where is the base voltage ?

Discussion in 'General Electronics Chat' started by Thevenin's Planet, Jun 4, 2009.

  1. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
    I recently performed a experiment on a voltage divider network as a front end to establish a small base biasing voltage. Placing a R1 as the resistor connected to the -Vcc, (for a PnP-pn2907 as Q1) and the base of Q1.Also the R2 was placed between the base and ground (which is positive in this case.
    The frist set of resistors is used for R1 and R2 is,R1 is one meg. ohms' and R2 220k ohms for the Q1 base bias. the collectro resistor (VRC is 1k ohms.VR2 as the base voltage is about .55 volts. IB=12.1 micro-amps. Ic=1 miliamps.VR1=15.5 volts.VCE=13.5 volts.VRC=2.7 volts.
    The second expertment is the same config grounded emitter directly to groundZ) a RC=1ooo ohms. The voltage divider is changed to R1=100k ohms and R2 (from base to ground)=220k ohms.According to the voltage divider formula there should be enough voltage across the base to be registered on the Volt Meter. Which says that the base is about 0.5 or 0.6 volts. VR1=14.5 volts. VR2=0.5 or 0.6 volt on the meter.VRC=11.4 volts
    In all serveral cases the bases seems to be locked to 0.5 or 0.6 volts and the R1's that is connected to -Vcc is almost carrying the whole -Vcc supply in each different rearrangement of the voltage divider network input circuits.
    What happen to the base voltage drop across the Rb1 ?

    Where is the Votage for the base ?
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    It sounds like what you are seeing is the forward biased base-emitter junction, The forward bias voltage is generally -0.7V which in this case is negative assuming that you are measuring the voltage relative to ground (0V).

    hgmjr
     
  3. Thevenin's Planet

    Thread Starter Active Member

    Nov 14, 2008
    183
    1
    Can it be understood that when emitter is grounded directly with external base resistor in parellel to internal emitter diode, the results are a base resistance voltage is locked to forward bias voltage.
     
  4. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    I think I have an idea of the circuit you are analyzing but I believe it is time for you to provide a diagram of the circuit to make sure that all is clear.

    hgmjr
     
  5. David Bridgen

    Senior Member

    Feb 10, 2005
    278
    0
    You talk about R1 and R2 forming a voltage divider and go on to say
    If there is no emitter resistor, R1 and R2 do not form a voltage divider. R2 is, as far as biasing the transistor goes, redundant.
    The bias current is simply the value of the supply voltage (minus half a volt) divided by the value of R1.

    In this case, the the voltage between base and V+ will always be Vbe.

    If you put a resistor in series with the emitter, the voltage between the base and V+ will be Vbe (call it half a volt) plus the voltage across the emitter resistor.

    At least one of those figures must be wrong.

    1mA through 1k is 1V,

    or 2.7V across 1k is 2.7mA
     
Loading...