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I have this common-emitter BJT amplifier:

schematic

simulate this circuit – Schematic created using CircuitLab

The text book says that with a bypass capacitor, the gain is -Rc/re, where Rc is resistor R6 and re is the internal resistance of the BJT. The internal resistance of the BJT is found by computing VE = VB - VBE (VB is found by voltage division, VBE is 0.7V), then IE = VE/RE = VE/R7, then re (internal resistance) = 25mV/Ie

Now I want to simulate this circuit and use simulated values to directly compute the gain by dividing the output by the input voltage, so I can compare with the theory, but I do not know exactly what is the input and what is the output. Are they taken right before and after the BJT, or right before and right after the caps, or...?

I have tried to take different combinations of output/input but the result does not even come close to the theoretical -Rc/re.

Any guidance will be appreciated. Thank you.

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Did you even look at this schematic? The transistor isn't being biased. It looks like that's what R5 and R4 are intended to do, but they are explicitly not connected to the base. What possible purpose do you imagine for R5 and R4? –  Olin Lathrop Jul 15 '13 at 11:41
    
Yeah, you are right. R4 and R5 share the same node as C1 and transistor base. I made a mistake while drawing the schematic but my question remains. Thank you. –  Douglas Edward Jul 15 '13 at 11:48
    
The schematic has been fixed. –  Douglas Edward Jul 15 '13 at 11:49
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2 Answers 2

There are several gains associated with voltage amplifiers. Consider the following model

enter image description here

In this model, the gain \$A_{VO} \$ is the open circuit voltage gain of the amplifier which, in your circuit, is given by \$R_C/r_e\$

But note that the output resistance of the amplifier (which is about \$R_C\$ in your circuit) forms a voltage divider with the load \$R_L\$.

So, the loaded voltage gain is:

\$A_{V} = A_{VO} \dfrac{R_L}{R_{out}+ R_L}\$

But, note that the input voltage \$V\$ is less than the source voltage due to the voltage division between the source resistance and the input resistance of your amplifier. Thus, the loaded gain with respect to the source is:

\$A_{VS} = A_V \dfrac{R_{in}}{R_S + R_{in}} = A_{VO} \dfrac{R_L}{R_{out}+ R_L} \dfrac{R_{in}}{R_S + R_{in}}\$

So, you cannot expect to measure anything close to the open circuit gain \$R_C/r_e \$

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Any guidance will be appreciated: -

It won't come near to \$ R_c/r_e \$ because you have an ac impedance connected to the collector which reduces the effective value of Rc down to about 700 ohms. Try making the 800 ohm output load more like 800kohm and see what happens.

You have also got a 12k input resistor that will reduce the signal appearing on the base and this will also give you the impression that gain is lower. Try making this a short.

Also, what frequency are you using in your simulation? The 1nF capacitor at the input will have an effective impedance of 159kohms at 1kHz.

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If I trace voltage after C3 divided by voltage before C1, I get this i.imgur.com/1TFHUr0.png so the gain is somewhere between -20 to 20 (shouldn't it be strictly negative??) Whereas when I compute -Rc/re... I get: VB = 7.3 V (voltage division), so VE = 6.6V, so IE = 6.6V/8kohm = 0.825 mA, so re (internal resistance) = 25mV/0.825mA = 30 ohm....so -Rc/re = -8kohm/30ohm = -266 I am using 3khz frequency. Surely there must be something wrong in my calculations, either the theoretical Rc/Re or in what I take as the output and as the input? –  Douglas Edward Jul 15 '13 at 12:37
    
@DouglasEdward to measure gain you have to establish the signal amplitudes as either RMS or peak-to-peak. Then you divide one number by the other. You can't hope to measure gain by numerically dividing one signal trace by the other because how would you account for instances of zero on the input that don't quite match a zero on the output - you'll get infinity! Show the input voltage trace and the corresponding output voltage trace on the same amplitude graph then let's go from here. Make the 12k 0ohms and the 1nF, 10uF too –  Andy aka Jul 15 '13 at 13:30
    
@DouglasEdward also read what I said about the output load of 800 ohms - the signal will see it shunting the 8k in the collector and therefore it will reduce your expected V_{gain} to less than one-tenth of what you expect –  Andy aka Jul 15 '13 at 13:33
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