Page 221 - Build Your Own Transistor Radios a Hobbyists Guide to High-Performance and Low-Powered Radio Circuits

From Table 13-4, this results in about 208 mV across the base of Q2. And because
the capacitive voltage divider network Cl and C2 attenuates the collector signal
voltage from Ql by 10, the output voltage at Ql 's collector must be 10 x 208 mV
peak, or about a 2-V peak sinusoidal waveform, and thus

()Ut
Actually, the output voltage can be calculated via Table 13-4 without directly
knowing the tail current lEE when the small-signal loop gain is chosen or
established. Since Q2 has a much higher impedance input resistance at the base of
Q2 compared to the input resistance at the emitter of the Colpitts osciUator, loading
effects from Q2 are ignored because the equivalent large-signal input resistance is

2~/Gm and this large signal resistance is further multiplied by rf or 100 in this
example. Note that 1/ n = Cl/(Cl 1 C2) = (180 pF)/(180 pF 1 1620 pF) = 1/10, or
n = 10, and rf = 100.

Just for "fun," if ~ = 100, then the small-signal input resistance is 2~/gmQl' Then

/ .02 1

where lEE = 1 mA. Thus 9mQl
So 2~/gmQl = 10,000 =

small signal-wise. Large signal-wise because gm/Gm = 3,

/11 X

The equivalent large-signal resistance loading the tank circuit across Ll then is

J X X
which is insignificant to Rl = 3 k

in terms of loading the Le tank circuit.

Unless inductor Ll is a lossless (ideal) inductor, when the oscililator circuit is built
based on the equations, a smaller-amplitude oscillation signal will be generated in
the range of 10 to 25 percent lower than the calculated amplitude. For example, if
the 15-~H Ll inductor is not ideal or lossless and has an unloaded Q of lOO, that

would be like paralleling a 30-k