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I've got the following small signal amplifier that I am simulating on multisim. The gain is around 5 and the setup works to amplify the given AC signal. The problem is that I have played around with the frequency of the AC source from 1kHz to 100MHz and haven't noticed any change in the gain of the amplifier. Is this supposed to happen ?

I would've thought the Gain would have been affected etc or that I would have seen something different ? I'm a bit puzzled and would like to know how the frequency of the input voltage affects the output signal in such an amplifierenter image description here

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  • \$\begingroup\$ Try to reduce the Cin capacitor value to10nF. What is a load of this amplifier? Also, is a BJT used in the simulation model includes any parasitic capacitance? \$\endgroup\$ – G36 Mar 11 at 7:46
  • \$\begingroup\$ There is no load. It's just connected to an oscilloscope at output to view signal. And no , there is no parasitic capacitance \$\endgroup\$ – user122343 Mar 11 at 7:47
  • \$\begingroup\$ If this is the case then add the Ccb capacitor (10pF) into a circuit. \$\endgroup\$ – G36 Mar 11 at 7:49
  • \$\begingroup\$ Also, add a load resistance 22k and reduce C3 to 6.8nF. \$\endgroup\$ – G36 Mar 11 at 7:50
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    \$\begingroup\$ Try using a real transistor model. \$\endgroup\$ – Andy aka Mar 11 at 8:02
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At the moment, you're directly applying a "hard" voltage to the base of Q1. That is unrealistic and can never be done in a real circuit. Try adding a resistor of 1 kohm in series with C2.

Normally in this circuit the collector-base capacitance of the transistor is the main factor that limits the bandwidth. This capacitor is often referred to as a Miller capacitance because the value of the actual collector-base capacitance is enlarged by the gain of the circuit, this is called the Miller effect.

When you apply a "hard" voltage at the base then this Miller effect is cancelled out so you will see an unrealistic high bandwidth value.

Other factors that influence the bandwidth are load capacitance, which is zero in your circuit which is also unrealistic. When you build this circuit and measure at the output with an oscilloscope probe you add a few pF of capacitance since the oscilloscope's probe has some input capacitance.

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  • \$\begingroup\$ Thank you. I appreciate the feedback you have given me. I was tryna see how the circuit responded but I understand what you said about it. Furthermore, can you provide me an explanation as to why the frequency of source affects the gain of the amplifier ? Is it due to the reactance of the coupling capacitor at input which now changes based on the frequency ? \$\endgroup\$ – user122343 Mar 11 at 9:23
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    \$\begingroup\$ why the frequency of source affects the gain of the amplifier ? You mean: Why does the gain of the amplifier change over frequency? That's because there are frequency dependent impedances present in the circuit: capacitors. To dive deeper into this, study how to do "small signal analysis" of such an amplifier. \$\endgroup\$ – Bimpelrekkie Mar 11 at 9:52
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A tale of two transistors

Generic NPN transistors often contain no parasitic capacitances, and often have other "ideal" properties. They are useful to see how parasitics affect frequency response. When a real transistor model is substituted, capacitances and other non-ideal effects become apparent.
In LTspice, the generic, default transistor (called NPN) yields an unrealistic response at the upper frequency end of a sweep (green). When a 2N2222 model is substituted, frequency response rolls off around 1 MHz (orange).At lower frequency, both transistors yield similar corner frequency (3Hz) due to the 10uf capacitor reacting against the bias resistors.
Mid-band gain of the 2N2222 may be higher than the generic NPN because 2N2222 \$ h_{fe} \$ is greater than generic.
generic NPN versus 2N2222

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That amplifier has two cascaded high pass filters (assuming you add a load resistor).

The filters cut-off frequencies, fc = 1/(2*piRC)

Where, for the input filter R = Rs + (((B+1)R4)//R1//R3) say B = 100 and Rs = source resistance = 0 Ohms.

And for the output filter R = RL + R2

I haven't done the maths but I would expect an overall high pass cutoff frequency of somewhere around 20Hz, way lower than the 1000Hz which you are injecting.

To put a low pass limit on the bandwidth you need to add a small cap across R2 or from the bottom of R2 to ground or across the load or from the top of the load to Vcc. Any one of these 4 positions for the cap will have a low pass effect limiting the bandwidth.

then low pass limit = fc = 1/(2*pi*RC) where R = RL//R2

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