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A transistor is a voltage-operated device (despite what many people/text books say) - voltage in, current out.
Thanks a lot for the help! i got you, you are saying the asymmetrical output is caused by the nonlinearity of the BJT.
Yes, absolutely.
This non-linearity is fundamental to how it operates. You'd get similar problems with any other active device (e.g. FET, MOST-FET, thermionic valve), as none of them have a linear relationship between voltage in, and current out.
However, that's why we love negative feedback
With 0.8v vs 1.2v, the difference is 0.4v, comparing to 0.8v that's 50%. Is that normal when a BJT is set at a reasonable operating point? Could it be that this rather large distortion is caused by the operating point being set on the shoulder or toe of the IV-curve? I don't have the IV-curve for 2N2222, so i just guessed maybe Ic about 3mA and Vc about 3v is ok, and i even reduced the signal level down to 1.0mV, but still asymmetrical output to the same degree.
When setting an operating point for a BJT, we think about things differently to this approach...
Firstly, we don't look up "IV curves" for specific BJTs, as they aren't really a device-specific thing.
It's a bit of a long story, this one, so I'll try to be concise...
Firstly, it's conventional to bias the BJT so the collector is in the centre of the available voltage swing. We do that on the assumption that signals are symmetrical, which is usually a fair assumption. YMMV. So for a 6V PSU, let's make Vc equal 3V...
Next, we need to decide on the collector current. For a small-signal amplifier, 1mA is a good starting point (more on this in a moment). So, we can make the collector resistor equal to 3k (Vsupply minus Vc).
Why 1mA? Basically to make the numbers easier!
Next,
If you explore the Ebers-Moll equation I mentioned in my previous post, it allows us to derive a simplified equation for gm.
What's gm? Simply, gm is a value of gain, but it's a little bit different to (voltage out) over (voltage in). Rather, it's (current out) over (voltage in). Remember, the BJT, just like FETs, MOS-FETs and valves/tubes are voltage-in, current-out devices... The "posh" word for this is "transconductance", FWIW... We use gm to express that. Unlike regular voltage gain, which is just a ratio, gm has a unit - it's Amps per Volt. Note that this is the reciprical of resistance (V/I), and the unit is Siemens.
In transistor circuits, it's convenient to work in milliamps rather than amps...
From Ebers-Moll:
For a BJT working at room temperature, gm=40.Ic
For a BJT working at a higher temperature, gm=35.Ic
I tend to use the latter, as gear often runs warm. But some textbooks will use the 40Ic version. Not a huge deal either way...
So, if we set the collector current to 1mA, we have a gm of 35mA per volt.
The next part of the story:
Your collector resistor is converting the current from the BJT into a voltage (simply Ohm's law). So the voltage at the collector is Ic times Rl. Now, Ic has already been set to 1mA to establish the correct DC conditions, and the signal itself is just a variation either side of 1mA. So,
...if gm is 35 times Ic, Ic is Vbe times gm
And
Vout is Ic times Rl
gain = gm times RlFor a typical circuit where Ic is 1mA, and Rl is 3k, then the overall gain should be 105.
In practice, it will be a little lower - there are always secondary effects (that can be considered later). But you'd probably get around 90 or so...
Note, BTW, that to calculate a value of gain, we didn't need to consider Hfe/Beta. That's lucky, as Hfe/Beta varies enormously from device to device. In a typical circuit, the biggest effect Hfe has is on input impedance, not the raw gain.
So what we see here is that getting large gain is very easy. OK, there is distortion that is readily visible on a 'scope, but that's where negative feedback comes in. Imagine if we had two of these stages - you'd have a gain of 90 times 90, or 8000 (simplifying a hugely). Far more gain than you'd actually need, but all that "excess gain" will, via the magic of negative feedback, be "thrown away" to produce a very linear amplifier indeed.
Many op-amps have just two gain stages. OK, they pull some "tricks" to get lots of open-loop gain, but essentially, all amplifiers start with the principles outlined above.
Personally, I feel that any time invested at this level is time well spent. The more you understand the very basics, the easier everything else becomes. But that much you already know
I do have some standard advice for people at this stage: spend some time learning how to separate
signals from
DC conditions. What we do here is establish the correct DC conditions, and then we
superimpose the AC signal on top of the DC conditions. Often, superposition is taught with no obvious "buy-in", but this is it - any signal can be regarded as a mixture of AC+DC, and in a simple transistor amplifier, that is very much the case - the DC conditions are an inconvenient pre-requisite of a practical amplifier. We're lucky today - we have thousands of great op-amps to choose from, and in most op-amp circuits, the DC components are usually 0 volts, so we're not even aware that we're dealing with them. Very different with discrete transistor circuits.
Hope this helps, and good luck,
Mark