555 monostable to trigger a 555 astable

elec_mech

Joined Nov 12, 2008
1,500
No, it sounds like you need a (latching?) circuit that activates after a complete clock cycle as opposed to one that activates on a high or low pulse. A D flip flop configured as a T flip flop mught work here, but I need to know more first.

I haven't read your entire post and it looks like there have been some changes, so let ask the following:

1) Is SW1 in your last schematic part of this two-switch activation or is it only to supply power to the circuit when it is being used? If the latter, I assume it is a toggle or rocker switch, correct?

2) Based on the last exchange between you and Bill, you want the following to occur?
a) User flips SW1 to apply power to circuit, but circuit does nothing yet.
b) User presses switch A and holds (nothing happens).
c) User presses switch B and holds (nothing happens).
d) User releases switch A and circuit begins.

3) Can user release switch B immediately after A is released or must they wait until circuit completes its mission, i.e., if user releases B before circuit finishes operation, the operation is stopped?

4) What is being turned on by the switch procedure? Only U2 or U1 & U2? My question being is U1 being used as part of the switch activation or is the switch activation used to turn on U1 (which later turns on U2)?
 

Thread Starter

thedude123

Joined Aug 29, 2011
39
No, it sounds like you need a (latching?) circuit that activates after a complete clock cycle as opposed to one that activates on a high or low pulse. A D flip flop configured as a T flip flop mught work here, but I need to know more first.

I haven't read your entire post and it looks like there have been some changes, so let ask the following:

1) Is SW1 in your last schematic part of this two-switch activation or is it only to supply power to the circuit when it is being used? If the latter, I assume it is a toggle or rocker switch, correct?

Yes, it is part of the two switch operation in the activation procedure.
It is a pushbutton switch, SPST on/off. I am using this switch in the breadboarding..

2) Based on the last exchange between you and Bill, you want the following to occur?
a) User flips SW1 to apply power to circuit, but circuit does nothing yet.
b) User presses switch A and holds (nothing happens).
c) User presses switch B and holds (nothing happens).
d) User releases switch A and circuit begins.

Almost,,, Let me reiterate to have it all in one place.
The proper procedure is this; Note that My test circuit uses a 4 sec delay vs a 20 sec delay to minimise test result times. Regardless,,, adjusting time delays and flashing cycles is easy once I can get them talking to eachother....
1.) Press and hold momentary switch (S3).
2.) Press SW1 to turn on the circuit.
3.) Release SW3 any time after that, (at users descretion)
4.) The circuit pauses for 4 sec. then LEDs flash (at U2 pin 3 output).
That is the circuit at its base level. If possible I would like to add the following features to it.
4a) Step 4 would continue for 4 sec... (blinking), then stop.

5.) If not to overly complicated,,, I would like to add a "reminder" action. After all steps have been completed flash the LEDs for 1 sec, every 5 min.
(I can imagine that this would be another 555 circuit. No problem, but I'm just having trouble figuring out how to connect them to control on another.... obviously...)

The user MUST follow this procedure or the circuit will start when they do not intend it to,, (While they are waiting to release SW3,,, SURPRISE! DOH!)
for example. If the user pushes SW1 without FIRST pressing and holding SW3,, then the delay time will run, then the LEDs will flash. The only way to reset the circuit within the delay time if this mistake is made is to
turn OFF SW1 before the delay times out and the LEDs flash. (there is also a buzzer in the output circuit, but I'll add that later. Buzzers are annoying during troubleshooting)


Sounds like a complicated procedure,, but it is not really.

"Hold the trigger, press the button, release the trigger and the circuit runs it's delay, then flashes".
(Do it backwards,,) ie. Press the button, THEN hold the trigger and it goes whithout you releasing the trigger"...


3) Can user release switch B immediately after A is released or must they wait until circuit completes its mission, i.e., if user releases B before circuit finishes operation, the operation is stopped?

I think I answered this part in the above...?

4) What is being turned on by the switch procedure? Only U2 or U1 & U2? My question being is U1 being used as part of the switch activation or is the switch activation used to turn on U1 (which later turns on U2)?

