Electronic – Help with 144Mhz small signal amplifier not amplifying

You don't need to forward bias the B-E junction. Whether you do depends on what you want the transistor to do. To keep it off, you want to not forward bias the B-E junction. As for the C-B junction, keeping that reverse biased is fundamental to how BJTs operate.

A BJT is basically a reverse biased junction that can be made selectively leaky. You apply a voltage accross C-E and with the base open nothing happens. The reverse biased junction doesn't allow any (except for small leakage we will ignore) current to flow. However, the special property of a BJT is that a little current thru the base messes up the insulating capability of the reverse biased junction. The gain, and hence the useful properties, of a BJT come from the fact that is only takes a little current to muck up the reverse biased junction such that it allows a lot more C-E current to flow. This ratio of C-E current to B-E current is the basic gain of the transistor. It can be as low as 5-10 in big mongo power transistors and 100s in high gain signal transistors.

The B-E junction also looks like a diode to the external circuit. It will have a forward voltage drop when conducting just like a regular diode. In silicon, this is 500-750 mV for most non-extreme applications.

If you want to use the transistor as a switch (either as full off or full on as you can make it) then you have to make sure there is no base current in the off case, and plenty enough to support the desired collector current in the on case. Driving the base to the emitter voltage is a good way to make sure the transistor is off. To turn it fully on, you need to provide at least 1/gain of the desired collector current.

In other cases, a BJT might be used in "linear" (it's often rather non-linear, but this is the term used to mean in-between mode or not-switch mode) mode, like a audio amplifier. In that case you want to always keep it somewhat on and have the input signal change its operating point. If done right, this can amplify the signal. Different configurations give you voltage gain, or current gain, or some combination. In these cases, biasing the transistor refers to keeping it somewhere in the middle of the operating range so that a little input signal can change the output both ways. Biasing is basically setting up the DC operating point.

The gain of your transistor circuit depends on what signal load the collector has to fight against. It sees \$R_C\$ which is either 166 ohms or 83 ohms (typo in original question) and it also sees \$R_L\$ which is 50 ohms.

Together they are in parallel and their signal impedance is somewhat less than 40 ohms.

To get gain from this type of circuit, the emitter impedance needs to be lower than the collector impedance (40 ohms max). As \$R_E\$ is (probably) 83 ohms, at low frequency audio the gain will be less than 1 because gain is \$\frac{Z_C}{Z_E}\$ and we've established that \$Z_C\$ is probably less than 40 ohms.

At higher frequencies the 1uF capacitor comes into play and the question is what frequency will it start to lower the emitter impedance and give positive gain. If we ask the question "what frequency does 1uF have an impedance of 83 ohms" the answer is about 1.9kHz.

At frequencies above 1.9kHz you will start to see gain gradually rise. Of course that's useless for your guitar (80Hz low E string).

The problem with this circuit is that you are wanting to drive a 50 ohm load and expect gain. That is unrealistic.

When it comes to using it to amplify an 810kHz signal you will get better results because the 1uF emitter cap has an impedance of less than an ohm and this means you'll have gains of over 50 probably.