Connect with us

Class A Amplifier design!!!!

Discussion in 'Electronics Homework Help' started by ishaan3731, Mar 12, 2012.

Scroll to continue with content
  1. ishaan3731


    Mar 12, 2012
    i am given assignment by my teacher to design class A power amplifier that gives 3W output....
    the circuit must be implemented both by series fed and transformer coupled techniques!!!!!

    i have to:
    1)implement the design problem on papers.......then find the efficiency also for both cases

    2)design the circuits on multisim

    Actually i dont know much about which transistor will be the best suited for this problem and will give output close to 3W....please suggest a transistor and also help me on how to start solving this problem as i am weak in designing !!!!

    Also maximum efficency of series fed is 25% and of transformer coupled is 50%......... from datasheet i could get the dc load line and ac load line so that i can get values of Icmax,Vce(sat) and Hfe and determine the values of resistors...i am using voltage divider biasing........ so i need values of R1,R2(voltage divider network),Re,Rc and main is the value of Vin........(Vin is input signal) . i assume Vcc to be 18 volts!!!!!
  2. GreenGiant


    Feb 9, 2012
  3. john monks

    john monks

    Mar 9, 2012
    For the series fed you have one resistor in series with a transistor. For simplicity we'll say that the transistor is in series with a resistive load. Now you want the transistor to be half on so as to have the collector or drain go from 0volts to 18volts. That will be 9volts. Now the trick is to select a resistance that will give you 3 watts with a sine wave that is 9 volts peak to peak or 9 volts divided by the square root of 2. Then you use the formula R=(E^2)/P or R=((9/square root of 2)^2)/3. Now you have the series resistor.
    For the transformer coupled amplifier your voltage at the collector or drain for your transistor will be 18 volts. Now you must find the reflected impedance that will give you 3 watts with a sine wave of 18 volts peak to peak. Note, your current in the transistor will be zero at 36 volts and twice normal current at zero volts. Now you use the formula R=(E^2)/P or R=(18 volts/square root of 2)/(3 watts). This R = the impedance of the transformer. Your steady state current = (36 volts - 18 volts)/impedance. Now you take your supply voltage 18 volts divided by your current and you will find that the efficiency of the amplifier is 50%.
    Of the series fed amplifier will come out to 25%.
    This is assuming that there is no emitter or source resistor and that the transformer is perfect. This is never the case but you can come close.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day