Shut-off Delay Timer for Dust Collector - 555?

Thread Starter

Crashcup

Joined Oct 11, 2017
17
Hi All, I'm pretty new to this forum, so be gentle.

I have what is probably a very simple application for most of you. I'm starting to build up a fixed dust collection system for my wood shop. My intention is to trigger the dust collector from micro switches on the blast gates.

From what I've read elsewhere, it may be a good idea to delay the shutoff of the DC to reduce the number of starts of the motor. I am thinking 3 to 5 minutes would be about right. That way if I work at one machine, move directly to another and open a different blast gate, the DC just keeps running, and shuts off a bit later when I'm truly done with it for a while.

So, the first thing that came to mind was to add a circuit using a 555 IC. I attached a first draft of a circuit, based on others I've seen online and modified to energize the relay all the time that one of the switches is on. Am I close with this? I'm not very experienced with electronics, but know the basics. And not much experience with the 555 other than playing with a few circuits on a breadboard years ago. (And now that I write that I realize that I should just hook this up and try).

I have an old wall wart with 9V output that I think I can use for the supply. The switches in the lower left would be a few switches to start with, and eventually up to 11 or 12, but always wired in parallel so any one of the switches can close the circuit. The relay shown at lower right will actually be a solid state relay; 3-32VDC in, 24-380VAC load. But I didn't know what the schematic symbol was, so... I think I've drawn the symbol for a regular coil type relay.

Other example circuits I've seen include a diode across the relay inputs - I don't quite remember the purpose... something about a current being induced when the input to the relay is shut off. But in the case of a solid state relay, this is probably not needed, right?

Also, I'm not sure of the need for a diode on the output of the 555. Not sure what the chances are of current flowing back into that pin and wrecking it. But it also shouldn't hurt to have it there I think.

Last question is: Does this make sense to use a 555? 5 minutes is a pretty long delay compared to examples I've seen. Playing around with the C1/R1 combinations, I'm getting values like 10 uF / 27 Mohms, or 5454 uF (5.4 millifarads?) and 50 kohms. I haven't shopped around yet to see if those are wild values that will be hard to find. Ideally I'd like to be able to use a pot to vary the resisitance to adjust the delay time.

Or is there a reason it would be better to use a microcontroller, like an Arduino Nano?

Sorry for the rambling post... and thanks in advance for any help!
Keith
 

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AnalogKid

Joined Aug 1, 2013
8,947
You're circuit will not do what you want, but you're pretty close. FIRST - there is no need for *anything* programmable. OK, I feel better now.

With a SSR, you do not need a diode across its input; you need the diode only if the device is an inductor like a relay coil.

You do not want a diode on the output of the 555. I see your idea, but there is a better way. Rather than have the switches trigger the 555 *and* drive the SSR directly, have the switches trigger the 555 only, and have only the 555 drive the SSR. A 555 is not a true monostable; it is called "retriggerable". In your case, this is a feature (it works in your favor), not a bug (another 555 personality quirk). The 555 output goes high when the trigger input goes low, but

<everything I said next was wrong> <see below>

Your concern about looooong timing periods is absolutely correct. A CMOS 555 has much higher impedances at its inputs, so you can get away with some longer timing periods (compared to a standard 555). Still, try to keep the resistors at or below 1 M. 270 uF and 1 M calculate out to 4.95 minutes. In real life this probably will be longer because part of the charging current passes through the capacitor as leakage current (a problem with large capacitors). The alternative is a circuit based on a counter. Not a big increase in complexity, but if you already have a CMOS 555 (LMC555) I'd start there.

Your switches are connected incorrectly. As shown in the 555 datasheet "monostable" circuit, the trigger input is pulled to GND to start the timer. Add a 1K resistor from pin 2 to the +9 V. All switches are in parallel from pin 2 to GND. Redraw and repost.

ak
 
Last edited:

eetech00

Joined Jun 8, 2013
2,280
Hi All, I'm pretty new to this forum, so be gentle.