SW1 will power the entire circuit, U1 and U2... Consider it the MAIN.. and the user action of the release of SW3 as the GO, (or start delay and then flash).
My Answers in BLUE above..I hope this helps....
 

Thread Starter

thedude123

Joined Aug 29, 2011
39
Fixing the circuit by removing the resistor on pin 4,, (it was leftover when I removed the RST switch, missed removing it,,,) And wiring both circuit pin 4 to Vcc, did not fix the issue. Not to say it wasn't the right thing to do, of coarse... They are wired to Vcc now.

Iv'e done some troubleshooting with some various results. With my two circuits, U1 being Monostable and U2 being Astable I have connected the output (pin3) of U1 to different pins on U2 and achieved different results.

Now for you experts this will be obvious.

Connecting U1 Pin 3 to U2 pin,,,
2- Pauses 4 sec then LED stays lit. (with or without U2 pin4 to Vcc)
4- Blinks for 4 sec, then turns off. (pin 4 jumper to Vcc removed)
5- Blinks slower for 4 sec, then turns off.

If I remove the jumper from U1 pin 3 to U2 pin 2 then U2 blinks, (ocillates) as desined.

So,,, the two circuits almost do what I want them too. They do work independantly as I need them to, but not together....

Note... When U1 pin3 is connected to U2 pin2...

Holding SW3, pressing SW1 and immediatly releasing SW3 gives a 4 sec delay, then the LED comes on,,, but stays on and does not flash (as the astable cirquit is designed to). If I repeat this and delay the release of SW3, the 4 sec delay still starts. ie,,, if I hold SW3 for more than 4 sec, then release, the LED comes on immediatly....

I thought I was close at this point,, but not quite...
 
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Wendy

Joined Mar 24, 2008
22,141
Sorry for my delay, I've been getting over some dental surgery.

I have redone my schematic to match what I have been troubleshooting and trial and error on the breadboard.

I am not using the reset momentary for the U1 monostable. It is not needed. Although, I just noticed that when I removed the switch, I left the 10k resistor wired on pin 4 to Vs...

Also, the pin4 to Vs on U2 was removed in the troubleshooting process.
Maybe It should be there,,,?

Thanks for you'alls patience.... and look forward to figuring this out.
Thanks. You are doing something astandard there, I'll get back with you. You aren't using pin 4 like I was describing.
 

elec_mech

Joined Nov 12, 2008
1,500
Almost,,, Let me reiterate to have it all in one place.
The proper procedure is this; Note that My test circuit uses a 4 sec delay vs a 20 sec delay to minimise test result times. Regardless,,, adjusting time delays and flashing cycles is easy once I can get them talking to eachother....
1.) Press and hold momentary switch (S3).
2.) Press SW1 to turn on the circuit.
3.) Release SW3 any time after that, (at users descretion)
4.) The circuit pauses for 4 sec. then LEDs flash (at U2 pin 3 output).
That is the circuit at its base level. If possible I would like to add the following features to it.
4a) Step 4 would continue for 4 sec... (blinking), then stop.
Okay, I think we can do this with a simple RC (resistor & capacitor) circuit in-line. Once the user presses SW3, the capacitor charges and four seconds later, U1 turns on then sends a 4 second pulse to the U2. In this way, you get your 4 second delay and four second blink.

5.) If not to overly complicated,,, I would like to add a "reminder" action. After all steps have been completed flash the LEDs for 1 sec, every 5 min.
(I can imagine that this would be another 555 circuit. No problem, but I'm just having trouble figuring out how to connect them to control on another.... obviously...)
Possible, but begs the question, will the user be holding SW1 continuously for five or more minutes to keep the circuit powered for this long?

