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Simple question about a 555 in one-shot (monostable) mode

A

AINTME

Jan 1, 1970
0
I have a 555 timer in monostable mode. I output a 45 second pulse. No
problem doing this. What I want to do though is to delay triggering this
pulse while another signal is high (12V) - what control do I use to do
this?

So: while input signal is 12v the output must stay high. When the input
signal goes low then I want the one-shot to trigger maintaining the output
(still) high for 45 seconds and then go low.

Can anyone help?

Thanks.
 
J

Jamie

Jan 1, 1970
0
AINTME said:
I have a 555 timer in monostable mode. I output a 45 second pulse. No
problem doing this. What I want to do though is to delay triggering this
pulse while another signal is high (12V) - what control do I use to do
this?

So: while input signal is 12v the output must stay high. When the input
signal goes low then I want the one-shot to trigger maintaining the output
(still) high for 45 seconds and then go low.

Can anyone help?

Thanks.
If I understand you correctly, You want a low transition or just a
low signal to start a One shot timer however, you don't want any other
transitions (resets) to effect the timer until it has expired it's
45 second count?
this is called a gate.

You have a couple of options here. use an OR gate where as the 555
timer out goes to one input and your Reset Line(trigger) is the
other input. the OR gate output would then control the RESET line for
example.
with this, when the 555 output is high, it will hold the GATE's output
on, this is because it's an OR gate, one or the other needs to be on for
the output to be on. This also means, a low on the other input will not
be seen at the RESET pin.

If you want to cut corners. place a NPN transistor in there instead.

Emitter to the 555 timer out, collector to the RESET Pin. use a pull
up resistor on the RESET line. and use a bias resistor on the base.
you don't need much current.
the difference here is, you'll now need s + level to reset it instead
of a - level because you'll be biasing the NPN.
when 555 output goes high. there will be now path to pull the RESET
pin to low once the timer starts.

I guess it's a matter of choice.

Btw, if you have multiple points of triggers, you can
isolate them via diodes with either method.
only the RESET pin directly will override.
 
A

AINTME

Jan 1, 1970
0
If I understand you correctly, You want a low transition or just a
low signal to start a One shot timer however, you don't want any other
transitions (resets) to effect the timer until it has expired it's
45 second count?
this is called a gate.


Sorry, that's not quite what I meant - I'll try to explain it better.
Under normal circumstances the circuit outputs a pulse of 45 seconds as
soon as it is energized (in a one shot configuration)
However there is this other line that goes between o and 12 volts.
Anytime this signal goes to 12 volts I want the whole process to freeze
and only start the 45 second pulse after that signal goes from 12 back to
0 volts. In other words if the signal goes high (no matter if the 555
output is already high)I want the output to go low 45 seconds AFTER the
12volts go back to 0. I think there may be a pin on the 555 that will do
this I just don't know which one will do it. Or maybe I need an external
gate.

Thanks for your help.
 
S

Sjouke Burry

Jan 1, 1970
0
AINTME said:
Sorry, that's not quite what I meant - I'll try to explain it better.
Under normal circumstances the circuit outputs a pulse of 45 seconds as
soon as it is energized (in a one shot configuration)
However there is this other line that goes between o and 12 volts.
Anytime this signal goes to 12 volts I want the whole process to freeze
and only start the 45 second pulse after that signal goes from 12 back to
0 volts. In other words if the signal goes high (no matter if the 555
output is already high)I want the output to go low 45 seconds AFTER the
12volts go back to 0. I think there may be a pin on the 555 that will do
this I just don't know which one will do it. Or maybe I need an external
gate.

Thanks for your help.
What you want is a "retriggerable oneshot",google gives as first result
http://stuff.mit.edu/~6.131/datasheets/74ls123.pdf
If you insist on using the 555, then use the same signal you use for
triggering, to keep the timing cap discharged.
Then on last removal of the trigger, the cap starts charging, and you
get 45 seconds.
Without extra components you wont succeed.
 
A

AINTME

Jan 1, 1970
0
What you want is a "retriggerable oneshot",google gives as first
result http://stuff.mit.edu/~6.131/datasheets/74ls123.pdf
If you insist on using the 555, then use the same signal you use for
triggering, to keep the timing cap discharged.
Then on last removal of the trigger, the cap starts charging, and you
get 45 seconds.
Without extra components you wont succeed.

