar 24, 2008
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    Generally it doesn't. You have to design accordingly. First check the voltage divider specs, as has been suggested earlier in this thread.

    A while back on another transistor question thread I made the point graphically. This was also mentioned earlier in another post on this thread verbally.

    simple BJT audio amp experiment (need help)

    [​IMG]
     
  19. EngIntoHW

    Thread Starter Member

    Apr 24, 2010
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    Hey,

    It was mentioned in this thread (by Marshall) that:
    However, it doesn't say how from the first place, without the voltage divider, the BE diode could be turned on by 0.5V.
     
  20. SgtWookie

    Expert

    Jul 17, 2007
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    The Vf of the BE junction is dependent upon the base current and junction temperature. As Ib increases, Vbe increases. The Vbe vs Ib plot is nearly a log function.
    Vbe vs junction temp for a given Ib is nearly linear.

    This is why people usually use base current rather than base voltage to control transistors; otherwise there are a lot more variables to worry about.

    If you are going to use a transistor in the linear mode, then you go by the hFE range.

    If you are going to use a transistor as a saturated switch (what you have shown in your first post), then you use a "forced" beta (hFE) of 10.

    If you wish, you can "fine tune" the base current after you get it working, but note that while using Ib=Ic/10 will work for 100% of transistors in a given batch, as you start decreasing the base current you will start experiencing failures due to high Vce(sat) (and thus high power dissipation in the transistor) due to variations in the transistors in the batch.

    Operating a transistor deeply saturated causes turn-off time to increase. However, it also minimizes the power dissipation. If turn-off times are not significant, you're better off to simply use Ib=Ic/10.
     
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