Maker Pro
Maker Pro

Designing a Time Delayed Relay

chopnhack

Apr 28, 2014
1,576
Joined
Apr 28, 2014
Messages
1,576
Hello all, I would like to say firstly that I have limited electronics background. I can solder and read schematics to some degree and on occasion complete a print to a project. I can sometimes troubleshoot, but generally need guidance to do this with any proficiency. Now with that out of the way, I want you to descend with me into lunacy as I attempt to make something!

Goal:
I am trying to switch a 110VAC (10.8A per spec sheet) inductive load "on" when triggered by a 220VAC (1.5 HP) inductive load. I would also like to have the 110V load run for about 30 seconds after the 220V load ceases. This is for hobby woodworking equipment that is used weekly.

What I Found:
I perused the SSR but found them to be cost prohibitive and I didn't find anything above 6A. In my searches I found that a NOTO mechanical relay would be my best option, allowing for both the load capacity and time delay features, however, the one I looked at was well over $300!

Next Step:
From here I am looking for some guidance in how I should approach this design. Should I first spec out a mechanical relay (which is much cheaper) and then build a timing circuit around it? I have attached a crude block diagram of what I am currently thinking of.

To clarify the diagram: The shaded lines are switched outputs from the relays and timer. L1 is my trigger input, when that machine is running, the 220v will flow through the N.O. relay energizing it. When that latches, L2, 110v will flow through the relay to the motor and the unit will run. The N.C. relay portion of the diagram is not in play as the relay is energized keeping the contacts open. When the 220v device is turned off, the N.C. relay actuates allowing the timer circuit to begin and keeps the motor running for the preset time.

Concerns:
Some immediate problems that came to mind are the N.C. relay lag and timer latency. I don't know if I can incorporate an RC circuit somewhere to buy me some time between the two circuits or not. Also, some protection from failure for when a relay malfunctions.

I hope you find my question clearly stated, but if I have omitted something or the questions I ask are wrong because of my limited electronics education, please explain in detail.

Thank you for your consideration!



Relay Design.jpg
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
I don't know what a NOTO relay is, but yes, an electromechanical relay will switch your inductive load. The contacts should have a "snubber" circuit across them to protect them from arcing when they switch OFF. I'll cover that in a later post.

Yes, commercial time delay relay units can be pretty expensive, and I'm sure you will be able to make something for a lot less than $300!

I'm not sure I understand your diagram, or perhaps the diagram won't do what you want. Can I explain your requirements back to you, and you tell me where I go wrong?

1. You have two motor-type loads, and two AC power sources: 220V and 110V.

2. The first motor, Motor1, is powered from 220VAC via some kind of switch, and the second one, Motor2, is powered from 110VAC.

3. While the switch is ON, 220VAC will be present at Motor1 so Motor1 will be running. During this time, you want Motor2 to run as well.

4. When the switch is turned OFF, and Motor1 stops, you want Motor2 to continue running for about 30 seconds, then shut off.

5. You want to insert a relay-type device or circuit between the 110VAC supply and Motor2 to control Motor2. This circuit will receive 220VAC (taken in parallel with Motor1) and will use this to control the relay that switches Motor2.

If that's all correct, I can design a circuit for you. It can be powered from a transformer (around 20VAC) connected to the 220VAC supply (in parallel with Motor1) and it can control a latching relay, whose contacts would be connected between the 110VAC supply and Motor2.

When the 220VAC supply appears, it would latch the relay ON, enabling Motor2. When the 220VAC supply disappears (corresponding to the switch being turned OFF), it would hold a charge in an electrolytic capacitor, and a second capacitor would discharge slowly. After about 30 seconds, when the second capacitor has discharged to a certain point, the circuit would trigger, and dump the charge from the first capacitor into the other coil of the latching relay, causing it to switch OFF.

This circuit would be built up on a piece of stripboard and would have around 20~30 small components. I can supply a schematic (circuit) diagram (free of course) but you would need to work out the component layout.

You would also have to take responsibility for the safety of the circuit - keeping the live parts properly insulated.

Component cost should be under USD 50.

Can you tell me where you're located, so I know which supplier to refer to for the components? Even better, put it in your profile. There's a field for it.

