# What is the use of R2 in this oscillation circuit?

Discussion in 'Electronic Basics' started by [email protected], Apr 29, 2005.

1. ### Guest

I have electronics hobby and just start learning how oscillator works,
like the loop gain must be >= 1 and the phase between the input and
output should be 0 degree. The problem is, when I see the actual (even
simple) oscillation circuit, I am at a loss how it exactly works.

"http://img13.imagevenue.com/loc70/a2a_osc123.jpg"

For example, in Fig.1, what's the use of resistor R2, from oscillation
point of view (not from transistor DC basing point of view)? I figure
that, since C3 is basically to short AC (i.e. RF in this case), it
seems to be equivalent to place one end of the crystal directly to the
emitter, as shown in Fig.2. Unfortunately, it does not oscillate this
way. Likewise, since C3 is to short the AC, then it should also be OK
to remove R2 and C3 completely, and place the crystal in a way shown in
Fig.3. However, it does not oscillate, either.

(1) So what is the function of R2 here?
(2) If it is a must for the oscillator to work, how to choose it?
(3) How to choose the value of C3?

2. ### colinGuest

the circuit in fig1 works by feeding back the signal at the emiter to the
base through the emiter base capacitance, a tune circuit is formed between
the crystal wich will be slightly inductive and the emiter capacitance in
series with c3 wich should be slightly higher than the load capacitance
specified for the crystal, the resultant signal at the base gives positive
feedback.

if you short out the emiter to ground you remove the signal, also the dc
curent would no longer be controled exept by the hfe.

Colin =^.^=