I think this will put me on the right track - it is a retriggerable
oneshot I'm after - I did not know what it was called - having a name for
it will help me with my search and also the fact that you say that I need
extra components other than the timer.

Thank you for your help!
 
R

Rich Grise

Jan 1, 1970
0
I think this will put me on the right track - it is a retriggerable
oneshot I'm after - I did not know what it was called - having a name for
it will help me with my search and also the fact that you say that I need
extra components other than the timer.

Thank you for your help!

Is that 12V-to-0V transition the _only_ thing that's clocking your
one-shot, or is it merely an inhibit and there's a different signal
triggering the one-shot?

Incidentally, it might be a convenient time for you to learn about
timing diagrams. :)

Cheers!
Rich
 
J

Jamie

Jan 1, 1970
0
AINTME said:
Sorry, that's not quite what I meant - I'll try to explain it better.
Under normal circumstances the circuit outputs a pulse of 45 seconds as
soon as it is energized (in a one shot configuration)
However there is this other line that goes between o and 12 volts.
Anytime this signal goes to 12 volts I want the whole process to freeze
and only start the 45 second pulse after that signal goes from 12 back to
0 volts. In other words if the signal goes high (no matter if the 555
output is already high)I want the output to go low 45 seconds AFTER the
12volts go back to 0. I think there may be a pin on the 555 that will do
this I just don't know which one will do it. Or maybe I need an external
gate.

Thanks for your help.
You want a Time off Delay that has it's time restarted when ever the
12 volts appears.

Just use a NPN transistor in common emitter to pull the Time Const cap
down to ground while the 12 volts is on. This will keep the cap
discharged and the output should remain on once the timer has been
triggered.
In the case where the timer is off when the 12 volts comes on. You
can put another NPN on the TRIGGER pin common emitter. this will make
sure it gets started.
You could if you wish. Use the same first transistor to pull the
TRIGGER to ground but you would need an isolation diode from that
transistor to the discharge pin with the anode to the discharge pin
and cathode to the collector, the TRIGGER could then be connected to
to the collector. If you don't use this diode, in turn what would happen
is it would not be a one shot any more. It would then start oscillating
because the discharge would come on internally when the Threshold is
reached. You don't want that to feed back to the trigger at that point:)

P.S.
You need to keep the Vcc on always to the timer. I don't know how your
doing it now but I suspect you are using initial power on ? Just use the
NPN transistor configuration as I explained. That is how most Delay off
timers I've seen are configured in the industrial world. It only makes
sense.
 
E

ehsjr

Jan 1, 1970
0
AINTME said:
I think this will put me on the right track - it is a retriggerable
oneshot I'm after - I did not know what it was called - having a name for
it will help me with my search and also the fact that you say that I need
extra components other than the timer.

Thank you for your help!


By combining your posts, I understand it as this:
You want a (+) output as long as the input is (+)
and it must stay (+) for 45 seconds after the
input drops. There must be no gap, no matter how
small, in the (+) output when the input signal
drops. You might be able to get away with oring
the 555 output with the input, but the brief
transition time of the 555 from non-triggered to
triggered when the input drops could create a gap.

My suggestion is that you add 3 transistors and support
components to the 45 second 555, like this:

PNP
+12 ---------------------+----------------- ----+
| e\ /c |
| --- Q2 |
| | |
[10K] [1K] |
| | |
+-------------------+ +---> Out
| | |
/c |c |
+----+-------+---| NPN / |
| | | \e Q1 +------| NPN |
[.1uF] [100K] [D1] | | \e Q3 |
| | |a | [3.3K] | [1K]
| | | | | | |
| | Input-+---- | --[555]--+ | |
| | | | |
Gnd --+----+-------------+-------------------+------+


(+) on the input will turn Q1 on through D1. That will turn
Q2 on, so you get a (+) out. When the input goes (-), the .1 uF
in parallel with the 100K will hold the base of Q1 (+) briefly,
keeping Q1 (and therefore Q2) on while the 555 triggers. That way,
there will be no dropout of the (+) output during transition. D1
prevents the (+) on the base of Q1 from interfereing with the
(-) trigger on the 555, so the 555 will trigger when the input
goes (-). The 555 output goes (+), which turns Q3 on. Q3 turns Q2
on, so the output stays (+). When the 555 times out 45 seconds
later, Q3 (and therefore Q2) turns off, and the output goes (-).

