C.E. Amp with Capacitor across BEJ

Discussion in 'Homework Help' started by jegues, May 12, 2011.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    Hello all,

    Does this circuit have a formal name to it? I've been trying to find it so I can read up on what this circuit actually does and how it differs from a typical C.E. Amp.

    What is the purpose of the capacitor across the BEJ? Certainly it's not big enough to serve as any type of coupling capacitor to eliminate DC for biasing purposes, correct?

    Any info on this would be greatly appreciated.

    Thanks again!
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    It's probably just getting you to think about the effects of base capacitance on frequency response.

    The cap is effectively in parallel with ac resistance (1+β)re and the 600k base bias resistance. The cut-off frequency will (or won't?) vary as the dynamic emitter resistance 're' varies, given 're' depends on Ie - which in turn relates to the value of VCE.

    Perhaps the role of the 100 ohm input resistance also needs some thinking about ...?

    Also need to remember the gain probably varies somewhat as the value of 're' changes. Gain will also influence the Miller effect (Ccb) so this is not quite as simple as one might think.
    Last edited: May 12, 2011
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    I just noticed it's not Vce that varies but Vcc, in relation to the effect on cut-off frequency. But I guess there will be changes in Ib, Ic & Ie as Vcc varies. Also the gain will change but presumably setting Cπ=0 removes the issue of the Miller effect...? Missed those conditions earlier.

    At Ie=0.3mA the effective current gain would be about 21. Which is very low.

    The effective AC base input resistance would be approx 1.82 kohm. The cut-off frequency would be about 16.8MHz(!!!) - does that make sense?

    If Vcc is doubled then Ie would approximately double changing the base input resistance to 910 ohm. The cut-off frequency would then be 17.66MHz. I'm not sure if this makes much sense.

    I decided to run a simulation using the 2N3904. One in fact needs an Rb of about 3.3M to get Ie down to ~0.3mA at Vcc=9V. Beta being much greater than 21. With Vcc=9V the upper 3dB point was about 7.8MHz (Ie=0.3mA, Av=30.6dB). With Vcc=18V the cut-off was about 5.7MHz (Ie=.83mA, Av=38.8dB). I suspect the simulation model was perhaps characterizing the Miller effect with the higher gain degrading the bandwidth.

    I wonder about the practical "value" of the question.
    Last edited: May 12, 2011