The user MUST follow this procedure or the circuit will start when they do not intend it to,, (While they are waiting to release SW3,,, SURPRISE! DOH!)
for example. If the user pushes SW1 without FIRST pressing and holding SW3,, then the delay time will run, then the LEDs will flash. The only way to reset the circuit within the delay time if this mistake is made is to
turn OFF SW1 before the delay times out and the LEDs flash. (there is also a buzzer in the output circuit, but I'll add that later. Buzzers are annoying during troubleshooting)
Okay, using a DPST switch for SW3 should allow us to achieve this. But, I'm confused. What happens if the user presses SW1 and SW3 out of sequence? It sounds like in either case the LED flashes. What is the difference, no time delay?

Sounds like a complicated procedure,, but it is not really.

"Hold the trigger, press the button, release the trigger and the circuit runs it's delay, then flashes".
(Do it backwards,,) ie. Press the button, THEN hold the trigger and it goes whithout you releasing the trigger"...
Really confused now.

Normal operation:
1) Press SW3 and hold.
2) Press SW1 and hold.
3) Release SW3 (is this optional or is the user required to release SW3 before the circuit starts)?
4) Wait 4 seconds then blink LED for 4 seconds
5) Keep SW1 pressed indefinitely and have (another?) LED blink once every five minutes.

Abnormal operation:
1) Press SW1 and hold.
2) Press SW3 and hold (or momentary press and release?).
3) ???

I'd like to see this numbered in order step-by-step as I have above with clear explanations on what happens when the user presses a button and when he/she releases the button at each stage.
 

Wendy

Joined Mar 24, 2008
22,141


OK, you have two distinct problems here.

Pin 4 is an input, it needs connected to Vcc if you aren't using it, but that isn't the thing that is killing your schematic.

The output of the second 555 is either ground or Vcc - 1.4V, the point the output is connected to will also be either ground or Vcc - 1.4V, in other words you are connecting a critical point on the first 555 oscillator and killing it.

Instead, try connecting the output of the 2nd 555 to pin 4 of the first, this will work.

I circled the resistor on pin 4 of the second 555 because it isn't needed, you can connect pin 4 directly to Vcc. It is a dunsel part, it isn't needed.




Take a look at the above schematic, which is a variation of what you are using. Pins 2 and 6 need to be allowed to float, put a hard voltage there and the oscillator dies.
 

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Thread Starter

thedude123

Joined Aug 29, 2011
39


OK, you have two distinct problems here.

Bill, Thanks for your help. I am understanding your descriptions of the two circuits as left to right in the schemtic;

"First" = U2 on the left
"Second" = U1 on the right
I labeled them U1 and U2 right to left as this was the way that I felt they would function. (U1 would trigger U2) Sorry for any confusion.


Pin 4 is an input, it needs connected to Vcc if you aren't using it, but that isn't the thing that is killing your schematic.

The output of the second 555 is either ground or Vcc - 1.4V, the point the output is connected to will also be either ground or Vcc - 1.4V, in other words you are connecting a critical point on the first 555 oscillator and killing it.

Instead, try connecting the output of the 2nd 555 to pin 4 of the first, this will work.

U1 outout connected to U2 pin 4. I did this and posted above post #23 as some troubleshooting and yes, the U2 oscillator blinks for 4 seconds then stops. This is getting close. But there is no delay before it oscillates.

I circled the resistor on pin 4 of the second 555 because it isn't needed, you can connect pin 4 directly to Vcc. It is a dunsel part, it isn't needed.

Corrected.




Take a look at the above schematic, which is a variation of what you are using. Pins 2 and 6 need to be allowed to float, put a hard voltage there and the oscillator dies.
New corrected schematic attached.
 

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Wendy

Joined Mar 24, 2008
22,141


You are getting warmer. There is still a potential chip eating problem.

U2 pin 7 is either open or switched to ground. That is its function, it is what it does. If R6 is adjusted to 0Ω then you have shorted this chip between the power supply rails. You avoided this on U1 with resistor R1

Add a 100Ω or larger in series with R6.

You are learning why people label their chips. I have to admit I'm guilty occasionally, usually when I start with one part, then add another later.

Have you tried this circuit yet? If you have and it still doesn't quite do what you want I'll tweak it, but as you've figured out I like to teach, and nothing teaches like doing it yourself.