Ed
 
A

AINTME

Jan 1, 1970
0
Is that 12V-to-0V transition the _only_ thing that's clocking your
one-shot, or is it merely an inhibit and there's a different signal
triggering the one-shot?

Incidentally, it might be a convenient time for you to learn about
timing diagrams. :)

Cheers!
Rich

I did not include a timing diagram because the ascii drawings look all
jumbled in my reader (Xnews) the only way I can see them is by emailing
the posting to myself - I can't however, reverse that process (write the
email and then post, it doesn't work).
Thank you for your input.
 
A

AINTME

Jan 1, 1970
0
ehsjr said:
AINTME said:
I think this will put me on the right track - it is a retriggerable
oneshot I'm after - I did not know what it was called - having a name
for it will help me with my search and also the fact that you say
that I need extra components other than the timer.

Thank you for your help!


By combining your posts, I understand it as this:
You want a (+) output as long as the input is (+)
and it must stay (+) for 45 seconds after the
input drops. There must be no gap, no matter how
small, in the (+) output when the input signal
drops. You might be able to get away with oring
the 555 output with the input, but the brief
transition time of the 555 from non-triggered to
triggered when the input drops could create a gap.

My suggestion is that you add 3 transistors and support
components to the 45 second 555, like this:

PNP
+12 ---------------------+----------------- ----+
| e\ /c |
| --- Q2 |
| | |
[10K] [1K] |
| | |
+-------------------+ +---> Out
| | |
/c |c |
+----+-------+---| NPN / |
| | | \e Q1 +------| NPN |
[.1uF] [100K] [D1] | | \e Q3 |
| | |a | [3.3K] | [1K]
| | | | | | |
| | Input-+---- | --[555]--+ | |
| | | | |
Gnd --+----+-------------+-------------------+------+


(+) on the input will turn Q1 on through D1. That will turn
Q2 on, so you get a (+) out. When the input goes (-), the .1 uF
in parallel with the 100K will hold the base of Q1 (+) briefly,
keeping Q1 (and therefore Q2) on while the 555 triggers. That way,
there will be no dropout of the (+) output during transition. D1
prevents the (+) on the base of Q1 from interfereing with the
(-) trigger on the 555, so the 555 will trigger when the input
goes (-). The 555 output goes (+), which turns Q3 on. Q3 turns Q2
on, so the output stays (+). When the 555 times out 45 seconds
later, Q3 (and therefore Q2) turns off, and the output goes (-).

Ed

Yes, thank you, this seems like it will work. Like another poster
suggested... what I need if a retriggerable monostable cct. Your cct here
seems pretty close to what I wanna do.
Thank you.
 
A

AINTME

Jan 1, 1970
0
Is that 12V-to-0V transition the _only_ thing that's clocking your
one-shot, or is it merely an inhibit and there's a different signal
triggering the one-shot?

The 45sec one shot triggers as soon as power is applied to the chip.The
output goes high right away.
But it will stay on as long as the other signal is 12volts after this
signal goes to 0 then the output goes low 45 seconds later. Everytime the
control signal goes hi it keeps the output hi until 45 seconds after the
signal goes back to 0v.
 
A

AINTME

Jan 1, 1970
0
You want a Time off Delay that has it's time restarted when ever the
12 volts appears.

Just use a NPN transistor in common emitter to pull the Time Const
cap
down to ground while the 12 volts is on. This will keep the cap
discharged and the output should remain on once the timer has been
triggered.
In the case where the timer is off when the 12 volts comes on. You
can put another NPN on the TRIGGER pin common emitter. this will make
sure it gets started.
You could if you wish. Use the same first transistor to pull the
TRIGGER to ground but you would need an isolation diode from that
transistor to the discharge pin with the anode to the discharge pin
and cathode to the collector, the TRIGGER could then be connected to
to the collector. If you don't use this diode, in turn what would
happen is it would not be a one shot any more. It would then start
oscillating because the discharge would come on internally when the
Threshold is reached. You don't want that to feed back to the trigger
at that point:)

P.S.
You need to keep the Vcc on always to the timer. I don't know how
your
doing it now but I suspect you are using initial power on ? Just use
the NPN transistor configuration as I explained. That is how most
Delay off timers I've seen are configured in the industrial world. It
only makes sense.
I'm getting a lot of good ideas from you guys - thank you very much.
 
