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If the load transistor were an NMOS, it would act as a common-drain amplifier and try to hold the output at a constant voltage. … A MOSFET driven with a constant \$V_{GS}\$ acts roughly like a constant current source, which is a good load for an amplifier. …

Your circuit will let you turn the volume down, but it won't let you amplify. If that's okay with you, your change shouldn't cause a problem. I'm not a hi-fi audio expert, but this circuit seems over …

Gain is a property of a circuit, not a signal. We normally consider the gain of an opamp to be independent of its inputs. For an ideal opamp, the gain is always infinite. You're probably confused bec …

For example: $$A_p = 0.01$$ $$A_{p,dB} = 10\cdot\log 0.01 = -20\:\mathrm{dB}$$ You can also have an actual negative gain, like what you get from an inverting amplifier. … To fully describe an amplifier, you usually need both the magnitude and the phase of the gain. …

As a rule of thumb, you should never have a negative \$V_{CE}\$ in your DC biasing. If you do, the emitter acts as a collector and the collector acts as an emitter. (They're not designed for that, so …

This page has an example of a fully-biased common emitter amplifier, including a schematic: This schematic features a resistor network to bias the base, as well as an emitter resistor. …

If the current through the resistors is approximately the same \$(I_C \approx I_E)\$, and you want the voltage drop across them to be the same \$(\frac {V_{CC}} 3)\$, then by Ohm's Law the resistors s …

You can do AC frequency analysis in your simulator to see the effect on the gain of the amplifier. What transistor are you using in your simulation? You could try a few different ones. …

As a general rule, all inputs of any kind need a DC path to ground. If you don't have a DC path to ground, the DC offset of the input can float to any voltage. You have capacitive coupling to the AC i …

I see a few problems. As Jippie says, C2 and C3 are reversed. The collector of Q1 will be biased somewhere between +50V and ground. The bases of your output transistors should be biased somewhere cl …

The output of your amplifier can be simplified to look like a voltage source in series with a resistance. (Just like VG and RG!) … In a common emitter amplifier, the input resistance comes from R1, R2, and the transistor base. The output resistance comes from RC and the transistor collector. …

It's a common-source amplifier. The input signal is AC-coupled directly to the gate. RG1 and RG2 provide a DC gate bias. MOSFETs act like voltage-controlled current sources. … So the circuit as a whole is a voltage amplifier, even though the MOSFET by itself acts like a transconductance amplifier. …

Look at the voltage scale of your graph. You're hitting the limits of your voltage supply. The output can never be higher than 15V (V2) or lower than about 1.1V (due to the R1/R2 voltage divider).

It's how an amplifier is supposed to behave. The basic transistor amplifier circuits are used because they behave approximately like 7.34(b). … \$(R_o)\$ Given reasonable sources and loads, what is the gain of the amplifier? …

Your electronics class has probably taught you the hybrid-pi model and given you some complex (yet accurate) formulas for gain, input resistance, and output resistance of the various amplifier topologies … As you suspected, a common collector amplifier will give you the output resistance you need. …