Hello again Nyame. OK, first things first. Your assessor's comments on your first circuit are valid (there is a 3.3V offset on your output - ie the output consists of your signal plus 3.1V DC) but as pointed out by the ifrst commenters including myself this was due to there being no DC path from the right hand side of the output capacitor to ground. In your simulator the voltmeter is ideal (ie is has infinite resistance). so the capacitor initially starts off with 0 (zero) volts across it. As there is no DC path to ground on the right hand side, the capacitor cannot change so it stays with 0 volts across it and it looks like you have an offset. In the real world this would not happen - even Digital voltmeters have an resistance of 10 Megohms or so, and would charge the capacitor and remove the offset. The 100K resistor that we recommended later did the same job. The 3.1V is the voltage on your collector for this circuit, which is far too low.The gain of that circuit was 4 because you had no capacitor across the emitter resistor. The emitter will tend to follow the base voltage (less 0.6 for the base-emitter junction, but for AC that doesn't matter). But the emitter and collector have the same current flowing through them (we can disregard the base current as it is tiny - 1/165 of the emitter current). So in the case above, you have the same AC current flowing through a 1K resistor in the emitter and a 3.9K resistor in the collector circuits. so the signal voltage across the collector resistor will be 3900/1000 = 3.9 - very near to 4.When you put the capacitor across the emitter resistor, it makes that resistor nearly zero ohms for AC (as above, it was around 10 ohms). The Emitter resistance of the transistor is now appreciable (26/IE in mA = 26/2.3 - 11 ohms). So the emitter resistance of AC purposes is the combination of the emitter resistance and the capacitor. Don't forget that capacitors are reactive so you can't just add them together. I estimated a total of about 15 ohms. So your AC gain is now 3900/15 = 260 - far more than you want. Thats 26V output for 100mV input - as it is powered from only 12V it is impossible for the circuit to deliver that much. If you look at the output graph in your original post it is very clipped - if you extend the waveform below where it is being clipped you'd have much more than the 6V that is there - so your gain would be at least 100. (I see your original circuit now has the 100K resistor on the output, you must have changed it?)OK, lets move to how we calculate the gain. an AC waveform can be expressed as the average value, the RMS value, the peak value, or the peak-to-peak value. I won't go into these here, but when you are dealing with waveforms on an oscilloscope then the peak-to-peak value is the easiest to use. Why? Because you have the lines on the 'scope display and the voltage settings to refer to. In your graph immediately above this comment, your simulator has kindly labelled the scales for you. We can see that the top peak of your input waveform is just hitting the 0.05V line, and the bottom peak is just hitting the -0.05V line. so that's 0.05+0.05 (if you like, 0.05 -(-0.05) = 0.1V - which is bang on what you set - 100mV input peak- to-peak. Now, your output has that DC offset on it, which complicates things. The top peak of your output is going just above the 3.3V line. The bottom peak is just above the 2.9V line. So the AC is 3.3-2.9 = 0.4V peak-to-peak. Which is 4 times more than your input, so great there, that's what you wanted.You said: from my Assessors calculation, Vo was 3.3, so he subtracted 3.3 minus base voltage 2.9 which gave him 0.4Don't get confused here. By "base voltage" he meant the voltage at the base of the waveform, NOT the voltage at the base of the transistor. I used "top" and "bottom" when referring to the waveform to avoid this confusion.
So in this case your circuit is working ok, but if you put much more input into that circuit is is going to clip again. As I said above, aim to have your collector voltage a bit more than half the power supply voltage - in the case I gave you it was almost at 8V. if your gain is out, split your emitter resistor and bypass the bit you don't want so you get your gain right. (in my example, I wanted an emitter resistor of around 450 ohms for the gain but the DC emitter resistance had to be 1K. So I used 390 ohms plus 620 ohms (total 1.01K) but bypassed the 620 with the capacitor so it would not contribute to the gain. In a circuit like this, your gain is (collector resistance / emitter resistance). By making at least some of your emitter resistance a physical resistor, you have far more control over the gain. But also make sure you have a good collector voltage as above. It's balancing act in a way, but it's not too difficult to get right.Cheers / David
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.This keeps the DC operating point as before (the total resistance is still ~ 1K) but uses a portion of it to control the AC gain to around 4.I have rounded to preferred values throughout (except the 16uF which would practically be 22uF).How does this work in your simulator?You should be able to put around 1V peak (2v p-p, x4 = 8V p-p output signal) without significant distortion.I might chuck this into my simulator later and see how it works.Cheers // David
