Delay on timer circuit

[Lazer]

Hi, I have a project that I am stuck on and would be really happy if I
had a solution.
I have two machines that both have footswitch controls (just SP/ST switches, no circuits, both pedals are the same) with 1/4" jacks.
The main unit turns on when the pedal is pressed and off when it's
released. The second unit turns on when the pedal is pressed and only turns off when it's pressed a second time.
At present I have to use seperate foot pedals for the two machines (a hassel). What I want is an adjustable splitter that accepts one footswitch and has leads that plug into both units. I want both units to turn on together when the footswitch is pressed. When the pedal is released I want the main unit to stop and the second unit to stop after a variable time delay (from 1-10 seconds). Also, if the pedal is pressed again before the second unit has stopped, I want both units to be on and the second unit's off-delay restarted when the pedal is released.
I've played with some circuits and relays, but can't figure this out. The voltage on the pedals is low (<24).
I need help!
Anyone know a circuit that could be a solution to my problems?

 

[dknguyen]

Sounds like the job for a microcontroller. You would need some amplification transistors or relays to amplify the 5V signal to 24V, and something (resistive divders maybe) to step down the pedal voltage from 24V to 5V. Then you can have the machines turn on and off even when you're not around!

 

[ljcox]

Alternatively you could do it with relays or CMOS devices.

Tell me which you prefer and I'll help.

 

[Lazer]

Delay off timer

I thought I would need a timer circuit with two relays and it's own power supply.

 

[ljcox]

You can do it with a relay and a capacitor. You will need a relay with at least 2 DPDT contacts.

Tells me the details of the relay that you have:- coil resistance, operate current (or voltage) release current (or voltage) if you can find these details. If not, tell me what you can and I'll guess the rest.

How do you want to vary the delay? With a pot?

As I understand it, this is what you want:-

1. press button, both start

2. release button, main stops second one runs for 1 to 10 s.

3. If the button is pressed before the delay has expired, Main is to turn on, delay is reset and the second one continues to run.

4. the second one only stops if the delay expires.

Please check this and I'll post a circuit tomorrow.

 

[Lazer]

Off delay timer

Yes Len, that is exactly what I want. Remember that the second machine turns off after a second press on the footswitch. That relay would have to open then close again after the time delay, wouldn't it? I'm sure that the footswitches both work by closing a relay circuit inside each machine, the only difference is that the second machine needs the pedal pressed a second time before it will stop (the second machine is a vacuum exhaust that needs to stay on after the first machine stops, for 1-10 seconds; because it's filters are expensive and it is very loud I would prefer that it's not on for long periods, if it's pedal is unplugged it will turn on; if the main machine's footswitch is unplugged it will not turn on). I was thinking of a pot to adjust the time delay. What has me stuck is the need for the second press on the pedal for the second machine to stop. Would you need that relay to open, close and open again after the time delay? I beleive that any power in the footswitch is <24 volts and is for the relays in the machines. They are just switches.
I hope this helps.

 

[ljcox]

I had missed the point that the second machine needs the pedal pressed to stop it and that it runs when the pedals are unplugged.

So it appears to me that the connection to the second machine must be open to make it run.

I thought about doing it with relays but decided that an electronic timer will be necessary given that the delay can be up to 10 sec.

Do you want to run the electronics from 24 Volt?

Is it +24V or - 24V?

EDIT. I have re-read your first post and need more clarification.

You said :-
"The main unit turns on when the pedal is pressed and off when it's released.

The second unit turns on when the pedal is pressed and only turns off when it's pressed a second time"

It is not clear to me which pedal you are referring to in the second sentence.

Call them Pedal 1 & Pedal 2.

Is this what you mean?

The main unit turns on when P1 is pressed and off when it's released.

The second unit turns on when P2 is pressed and only turns off when it's pressed a second time

It would help me if you could draw a timing diagram.

Show the voltage across the contacts of each pedal. See attachment for an example,

 

[ljcox]

Following on from my previous post, I have an additional question.

In the first post you wrote:-

"I have two machines that both have footswitch controls (just SP/ST switches, no circuits, both pedals are the same)"

But in a later post you wrote:-

"the second machine is a vacuum exhaust that needs to stay on after the first machine stops, for 1-10 seconds; ..., if it's pedal is unplugged it will turn on; if the main machine's footswitch is unplugged it will not turn on".

