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Q:Triac Switch

A

Andrew Holme

Jan 1, 1970
0
Chris said:
http://www.chrisncarol.com/switch.jpg

would this be ok, basically when the 555 is high load1 is active when
it's low load2 is active, is this correct?

Thanks

No.

1. The +5V rail is shorted to ground through the base-emitter junctions of
Q1 and T3
2. T2, R2 and T5 are redundant

Why is the Q output AC-coupled? Is it to pulse the opto-triacs to save
power?
 
A

Andrew Holme

Jan 1, 1970
0
Andrew said:

You can just connect the opto-LEDs, through current-limiting resistors,
directly to the Q output.
 
J

Jasen Betts

Jan 1, 1970
0
http://www.chrisncarol.com/switch.jpg

would this be ok, basically when the 555 is high load1 is active when
it's low load2 is active, is this correct?

Thanks

it's an unusual looking circuit (there's a short circuit through the two
first output transistors) what's the intended purpose of C3

you can simpply connect the optocouplers (and resistors) above and below
the output of an NE555 to have them firing alternately.

you have the 555 configured to be on for upto about 4 seconds (adjustable)
and then off for about 15 miliseconds (or approx one mains cycle) - is
that what you want?

if you want an even mark-space (on and off for the same length of time)
connect the 250K pot between pins 3 and 2 and disconnect pin 7 and remove
R10

something like this

0V +--------------+--+5V
| | . . . . . | . . . . .
+------(---.1 . . 8.--+-[220R]---.1 6.
| 22u | . ~ . .MOC3022.
+--||--(-+-.2 7. +--------.2 5.
| | . NE555 . | . .
+------(-(-.3 6.--+ | .3 4.
| | | . . | | . . . . .
+---+ +-(-.4 5. | |
| | | . . . . . | | . . . . .
| v | | +--------.1 6.
+-[250K]-+------------+ | .MOC3022.
| | +--.2 5.
+-----------------------+ | . .
| .3 4.
0V--[220R]--+ . . . . .

The stuff on the other side of the optocoupler seems needlessly complex
but I guess that depends on what the load is you intend to drive. doing
that part the way you plan won't cause any problems.


It appears that you copied the a "solid state relay" schematic from
somewhere (and got atleast one copy wrong)

You don't need the whole solid state relay input stage if the input
has a predictable voltage range. a simple resistor is sufficient...
the 3022 has a 10mV input
sensitivity and Fairchild suggests feediing it about 15mA in case it
degrades a little, the internal LED has a voltage drop of about 1.5V
so 15ma at 3.5V comes to 233 ohms so 220 is close enough.

Bye.
Jasen
 
B

Bob Monsen

Jan 1, 1970
0
http://www.chrisncarol.com/switch.jpg

would this be ok, basically when the 555 is high load1 is active when
it's low load2 is active, is this correct?

Thanks

You can probably get by with fewer transistors and lower operating
current.

The MOC3022 needs at least 10mA to trigger, so you work backwards
from there. Its input forward voltage is less than 1.5V, and a saturated
2N4401 is something like 0.3V. Thus, you will have 3.2V, and you want to
have 10mA through it, so you need at most 320 ohms of resistance. Make it
300 ohms. Passing 10mA through a signal NPN will probably require 500uA
into the base, so use an 8.2k resistor from Q to the base. This will
reliably turn on a MOC3022 when Q is high, and turn it off when Q is low.

Now, the other side of the coin is turning one on when Q is low. To do
that, you use a PNP transistor in an inverted configuration, emitter to
5V, collector to the MOC3022, and the other MOC input terminal to a 300
ohm resistor which then goes to ground. Again, connect the base to Q using
an 8.2k resistor, and your MOC3022 will turn on when Q is low, and turn
off when Q is high.

When you turn off the 555, both branches will go off. If you want it to
stay on after the 555 goes away, use what is called a latching relay.
There are two coils, and they latch the relay one way or the other.

Good luck.
 
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