It may be the circuit is inverted on pin 4. I use a simple 2 resistor 1 transistor inverter when this happens, something like what I did in this circuit.

CMOS 555 Long Duration LED Flyback Flasher

I have several 555 design resources if you are interested, cheat sheets and the like. The 555 schematic I offered up came from my cookbook. I like to draw, but I use my own package to do so.
 
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Thread Starter

thedude123

Joined Aug 29, 2011
39
Okay, I think we can do this with a simple RC (resistor & capacitor) circuit in-line. Once the user presses,, (releases) SW3, the capacitor charges and four seconds later, U1 turns on then sends a 4 second pulse to the U2. In this way, you get your 4 second delay and four second blink.

"U1 turns on then sends a 4 second pulse to the U2."
Exactly. It does work this way as is now with a few flaws.
After the 4 sec delay the LED just stays on, doesn't blink. (If I remove the U1 pin3 to U2 pin2 the LED will blink. Bill described and corrected this in his post below. The key I am missing in connecting these two is as you stated.
Also, The U1 4 sec delay starts when the SW3 is pressed and held, not when it is released. This could be due to wrong SW3 wiring. So what triggers a monostable? The momentary or permanant grounding of pin2?



Possible, but begs the question, will the user be holding SW1 continuously for five or more minutes to keep the circuit powered for this long?

SW1 is a pushbutton ON/OFF switch



Okay, using a DPST switch for SW3 should allow us to achieve this. But, I'm confused. What happens if the user presses SW1 and SW3 out of sequence? It sounds like in either case the LED flashes. What is the difference, no time delay?

There is no difference in circuit operation, either way.
If the switches are pressed out of sequence then the circuit will start as designed. The user will think that they start the circuit when releasing SW3 as planned, but since they operated the switches out of sequence, the 4 sec delay occurs when they did not expect it too. (while they are still holding and waiting to release SW3).

See, when the user is holding SW3 thinking that they have control of when the circuit starts. Then 4 sec later, LED flash proving that they performed the sequence wrong. If the user realizes they have done it wrong they can ONLY operate SW1 and turn it off before the 4 sec delay triggers the LED. This is an acceptable correction by the user of the botched sequence.

The key to operating this circuit properly is the that the user releases SW3 whenever they are ready to start the 4 sec delay. Operating SW1 BEFORE holding SW3 the user THINKS they have control of starting the circuit, when in fact it has already started because they pressed them out of sequence. This causes quite a suprise when the LED, (or Alarm) goes off "unexpectadly".........

Really confused now.

Normal operation:
1) Press SW3 and hold. YES
2) Press SW1 and hold.
NO. press and release. It is an ON/OFF button. the one I linked to in the obove posts.
3) Release SW3 YES, This will start the circuits.
(is this optional or is the user required to release SW3 before the circuit starts)? They are required to release SW3 to start the circuit.
4) Wait 4 seconds then blink LED for 4 seconds YES
5) Keep SW1 pressed indefinitely and have (another?) LED blink once every five minutes. Read Above... correction to step2

Abnormal operation:
1) Press SW1 and hold. Read Above... correction to step2
2) Press SW3 and hold (or momentary press and release?).
3) ???

I'd like to see this numbered in order step-by-step as I have above with clear explanations on what happens when the user presses a button and when he/she releases the button at each stage.
There are two switches. SW1 and SW3,
SW1 is this If it helps, just consider SW1 a toggle ON/OFF switch
http://www.radioshack.com/product/index.jsp?productId=3159591&filterName=Type&filterValue=SPST

SW3 is this
http://www.radioshack.com/product/index.jsp?productId=2049718&filterName=Type&filterValue=SPDT

If these are not the correct switches, please advise.



Proper operation;
1.) Press and hold SW3
2.) Press and release pushbutton on/off switch, SW1
-- Nothing happens because the user is still holding SW3.
3.) When the user releases SW3 the circuit starts;
action- A delay of 4 sec. then LED flashes for approx 4 sec then turns off.
4.) After about 5 min the LED will flash to remind the user to turn off the circuit by pressing and releasing SW1. (this step may not be required but it would be nice to have as a reminder to turn off the cirquit with SW1)

Improper operation.
1.) Pressing and releasing the on/off SW1 without FIRST holding SW3.
The circuit will start and perform as normal. This would be a mistake by the user in the procedure and the LED would go off.

