I am trying to simulate a circuit from an exercise. I am confident that I have calculated everything correctly, but unfortunately, the simulation does not match the calculation. The amplification in the upper half of the waveform is weaker than in the lower half. I suspect that the cause of this issue is the parameters of the transistor I selected. However, I don't have enough knowledge to adjust the SPICE model of the transistor to make the simulation match the calculations. Could you please help me with adjusting the model? Or is there another reason why the amplification is not symmetric in the simulation?

At first, consider this simple circuit:


Given:

1. Current gain if the transistor is B = β = 200
2. Early voltage \( |U_A| \) = 50 V
3. Supply voltage \( U_b \) = 4 V
4. Load resistor \( R_L \)= 50 Ω
5. Collector resistor \( R_C \) = 250 Ω
6. Thermal voltage \( U_T \) = 26 mV

The input characteristic figure of the transistor is given as well. I had to determine the value of the resistor \( R_B \) that sets the base-emitter voltage \( U_{BE} \) of 0.7 V at the operating point. To do this, I wrote the following equation and drew a load line:
\( U_b = I_B R_B + U_{BE} \implies I_B = (U_b - U_{BE}) / R_B = \frac{U_b}{R_B} - \frac{U_{BE}}{R_B} \)
1. \( U_b = U_{BE} \implies I_B = 0\)
2. \( U_{BE} = 0.7 V \implies I_B = 40 \mu A \)


With this, I was able to calculate the value of \( R_B \):
\( R_B = \frac{\Delta U_{BE}}{\Delta I_B} = - \frac{0.7 V - 4 V}{40 \mu A - 0 \mu A} = 82.5 k\Omega \)

The next figure shows the output characteristic curves of the bipolar transistor from the circuit. I had to draw the corresponding load line for the resistor \( R_C \) and to mark the operating point. According to the diagram, the collector-emitter voltage \( U_{CE} \) is 1.9 V at this operating point:
1. \( U_b = I_C R_C \implies I_C = \frac{U_b}{R_C} = \frac{4 V}{250 \Omega} = 16 mA \) (if \( U_CE = 0 V \))
2. \(I_C = 0 A\) if \( U_{CE} = 4 V \) (= \( U_b \))


According to the given circuit, it is a common-emitter amplifier. The gain of this amplifier can be calculated as follows:
\( A = - S r_a = -13.22 \)
where
\( S = \frac{I_C}{U_T} (1 + \frac{U_{CE}}{|{U_A}|}) = \frac{8 mA}{26 mV} (1 + \frac{1.9 V}{50 V}) = 319.385 mS \)
\( r_a = (r_{CE} || R_C || R_L) = 41.4 \Omega \)

Now to the simulation. I took the bipolar transistor 2N2222 and adjusted its model as follows (bold marked values):
.model own2N2222 NPN(IS=1E-14 VAF=50 BF=200 IKF=0.3 XTB=1.5 BR=3 CJC=8E-12 CJE=25E-12 TR=100E-9 TF=400E-12 ITF=1 VTF=2 XTF=3 RB=10 RC=.3 RE=.2 Vceo=1.9 Icrating=800m mfg=NXP).

The amplified signal is blue. Its maximal positive value is around 100 mV but I expected to see the value around +132.22 mV (according to the gain A calculated above). Its minimal negative value is around -135 mV (almost as expected).

 

Notice the use of an emitter resistor to stabilize the gain as emitter current changes. The would preserve the symmetry of a symmetric waveform applied to the base,

 

For a grounded emitter circuit you use the transconductance of the BJT, not its current-gain, to determine the AC voltage gain.
The transconductance is Ice / Vt where Ice is the collector-emitter current and Vt is the thermal voltage (≈26mV@ 25°C), thus the gain is very sensitive to temperature.
So for example, for a 1mA emitter current, the gm is 1mA / 26mV = 38mS (S = A/V where A is the collector current and V is the AC base voltage).

If you want to use the current-gain (Beta) to determine the gain, the you need to put a resistor in series with the base to convert input voltage into base current.

Adding an emitter resistor as sparky did, adds negative feedback, reducing the effect of Beta variations to stabilize the gain, which is then determined by the collector resistor value versus the emitter resistor value.

 

yes agree The circuit built for AAC It was 3.3V using arduino supply, is now modified for 4V.
R3 establishes a feed back path. The R1/R3 divider center node is set voltage above 800mV while the resistor values
that will give enough current 3k and 1.4k for 4V operation. Your operating point will be helpful in correcting this.
3k + 1.4k and Voltage is 4V, my simulation measures 75uA DC for Base current when C3 is disconnected.
The real improvement will be adjusting R2 and R6 400KHz wave shape can approach resonance and %THD reduce harmonics.

