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Transistor help

Discussion in 'Electronic Basics' started by Wong, Nov 13, 2003.

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  1. Wong

    Wong Guest

    I just read a tutorial about biasing a transistor. It says that we
    cant use the formula Ic=Beta.Ib when it is in saturation region but in
    active region we can use that. But in another tutorial, I found that
    the transistor is in saturation region as a switch but it applied the
    above formula, so which one is true ?
  2. They might both be, depending on the definition of beta. If beta is
    defined as the ratio of collector to base current, and you approximate
    beta by a constant, then the first one is better, If you include in
    the definition that beta changes as collector voltage approaches and
    falls below base voltage, then the second one can still be right.
    Transistors in saturation are often driven to a forced bets that is
    well below the beta you see when there is plenty of collector
    voltage. For instance, if a transistor exhibits a beta of about 100
    over a large range of collector voltage, but you saturate it by
    driving it with a base current equal to 1/10 of the collector current,
    you are forcing it to the beta = 10 operating point (forcing the beta
    down to 10, 1/10th of its normal beta). This is a measure of how
    thoroughly saturated the transistor is.
  3. Mac

    Mac Guest

    When you use a transistor as a switch, you normally want to saturate it.
    So you calculate the maximum base current you could possibly need, then
    multiply by 10 and design for that current.

    For example, suppose I have a transistor with a gauranteed minimum DC
    current gain of 100. I am using it as an inverter, and I have a 1k pullup
    on the collector. I know that I need to drive it down to 0.8V to get a
    valid LVTTL low.

    So, I can go through the calculations to find the minimum collector current:

    (3.3V - 0.8V) / 1k = 2.5 mA

    This means that the minimum base current is:

    2.5mA / 100 = 0.025mA

    So, in theory, I would be safe if I can put 0.025 mA into the base. But I
    want to be really safe, so I'll shoot for 0.25mA into the base. So that is
    why I use the formula, to find the minimum current, which I then multiply
    by 10.

    Anyway, to finish, since the base will be at about 0.8 volts, and my input
    voltage will probably be 3.3 volts, I need (3.3 - 0.8) / 0.25mA = 10k in
    series with the base. Since I have a ten-fold safety margin, I don't need
    to worry about 5% resistors or inputs that might be substantially less
    than 3.3V.


  4. So put a milliammeter on the collector of a transistor and a
    microammeter on the base. You will see that as you get closer to
    saturation, it takes more and more increase in base current for the
    same increase in collector current. I.e. the Ic to Ib ratio drops.
    Instead of 160 for example, it drops to as low as 10 when fully

    You don't have to use a milliammeter. You can measure the voltage
    across the resistors, such as a 1k which gives 1 milliamp per volt.

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  5. GPG

    GPG Guest

    2SD1071, Darlington power switch ,300V, 6A, 40W, B1500, TO220
  6. Wong

    Wong Guest

    OK, one more question. Why we need to use as switch in saturation
    region and then, amplifier in active region ? Is this related to the
    "Beta Stabilization" ?
  7. If you try to set the base current to turn a switch on to 1 volt
    collector to emitter, you will have a hard time making two units that
    act the same. Some will drop 1/2 volt, and some will drop volts.
    Even a change in temperature will alter the drop when the switch is
    on. Double or quadruple that base current, and thousands of units
    will saturate with the same (smaller) drop to within 100 mv or so of
    each other. If that switch is delivering power (as in a switching
    supply) it will also produce only a fraction of the heat because of
    the lower voltage drop. All that for a bit more base drive.
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