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I'm having some trouble understanding the behaviour of the basic common emitter fixed bias amplifier. I've been looking at this for awhile now and I'm starting to think there is a foundational problem with my understanding.

I understand that I need at least 0.7V applied to the base to turn on the base emitter junction, and also that I want to choose R1 so the base current biases the transistor in the middle of it's load line.

My confusion starts when I inject a signal to be amplified. The signal will be superimposed on the bias, however wouldn't that super imposed signal now be varying slightly above and below 0.7V thus turning the transistor off and on, creating a very distorted output signal?

I've tried to think this through and I suspect the current in R1 must start to vary slightly in relation to the input signal?

Another possibility is that I know the base emitter junction doesn't exactly turn off at 0.7V but it does become nonlinear. Would that mean the outputted signal is distored on the negative part of the input signal?

Is this correct?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Welcome to EE.SE. I've taken the liberty of turning your voltage source the right way up (positive on top) and making the ground symbol point to, um, ground. I also deleted the 1 V parameter on V1 as I reckoned you weren't trying to run the circuit at the default value of 1 V. \$\endgroup\$ – Transistor Feb 19 at 19:06
  • \$\begingroup\$ Try simulating it. Superposition means you can add up the contributions of each source seperately, while other sources are zero. So, check what V1 does by shorting V2 (setting it to zero volts), next, check the contributions of V2 by shorting V1. \$\endgroup\$ – Huisman Feb 19 at 19:24
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The transistor does not just go from hard on to hard off. Rather, the emitter current of the transistor follows the diode equation. Roughly, \$i_e = I_{SS}\left(e^\frac{v_{be}}{V_T} - 1\right)\$ (Google for the diode equation for the definitions of \$I_{SS}\$ and \$V_T\$). At room temperature, \$V_T \simeq 26\mathrm{mV}\$.

So in your simple circuit, a small (like, less than 1mV or so peak-peak) signal will act to increase and decrease the emitter current in an amount proportional to the signal voltage.

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  • \$\begingroup\$ I think I may have thought of a better way of asking my question. In the basic circuit I've drawn, would the output signal be distorted on the negative peak of the input signal? When I look at the graph of the diode equation, I believe the answer is yes. \$\endgroup\$ – lowteq Feb 19 at 21:55
  • \$\begingroup\$ It depends on how small your input signal is. The percentage distortion will get arbitrarily small as you reduced the input signal amplitude. \$\endgroup\$ – TimWescott Feb 19 at 21:57
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I understand that I need at least 0.7V applied to the base to turn on the base emitter junction,

Not exactly. A base-emitter junction can be considered forward-biased with larger or smaller magnitudes than \$700\:\text{mV}\$. It's just that the collector current (speaking now about common small-signal BJTs) will change by roughly a factor of 10 for every \$60\:\text{mV}\$ change in the base-emitter voltage. So if \$I_\text{C}=1\:\text{mA}\$ when \$V_\text{BE}=670\:\text{mV}\$, then you might expect that \$I_\text{C}=100\:\mu\text{A}\$ when \$V_\text{BE}=610\:\text{mV}\$ (assuming the temperature didn't change and you are using the same BJT as before.)

and also that I want to choose R1 so the base current biases the transistor in the middle of it's load line.

Again, that depends. But yes, you want to bias the BJT somewhere useful. How you make that decision may involve a little more than just "middle of...."

My confusion starts when I inject a signal to be amplified. The signal will be superimposed on the bias, however wouldn't that super imposed signal now be varying slightly above and below 0.7V thus turning the transistor off and on, creating a very distorted output signal?

Actually, for your circuit it's quite true that the output signal will be distorted. It's one of the reasons that it's not used (much) without some kind of global NFB included. You are looking at a grounded emitter, so the only AC impedance will be \$r_e=\frac{V_T}{I_\text{C}}\$, where \$V_T\$ is the thermal voltage.

