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I am a unskilled in this domain, so I hope you can bear with me a little. I am doing a project for my teacher and I chose to do this

I tried to search for the best suitable values for the components and I put them on Pspice. enter image description here

V5.Vamp=0.8v,fre=1k,Voff=0v.

After stimulating using

I get this graph

enter image description here

I am not sure if this is how it supposed to be. Is there anything wrong that I did?

Thanks in advance.

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  • \$\begingroup\$ Looks good BUT reduce drive voltage of C input to about 0.8V an see what happens :-) \$\endgroup\$
    – Russell McMahon ♦
    Commented Sep 3, 2015 at 6:21
  • \$\begingroup\$ Are you STILL getting the output waveform shown with Vin = 0.8V. If so something is "wrong" \$\endgroup\$
    – Russell McMahon ♦
    Commented Sep 3, 2015 at 6:57
  • \$\begingroup\$ 3rd time lucky. As Dave Tweed pointed out I made a silly mistake in overlooking the effect of C2. I suggest you try the 20 mV input as per my amended answer and THEN remove C2 and use 0.8V and see what happens in each case. Should be educational. \$\endgroup\$
    – Russell McMahon ♦
    Commented Sep 4, 2015 at 2:49

2 Answers 2

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Due to C2 the amplification of the circuit is very high. Russell's comment about the gain being 4.7 is correctly based on the ratio of collector resistor to emitter resistor BUT the emitter resistor is shunted with 20 uF and that has an AC impedance of 8 ohms at 1kHz hence your gain is through the roof. Remove C2 and check that the gain is about 4.7.

What you do next depends on what you want the circuit amplification to be and where the low frequency gain starts to reduce but, needless to say it usually involves restoring C2 to its former position but placing a resistor in series with it.

I estimate that you should be able to get a clean 7 or 8 volts p-p on the output and, with a suitable resistor in series with C2 you could achieve gains of up to 20 without much distortion. Without C2 the gain is approximately 4.7.

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  • \$\begingroup\$ However, note the very low values used in the bias network. This low input impedance, coupled with the relatively small value for C3, is going to both attenuate the input signal and introduce a significant phase shift. The latter is quite evident in the plot. \$\endgroup\$
    – Dave Tweed
    Commented Sep 3, 2015 at 11:14
  • \$\begingroup\$ @DaveTweed It's all down to where the OP pitches the signal input frequency. At 1kHz, C3 is 159 ohms and will be a little bit attenuated by the bias. At 20Hz, C3 is about 8k and will be significantly attenuated. \$\endgroup\$
    – Andy aka
    Commented Sep 3, 2015 at 11:32
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Your choices of components look good.
Ideally Vc_DC shouild Vbattery/2 and the 4.635V you see is entirely close enough to Vbattery/2 = 9V/2 = 4.5V/2 for this purpose.

(1) As shown the gain is very high. It is ABOUT Rc/Reinternal
where Reinternial is the effective emitter resistamce of the transistor and is approximately equal to 26/Ie_ma.
The reason for this formula is based on the behaviour of a silicon junction with current - just accept it for now. Here Ic ~= Ie~= V_r6/R6 = 0.93 mA.
So Reinternal = 26/0.3 = 28 Ohms.
Gain ~= Rc/Reinternal = 4.7k/28 =~ 168

For 4.5V max swing on collector Vin_half_cycle_peak = 4.5/168 = 27 mV So an input of about 20 mV AC 1/2 cycle peak will about give maximum allowable Vc.
As your simulator gives 5V peak for a 5V signal try Vin = 20 mV.

(2) Next, remove C2.

The gain, as you probably know is ~= R4/R6 = 4.7k/1k = 4.7 ~= 5
so with ~- 4.5V max swing at the collector you can expect the transistor to be driven into on or off modes by some of the saignal. c With a gain of 5 and swing of 4.5V the max Von you can tolerate is ABOUT 4.5/5 = 0.9V.
If you drive it with say 0.8V 1/2 peak or less the output will "make sense"

At present when Vin is too high the transistor turns hard on.

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  • \$\begingroup\$ I changed Vamp to 0.8 as you recommended, and edited the post. \$\endgroup\$
    – TU1
    Commented Sep 3, 2015 at 6:54
  • \$\begingroup\$ You forgot about the emitter bypass capacitor. \$\endgroup\$
    – Dave Tweed
    Commented Sep 3, 2015 at 11:10
  • \$\begingroup\$ @DaveTweed Um. How did that happen. Yes. I was completely aware of it being there and (of course) it's effect, but somehow ... Must have been late :-). || Answer amended and added to. \$\endgroup\$
    – Russell McMahon ♦
    Commented Sep 4, 2015 at 2:49

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