# Common-emitter amplifier simulation is giving me the wrong result

I'm trying to simulate a common-emitter amplifier with a BJT and check that the results match with the equations I have.

The circuit I have is basically this:

This is the simulation: https://easyeda.com/editor#mode=sim,id=e10ea3e41e3d46f0b3a0bdef250a14ac. (you can simulate it by pressing F8)

I input 1.2 V and I'm expecting an output of 6.33 V, but I'm getting 0.1 mV approx. according to the simulator. I suspect that I have misinterpreted the BJT datasheet, let me explain:

The BJT that is supposed to be used by the simulator is 2N2222 whose datasheet is here: https://datasheet.lcsc.com/szlcsc/1809200018_ST-Semtech-2N2222A_C118536.pdf

According to the above, the expected DC gain (beta) for this BJT is 75 when iC is 10 mA. So I have found an RI that causes iC to be 11 mA just in case:

iC = 11 mA
Vi = 1.2 V
iC = beta * (Vi - 0.6) / RI


which gives me RI = 4k ohms approx.

I calculate the expected output voltage with the following formula:

vOUT = Vd - ( ((vIN - 0.6) / RIN) * beta * RL )


Given that RI = 4k ohms, Vd = 10 V, vIN = 1.2 V beta = 75, and choosing an RL = 500 ohms, the above equation gives 6.3325 V.

However, as explained, the simulator gives me 0.1 mV approx. What am I doing wrong?

PD.: you can check my formulas here

• The minimum $\beta$ is 75, but the simulator model probably uses a "typical" value instead of the minimum value. Commented Oct 30, 2020 at 18:24
• I have tried it with many betas. I have been able to verify that the beta used by this simulator using a common-collector config. with the same BJT is 100, however the BJT is working under different conditions (current) in that schema. Trying to estimate the actual beta the BJT is using in my circuit gives me something like 197, it makes no sense at all looking at the datasheet. Commented Oct 30, 2020 at 18:28
• This isn't the way to design common emitter stages when there is no emitter resistor. Even a simulator can be a mile off. Commented Oct 30, 2020 at 18:30
• So I should add an emitter resistor? Commented Oct 30, 2020 at 18:30
• It depends on the rest of the circuit that might surround this single stage. Commented Oct 30, 2020 at 18:31

Trying to estimate the actual beta the BJT is using in my circuit gives me something like 197, it makes no sense at all looking at the datasheet.

Using the data sheet graph at 11 mA collector current: -

According to the graph β is a shade under 200. Using a minimum figure is bound to be creating a discrepancy with a simulator that will use typical values.

However, in this particular circuit it's just pot-luck that it turns out to be about right - without an emitter resistor you at at the peril of the vagaries of temperature and forward diode drop characteristic changes. And, you'll get too much gain variation along with significant signal amplification distortion.

I mean - just look at the graph from your linked data sheet and how much β changes with temperature and collector current. Nobody can design a single transistor stage like this without an emitter resistor.

• How is that the graph says something different than the table? Commented Oct 30, 2020 at 18:44
• @Martel, the table gives a minimum value, the graph gives the typical behavior. Commented Oct 30, 2020 at 18:46
• @Martel the table is giving you minimum values and not typical values. Commented Oct 30, 2020 at 18:46

The minimum β is 75, but the simulator model probably uses a "typical" value instead of the minimum value.

In fact, your simulator says it is "powered by LTSpice", and the LTSPice 2N2222 model has a forward $$\\beta\$$ of 200:

In the real world, the $$\\beta\$$ parameter is not very well controlled from device to device, so we usually try to design our amplifiers so their gain does not depend strongly on $$\\beta\$$. That means the simple common-emitter stage you're learning about now isn't usually a good choice for a real world design.

• How can you design an amplifier does is not dependant on the BJT gain? I have tried with a gain of 200 and changing RI to 10k ohms. This gives me vOUT = 4 V but the simulator still gives me 5.1 V Commented Oct 30, 2020 at 18:35
• @Martel, An emitter resistor as you mentioned in comments on the main post is the simplest way. Building up an op-amp with a half dozen or a dozen transistors and putting negative feedback around it is a more complicated way. Commented Oct 30, 2020 at 18:41

Here there is a file inside the LTspice directory which is called "standard.bjt". Here is a code of 2N2222:

.model 2N2222 NPN(IS=1E-14 VAF=100 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=30 Icrating=800m mfg=NXP)

After modifying your values, then you can paste it with a spice command on your LTspice circuit.

PS: Do not forget also you can change the name of 2N2222 whatever you want as a name.