# What causes signal distortion in the BJT circuit I designed?

I set up the circuit as above and ran a simulation. Then, I noticed that the negative part of the signal was flattened, almost like a rectifier. However, when I reduced the capacitance of the capacitor, or increase the resistance of R3, the distortion disappeared. Why is this phenomenon occurring?

• However, when I reduced the capacitance of the capacitor, or increase the resistance of R3, the distortion disappeared. This should tell you that the effect is dependent upon the load current, which is the current flowing through C1. That should give you a clue regarding what is happening. Commented May 20 at 1:06

Your biasing is too close to ground.

simulate this circuit – Schematic created using CircuitLab

Here is a simulation when Vbias is 1 V vs when Vbias is 6 V.

What happens when Vbias is 1 V, is that when the input signal is low, $$\V_{BE}\$$ falls down to 0.6 V and even lower, causing the valleys to be wide and not as deep as they would be otherwise. Here is a simulation showing the base voltage, the emitter voltage and the base-emitter voltage.

Alternatively to raising the base bias voltage, you could reduce the emitter resistance to about 20 $$\\Omega\$$ per periblepsis's answer. Both techniques increase the quiescent emitter current, which will give you more linearity.

You are showing us an emitter follower (or common collector) topology. You've included a DC bias to your base source, which is good to do. But the problem is that while $$\Q_1\$$ can source current to the $$\C_1+R_3\$$ branch, it cannot sink current. So the only circuit element that is able to sink current from the $$\C_1+R_3\$$ branch is $$\R_2\$$. And it can't sink much, or fast. It's not an active sink.

If you refuse to change the topology to make it possess both an active source and an active sink, then you must greatly reduce the value of $$\R_2\$$. There is no possible "good" value for $$\R_2\$$. But if you make it less than the value of $$\R_3\$$ then the output may look better to you.

The real answer, though, is to use a different structure to drive the $$\C_1+R_3\$$ branch.

The comments and questions are all raising valid points and solutions. 1) Your bias voltage at at VB=1VDC gives you 300uA emitter current when your AC input of 100mV (peak,p-p,RMS?). Which ever the case for the AC input the current swing in 50 ohm resitor is many times your DC bias current. Raising the base bias voltage will help fix the problem while keeping the load constant. 1a) change your bias current by reducing the 1.5k resistor will have a similar effect 2) If you keep the same DC bias (redundant?) point, you need to increase the load impedance. Increasing the 50 ohm resistor or reducing the coupling capacitor value to where it's reactance will increase the load. 3) Replace the 2N2222 with a device/circuit with a higher effective gm (hint... .not another bipolar)