1. sneha sini

    Thread Starter New Member

    Jun 10, 2013
    5
    1
    output of BFSK is high frequency for high modulating input and low frequency for modulating input 0.
    but when we do in practical, we got high frequency for modulating input 1 but same frequency for input 0. iam attaching the circuit diagram with this. instead of pnp transistor im using npn. we design for frequencies 1000- 1200 Hz.

    npn, when the transistor (Q1) is ON, Rc || Ra ; ie frequency increases.
    when Q1 off, Rb C plays ie, frequency decreases. (it acts as astable)

    for low modulating input i cant obtain low frequencies. (we obtain same frequency as input)
     
  2. sneha sini

    Thread Starter New Member

    Jun 10, 2013
    5
    1
    BFSK is Binary frequency shift keying
     
    PackratKing likes this.
  3. WBahn

    Moderator

    Mar 31, 2012
    17,768
    4,804
    Why did you replace the PNP transistor with an NPN transistor?

    What do you mean that you obtain the same frequency at the output as the input? Do mean that you get about 1270Hz when the input is HI and 150Hz when the input is LO?

    If Q1 is being operated as a switch, your timing resistor values would be (Ra||Rc)+Rb and Ra+Rb
     
  4. vk6zgo

    Active Member

    Jul 21, 2012
    677
    85
    When Q1 is turned on,Rc is in parallel with Ra ,reducing the value ot t=CR,& hence increasing the frequency.

    In this circuit,as it stands,you must use a PNP transistor,as an NPN will not turn on .

    Even if you invert the NPN,the circuit conditions are not correct.

    Either do it the easy way & use a PNP,or redesign the circuit so it works with an NPN.

    P.S. Actually,I'm not quite sure it will work properly with a PNP,as it all depends on the amplitude of the digital signal driving Q1 base.
     
    Last edited: Jul 22, 2013
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