how can this circuit be adjusted to operate smoothly and correctly, the transistor hfe is 165

how can this circuit be adjusted to operate smoothly and correctly, the transistor hfe is 165

Nyame... a well designed circuit (and this one is not bad) should be able to cope with fairly wide variations in Hfe without altering the operating conditions too much.  It just needs a minor adjustmentIf the transistor Hfe is 165, the the emitter resistor of 1K will appear at the base as a resistance of 165 x 1k = 165K at the base, ie across the Base to ground resistor of 2.6K.  This is a LOT more than 2.6K so we can disregard it.So a bit of calculation shows that the base voltage will be around 2.6 / (2.6+8) * 12 = 2.94V.  The base-emitter junction will have 0.6V across it, so the voltage at the emitter will be around 2.3V.  That means the emitter will have around 2.3 mA current 2.3V across a 1K resistor).The same current is flowing in the collector circuit.  That means the 3.9K collector resistor will have 3900 x 2.3 =8.97V across it.  So the collector is sitting at only about 3V above ground.  So the collector can move 9V positive (hence your output positive half cycles are good) but only about a couple of volts negative (which is why your negative half cycles are cut off).In this type of circuit you should aim to get a bit more than half the supply voltage at your collector - around 6.5 to 7 volts.  So again do the calculations - we have 2.3mA of current and we want around 5.5 volts in the collector resistor: R = V/I = 5.5 / 2.3 - 2.39K.  So I'd try a 2.2K or a 2.7K resistor in place of the 3.9K collector resistor.  Your output swings should be more symmetrical before clipping sets in, and it should appear on both half cycles  at around the same increase in input voltage.  That 8K base bias resistor you could change to an 8.2K, and the 2.6K to 2.7K, which are preferred values - those small changes won't make a lot of difference.I'm about to go out so don't have time to either put this into a simulator or breadboard it.  But I'm fairly confident this will help.  Note that my calculations above are a simplification and are probably not the way you've been taught - you'd be calculating more precisely, taking into account the emitter resistance of the transistor, etc.  But I got taught this quick and dirty way in a series of articles in the old Practical Electronics MANY years ago and it works well for most things (Alan W, any chance of finding those articles??)Let me know how you go / cheers / David

Thank Sir,The required voltage  gain for this Circuit has to be 4.the A.C Signal is 10 mili volts, 1 KhzSo changing the collector resistor will change the gain target Sir,my regards Nyame

There is another issues as well, the  16uF capacitor shunting the 1K emitter resistor at 1KHz has a low impedance, the AC gain is going to be high, and the amplifier will likely saturate with 100mV p-p base drive. It is common to split the emitter resistor and put the capacitor across a portion of it to tailor the frequency response and gain. You may need to reduce the input swing with the given circuit and values. Too much gain and the distortion will be high, there is a trade off.Cheers,Richard

Thanks Sir the voltage gain target is 4the A.C signal is 10 mili volts 1KHz,  also changing the collector will change the gain target sirCheers Nyame 

OK,  >The required voltage  gain for this Circuit has to be 4.That gives us extra information to work with.The emitter resistance of the transistor is roughly 26/IE (mA) which in this case is going to be around 11 ohms.  At 1KHz the 16uf (another non preferred value!) emitter capacitor's impedance is just under 10 ohms.  The total impedance is thus around 15 ohms.  The gain of the circuit originally would have been 3900 (collector resistor) / 15 (emitter impedance) or around 260.  No wonder the poor thing was clipping!In your original case you could have changed this by just omitting that 16uF cap on the emitter.  The gain then would have been  3900 / 1011 (collector resistance / emitter resistance = 1K + 11 ohms from transistor) or just under 4.No let's say you used a 2.2 K resistor in the collector circuit as I suggested above.  To get a gain of 4, you need an emitter resistance of 1/4 of 2.2K, IE about 550 Ohms.  To get this, you could split the emitter resistance of 1K up into:560 ohms from the emitter to a mid point470 ohms from the midpoint to ground, bypassed with the 16 uF capacitorThis is as suggested by Richard Gabric above (great minds think alike .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

Nyame...later....just spent a happy hour or so playing with Multisim (most of it trying to remember how it works as I have not used it for yonks I found I had to reduce the collector resistor a bit for ideal symmetry.  I used a 1.8K collector resistor, and on the emitter a 390 Ohm then a 620 Ohm with a 22uF cap across it to ground.  This gave me a gain in the region of 4.3 and an output signal around 8v p-p before clipping.  At your stated input (10 or 100 mv I think) you'd have no problems with clipping!You could use a preset on either the 390 ohm (470 ohm preset) or the 1.8K (2.2K preset) if required to get your gain to exactly 4.Job done!!  Let me know how you go.  Cheers // DavidHere's a screenshot but it won't be much good if it can't be bigger than this.I tried to post a scope shot but it came out too small to be of any use.
The green blocks are (from left) function gen, DVM (collector) , DVM (emitter), oscilloscope.I got 7.9V on collector, 2.3 on emitter with no signal.

thanks a million for that effort Sir please find what i got in my own simulation am not to clear with the results please find attached down side how a previous gain was calculated 
cheers Nyame 

NyameAs with my pics, they are very small (I am trying to get that fixed) but your input is 100mV Peak-to-peak, and your output is about 400 mV peak-peak, so you're spot on.  And no distortion showing.  The bottom pic is too small to read..  Your time scales for input and output are different, that's why the output waveform looks like it is higher frequency.  Is anything else not clear?  Are you ok with the calculations?>thanks a million for that effort Sir 
My name's David!!It's a great pleasure Nyame, you would be surprised how much advice we put in that never gets a reply, let alone any thanks.  So you are very welcome.Cheers // David

Thanks David !    Please to make things  more clearer for me,  kindly  explain to me how you spotted the gain of 4 in my Vo Simulation ?because from the simulation i can only spot 0.3 at the top of Vo
cheers Nyame 

David !i have tried to capture the calculation how the gain was calculated by my assessor, from graph, please am trying to be-careful with my correction in other to learn more, so am sorry for any inconveniences that will cause you
please find below attached graph  

from my Assessors calculation,  Vo was 3.3 so he subtracted 3.3 minus base voltage 2.9 which gave him 0.4then he said since Vi is 0.1 the gain is 4, that is exactly were am not clearplease kindly put me through cheers Nyame 

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

David !           Thanks a million,this task has been a complex one for me you have really elaborated everything i needed to know with this stage i most say. ! now i have two more taskfirst how can that circuit be adjusted to achieve again of 10 !using A.C requirements as 10mili volts, and 1 KHz,Vcc 12, transistor hfe 165, C1, C2 Capacitors uf
the last task is achieving open gain ! 
 A.C sign 10mili volts, 1Khz,Vcc 12, transistor hfe 165C1, C2 Capacitors 1 uf

cheers Nyame 

David !please does this amplifier meet the gain of 10 ?cheers Nyame 

Please find the graph