how to think values of components???

Discussion in 'General Electronics Chat' started by indianhits, Aug 22, 2009.

  1. indianhits

    Thread Starter Active Member

    Jul 25, 2009
    hello guys i want ti know how do you guys think of components values like resistors and capacitors value and all and how do you think or make your own circuit diagrams

    and in CE amplifier circuit we use a resistor at the emitter to ground side what is the use if it?????????

    Thank you!!!
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    The ability to design a circuit is the result of some time spent studying electronics. Some instruction in a classroom setting is good for a structured approach. With a solid grounding in the basics, it is possible to move on to more complex circuits, and learn how and why they operate as they do. With enough of that, one may start designing circuits.

    Like everything else, it takes time and study to master the subject.
  3. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    If you try to put a linear bias on a transistor without an emitter resistor (or some form of feedback stabilisation), what voltage do you use?

    The base-emitter junction starts to conduct somewhere round 0.6 - 0.7V and is dependent on temperature. You will not easily get a stable bias.

    If you add an emitter resistor, you can bias the base rather higher (say 1V) and the excess 0.3 - 0.4V is dropped across the emitter resistor. For example say the resistor is 100 Ohm, the emitter current range is now 3 -4 mA and the collector current is about the same as it's effectively a series circuit.

    Without a capacitor across the emitter resistor, the circuit gain is the ratio of the emitter resistor to the collector resistor. eg. with (100 Ohm emitter and) a 1K collector resistor, which would have 3 to 4 volts across it (assuming a high enough supply voltage), the amp would have a gain of 10.

    The amp will also be linear over a larger range as the varying input voltage to the base (from whatever signal you are amplifying) can be larger amplitude without risk of either saturating or completely turning off the transistor.

    Overall, it's more stable and more flexible to component and temperature variations.
  4. rspuzio

    Active Member

    Jan 19, 2009
    In addition to what R. Jenkins pointed out, here is another
    aspect of the issue:

    With the emitter connected directly to ground, the the output of
    the amplifier would vary as the exponential of the input voltage.
    For most applications, this is not good --- we want the output of
    the amplifier to be proportional to the input, not to the exponential
    of the input. To be sure, decreasing the range of the input helps,
    but it only goes so far.

    Placing a resistor between the emitter and ground goes a long way
    to solving this problem --- if you put a big enough resistor between
    the emitter and ground, the relation between voltage of the base
    and collector current is dominated by the resistor, which behaves
    according to the linear Ohm's law with the non-linearity of the
    transistor only having a small effect. For some applications, such
    as an amplifier in a simple radio, this remaining non-linearity is
    small enough that we can safely ignore it. If we are interested in
    higher fidelity, then, since the deviation from linearity is small, we
    can easily enough make it even smaller by some technique such as
    canceling it against an opposite deviation (as in a push-pull design)
    or by introducing a bit of negative feedback.

    Since you asked about computing values, let's make a
    little computation here. The exponential constant for a PN junction
    is around 25 mV. Let's say that, as in the example R. Jenkins
    gave, the emitter current varies between 3 and 4 mA. Then the
    ratio of the currents is 4/3 = 1.33..., we have 25 ln 1.33 = 7.2,
    which means that the base-emitter voltage drop only varies by
    7.2 mV. By Ohm's law, the voltage across the 100 Ω resistor will
    vary by 100 mV for the same range of output currents, so we see
    that the resistor is responsible for 86% of the change in base
    voltage over that range. I don't want to bore you with more
    math, but if you do the algebra with the logarithms, you find
    that the plot of current versus voltage differs from a straight
    line by much less than a percent. Looking at a plot, you would be
    hard pressed to distinguish it from a straight line; for not too
    demanding purposes, this will do just fine.
  5. indianhits

    Thread Starter Active Member

    Jul 25, 2009
    i am having hard time understanding electronics now i am completely dependent on this site cause our college madam is not teaching this stuff properly and i want to master this subject

    Thanks guys.Thanks!!!