To a first order approximation, \$I_\text{C}\$ follows the Shockley equation. And that means that approximately a 10-fold change in \$I_\text{C}\$ for a \$60\:\text{mV}\$ change in \$V_\text{BE}\$. So once again, suppose \$I_\text{C}=1\:\text{mA}\$ and that you've biased things so that \$V_\text{BE}=670\:\text{mV}\$. From this, the voltage gain will be hovering at about \$A_V=\frac{R_\text{C}=1\:\text{k}\Omega}{r_e\approx 26\:\Omega}\approx 38.5\$. Suppose your AC signal is \$v_{ac}=\pm25\:\text{mV}\$. Near the top of the positive-going swing, it pulls upward on the base by \$25\:\text{mV}\$ and this causes the collector current to increase by a factor of about 2.6. So the gain raises to about \$A_V=\frac{R_\text{C}=1\:\text{k}\Omega}{r_e\approx 10\:\Omega}\approx 100\$. Near the bottom of the negative-going swing, it pulls downward on the base by \$25\:\text{mV}\$ and this causes the collector current to decrease by a factor of about 2.6. So the gain falls to about \$A_V=\frac{R_\text{C}=1\:\text{k}\Omega}{r_e\approx 68\:\Omega}\approx 14.8\$. In short, the voltage gain is varying with the signal, everywhere from about 15 to about 100.

So yes, the signal at the collector is usually distorted. If you keep the input signal swing small enough, say below a millivolt or so, then it's not so bad. But this kind of stage isn't often used as a single stage. Instead, global NFB as part of a multi-stage system is often applied to help straighten things out. (Or an emitter resistor is added to provide local NFB.)


So far, I've just addressed your questions. But the circuit isn't very good. Different BJTs from the same family of parts will vary significantly in a variety of ways. They will have different saturation currents which affect their base-emitter voltage for a given collector current; they will exhibit different values of \$\beta\$; they will vary somewhat differently as the ambient temperature rises and falls during the day; etc. If you do know how to calculate (in theory) the operating point given some varying parameters, try it out and see how differently that circuit biases itself with slight variations in these three parameters. If not, use Spice and see what it says. Most especially, play around with \$\beta\$ and see what happens. (Assuming you can get it into active mode and not saturated, of course.)

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When you are superimposing the signal on top of the bias voltage, you have to make sure that the signal is "well-behaved". You cannot expect the circuit to work for all kinds of possible inputs. The definition of "well-behaved" depends on what you are trying to do with the circuit. Since in the question you are trying to make a linear amplifier, you have to assume that the signal follows the small-signal discipline/approximation and it needs to be ensured that the BJT is in forward-active region.
For that, as you said, the \$V_{BE}\$ needs to higher than 0.6V or 0.7V (This is also an approximation) and the \$V_{CE}\$ needs to be higher than ~0.2V. If you assume that 1mA current flows through the collector, then the collector voltage becomes 4V (assuming 5V supply). In this case you have maximum output swing of \$min(1V,3.8V) = 1V\$. Since small signal gain comes around 40, the allowed input swing becomes 25mV (=1/40).
But remember that this gain is derived using small signal approximation. So indeed as you said, in reality you will have some distortion. In this particular case, since, $$I_c = I_se^\frac{V_{be}}{V_t}$$ $$I_C+i_c = I_se^\frac{V_{BE}+v_{be}}{V_t} = I_Ce^\frac{v_{be}}{V_t}$$ $$I_C +i_c = I_C(1+\frac{v_{be}}{V_t}+\frac{1}{2}(\frac{v_{be}}{V_t})^2+...)$$ $$i_c = \frac{I_C}{V_t}v_{be}+ \frac{I_C}{2}(\frac{v_{be}}{V_t})^2...$$ Here the capital letters refer to the bias voltages and the small letters refer to the variations. Clearly, the first term is what you get from the small signal approximation, where you assume that the higher order terms do not matter as they are very small.
In reality, you will get some distortion due to these higher order terms and to ensure that your amplifier is sufficiently linear, you would need to ensure that \$\frac{V_t}{v_{be}}\$ (ratio of the first two terms called harmonic distortion) is high accordingly. So for higher linearity, you need to reduce your input swing.

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