Technics QUARTZ SP-15 - Page 6

e Switching circuit operation
@ Commerical power (50/60Hz) is rectified by D1 and D4 through line filter L4 to obtain DC 168V. The DC 168V
is divided by R2 and R3 so that TP. 0 = OV, TP. 2 = 84V and TP. 3 = 168V.
@ C7 is charged through R4. When trigger diode D2 reaches ‘‘on” level, the trigger current flows from D2 to N1 via
R5. Since the coils of N2 and N3 are reverse in phase, when a current flows into N1 in the direction of ¥ , the
base current is generated at N2 and N3 so that Q1 and Q2 are respectively reverse- and forward-biased.
Therefore, Q1 turns off and Q@2 turns on, and the primary current of T2 flows from N1 > N5> Q2 (C-E) > TP. 0.
At that time, a feedback current is generated at N4 and it flows into N1 through R6. The current flowing into
N1 becomes of positive feedback, then Q1 and Q2 keep turning on, and the collector current of @2 keeps
increasing in accordance with the inductance of N5.
® However, N3 is directly connected between base and emitter of Q2, proximately being short-circuited, therefore,
the feedback current is saturated at a given value determined by R6. Accordingly, the base current of Q2
decreases in accordance with the inductance of N3. When the base current of Q2 is about to become smaller, the
counter electromotive force is generated at N1, causing the feedback current to flow in the reverse direction, and
then a positive feedback line is formed for Q1. Consequently, Q1 turns on and Q2 turns off, and the primary
current of T2 flows from TP. 1 + Q1 (C-E) > N5> TP. 2. Similarly, when the current is saturated, a positive
feedback line is formed for Q2.
O Through repetition of the above procedure, the oscillation is continued and 20 kHz pulse voltage can be obtained
at N6 on the secondary side of T2.
Fig. 16 Switching circuit
(e-mark shows the polarity of coil.)
Fuse
600mA
AC
50/60H2225 Cl