I have what is probably a very simple application for most of you. I'm starting to build up a fixed dust collection system for my wood shop. My intention is to trigger the dust collector from micro switches on the blast gates.

From what I've read elsewhere, it may be a good idea to delay the shutoff of the DC to reduce the number of starts of the motor. I am thinking 3 to 5 minutes would be about right. That way if I work at one machine, move directly to another and open a different blast gate, the DC just keeps running, and shuts off a bit later when I'm truly done with it for a while.

So, the first thing that came to mind was to add a circuit using a 555 IC. I attached a first draft of a circuit, based on others I've seen online and modified to energize the relay all the time that one of the switches is on. Am I close with this? I'm not very experienced with electronics, but know the basics. And not much experience with the 555 other than playing with a few circuits on a breadboard years ago. (And now that I write that I realize that I should just hook this up and try).

I have an old wall wart with 9V output that I think I can use for the supply. The switches in the lower left would be a few switches to start with, and eventually up to 11 or 12, but always wired in parallel so any one of the switches can close the circuit. The relay shown at lower right will actually be a solid state relay; 3-32VDC in, 24-380VAC load. But I didn't know what the schematic symbol was, so... I think I've drawn the symbol for a regular coil type relay.

Other example circuits I've seen include a diode across the relay inputs - I don't quite remember the purpose... something about a current being induced when the input to the relay is shut off. But in the case of a solid state relay, this is probably not needed, right?

Also, I'm not sure of the need for a diode on the output of the 555. Not sure what the chances are of current flowing back into that pin and wrecking it. But it also shouldn't hurt to have it there I think.

Last question is: Does this make sense to use a 555? 5 minutes is a pretty long delay compared to examples I've seen. Playing around with the C1/R1 combinations, I'm getting values like 10 uF / 27 Mohms, or 5454 uF (5.4 millifarads?) and 50 kohms. I haven't shopped around yet to see if those are wild values that will be hard to find. Ideally I'd like to be able to use a pot to vary the resisitance to adjust the delay time.

Or is there a reason it would be better to use a microcontroller, like an Arduino Nano?

Sorry for the rambling post... and thanks in advance for any help!
Keith
Hi

In what type of environment will this timer be operating? Lots of electrical noise?
Will the switches require long wire runs to the timer? A few feet maybe?

eT
 

Thread Starter

Crashcup

Joined Oct 11, 2017
17
You're circuit will not do what you want, but you're pretty close....

I see your idea, but there is a better way. Rather than have the switches trigger the 555 *and* drive the SSR directly, have the switches trigger the 555 only, and have only the 555 drive the SSR. A 555 is not a true monostable; it is called "retriggerable". In your case, this is a feature (it works in your favor), not a bug (another 555 personality quirk).
ak
Thanks ak. I think you answered a question I forgot to ask. Wasn't sure if pin 2 was held "high" if the output stayed high or needed the triggering of pin 2 going "low" to send the output high. So I believe what you're saying is that as long as trigger pin 2 is high, the output is high? Then, when the trigger goes low, output will stay high for the delay time and then go low.

"Your switches are connected incorrectly. As shown in the 555 datasheet "monostable" circuit, the trigger input is pulled to GND to start the timer. Add a 1K resistor from pin 2 to the +9 V. All switches are in parallel from pin 2 to GND".

So the thing to do is have the trigger pin 2 connected to Vcc through the resistor, and wire the switches so they connect pin 2 to Gnd when they're closed? I think I get it.

I'll either change my drawing and repost or... see if I can sketch it in Eagle and post. Will have to watch a couple videos on Eagle to get started.
 

Thread Starter

Crashcup

Joined Oct 11, 2017
17
Hi

In what type of environment will this timer be operating? Lots of electrical noise?
Will the switches require long wire runs to the timer? A few feet maybe?

eT
Hi et, thank you. I think you're right on both counts. The wire runs for the switches for sure will be pretty long - up to maybe 70 ft from the DC and switching circuit.

