Class A Amp Load and Bias Problem

Discussion in 'The Projects Forum' started by logans-electronics, Sep 6, 2009.

  1. logans-electronics

    Thread Starter Member

    Sep 1, 2009
    I am looking for formulas and a model circuit to drive a 8 ohms speaker using a TIP35C or TIP41C Transistor....Basic Class A amp.. I will be using an LM833 op amp to send the audio signal in. I can adjust my rail voltage higher or lower as you recommend.
    See attached.....
    Max wattage is not important....may just 10-20 for now. I did check out the LM386 and looked at using output IC's. but I am just trying to understand how amplifiers work.

  2. hobbyist

    Distinguished Member

    Aug 10, 2008
    At 30v. RC would have been around 20 watts, so I adjusted the supply down to 15v. to give a lower wattage resistor of around 10 watts for RC.

    I also used a 2n3055 pwr. transistor, because that's all my simulator has available in the power range.

    The input Vpk. should be around 2v. to avoid distortion on the output.

    No gain to be had but it supplies the needed power to output to a 8 ohm load.

    Very inefficient due to it's inherent characteristics of class A amp.

    class A pwr. 2n3055.jpg
  3. logans-electronics

    Thread Starter Member

    Sep 1, 2009
    Thanks for doing that... I will have to look in to the Circuit Maker Program.
    I am using Multisim....

    Could you show me some of the formulas to develop the 4 resistors base on your selected transistor?

  4. Audioguru

    New Member

    Dec 20, 2007
    You do not have the transistor biased correctly again. Its base current is so low that its emitter voltage is almost nothing.

    Your circuit is missing an important supply bypass capacitor again.

    Your input capacitor has backwards polarity again. Its value is so low that the transistor will pass only very high frequencies.

    You should never have DC current in a speaker like you have here.
  5. hobbyist

    Distinguished Member

    Aug 10, 2008

    Power amp is to supply power to a low load resistance.

    So sometimes it has no voltage gain, meaning what goes in comes out,
    however the samae voltage going out as coming in has the proper current needed to power the load where the signal input circuit would be bogged down by such a low load.

    with that in mind this amp is designed with no voltage gain. Just power output.

    1) chose to make RC (7.5 ohms) at least close to the load value, of 8 ohms.

    2) with Rc in parrallel with the load the total output res. is around 4 ohms.
    so I chose RE to be 3.9 ohm resistor.
    this now makes a voltage gain of unity (1) but gives a DC bias at the emitter to keep the transistor in it's linear region. Puts about 3V. across the collector emitter of the transistor.

    4) calculate IC ~= 1A.

    5) RE x IC = VE = (1A x 3.9ohms = 3.9V.)

    6) VE + Vbe = VB (3.9V. + 0.7V. = 4.6V.)

    7) choose RB1 (res. base to ground) to be about( 20 x RE ) = ~ 82 ohm resistor.

    8) calculate IRB1 (current thru RB1) = (VB / RB1) = 56mA.

    9) RB2 (the top base res.) = (VCC - VB) / IRB1 = (15V. - 4.6V.) / 56mA. = 185 ohms.

    adjust resitor values to nominal values
    RC = 7.5 ohms
    RE = 3.9 ohms
    RB1 = 82 ohms
    RB2 = 180 ohms.

    Build test (simulator) and adjust values as needed.