[SOLVED] How to determine the small-signal voltage gain of Common mode BJT amplifier?

[Max_JVC]

Hello Guys.

I have tried to calculate the voltage gain of the following circuit without measuring, only with the values that I have of the components. There are some formulas in texts like Sedra and Boylestad, but they are nowhere near what I actually measure with the multimeter. The circuit is as follows:

The expression for the calculation with load that appears in the mentioned texts is the following:

\[ A_{v_L}=\frac{V_o}{V_i}=-\frac{R_C//R_L}{r_e} \]


I hope you can help me greetings! (Sorry, my English is very poor!)

 

[FvM]

You didn't tell your calculation or measurement results, thus we can't see which are wrong. I can only guess that you probably don't know what re in the expression means and how it's determined. It's 1/gm and can be estimated with 26 mV/Ic at 25 °C.

 

[KlausST]

Hi,

C2 polarity seems to be wrong.

There are some formulas in texts like Sedra and Boylestad, but they are nowhere near what I actually measure with the multimeter.

.. then it's a good idea to tell us
* the simulation results and the
* the measured results

And how you measured them.
AC measurement is not a simple as DC measurement. There are different AC measurement and calculation methods with different results.

Klaus

 

[Max_JVC]

You didn't tell your calculation or measurement results, thus we can't see which are wrong. I can only guess that you probably don't know what re in the expression means and how it's determined. It's 1/gm and can be estimated with 26 mV/Ic at 25 °C.

Hello.
Simulation results:
IC=508mA; IE=518mA; IB=9.97mA
VCE=19.89V; VCB=-18.92V; VBE=0.97V
VRL(rms)=289.236mV

I know what is re in the expression.

Hi,

C2 polarity seems to be wrong.


.. then it's a good idea to tell us
* the simulation results and the
* the measured results

And how you measured them.
AC measurement is not a simple as DC measurement. There are different AC measurement and calculation methods with different results.

Klaus

Why is the C2 polarity wrong? The voltage at that point never reaches 0, it is always positive.

I did the measurements like this:

Oscilloscope shows:

Thanks!

 

[Max_JVC]

I can only guess that you probably don't know what re in the expression

I did the calculation as follows:

\[ A_{v_L}=\frac{R_L//R_C}{r_e}=I_C\frac{R_L//R_C}{V_T}=508mA\cdot\frac{5\Omega}{26mV}=97.69 \hspace{2mm}V/V \]

According to measurements:

\[ A_{v_L}=\frac{289mV}{40mV}=7.25 \hspace{2mm}V/V \]

I increased the signal frequency and the capacitors value to mitigate the low frequency gain attenuation but the latter value goes up to 11 V/V hopefully.

 

[FvM]

You still didn't report the measured voltage gain.

I see now, that "what I actually measure with the multimeter" actually means simulation. A real 2N3904 wold immediately burn with about 10W power dissipation.

The circuit has several issues:
1. C2 capacitance is far to small, causing a large voltage drop
2. 2N3904 isn't suited for the intended current and power level. Emitter and base bulk resistance has a large contribution to effective re. Respectively the estimation 26 mV/Ic isn't applicable.

C2 polarity is still wrong (you may want to visualize the node voltages in your simulation), but that's irrelevant for the present discussion. Multisim doesn't care for capacitor polarity.

 

[KlausST]

Hi,

C2 polarity is still wrong (you may want to visualize the node voltages in your simulation), but that's irrelevant for the present discussion. Multisim doesn't care for capacitor polarity.

The discussion is "measured results vs simualted results".

While the simulation does not care about capacitor polarity, in the real circuit a wrongly oriented capacitor may shift DC operating point of the BJT.... modifying the AC measurement results.

Why is the C2 polarity wrong? The voltage at that point never reaches 0, it is always positive.

You have a simulation tool, so use it.
A capacitor has two connections. There is no single "at that point". You always need to take both nodes into account.

So either directly measure the DC voltage across the capacitor....
or measure both nodes WRT GROUND and decide which is the more positive one. (It should be the one marked with a "+")

Klaus

 

[FvM]

You can drop in a power transistor like 2N3055 and see a much higher gain. Don't forget to increase C2.

 

[LvW]

Typing error? Rc and RE are 10 resp. 4.5 Ohms ?