Transistor Amplifier Advice - Page 1
Forum section: electronics / beginners

author: Mint.
author_rank: Frequent Contributor
timestamp: June 25, 2012, 08:14:26 am
content: So I am playing around with transistors and I want to see the current gain characteristics of the transistor. I have got in the book I'm learning with right now a circuit similar to this:However it does not give me much values or where to start if I were to build this. If I were to build this, would it amplify a square wave? Can somebody please direct me through the steps I have to undertake to make a transistor amplifier or to simply view the characteristics of the transistor in a more practical way than simulations.-Thanks in advance

author: Kremmen
author_rank: Super Contributor
timestamp: June 25, 2012, 11:50:34 am
content: That is the simplest possible common emitter amplifier stage. The resistors R1 and R2 form a voltage divider that biases the transistor into the "linear" region. The transfer curve is not exactly linear anywhere, but it is most distorted at the ends, so you get best results by using the middle part of the transfer curve. The resistor values should be reasonable in relation to the base current needed for the biasing, offhand we are talking kilo-ohms or tens of kohms. The cap is there to isolate this DC bias from the signal source, a value on the order of 1 uF should work, depending on what are the limits of your frequency range.With no emitter degeneration you will get a voltage amplification G = Rc / re', where re' is the intrinsic emitter resistance of the transistor, say ~25 ohms at ~1mA Ic (YMMV). So if Rc is say 1k, your voltage gain would be roughly 40.There are issues with this simple circuit: the resistor divider is not a true current source, and the intrinsic emitter resistance is nonlinear subject to collector current.This nonlinearity will have a significant impact as the collector current varies as a function of the input signal. Assume for the sake of argument that your Ic varies btw 0.5 and 2 mA , i.e. btw 50% and 200% of the idle current. That could mean the intrinsic emitter resiostance varies from ~12,5 ohms to 50 ohms. The gain varies accordingly, producing a gruesome amount of 2nd harmonic distortion.Adding emitter degeneration in the form of a resistor will help remedy that. So if you add a say 225 ohm emitter resistor to ground you create a 250 ohm emitter total resistance. Increasing Rc to 10k will keep the voltage gain at 40, although with significant increase in the source impedance of the amplifier. However when you earlier had a gain variation of 80 to 20, you now have only 42.1 to 36.4. Not exactly Hi-Fi but much better anyway.Addition: Just for practical experimentation, you may want to replace one (or even both) of the resistor with a trimmer. This way it is easy to test the voltage amplification of the circuit and see what effect the input signal will have.

author: vk6zgo
author_rank: Super Contributor
timestamp: June 26, 2012, 01:04:14 am
content: Quote from: Mint. on June 25, 2012, 08:14:26 amIt would not be ideal for a square wave,because such signals have a DC component which you would lose due to the input coupling capacitor.If you make the cap large enough,& can tolerate a bit of "tilt" in your square wave,it will be reasonable.As an experiment,try using various values of coupling cap,& see what happens to your square wave.Kremmen's posting is full of good stuff---read it several times!