Electrical noise I'm not sure how to evaluate, but it's a room full of tools with induction motors, so I'm guessing there will be a fair amount of EMI. I was hoping a 9V DC signal for switching wouldn't have any problems with that, but do you think some steps need to be taken?
 

AnalogKid

Joined Aug 1, 2013
8,947
Thanks ak. I think you answered a question I forgot to ask. Wasn't sure if pin 2 was held "high" if the output stayed high or needed the triggering of pin 2 going "low" to send the output high. So I believe what you're saying is that as long as trigger pin 2 is high, the output is high? Then, when the trigger goes low, output will stay high for the delay time and then go low.
Nope. When Triger (2) goes low, Output (3) goes High. That's why the switches are connected to GND; they pull the Trigger input Low.

BUT, my explanation above also has issues. More later.

ak
 

AnalogKid

Joined Aug 1, 2013
8,947
Been a while since I've used a 555 in a non-standard way. The standard 555 monostable will *not* do what you want (a true retriggerable monostable), but it will if we move the switches.

Inside the 555 is a flipflop which is set by the Trigger and reset by the Threshold. The important thing is that the Trigger input overrides the Threshold input, so if the Trigger is held low longer than the timing period, the output stays high. When the Trigger eventually does go high, the output immediately goes low and sits until the next Trigger negative transition. That is not what you want. Here is how to fix it.

1. Connect the Trigger (2) to pin 6 and the top of the timing capacitor. Disconnect pin 7.
2. Connect the paralleled switches directly across the timing capacitor, from 2-6 to GND.

Here is what happens:

1. When a switch closes, it discharges the timing cap and pulls the Trigger Low. This sets the internal ff and drives the output High.
2. As long as the cap and the trigger input are held at GND by the switch, the output stays high and the timing period is *not* started.
3. When the switch opens, the cap starts to charge up toward Vcc.
4. As the cap charges up through the timing resistor, the voltage on the Trigger input also rises.
5. When the cap reaches 1/3 Vcc, the trigger input releases the Set command to the ff.
6. The voltage across the cap continues to rise.
7. When the cap reaches 2/3 Vcc, the threshold input Resets the internal ff. This drives the output low.

If a switch closes in the middle of the timing period, it discharges the timing cap and holds it low. This is the same as step 1.

The 555 is not doing much of what makes a 555 special, and can be replaced by one transistor.

I'll whip up a schematic later.

ak
 
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AnalogKid

Joined Aug 1, 2013
8,947
OK, here we go. First, the retriggerable circuit described above, based on a CMOS 555. In this circuit, the 555 is acting as a single Schmitt Trigger inverter (like 1/6 of a CD40106) with a higher-current output stage. Hey ... there's a thought ...

Note: there is no need for a 1% resistor. 825K is the closest thing to 820K in my library.

ak
555-Retrig-1-1-c.gif
 
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AnalogKid

Joined Aug 1, 2013
8,947
Third, the same function with no active components. This circuit assumes that the SSR has a 3-32 V input at around 3 mA. Note the 24 V source; this greatly reduces the size of the capacitor.

ak
555-Retrig-1-3-c.gif
 

Thread Starter

Crashcup

Joined Oct 11, 2017
17
Wow, thanks very much for doing this ak. I took some time to thing through each of these, except the last one. Hopefully it makes sense for me to comment by replying to each of those separate replies.

I'll have to read about Schmitt triggers later when I get another break. I'm just not familiar with the general functionality.

Learning a lot today!
 

Thread Starter

Crashcup

Joined Oct 11, 2017
17
Third, the same function with no active components. This circuit assumes that the SSR has a 3-32 V input at around 3 mA. Note the 24 V source; this greatly reduces the size of the capacitor.

ak
View attachment 146909
This is very cool in its simplicity. At first glance, I didn't think there was a time delay built in, but apparently that's what C1 does. When a switch is closed, it activates the SSR, and also charges C1. When the switch opens again, C1 begins discharging, maintaining a high enough voltage to keep the SSR active for the delay.
The 24V must be to start at a higher voltage point to give more time before it gets down to 3V. This wouldn't be a problem, I probably have a furnace transformer around that's 24VAC, then would need to include a rectifier.

