AMP: Simple RF Amplifer

Simplest Class A RF Amplifier

Oct/20 2016

Insert Gain: 10 dB up to at least 100 Mhz.

BJT: 2sc9018, 800 Mhz Ft

Output transformer: 6T:2T on TV core

low output:(VCC 8V ~ 11V)

bias 6.8k--+--1k, 43-51ohm,8-10mA or (8.2k--+--1.2k 43-51ohm, 10-13mA), (1.3k ---+---9.1k, 43-51 ohm, 10-13mA)

47 ohm degeneration

output level > 10 dbm:

13.8 Volt

bias: bias 6.8k ,1k

degeneration 68, bias current 15 ma, max output 15 dbm

Simple BJT Follower

suppose final output signal peak voltage is: Vo volt. The DC quiescent point, suppose is Vd.

RL=R4, Re=R6

in negative peaking point all current in loop R4/C2/R6, pumped by C2, and suppose C2 is big enough, then DC voltage(VD) on C2 almost does not change, then:

Vo = ( Vd/(RL+Re) )*RL = VD/(1+Re/RL)

VD=(1+Re/RL)*Vo

this request quiescent current :

ID = (1/RL + 1/Re)*Vo

this tell in given Vo, RL, request Re to a small value to get a higher output voltage, but will increase quiescent current, large average transistor power, ,obviously.

given a power supply Vcc, we could expressed the power dissipation of transistor as:

P=(Vcc-VD)*ID= (Vcc-(1+Re/RL)*Vo)*( (1/RL + 1/Re)*Vo )

if Vcc=12V, Vo=1Vpeak, RL=50 we got a power curve:

As Re is 50ohm, power dissipation around 300 mW. this tell the Re should around 1*R to 5*R. so raise VD, and Re help to reduce power dissipation. but there are another limitation arise. as a buffer the input Vin suppose equal Vo, the BJT Vce at least Vo to make sure the positive peak does not cutoff. further more, it's better to give a 2*Vin voltage drop on Vce, don't stress the transistor to much (then more distortion).

Conclusion:

Choose Vce starting from Vcc- 2*Vo - 0.7, or even higher value, then reduce the Re enable output could get design goal:Vo. if you get Re is reach to 3*RL or less, watch the transistor dissipation.

such a simple buffer suitable to a light load, otherwise we must pay Power for heavy load.