Using Transistor Datasheets: Transitional Frequency and Gain

Discussion in 'The Projects Forum' started by jh919, Mar 25, 2014.

  1. jh919

    Thread Starter New Member

    Mar 24, 2014
    Hello all,

    I am currently trying to give current gain to a signal which has spectral content primarily at 40KHz and about 167Hz. (The waveform is a series of 1ms pulses of 40KHz square wave spaced apart by 5ms).

    I am trying to use a common emitter transistor amplifier for this, which is shown in this diagram i've attached.

    Using a oscilloscope, it seems to work ok in amplifying the signal except the 1ms pulses almost completely their 40kHz content, and just become 1ms pulses of logic high voltage.

    I tried checking the datasheet of the transistor (CDIL BC108), and the transitional frequency is given as 150MHz.

    I then measured the voltage at my Rb, and Rc, and using V=IR, found out the Ib was 3.7μA, and the Ic was 0.26mA, so the gain should be about 59.

    According to these measurements then, and the equation Transitional frequency = Gain x frequency at which gain is at -3dB of low f gain,
    150 000 000 / 59,

    the -dB frequency should be 2.5MHz, and the transistor should have no problem in amplifying 40kHz content.

    Following this I measured the frequency response of the transistor myself, and found out amplification dropped to nothing after just after 10kHz.

    Can anyone tell me why my findings are diffrerent to the datasheet, where and how I'm going wrong with my understanding? Thanks in advanced!!!

  2. ericgibbs

    Senior Member

    Jan 29, 2010
    What is the driving source.?

    A 380K Base resistor is a very high value.!

  3. jh919

    Thread Starter New Member

    Mar 24, 2014
    Hi, I'm sorry to say I'd forgotten that capacitors will affect the frequency response. Just reconsidering that now
  4. jh919

    Thread Starter New Member

    Mar 24, 2014
    Oh, the driving source is the output of a PIC18, with logic high at about 3.5 volts
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    Are you trying to drive a crystal?

    380K in the base is a large resistance: what is the source impedance?

    An inverter such as this is not really a current amplifier either, but it should be able to pass 40KHz. It is working in the cutoff and saturation regions, it is not a linear amp as used here.
  6. ericgibbs

    Senior Member

    Jan 29, 2010
    With a PIC pin Hi of 3.5V, thats 3.5-.7= 2.8V/380K = 7micro amps Base current, the circuit as designed will not work as you expect.

    Also you say, you want current gain, that transistor configuration will not give a current gain.
  7. RichardO

    Well-Known Member

    May 4, 2013
    A couple of questions:
    Is the crystal a piezoelectric transducer?
    Did you do your measurement with the crystal in the circuit?

    If the answer to these questions is yes then your problem is caused by the capacitance of the transducer. The 40 KHz transducer I looked at (Kobitone 255-400ST16-ROX) has about 2400 pF of capacitance.

    The upper 3 dB frequency of your circuit is: 1/ (2 * pi * 38 K ohms * 2400 pF). This is only about 1.7 KHz!

    You need a much smaller collector resistor to get your bandwidth up. I would use a 1 K ohm collector resistor and a 4.7 K ohm base resistor. I would also increase the 0.01 uf coupling capacitor to 0.1 uF.

    Note that with these values the circuit will not be very power efficient but it will work good enough to get you going. You will want to use some kind of push-pull driver in the future.