Which Sensor to Use ... Have You
The Electret microphone has a small FET inside it that
needs to be powered. This is supplied via R5, a 22 kilohm
resistor. When sound is detected by the microphones element (a tiny capacitive-plate-charged disk inside), the built-in FET amplifies the tiny signal and it is modulated against
the supply voltage from R5. Capacitor C2 (0.22 µF) picks
up this tiny modulated voltage and passes it to the
LM358’s non-inverting positive input.
The sound-activated section has very little output
swing, so R7 is used to tie the non-inverting input very
loosely low to prevent the LM358 from toggling when no
input signal is present (but with minimal loading on C2).
Using sound as the triggering source for the op-amp has
some drawbacks because it is not very constant. In reality,
the positive input detects the audio peaks, which can move
in the kHz range and is not good for switching any relay on
or off. Hence, C1 is used. Also, because the incoming microphone voltage has a very small swing, R8 (at one megohm)
is used to effectively stretch the voltage scale in the window
that the negative, inverting input of the LM358 is referencing.
D1 prevents reverse voltage when the relay disengages
from destroying the FET. The LED has a 1.2 kilohm cur-rent-dropping resistor.
To select the LDR input for decreasing light, toggle
“ON.” For the thermistor to sense less heat, toggle “ON.”
The opposing resistor is tied to high, positive, and the
sensor is tied to low, negative. Here on R5, a 100 kilohm
resister is used. R6, C2, and R8 are wire links, and the dotted line under C2 is not used here. R7 is omitted.
The microphone input, R5 (at 22 kilohms) and R6, is a
wire link. The dotted link under C2 (0.22 µF) is not used, and
R8 is one megohm. For R7, 10 megohms is only used here.
To use five-volt logic to turn on the relay, attach a 10 kilohm resistor to -R6 and a 100 kilohm resistor to +R5. The junction of these should produce just over one volt to ground,
which will hold the positive input of the IC1 logic low until the
five-volt logic arrives. If you need a higher impedance, try a
ratio of 100KΩ/1MΩ. To use an inverted logic voltage, just
swap the high and low resistors or try R5 = 56KΩ, R6 = 39KΩ.
This will hold the LM358 non-inverting, positive input high
until the logic low arrives. Now, use a wire link instead of C2
and make (also) the dotted line below it, R8, a wire link. Both
input terminals now become the positive input signal. You’ll
need a ground point for this input, and there is a spare pad
available just below R6. R7 is omitted. Adjust VR1 to set the
exact voltage level needed. (Both of the usual input terminals
become linked to provide a junction point for R5 and R6.)
Both C2 and the dotted line under it need to be wire links here.
Using the Voltage Divider
Option as the Input
To activate the relay using any input voltage from zero
to 12 volts (user defined), select the input voltage to be
just above for ON or just below for OFF. R5 and R6 then
hold the positive input of the LM358 slightly higher or
lower than the incoming, user-defined input-voltage signal,
allowing the relay to toggle on or off when it does. Fine
tune this using VR1, which may end up being very offset
the closer you get to the positive or negative supply rails.
The Standard for checking
Good enough to be the
choice of Panasonic,
Pioneer, NBC, ABC, Ford,
JVC, NASA and thousands
of independent service
Voltage Divider Definition:
Two resistors in series connected
across a voltage source. The voltage
at the common junction is a fraction
of the total applied voltage, determined by the resistor values.
Locate shorted or leaky
components or conditions
to the exact spot in-circuit
Still cutting up the pcb,
and unsoldering every
part trying to guess at
where the short is?
eg: Pos Neg = < side connected to
10KΩ + 10KΩ = 1/2-supply voltage
2KΩ + 8KΩ = 3/4-supply voltage
8KΩ + 2KΩ = 1/4-supply voltage
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All are measured between the
junction and the Neg/Zero supply
rail. You can have more than two
resistors. Three resistors, for example
will produce a window voltage; eg., 20
volts supply, 10KΩ + 5KΩ + 10KΩ
will produce “four volts” across the
five kilohm resistor — 2/5 + “1/5” +
2/5 of the total supply. But this voltage has no direct reference to either
supply rail, as it is floating somewhere in between.