I hope this helps describe the needed procedure.
 

Thread Starter

thedude123

Joined Aug 29, 2011
39


You are getting warmer. There is still a potential chip eating problem.

U2 pin 7 is either open or switched to ground. That is its function, it is what it does. If R6 is adjusted to 0Ω then you have shorted this chip between the power supply rails. You avoided this on U1 with resistor R1

Add a 100Ω or larger in series with R6.

Will do. Thanks.

You are learning why people label their chips. I have to admit I'm guilty occasionally, usually when I start with one part, then add another later.

? I have labeled them,, U1 and U2. I have moved the labels to be on top of the 555s for better clarity.

Have you tried this circuit yet? If you have and it still doesn't quite do what you want I'll tweak it, but as you've figured out I like to teach, and nothing teaches like doing it yourself.

Yes,, in post #23 I noted it in troubleshooting. Although,, this could have been back when I still had the resistor on pin4. I will retry tonight.
When I tried it, triggering SW3 would make the U2 flash the LED for 4 seconds time, then turn off. Close,,, but not quite there yet. There was no delay.

It may be the circuit is inverted on pin ""for"",,, (pin 4??). I use a simple 2 resistor 1 transistor inverter when this happens, something like what I did in this circuit.

CMOS 555 Long Duration LED Flyback Flasher

I will try adding Res / Trans as you have.
1.) Is the L1 choke critical to my application?
2.) Should I add both Q1 and Q2? Seems that just Q1 is needed at my U1 pin3?

I have several 555 design resources if you are interested, cheat sheets and the like. The 555 schematic I offered up came from my cookbook. I like to draw, but I use my own package to do so.

I have spent some time breowsing your blog, yes. One of my TOP resources to get to where I am now and still learning.
Comments above.

Made changes to my schematic and O-Scoped.. pic attached
 

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SgtWookie

Joined Jul 17, 2007
22,210
I do not see a ground reference anywhere in your schematic. Place one on the bottom supply "rail". Otherwise, you will get strange readings.

Also, I suggest that you move your power switch to the positive battery terminal.

There are 555 SPICE models out there that send current to node 0 (ground) instead of through the timer's ground pin; this can cause some very strange readings if the 555's ground pin is not connected to ground (node 0).

I discovered that when I tried to use a 555 model with a negative voltage for ground, and ground for Vcc. The circuit would not work at all until I connected the ground pin to ground.
Your entire circuit is floating with no ground reference - unless the 555's have an internal connection to node 0.
 
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Thread Starter

thedude123

Joined Aug 29, 2011
39
I do not see a ground reference anywhere in your schematic. Place one on the bottom supply "rail". Otherwise, you will get strange readings.

Also, I suggest that you move your power switch to the positive battery terminal.

There are 555 SPICE models out there that send current to node 0 (ground) instead of through the timer's ground pin; this can cause some very strange readings if the 555's ground pin is not connected to ground (node 0).

I discovered that when I tried to use a 555 model with a negative voltage for ground, and ground for Vcc. The circuit would not work at all until I connected the ground pin to ground.
Your entire circuit is floating with no ground reference - unless the 555's have an internal connection to node 0.
I would have thought that the ground terminal at the 9V source would be acceptble.. I've added one on the ground rail anyway.

I moved the SW1 power switch to the positive side of the source.

Thanks for the tips on good practice.
 

Wendy

Joined Mar 24, 2008
22,141
I would listen well to Wook, when it comes to simulators he leaves me in the dust. I don't simulate circuits at all (except between my ears), I build, verify, remember, and document. Both have strong and weak points.

I'm not sure what you are talking about for C3. If you are referring to the odd startup shown below this is very characteristics for all 555 circuits, enough so I have developed techniques to prevent it if need be.



