textbook era?

Discussion in 'Homework Help' started by Neil Groves, Sep 29, 2014.

  1. Neil Groves

    Thread Starter Member

    Sep 14, 2011
    I am building a small trannie circuit to play with and my calculations tell me that the bias current for the base of the transistor is 500+ micro amps, the circuit however tells me different, I havn't done a practical reading yet, just want to know what is going on here, is it me or them?

  2. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    I think your calculations are off by a factor of ten? How did you arrive at 500+uA?
    BTW your schem mentions 'Vce(sat)=0', but the transistor isn't saturated when used as shown.
  3. Neil Groves

    Thread Starter Member

    Sep 14, 2011
    let me recalculate......

  4. Neil Groves

    Thread Starter Member

    Sep 14, 2011
    20k + 3.6k = 23600

    I divide that by 12v which gives me 0.000508 Amps = 508 micro amps right?

  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    That's incorrect.
    Your calculation is only a rough estimate of the current in the voltage divider formed by the two resistors. It doesn't give you the base current. Are you familiar with BJT voltage divider bias calculations?
    Last edited: Sep 30, 2014
  6. b1u3sf4n09


    May 23, 2014
    Use the voltage divider formed by the two resistors to find your transistor's base voltage.
  7. Neil Groves

    Thread Starter Member

    Sep 14, 2011
    BJT voltage divider bias? No sir I am not, in fact never heard of such a thing, I'll have to Google it...... The plot thickens :)
  8. crutschow


    Mar 14, 2008
    The voltage divider generates a voltage at the transistor base. This voltage cause the transistor to turn on and generate a voltage at the transistor emitter one Vbe below the base voltage (about 0.7V) which determines the emitter current through the 220 ohm emitter resistor. This emitter current is reduced by the transistor Beta value (100) to give the base current.

    Note that the above assumes the base current is small enough so as to have only a small effect on the base bias voltage due to the base current flowing through the equivalent bias divider resistance.

    Make sense?
    Neil Groves likes this.
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
  10. Neil Groves

    Thread Starter Member

    Sep 14, 2011
    Thanks guys, lots of info to sift through, job for the weekend me thinks :)

  11. hobbyist

    Distinguished Member

    Aug 10, 2008
    with the beta value given, you can just use an assumption of taking the collector current given and dividing it by beta, (4.58mA / 100) to give the approximate base current.

    However for future reference:

    If the collector current or the emitter current or any voltage values were not given in the schematic, then a first order approximation can be done as follows.

    First do the voltage divider equation (VCC x 3.6K / (3.6K + 20K)) to give the voltage at the base, VB.
    then take VB and subtract Vbe (0.7v) from it to arrive at the voltage at the emitter, VE.
    then divide the emitter resistor into VE, to solve for emitter current, from there divide the emitter current by the value of Beta given in this case 100, to solve for base current.
    This will allow you to make a quick check on voltages and currents calculated, before you breadboard the actual circuit, once it is breadboarded, you can take emperial measurements to see if the actual meter readings are reletively close to your calculated values
    Neil Groves likes this.
  12. Neil Groves

    Thread Starter Member

    Sep 14, 2011
    That's excellently clear hobbyist, much fun to be had here, thankyou very much.

  13. MrAl

    Well-Known Member

    Jun 17, 2014

    It could be that you did not consider the fact that the emitter current can raise the voltage across the emitter resistor and thus lower the base current, which would be the same as ignoring the collector current and just calculating the base current without that influence. The current would then be much higher like around 350ua which is much higher than 46ua.
    To do this right the transistor has to be looked at as a current controlled current source and all resistors except for the 1.2k must be in the calculation.

    Not sure why they quote Vsat=0v though.

    Here is the solution for the base current given an ideal transistor where the beta is always the same and base emitter voltage drop is always the same:
    E1 is the source voltage (12v in your case)
    E2 is the base emitter voltage drop (usually taken to be 0.7v for these approximations)
    R1 is the upper bias resistor (20k),
    R2 is the lower bias resistor (3.6k),
    R3 is the emitter resistor (220 ohms),
    B is the beta.

    For the values given in the schematic, we get:
    44.735677 microamps.
    Last edited: Oct 1, 2014