D

Dleer

Jan 1, 1970
0
"and" the 2 inputs, you'll have to get levels compatible if using ttl
 
J

Jamie

Jan 1, 1970
0
AINTME said:
I did not include a timing diagram because the ascii drawings look all
jumbled in my reader (Xnews) the only way I can see them is by emailing
the posting to myself - I can't however, reverse that process (write the
email and then post, it doesn't work).
Thank you for your input.




TRIGGER_____________
o-------| |
: | |
: | |
: | 555 |
: | |
: | |
: | | D1 small signal type.
: | |Discharge
: | |->|---+---- (C)
: |___________| : \
: : |-(B)-[4.7k]--- 12V signal.
: : /
: : : (E) NPN switching
:--------------------------- :
:
GRD

I don't know how thats going to appear ..that was done in note pad.
add that to the circuit..
when the 12V comes on, that will hit the trigger and hold the charge
state low while it's on. This will keep the output on.
when you remove the 12 volts, the timer will start charging the cap,
output will remain on. if the 12 volts comes in before the cap gets
charged enought to hit the threshold, the NPN transistor will discharge
it again to reset its time back to the start. pulling the trigger this
time will make no difference.
When the charge reaches 2/3 Vcc voltage, the threshold will trip
which will then put the internal flip flop Q output back to low again
that reflects on the 555 output. etc.. This was out of my head.
if you're not using a low ESR Cap, D1 should be ok with a simple
200 ma type diode., it also depends on the slew rate of the 12V control
signal.
 
J

John Fields

Jan 1, 1970
0
The 45sec one shot triggers as soon as power is applied to the chip.The
output goes high right away.
But it will stay on as long as the other signal is 12volts after this
signal goes to 0 then the output goes low 45 seconds later. Everytime the
control signal goes hi it keeps the output hi until 45 seconds after the
signal goes back to 0v.

---

View in Courier:

+12V>----------------+------+---------+--------+
|R1 | | |R4
[10M] | | [3K3]
| | | |
Q1 | | +---|--[1M]--+
2N3904 | |R2 | | R3 |
C----+ [10K]<--+--|+\ |
VIN>--[10K]---B | | | >------+-->VOUT
E +------|--------|-/ U1A
| |+ | | LM393
| [4.7µF] | |
| |C1 | |
GND>------------+----+------+---------+----------->GND


With no +12V, C1 will be discharged and 0V will appear on U1A-.

Then, when power comes up, U1A+ will go more positive than U1A-,
causing VOUT to rise to +12V and stay there until C1 charges to a
voltage higher than that on U1A+. At that time VOUT will fall to
about 0.4V worst case.

Any time VIN goes high it will turn on Q1, which will discharge C1
quickly, making the voltage on U1A- more negative than the voltage
on U1A+, forcing VOUT high until VIN goes low and allows C1 to
charge up again.

R2 should be set somewhere in the vicinity of 8V for a 45 second
timeout and it would be a good idea to use a polyester (Mylar) cap
for C1 in order to get low leakage and a reasonable tempco. If you
want to/have to use an aluminum electrolytic, observe polarity.

Also, connect all of the unused pins of the second comparator to
ground.
 
J

John Fields

Jan 1, 1970
0
The 45sec one shot triggers as soon as power is applied to the chip.The
output goes high right away.
But it will stay on as long as the other signal is 12volts after this
signal goes to 0 then the output goes low 45 seconds later. Everytime the
control signal goes hi it keeps the output hi until 45 seconds after the
signal goes back to 0v.

---

View in Courier:

+12V>----------------+------+---------+--------+
|R1 | | |R4
[10M] | | [3K3]
| | | |
Q1 | | +---|--[1M]--+
2N3904 | |R2 | | R3 |
C----+ [10K]<--+--|+\ |
VIN>--[10K]---B | | | >------+-->VOUT
E +------|--------|-/ U1A
| |+ | | LM393
| [4.7µF] | |
| |C1 | |
GND>------------+----+------+---------+----------->GND


With no +12V, C1 will be discharged and 0V will appear on U1A-.

Then, when power comes up, U1A+ will go more positive than U1A-,
causing VOUT to rise to +12V and stay there until C1 charges to a
voltage higher than that on U1A+. At that time VOUT will fall to
about 0.4V worst case.

Any time VIN goes high it will turn on Q1, which will discharge C1
quickly, making the voltage on U1A- more negative than the voltage
on U1A+, forcing VOUT high until VIN goes low and allows C1 to
charge up again.