Two things I am unsure of:
Most of the delay circuits use a capacitor discharging through a resistor, but is that not required? I realize that a capacitor always has leakage so it will always bleed down when there isn't voltage on the + side, so it will slowly discharge down to 0..... but why is a resistor usually used in these circuits?

An SSR like I have, if it's labeled 3-32VDC on the input, will that generally mean that it has a threshold right around the 3 V?
 

Thread Starter

Crashcup

Joined Oct 11, 2017
17
OK, here we go. First, the retriggerable circuit described above, based on a CMOS 555. In this circuit, the 555 is acting as a single Schmitt Trigger inverter (like 1/6 of a CD40106) with a higher-current output stage. Hey ... there's a thought ...

Note: there is no need for a 1% resistor. 825K is the closest thing to 820K in my library.

ak
View attachment 146907
This one took me a while to think through, and I had to look up the 555 block diagram and comparators and flip-flops.

So in this, it looks like nothing would ever change for pin 5, Cv, and pin 7, Discharge. Apparently 7 would get connected to Gnd when the output goes high, but I don't see that it does anything.
5 looks like it would stay at 2/3 Vcc all the time. Is there a purpose to C3?

Also 4, the reset, is always at 9V -I did read somewhere that if this isn't being used as an active control, it should be kept at Vcc. Not exactly sure why. The block diagram I looked at wasn't exactly clear on what the reset does in terms of the flip-flop output value.

When all switches are open, 6 and 2 will be at Vcc. 6 drives comparator #2 output high; 2 drives comparator #1 output low. The block diagram tells me Comp#1 is Set and Comp#2 is Reset on the flip-flop, so in this state, flip-flop output Q-bar (whatever it's called, opposite of Q) is low, so output pin 3 is low.

When a switch is closed, 2 and 6 go low. C1 discharges to ground. 2 low drives comp#1 high (Set). 6 low drives comp#2 low (Reset). Result from flip-flop is Q-bar goes high, setting output pin 3 high.

Now the interesting part. When the switch is opened again, are 2 and 6 delayed in reaching Vcc again by the charging of C1? It's kind of hard to picture them not getting the full Vcc right away, but I guess it makes sense if C1 and R1 are big enough. The current through R1 must be small, so we get a long delay before pins 2 and 6 get a high enough voltage to change the output state again. (Also, for a moment I thought that closing a switch would short Vcc to Gnd, but the large resistor prevents that).

In this phase, where a switch is opened again, it's interesting what happens with the comparators and the FF. If I understand right, comparator #1 would switch first, when pin 2 reaches 1/3 Vcc. Comp #2 won't switch until pin 6 reaches 2/3 Vcc. In between that, the state of both comparators will be low. A website where I looked this up describes that state of S and R at 0 on the FF as being an invalid state. But, it also seems to show that if it switches from S=1/R=0 to S=0/R=0, the output Qbar will stay high. When R finally goes high (pin 6 reaches 2/3 Vcc), Qbar finally goes low, turning off the pin 3 output. Very confusing. Does that sound about right?
 

AnalogKid

Joined Aug 1, 2013
8,947
Two things I am unsure of:
Yes to everything. You continue to analyze correctly.

Yes, all capacitors have leakage current, so they all self-discharge. But usually that current is very much lower than whatever signal current is going in/out of the cap. In a timing circuit, the external resistor sets the charge or discharge current only because its resistance is much lower than the capacitor's internal leakage equivalent impedance. This is why long time circuits with a 555 are not recommended. A 12 V circuit with a 1 M resistor has a charging current that peaks at 12 uA and averages less than 4 uA over a cycle. If the leakage current for a multi-hundred uF aluminum electrolytic is 400 nA, that's a 10% error. As a one-off for a home project that's not a problem, just change the resistor value to compensate. But in a production product, no way. Also, the leakage current varies significantly from one cap to another, and with temperature.