C3 bypass electrolytic capacitor between collector and ground can further improve a low voltage circuits.
Measure the base current when C3 is again connected. The explanation in algebra called polynomial expression.
The algebra procedure is more intuitive when you measure the feedback funny business which agrees with Kirchhoff.

Feedback was used way back but in 1927 while crossing the Manhattan river by ferry Harold Steven Black conceptualized negative feedback for amplifiers. Video below, In an interview Black explains "The flash of recognition" he says he does'nt know how it happened.

 

The gain is going to be affected by the load.
Make the load 10 times the value of Rc so that it has minimum effect.

The output is not symmetrical (positive side is saturating) because the base current is too low. Try a lower value of RB.
Also try reducing the amplitude of the input signal.

 

Revision (10 times or more using potentiometer) and a few adjustments. At 50Ω the amplitude drops 50% the THD remains 0.67% THD
A low noise 2N5089 would further reduce distortion. A single npn amplifier circuit has more potential than I realized,
Because the circuit draws 15mA. I think possibly giving up some gain might be necessary. R2 warmest 42mW, possibly needs capacitor at V1.

 

Thank you for your many answers. Although you suggested several solutions, I’m not sure I fully understood the cause of the asymmetry in the output. So, I would like to summarize it in simple words for myself. Could you please validate my summary?

The asymmetry in the output occurs because the base current becomes too low at certain points in time. To make the output signal symmetrical, a more sophisticated circuit is needed (for instance, as @Sparky1 suggested).

 

Ah. If it is symmetry you are after, then you need to apply negative feedback and sacrifice some gain.

Take the base bias voltage from the collector.
Bias the transistor so that the collector is at half the supply voltage.
Reduce the input signal amplitude.

Have a look at the first portion of this post.
https://forum.allaboutcircuits.com/threads/push-pull-amplifier-biasing.202590/page-3#post-1931565

 

I’m not sure I fully understood the cause of the asymmetry in the output.

The base current is a log function of the base voltage (the base-emitter junction looks like a forward-biased diode), thus the current on the positive peak of a sine-wave input is proportional larger than the negative peak.
That causes the asymmetry in the output you observed.
If you use a current signal input to the base, or add negative feedback as suggested, you can significantly reduce it.

 

The base current is a log function of the base voltage (the base-emitter junction looks like a forward-biased diode)

Did you perhaps mean "The base-emitter current is an exponential function of the base-emitter voltage"?
\( I_B = I_S (e^{\frac{U_{BE}}{U_T}} - 1) \)

I believe I now understand the cause of the problem well enough.

From the following points:

• The formula for the base current: \( I_B = I_S (e^{\frac{U_{BE}}{U_T}} - 1) \)
• The relationship between the collector current and the base current: \( I_C = \beta I_B \)
• The gain formula for a common-emitter circuit: \( A = - S r_a\), (where \( S \) depends on \( I_C \))

I can see how the alternating input voltage \( U_e \) , combined with the constant voltage \( U_b - I_B R_B \) affects the gain.

Indeed, with a larger \( U_e \), I get a larger \( I_B \), which leads to a larger \( I_C \). This increase in \( I_C \) results in a larger gain, which causes the output signal to become non-symmetric.

Now, the suggestion to reduce the amplitude of the input voltage makes perfect sense. By doing so, I decrease the difference in base current between the maximum and minimum values of \( U_{BE} \).

Background: Almost a year ago, I took an exam on "Electrical Circuits." This exam covered the basics of analog electronics, especially amplifiers. I prepared mainly from a theoretical perspective, focusing on calculations for given values. Now, I have some time to finally combine that theory with practical experience. The circuit from my first message in this topic was the first exercise involving BJTs.

 

The underlying reason for the asymmetry is that a transistor is a very nonlinear device. In order to get (reasonably) linear behavior out of it, you need to do one of two things: Operate it over a sufficiently small portion of it's charactersitic so that if behaves acceptably linear over that span, or add a feedback mechanism that servos the circuit so as to make it linear. In the first case, you are primarily trading range (of input and/or output) for linearity, and in the second you are primarily trading gain for linearity.

 

Did you perhaps mean "The base-emitter current is an exponential function of the base-emitter voltage"?

Yes.

 

The underlying reason for the asymmetry is that a transistor is a very nonlinear device.

That's a general statement with no explanation.
The non=linear relation between base-emitter voltage and base current is the primary reason for that non-linearity.