R2 should be set somewhere in the vicinity of 8V for a 45 second
timeout and it would be a good idea to use a polyester (Mylar) cap
for C1 in order to get low leakage and a reasonable tempco. If you
want to/have to use an aluminum electrolytic, observe polarity.

Also, connect all of the unused pins of the second comparator to
ground.

---
Slow day...

Version 4
SHEET 1 880 680
WIRE 0 96 -416 96
WIRE 128 96 0 96
WIRE 320 96 128 96
WIRE 448 96 320 96
WIRE 0 128 0 96
WIRE 128 128 128 96
WIRE 448 128 448 96
WIRE 320 224 320 96
WIRE 0 240 0 208
WIRE 288 240 0 240
WIRE 0 256 0 240
WIRE 0 256 -96 256
WIRE 448 256 448 208
WIRE 448 256 352 256
WIRE 128 272 128 208
WIRE 240 272 128 272
WIRE 288 272 240 272
WIRE -272 304 -320 304
WIRE -160 304 -192 304
WIRE 0 304 0 256
WIRE 128 304 128 272
WIRE 320 320 320 288
WIRE -416 336 -416 96
WIRE -320 336 -320 304
WIRE 240 384 240 272
WIRE 288 384 240 384
WIRE 448 384 448 256
WIRE 448 384 368 384
WIRE -416 448 -416 416
WIRE -320 448 -320 416
WIRE -320 448 -416 448
WIRE -96 448 -96 352
WIRE -96 448 -320 448
WIRE 0 448 0 368
WIRE 0 448 -96 448
WIRE 128 448 128 384
WIRE 128 448 0 448
WIRE -416 496 -416 448
FLAG 320 320 0
FLAG -416 496 0
SYMBOL Comparators\\LT1017 320 256 R0
SYMATTR InstName U1
SYMBOL res 384 368 R90
WINDOW 0 66 54 VBottom 0
WINDOW 3 72 58 VTop 0
SYMATTR InstName R1
SYMATTR Value 1e6
SYMBOL res 432 112 R0
WINDOW 0 48 44 Left 0
SYMATTR InstName R2
SYMATTR Value 10k
SYMBOL res 112 112 R0
WINDOW 0 38 42 Left 0
SYMATTR InstName R3
SYMATTR Value 10k
SYMBOL res 112 288 R0
SYMATTR InstName R4
SYMATTR Value 20k
SYMBOL res -16 112 R0
WINDOW 0 45 40 Left 0
SYMATTR InstName R5
SYMATTR Value 10e6
SYMBOL cap -16 304 R0
SYMATTR InstName C1
SYMATTR Value 4.7e-6
SYMBOL voltage -416 320 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value PULSE(0 12 1 1e-6 1e-6)
SYMBOL npn -160 256 R0
WINDOW 0 62 20 Left 0
WINDOW 3 30 47 Left 0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL voltage -320 320 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value PULSE(0 12 60 1e-6 1e-6 10)
SYMBOL res -176 288 R90
WINDOW 0 -39 56 VBottom 0
WINDOW 3 -38 62 VTop 0
SYMATTR InstName R6
SYMATTR Value 10k
TEXT -392 472 Left 0 !.tran 240
 
R

Rich Grise

Jan 1, 1970
0
Also, connect all of the unused pins of the second comparator to
ground.

Just the inputs. Leave the output floating.

Cheers!
Rich
 
J

John Fields

Jan 1, 1970
0
Just the inputs. Leave the output floating.

---
Interesting...

Their 1989 General Purpose Linear Devices databook, 5-65 states:
"All pins of any unused comparators should be grounded", while page
8 of the datasheet at:

http://cache.national.com/ds/LM/LM193.pdf


states: "All input pins of any unused comparators should be tied to
the negative supply."
 
R

Rich Grise

Jan 1, 1970
0
Interesting...

Their 1989 General Purpose Linear Devices databook, 5-65 states:
"All pins of any unused comparators should be grounded", while page
8 of the datasheet at:

http://cache.national.com/ds/LM/LM193.pdf

states: "All input pins of any unused comparators should be tied to
the negative supply."

I will grant you that grounding an open-collector output would do
no harm, but I wonder if it's really beneficial, i.e., would it
make any actual difference in the operation overall?

Thanks,
Rich
 
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