How can you know how much gain a particular transistor will be giving? I'm familar with how op-amp's gain can be set using resistors, I assume something similar is going on with transistors, but I haven't been able to find anything explaining it.

http://www.electronics-tutorials.com/amplifiers/small-signal-amplifiers.htm

Here - check it out... it'll explain

Assuming that the input bias network and base input impedance do not load the signal source down too much, the gain of a transistor circuit is -Rc/Re, where Rc = colletor resistor, Re = *unbypassed* emitter resistor.

There is an internal emitter resistor equal to 25mv/Ie that is always there and cannot be bypassed like an external resistor can.

The usual caveats you'd imagine apply - you have to bias it correctly, the signal has to be small enough to not make the thing clip, etc., etc.

Most low gain opamp circuits can be adapted to work with transistors when the transistors are set up with the proper bias. The emitter of a bipolar junction transistor functions perfectly well as a low impedance noninverting input.

RC=10k
 Re=1k
 Rc/Re=10
 Gain of 10
 I'm attempting to gain confidence in my calculating, reading, writing, and comprehension of equations...could someone grade my figures?
 [no clipping or signal loading, input biased]

QuoteRC=10k
Re=1k
Rc/Re=10
Gain of 10
I'm attempting to gain confidence in my calculating, reading, writing, and comprehension of equations...could someone grade my figures?

You forgot the negative sign :wink: the gain is negative 10. The equation is ( -Rc ) /  Re  
The resulting signal has a gain of â€"10, it is ten times larger and inverted when compared to the input.

QuoteRC=10k
Re=1k
Rc/Re=10
Gain of 10

To a first approximation, yes. Remember that the true story always includes that internal emitter resistor that's a part of the transistor, 25mv/Ie. That will affect the 1K emitter only slightly, but the gain will be slightly less than 10.

One thing I've never been too clear about is the max gain of a resistive loaded single transistor. A BJT usually maxes at about 46dB. JFETs seem to max at much less.  Why these limitations?

Let's take the next step. NPN transistor, high gain (>200), biased through non-interfering means to half the power supply voltage on the collector. Emitter grounded. 10K collector resistor, and Ie= 4.5ma (that is, Vsuppy = 9V).

A) what's the value of the internal emitter resistor?

B) what's the small signal voltage gain?

C) I add a 100K resistor in series with a huge capacitance from collector to base, and a 10K resistor in series with the base, and apply signal to the 10K at the base. What's the gain from input to collector?

I'm lost here. What's Ie? And how does it = 4.5mA at 9V?

Ie is emitter current, Ic is collector current, r'e is internal emitter ac resistance, Vb is base voltage:

Ie ≈ Ic = Vb/ r’e