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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? …

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. …

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. …

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. …

But a common-emitter amplifier is supposed to be a voltage amplifier. So what do we do? … The resistor voltage is proportional to the current, so your linear voltage-controlled current source becomes a linear voltage-controlled voltage source -- an amplifier! …

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 …

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 …

That makes it a common-collector amplifier (emitter follower). … If I take the view that I'm controlling \$V_{BE}\$ with the load on the collector, then it starts to sound more like a transconductance amplifier than any of the usual voltage amplifier topologies. …

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. …

There are reasonably priced op amps that allow 90V between the supplies (example). There are also ridiculously expensive power op amps that can handle hundreds of volts (example). But lower voltage op …

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 …

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 …

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 …

The non-inverting op amp input should be connected to ground, not +15V. Otherwise you saturate the output because V+ is always greater than V-.

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. …