I just checked. A low cost 330 uF 25 V cap at Digi-Key has a leakage current of 3 uA (!) or 0.01CV, which ever is greater. 0.01 x 330 Uf x 9 V = 30 uA (!!!). A $1 part brings this down to 16 uA. So, the timing period probably will be noticeably shorter than calculated. That is worse than I thought. Other than forum discussions, I've never actually designed an RC timer for anything over one minute. I've always thought they would suck, but I never ran the numbers to see how badly. Still, others around here swear by them.

Many "hockey-puck" SSR's have a constant-current circuit at the input, so the device draws the same 3 mA (a typical value from memory) for any voltage in the operating range. Thus, 3 mA is the discharge current for the timing cap, and is large enough that no additional resistor is needed. In fact, it is so large that the timing cap is relatively huge. Again, I never would propose this for a commercial product, but for home it is an interesting variation on the theme.

A 24 V transformer has a calculated peak voltage after rectification of over 33 V. Plus your line voltage might be a bit high. Plus the rating is at full load, and this circuit is nowhere near that. The idea is fine, but the transformer secondary should be *rated* at 20 v or less. The circuit already is a large filter capacitor, so all that is needed is one diode.

ak
 
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eetech00

Joined Jun 8, 2013
2,280
Hi et, thank you. I think you're right on both counts. The wire runs for the switches for sure will be pretty long - up to maybe 70 ft from the DC and switching circuit.

Electrical noise I'm not sure how to evaluate, but it's a room full of tools with induction motors, so I'm guessing there will be a fair amount of EMI. I was hoping a 9V DC signal for switching wouldn't have any problems with that, but do you think some steps need to be taken?
Well..a little concerned about using cmos in that environment because of its sensitivity but I guess we'll see.
Could always add input protection later and/or use the NE555 instead....but might be ok as is.

eT
 

Thread Starter

Crashcup

Joined Oct 11, 2017
17
Well..a little concerned about using cmos in that environment because of its sensitivity but I guess we'll see.
Could always add input protection later and/or use the NE555 instead....but might be ok as is.

eT
What would input protection be? Would there be any disadvantage to adding it "just in case"?
 

AnalogKid

Joined Aug 1, 2013
8,947
Well..a little concerned about using cmos in that environment because of its sensitivity but I guess we'll see.
Could always add input protection later and/or use the NE555 instead...
In each circuit, the only pins exposed to external wires have a 330 uF capacitor to one rail. That is better protection than diodes.

ak
 
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AnalogKid

Joined Aug 1, 2013
8,947
This a counter-based solution mentioned before, based on something for another thread. The CD4060 is an CMOS oscillator and 14-stage divider. This means that the square wave at Q14 is 16,384 times slower than the oscillator frequency. This means that the for a time delay equivalent to the other circuits, the timing components can be 8192 times smaller. The 4060 is one of the greatest CMOS parts ever. However, it is standard CMOS so its output stage is good for only 1-2 mA and it cannot drive the SSR directly.

When the Q14 output is low, transistor Q1 is on and the SSR is on. I added an LED so you can see that the part is in the timing phase.

A closed switch holds the device in reset, so all outputs are low, the SSR is on, and the LED is off. When all switches are open, the oscillator starts and the LED starts blinking at about 1 Hz. After 8192 oscillator cycles, Q14 goes high. This turns off Q1 and the SSR, and freezes the oscillator through D1 (the 4060 does not have an enable/disable pin). When any switch closes it resets the counter, turning on the SSR.

C2 and R3 form a power-on-reset. When the circuit is powered on it runs through a 5 minute fan cycle even if all switches are open.

ak
Fan-Off-Delay-1.gif
 
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