Does this amplifier model achieve a voltage gain of 10 as specified?

i wish to know if this amplifier meet the gain of 10 ?

Hi Nyame-The DC gain will be somewhere around 10 (ignoring the input and output blocking capacitors), but with the capacitor across the emitter resistor the AC gain will be much higher than 10.  In this case the AC gain will depend on the Hfe of the transistor.-Rick

Hi again Nyame.  I am as usual about to go to work so no time for simulations my end.  However quick calculations show that your DC design is just about spot on - your collector should be sitting at 6V, well done.As Rick says, your gain is huge.  Your input is (look at the peaks of your sine waves on the scales) about 0.02V (20mV) peak to peak.  Your output is just over a volt peak to peak.  so your gain is 1000(mV) / 20(mV) = 50.As Rick says, take that capacitor off the emitter resistor and you should be nearer the mark.  try that in your simulation and I'll get back to this later.Cheers / David

Hi David, thanks for your recommendation, when i pulled off the emitter capacitor, i witness DC offset on the simulator, 
i dont really know how to manage that, pleaes i need your help

cheersNyame 

Hi Rick, thanks for that recommendation, first the transistor gain is 165, A.C 10mili volts, 1KHz, there is an issue once the emitter capacitor is pull off, i witness DC offset in the simulator, please how can we avoid the DC offset ?  cheers Nyame 

please just a quick reminder, my circuit properties are transistor hfe 165, a.c 10 mili, 1KHz, vcc 12
cheers Nyame 

Nyame sorry, very busy, will reply in detail tonight.  Best // David

Hi Nyame-The answer about the offset is the same as it was the last time you asked HERE -Rick

Hi Nyame,I am a bit concerned that you seem to be struggling with what have been well explained principles in determining how to set up the amplifier with the required characteristics.  Have you actually got all the theory behind you to understand the basic circuit concepts (resistance, capacitance, impedance and transistors), or have you been plugging in values somewhat randomly into the simulator without doing the maths necessary to come up with values needed.Please understand that I am not being judgmental here, but you appear to be doing a course in electronics, and having some problems understanding  what is needed to achieve what you want. It could be that you need to do a bit of reading (there is a lot of theory accessible on the web), and perhaps if there are other students around, a bit of collaborative work may be of help, it's a good way of learning (aside from posting to this forum).  I wish you well in your endeavors,cheers,Richard

Hi Richard,  well said, but please is more of self learning process, and as you can see i always come up with something that needs a final touch not like am giving the assessment for the forum to do all for me, i always try to build 75% of my tasks ,and as a new student in electronics i take pride in learning by any means. 
please remember you cant get all the theory in a dayNo offence !cheers Nyame

Hi David !please take a quick look at what i did again to target the gain of 10cheers Nyame 

Hi Rick,please take a quick look at the circuit now cheers Nyame 

Nyame...job done! You have 0.02V peak-to-peak input, 0.2v peak-to-peak output, that looks like a gain of 10 to me.  And no clipping.  Nice work.Couple of points:- I see you increased your collector resistor from 3.7K to 3.8K.  Why did you have to do this?  The gain is effectively the collector resistance over the emitter resistance.   Apart from the actual emitter resistance (370 Ohms) you have the transistor emitter resistance (26/IE (mA) = 26/1.3 = 20 Ohms, so your actual emitter resistance is 370+20=390 Ohms.  so you could have used a 3.9K resistor in the collector and your gain should have been exactly 10.- you should get used to using preferred values where possible.  There is a table of preferred values here on EEWeb.  you should aim to use the E12 series where possible.  Yes, these values were intended for 10% tolerance resistors, and most resistors these days are 1% tolerance, but the E12 values are very standard and you will find them in ANY store or electronics shop, whereas other values you will probably have to special order or get online.  - If preferred values will not fit the precise values you need, by all means go up to the E24 or E48 series, or use a preset resistor so you can advust for the precise gain or output you need.- finally it may be possible to get your simulator to calculate the exact gain for you.  This varies form simulator to simulator, I think mine (multisim) can do it and must try to make it work sometime.  I grew up in the days when you would put the circuit on a breadboard and test it with real components and meters and oscilloscopes.  It always seems like more fun :-)Cheers / DavidSorry one final final thing. The meters you have on your outputs seem to be indicating the RMS value of your AC waveform, not the DC offset. Take the values indicated on your meters, multiply them by the square root of 2 (1.414) and you get the peak value which is exactly what is displayed on your scope waveforms. .  Are you sure that meter is set to DC, not AC??

Hi Nyame-I apologize.  I did not see that you added the load resistor across the output as suggested in the other post, but many of the comments from the other post are still valid.  With an output blocking capacitor you should not have any DC offset.Simulators are wonderful things, but at some point it's best to just build the circuit and see what it does in the "real world".I'm not the expert on circuit simulation.  I think your circuit is fine and the DC offset is an artifact of the simulator.-Rick

Hi David ! thanks  very much for the aid, my last task is to achieve open gain for the same circuit !  
please kindly go through the attached circuit and see if it meet the requirement of open gain cheersNyame  

Nyame...you should not need to ask me now.  Your output waveform is good, and your gain is now 1400mV / 10mV  = 140. I have taken peak values here instead of peak-to-peak.  996mV RMS on your output meter = 996 * 1.414  = 1408mV peak.All Good!

