Now,
let’s come to the COLD side [secondary side] of the SMPS power supply. Please refer the previous post here for more
details.
The secondary side has the
output voltages to main board components.
There
are 6 voltage outputs:
+14VDC
> to audio output IC.
+5VDC
> to CPU side of the syscon chroma IC
An
other +5VDC > to tuner stage
+B110VDC
> to line output transformer [LOT] section.
+27VDC
> to vertical scan output IC
+8VDC
> to syscon-chroma IC, pins 17 & 49 of IC8873
Let’s
discuss all these supply voltages in detail.
+B110VDC: this voltage should be there all the time
when the set is powered ON; provided there is no fault to primary [HOT] side of
the SMPS.
+27VDC: this voltage too will be there all the time
when the set is powered ON; provided there is no fault to primary {HOT] side of
the SMPS
+14VDC:
this voltage too will be there, when the set is powered ON.
Then
comes +5VDC [CPU]: this voltage is the
output from a voltage regulator IC; 78L05.
It too will be, and should be there, as the input to this regulator IC
will be present when the set is powered ON.
Refer to the circuit diagram.
Input voltage to this regulator IC comes from pin-14 of the SMPS
transformer [after rectification].
The
+8VDC 7 +5VDC< {the voltage goes to pin36}are switched type, means these tow
voltages will be absent when the set it powered On. These voltages will be present only after we
press either the power button on the remote control handset or at the front
pane control of the TV set. Here more
description is needed I thinks.
Take
the case of these two voltages only.
Refer
the circuit diagram.
Consider
all these 3 voltages [+5V CPU, +5VDC, +8VDC]
The
source of all these voltages are generated by pin-14 of the SMPS transformer,
and then rectified by (D604-FR104). The
rectified voltage is then fed to input of the regulator IC 67L05, which in turn
outputs +5VDC CPU. It will be present
all the time, because this voltage is needed to CPU section circuit inside the
syscon-chroma IC. Without this voltage,
the CPU section circuit within the IC wont work and there by the set too will
not response to any external or remote commands. In short, the set will be dead. This voltage is not switched type. Means it will always be present. The rectified voltage is also fed to collector
of a NPN transistor [V507-C8080] and the collector of another NPN transistor
[V508-C8050]; switches the +5VDC. Refer
the circuit diagram and try to understand what happens when we Switch ON the set
first, after it is plugged to AC mains, and then we press the power button; either
at the remote control, or at the front panel of the TV set.
When
we first Switch ON on the set:
+B110VDC
is OK - present
+27VDC
is OK – present
+5VDC
(CPU) is OK – present.
+14VDC
is OK – present.
But;
there will be no 8VDC & 5VDC. Both these voltages are by-passed by transistors V607
& V508 respectively.
Why? Here comes the point.
Note
the circuit with NPN transistor V507-C8080.
Its
base terminal is pulled down to 8.7VDC by using a Zener diode [VD8]. By no means the voltage at this pont will go
up, than the break-down voltage of this zener diode; ie 8.7V.
But,
from where this voltage comes from?
There
is 3.3KOms resistor connected to the cathode of this Zener diode; from the
27VDC voltage output.
Now
what happens?
When
we press Power button either on the remote control or at the front panel control of the set; to Power On; there should be 8VDC to the base terminals of transistor
V507 . But, it won’t be there now.
Why?
Refer
the circuit diagram once again.
The
base of this transistor is connected to [Power On] pin-64 of the syscon
IC. The voltage level of this pin will
be hight [=H]; when we switch On the set.
Just see, this high voltage is fed to the base of a NPN transistor
[Q210-2SC1815]. The collector voltage to
this transistor is fed from (+5V CPU), via a serial resistor CR292, and then a
LED (power indicator) and an other resistor R291.
As
we know, when the base bias of an NPN transistor is at high [H] level, it will
conduct and its collector voltage will drop considerably due to load. Here the emitter of this transistor is
grounded. So, when its base voltage is
high [H], its collector voltage sill be low.
In
fact, when we switch On the set by its AC main Power Switch, the system control
section voltages to the IC will be present, the voltage at pin number 64 of
IC8873 will be high[H], and the voltage at the collector of Q210 will be low,
and in turn the base voltage of +8V regulator transistor too will be low. So this transistor won’t conduct, and there
will be no +8V.
Let
it be.
Now
come over to the base of V508; the 5VDC controller transistor C8050.
Refer
the circuit diagram again.
Its
collector terminal get voltage from the cathode of D604 itself; through
resistors R572 & R571. The base of
this transistor is pulled down to 5V by using a 5V6 zener diode, and the base
bias comes from the emitter of V507 [+8VDC].
It
is clear that the base bias voltage to this transistor will be present only
after the +8VDC has been bypassed by V507.
In short, there will be no +5VDc without +8VDC.
When
+8VDC is present, +5VDC too will be there, and should be.
In
short; the Main Switching Point to these tow voltages are of pin-64 of the
system control IC.
When
the voltage level of this pin is high [H], there will be no +8VDC &
+5VDC. When we press the power button,
the voltage at pin-64 of the IC will go low[L], the collector terminal voltage
of Q210 will go up; and the base voltage of V507 too.
Now
the transistor V507 will conduct (bypass) +8VDC. The base bias of V508 will go up and it too
will start to bypass the +5VDc.
This
is the main switching process takes place, when we press the Power button on
the remote control.
Please
do not mind my language. I’m not a
person belongs to an English speaking country.
Please try to understand the idea I’ve tried to express.
