How to implement an Upper Cut Off Frequency?

Discussion in 'Homework Help' started by Denny1234, Apr 17, 2008.

  1. Denny1234

    Thread Starter Member

    Feb 17, 2008

    Im designing an common emitter amplifier and I have the following specification.

    Frequency response:
    Lower cut-off frequency: 40 Hz
    Upper cut-off frequency: 1 MHz

    I have worked out a suitable value for a coupling capacitor to give me that lower cut off but I have never covered how to implement an upper cut off frequency before?

    I would know how to calculate the desired value fine but where in a common emitter amplifier would you place a capacitor to give you an upper cut off frequency?

    Thanks for any help.
  2. Caveman

    Senior Member

    Apr 15, 2008
    You can figure this out pretty intuitively. Basically, the gain of a common emitter amplifier is the collector resistance divided by the emitter resistance. You can substitute impedance for resistance if you like.
    G = Zc/Ze
    If you want a low pass filter (which you do), you want the gain to go down as frequency goes up. From the gain equation, you want Zc to go down or Ze to go up as frequency goes up.
    Zc goes down if you put a capacitor in parallel with Rc. Technically, you could also put this capacitor from the collector to ground.
    Ze goes up if you put an inductor in series.

    One more thing. If you put the capacitor from the collector to the base, you will get an effectively bigger capacitor via the miller effect. Look it up.
  3. mik3

    Senior Member

    Feb 4, 2008
    make a band pass filter by using an op-amp and put it before the base of your amplifier. Search in google to find this kind of filters, they are called active filters. If you want you can make a band pass filter by using inductors and capacitors but active filters are better because you dont have to use a bulky inductor on your PCB.
  4. Denny1234

    Thread Starter Member

    Feb 17, 2008
    Thanks for your help guys.

    Ive had to split my emitter resistor into 2 (bypassing one of them with my parallel capacitor to ground at my cut off frequency), to give me the midband gain i require but I think my circuit is fine.

    One really stupid question though, this is my first ever circuit and i have never used multisim. How what i measure to see if my gain and frequency responses work, i think i need to use the bode plotter but im not sure how i would connect it up?. Sorry for the newbie question but this work is a year ahead of where I am at (ive just started my course but i think i get the basics.
  5. mik3ca

    Active Member

    Feb 11, 2007
    an inductor acts as a high resistance at high frequencies and acts as a low resistance at low frequencies.
    a capacitor acts as a low resistance at high frequencies and acts as a high resistance at low frequencies.
    the resistances (reactance) of a capacitor and inductor are equal at their resonant frequency which is found using: 1 / (2 * pi * sqr(L * C))
    sqr = square root, and pi = the pie function on your calculator.

    What you need is a low-pass and a high-pass filter. You can use the equation above to determine the cut-off point.

    you could connect the signal in question to ground through a capacitor, or you could add an inductor in series with your signal.