Hi David its been a while, trust you are doing good, please help me assess the attachment below to see if i solved it the right way cheers Ephraim 

Hi Nyame...trust you're well...could you post a bigger version of that, I cannot get it readable.  Put it in Paint and resize it to 200%, then post again.  Sorry, the pic posting in this forum is deplorable.....

Hi David i think is somehow clear here 

here is the results  i got 

on my breadboard 4.7k , 22k and 47k  are the valued  used  

 Hi David i also need to build this circuit and simulate  it  using NOR gate for the  process and switch/ NAND gate to create the correct input point, and build it truth table 

Nyame hi, thanks, I see you did post the diagram in another post as well.  I really don't like the way they are doing this.  Your results are correct in that your ratios are correct.  Let's look at what's going on here.in the first case in your table, you only have the 40K resistor in circuit.  This is actually 47K you say.  Now the feedback resistor across the opamp (you have shown 450 ohms) is probably 470 ohms?  So the gain of this circuit would be -(470/47000) = 0,01.  if this is the case then since your output was -0.083 mV your battery voltage would be 8.3 volts.Now lets take the case of your second line.  The gain of the circuit would be -(470/22000) = 0.1236.  Multiply this by your battery voltage of 8.3 and this gives us 177.3 mV - within half a percent of your measured 178 mV.  (I'd guess you were using 1% resistors?)You can do the same calculations for the other cases and your results should be accurate.  So how would I teach this better?Well for starters this is a digital to analog converter.  Each switch will have a "weighting" showing how much it will affect the output.  And each resistor should be an exact multiple of the feedback resistor - 100x, 200x, etc.  You can do this easily.  Make your feedback resistor 400 ohms exactly.  You can do this easily with  a 390 ohm and a 10 ohm resistor in series.   Then make your switch resistors multiples of this as follows
40K - 39K + 1K in series.  Weighting = 1
20K - 2 x 10K resistors in series    Weighting = 2
10K - one 10K resistor      Weighting = 4
5K - 2 x 10K  resistors in parallel  Weighting = 8
How do we get the Weightings?  Well the 40K resistor will let one "unit" of current through.  The 20K resistor is half of that so will let through twice as much current - 2 "units", the 10K will let through 4 times as much and the 5 k will let through 8 times as much current.So we can now calculate the gain of the circuit very precisely.  we jsut add up the weigntings of the resistors that are switched in and multiply it by 0.01 (because our 400 ohm feedback resistor is 0.01 of the  40K resistor.  So our table will now look as follows:

The weighting is a very important concept with Digital to analog converters (DACs) because it lets you calculate your output very easily.   As above it will be - (sum of weightings) x 0.01 x input voltage (in this case the battery voltage).You were asked to ignore the sign of the voltage (they will be negative) so I have too.if your weighting is 1, then your output will be  1 x 0,1 x 9V = 90 mVin your third row the weighting is 4, so  output - 4 x 0.1 x 9V = 360 mVin your 4th row you have weightings of 8 and 2, the sum of these is 10, sooutput = (8+2) x 0.01 x 9V = 900 mv
Another improvement I would make would be to make the input a precise voltage - say 5 volts which you could get from a 7805 or 78L05 voltage regulator IC.  The your results would be more predictable because they would not depend on a battery voltage which might not be exactly 9V - in your table your outputs would then be 50, 100, 200 and 500 mV.  In real life DACs a voltage reference is used like this to ensure that the output is very precise.Let me know if all this is clear or if you need more explanation.Cheers // David

Hi Davidthank you very much, i most say  your a good teacher they way you explain things to me i really understand them and find it easy to explore, for that am very grateful. 
there is a logic task i also sent you i guest you did not see it ,let me post it again cheers Ephraim 

 Hi David i also need to build this circuit and simulate  it  using NOR gate for the  process and switch/ NAND gate to create the correct input point, and build it truth table 

Nyame I did see the logic problem - just have not had time to reply.  However - I am not here to do your work for you - you won't learn anything that way!   I'll give you some hints.  Fill the table out in the ABC columns, starting from 000, 001 ....etc up to 110, 111 at the bottom.Now you have to  fill in the D column, whether it is a 1 or 0 for each combination of ABC.You have three AND gates which will only output a 1 for a particular combination of ABC. they will be 0 at all other times.  The three AND gates feed into an OR gate.  So if any or the three AND gates outputs a 1, the OR gate will output a 1 to D.So what you have to do is decide which combinations of ABC will cause the three AND gates to output a 1, and then fill in 1 opposite that combination in your table.A on its own means A must be 1 to be True, if A is 0 then it is False.A with a bar over it means "NOT A" which means if A is 1 is it False, if A is ) it is True.An AND gate requires True or "1"  conditions at all its inputs to output a 1And that's given you way enough information to complete this yourself.  do so and post your completed table, and I'll tell you if you're right.> I must say  you're a good teacherThank you!  You're a good pupil in that you do reply and thank me, most posters on this forum do not bother.  If you do that I'm happy to give you my time.  Talk soon.