# VHF Amplifier Design

I am an RF enthusiast that wants to learn how to think like an engineer. I am trying to build a FM receiver that consists of a wideband FM amplifier and a slope detector. I have so far created an amplifier. It does seem to work but I would like to revise my thought process.

Assumptions that I have used for biasing:

• In a CE amplifier, current through emitter resistor equals current through collector resistor.
• Emitter resistor current is 100x the base current (Beta=100)
• Biasing current should be at least 10x base current for stability.
• Current flowing through both bias resistors is equal
• Adding a emitter capacitor to ground increases gain via creating a lower resistance AC path to ground
• Using an inductor instead of collector resistor lets you set DC bias and gain is proportional to frequency because of inductive impedance.

My arbitrary design ideas:

• Emitter resistor should drop 1 volt, base voltage should be 1.6V
• We have a 5V supply
• datasheet for 2n3904 says B=100 if 10mA current flows through emitter. 1V/10mA=100 ohm emitter resistor.

Biasing TR1:

• Emitter resistor current= 0.010A
• Base current is 0.01A/100 = 0.0001A
• Bias current should be 10*0.0001A=1mA
• Larger bias resistor: 5V-1.6Vbase = 3.4V with 1mA through it, so 3.4K resistor.
• Smaller bias resistor: 1.6Vbase/1mA = 1.6k resistor.
• Add a bypass cap and simulate in LTSPICE for a good value, experiment with load inductor values.
• Question: I grabbed wrong resistors, 34k and 16k which should violate stability rule of thumb. How is this stage still functioning?

Biasing TR2:

• Base current is 10mA through emitter resistor/100 = 0.0001A base current; Biasing current should be 1mA, 10x larger than base current for stability.
• Larger biasing resistor = 5V-1.6Vbase / 1mA = 3.4k (I have 3.3k on hand)
• Smaller biasing resistor= 1.6V/1mA = 1.6k (I have 2k on hand)
• Same inductor load because I know it works

How can I think of amplifier design more formally? My only education on the topic is various youtube videos and internet tutorials on the subject. Thank you for your input.

• With a β of 100, the emitter current will be 101× the base current; β is the ratio of collector current to base current, and emitter current is the sum of the two. This does not actually matter, however, as β varies by quite a large amount for any given transistor, both between transistors of the same part number and for a single transistor at different temperatures. Commented Feb 5, 2022 at 6:09
• I apply generally the rule Rb low (as R3) = 10 or 20x the emitter resistor (R4). R4 seems a bit too low, but it can be ok anyway (emitter current will be higher). One can now do a simulation "test" also to verify that DC analysis is ok. Commented Feb 5, 2022 at 9:36
• NB: no information about generator impedance and probe used (x1, x10)? Commented Feb 5, 2022 at 9:59
• Sorry, probe used was 10x Commented Feb 5, 2022 at 14:06