Generally this one time extra long pulse is no big deal for oscillators.
 

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elec_mech

Joined Nov 12, 2008
1,500
Okay, I completely missed the link to the toggle switch - that makes a lot more sense now.

Bill and SgtWookie and other people smarter than myself can comment and correct, but you can add the 4 second startup delay in one of two ways - add another 555 circuit or use an RC circuit.

Now, I haven't played with these much myself just yet, but I've attached a few ideas that may or may not work based on the RC circuit. I don't know if R1 in circuit B or Rb in circuit C is necessary or possibly detrimental, but I added the former to keep pin 4 pulled low (in reset) and the latter to limit current to the transistor.

You'll need to figure out the value for Rt and Ct to get 4 seconds. You can use this to help you: http://www.bowdenshobbycircuits.info/rc.htm.

Note that the 555 operates well below 9V, so you may need to adjust your R & C values to give you a longer time (4 seconds to reach 5V instead of 9V for example, in case 5V is enough to trigger the reset or trigger pins). Alternately, you may elect to add a zener diode so you're sure the output does not go high until the zener voltage is reached. Since you're using 9V, you may have to go with a 8.2V zener such as 1N4738.
 

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Wendy

Joined Mar 24, 2008
22,141
Certain specific 555 circuits will not work at the low end of the voltage spec. This is because a 555 outputting a high is Vcc - 1.4VDC. You really look at 555 it reads as if it were designed for TTL (other than the wide power supply voltage range).

This doesn't apply for the OPs needs though.

I have to admit to being a fan of Bill Bowden's website. His drawings look a lot like mine, I suspect we both came up with the same idea of using MS Paint independently, except I offer my templates publically.
 

Thread Starter

thedude123

Joined Aug 29, 2011
39
I would listen well to Wook, when it comes to simulators he leaves me in the dust. I don't simulate circuits at all (except between my ears), I build, verify, remember, and document. Both have strong and weak points.

I'm not sure what you are talking about for C3. If you are referring to the odd startup shown below this is very characteristics for all 555 circuits, enough so I have developed techniques to prevent it if need be.



Generally this one time extra long pulse is no big deal for oscillators.
Bill, I understand the startup variance in the oscillations. It is acceptable. I was referring to the oscillation band increasing over time. It is hard to tell from the graph but the Red oscillation line rises.

Since I added the transistor as the inverter as you suggest, naturally the oscillation voltage band now decreases. This could be a "sim" thing and my lack of knowledge to use them. So far they have been ok,,, but I dont trust them completely... I breadboard it when I believe it to be what I want.
 

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thedude123

Joined Aug 29, 2011
39
Certain specific 555 circuits will not work at the low end of the voltage spec. This is because a 555 outputting a high is Vcc - 1.4VDC. You really look at 555 it reads as if it were designed for TTL (other than the wide power supply voltage range).

This doesn't apply for the OPs needs though.

I have to admit to being a fan of Bill Bowden's website. His drawings look a lot like mine, I suspect we both came up with the same idea of using MS Paint independently, except I offer my templates publically.

Fan as well,,,, but on the circuit of his that I was building from,,, he does not label the diodes on his schematic. Should we assume they are the common 1N4148 ??
 

Wendy

Joined Mar 24, 2008
22,141
Generally diodes are not critical, so they would probably work. The main thing is current, a 1N4001 will handle up to an amp. I bought several thousands of 1N4007 for $1, wanna guess what my first choice is?

The other thing about diodes is their speed. The 1N4148 (as well as the 1N4454 and 1N918) are digital low power diodes, they don't have much current but are pretty fast.

Most cases though it simply doesn't matter. I have heard the term audition a part, it is very descriptive of what I do in many cases.

When you refer to an oscillation band are you referring to the P-P voltage of the triangle wave, or the frequency of the oscillation?
 

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thedude123

Joined Aug 29, 2011
39
That's good to know. I bought about 10 of the 4148s just to have them on hand.
See latest attachment. Seemed odd that the oscillation is dropping like that.

I am home now in front of my breadboard.... willing to try anything at this point.
 
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