These statements appear to be contradictory.

The first indicates that both switches are SPST and are normally open (N/O).

The second implies that the second pedal is normally closed (N/C).

Perhaps they are SPDT and one is wired to the N/O contact and the other to the N/C contact?

I expect that what you need will be a very simple circuit to design, but I need to know exactly how the existing pedals work.

 

[ljcox]

Here is a suggestion. I am assuming that you can derive +24 Volt from the machine.

The +12 Volt can be derived from the +24 V with a resistor and Zener diode.

Circuit Description.

Initial state:- relays A & B are released, capacitor C1 is discharged thus the output of U1a is high so C2 is charged to 12 Volt and the outputs of U1b & U2a are low.

Relay A operates when the Pedal is pressed, it starts the Main machine and charges C1 rapidly to 12 Volt.

The output of U1a goes low and thus the output of U2a goes high which turns Q1 on operating relay B.

Relay B starts the Secondary machine.

C2 starts discharging via R3 so the output of U1b goes low after about 0.5 second and thus B releases.

Relay A releases when the pedal is released. C1 starts discharging via R1 & P1.

When the voltage across C1 decays to the lower threshold level of U1a, the output of U1a goes high which causes the output of U2a to go high thus turning on Q1 and operating B and also C2 starts to charge.

When the voltage across C2 reaches the upper threshold level of U1b, the output of U1b goes low. thus Q1 turns off and B releases.

Setting up the timing.

The 40106 Schmitt Trigger IC has a wide threshold spread, so it is not possible to calculate the precise value needed for C1.

Also, electrolytic capacitors have a +/- 10% tolerance, so a 10 uF cap can be anywhere between 9 uF and 11 uF.

Therefore, you need to experiment with the value of C1 to give the correct timing.

I suggest that you buy three 10 uF, one each of 6.8 uF, 3.3 uF, 2.2 uF and 1 uF.

Then you can (if necessary) connect these caps in parallel, eg. 6.8 uF in parallel with a 2.2 uF will give you NOMINALLY 9.0 uF.

So I suggest that you start with one of the 10 uF caps temporarily connected as C1, set the pot to its maximum resistance and then measure the time from when the pedal is released until relay B operates for the second time.

This time needs to be about 11 sec. so that the minimum time is about 1 sec (ie. when the pot is set to its minimum).

So try each 10 uF in turn, and if none of them give you about 11 sec. then:-

1. if the time is <11 sec. you will have to put one of the other caps in parallel with the 10 uF

2. if the time is >11s remove the 10 uF and replace it with the 6.8 in parallel with say the 2.2.

EDITED to correct errors.

 

[ljcox]

EDIT to include improved diagram and circuit description.
Circuit Description.

Initial state:- relays A & B are released, capacitor C1 is discharged thus X is high so C2 is discharged.
X high makes Y low thus C3 is charged to 12 V.
Therefore, both inputs of U1c are high so Z is low and thus Q1 is off.

When the Pedal is pressed, relay A operates & starts the Main machine.
The other contact of A charges C1 rapidly to 12 Volt.

X goes low so C2 pulls pin 9 low thus Z goes high which turns Q1 on & this operates relay B.
Note that Z is high if either input of U1c is low.

Relay B starts the Evac machine.

C2 starts charging via R3. When the voltage at U1/9 reaches the upper threshold level (after about 0.5 second) Z goes low so B releases.

When the pedal is released, relay A releases. C1 starts discharging via R1 & P1.

When the voltage across C1 decays to the lower threshold level of U1a, X goes high which causes Y to go low so C3 pulls pin 8 low thus Z goes high which turns on Q1 & this operates B. C3 charges via R4 and C2 discharges via R3.

When the voltage at U1/8 reaches the upper threshold level (after about 0.5 second), Z goes low thus Q1 turns off and B releases.

The circuit is now back to the initial state.

 

[ljcox]

Lazer, I replaced the circuit in the above post to fit in with your latest PM.

This post is mainly to bring it to the top of the queue.