Edit: Oh, and welcome to Electronics Point!
 

hevans1944

Hop - AC8NS
Jun 21, 2012
4,886
Joined
Jun 21, 2012
Messages
4,886
There are commercially available products that do what you want to do. The inductive sensor relay contacts are connected to operate a 110 VAC delay-on-release relay, also a commercial product. The delay-on-release relay closes its normally-open contacts as soon as it is energized by the inductive sensor relay, but it has an adjustable delay after it is de-energized before its normally-open contacts resume the open state.
 

chopnhack

Apr 28, 2014
1,576
Joined
Apr 28, 2014
Messages
1,576
Thanks to both of you for your quick replies!

Kris, Wow.. Amazingly you have nailed it! And with clarity and fewer parts than I thought!! I was sure that a timer IC would be necessary, but if I read you correctly you are relying on a simple RC discharge cycle coupled with another cap to fire off the other relay leg. Very elegant. Not that I care too much if the RC time varies by more than a few seconds since its not mission critical - but for scholarly purposes - aren't RC discharge times quite variable based on the component's quality? The NOTO is short hand for normally open timed off - at least on one website. I'd prefer to run the circuit off of 110v, but either way I would love to see the circuitry and how you mapped out the flow.

Thanks ever so much!
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Thanks :) Yes you're right, an RC circuit will not be very accurate. Capacitors that can produce delays around 30 seconds are generally aluminium electrolytic types, which have an accuracy of ±20% (unless you special-order them from the manufacturer in big quantities), and that's only the initial tolerance at room temperature, when they're new!

The initial tolerance can be adjusted out by varying the R of the RC circuit (i.e. make it a trimpot) but capacitance will drop around 10% with a temperature change from 20°C to -20°C. If this is a problem, there are other more accurate timing methods available.

Regarding powering the circuit. My original thought was to power it from the 220VAC source, using a transformer, so that all the circuitry (apart from the transformer primary, and the relay contacts) would be low-voltage, for safety. Also, the transformer could do double duty - it would supply the power for the circuit, and it would provide an easy way for the circuit to detect when the 220VAC has disappeared.

Powering the circuit from 110VAC is also possible, but this would mean that it would always be powered up, and always wasting a small amount of power. It would simplify the circuit somewhat though, because it would always have power available, so it wouldn't need to store charge in a capacitor so it could turn the relay OFF after its own power source has disappeared. This would also give us more options for measuring the timeout period - we could use a frequency divider, clocked from the 110VAC mains frequency, to get very accurate timing.

If the circuit is powered from 110VAC, it needs a way to monitor the 220VAC at Motor1. This could be done pretty cheaply with an optocoupler.

In either case (220VAC- or 110VAC-powered), there's also the option of using a non-isolated, line-powered circuit, which avoids the transformer. This trick is used when cost and/or space is at a premium. The transformer is replaced by a capacitor, which drops the AC voltage down to a usable DC voltage. With this arrangement though, the whole circuit is potentially live, and must be fully enclosed - in potting resin, or perhaps hot melt glue, or at the very least, a plastic enclosure. The time adjustment could be made through a small hole, or the trimpot could poke through the potting resin or the enclosure.

I would go for this option if (a) you want to do it that way, and (b) you can guarantee me that you will keep the board safely isolated, add clear warning signs, and take responsibility for ensuring that it stays safe.

No matter which approach we use, I envisage a unit with six wires: two from the 110VAC supply, two from the 220VAC supply that's active when Motor1 is running, and two going out to Motor2. And one time period adjustment using a screwdriver, unless the circuit is powered from 110VAC and uses the mains frequency for timing.

What do you think?

Please add your location to your profile, and tell me what your mains frequency is.
 

chopnhack

Apr 28, 2014
1,576
Joined
Apr 28, 2014
Messages
1,576
Thank you again Kris! To answer your last question first, 60hz, 110/120v and I think both ideas are brilliant. I like option 3 as well, but safety first.

Either option 110 or 220 would work. Using 110V would not waste any significant power because the entire circuit would only be live when the power is turned on for motor 1 which I realize I did not show! Motor 1 would be at the end of the blue arrow - the 220V Line 1. What you describe in the 110V option is basically what I blocked out in my diagram, except you mention using an optocoupler whereas I mention using a relay. The 110V option would be my preference as I believe it will be more compact in size.

I am guessing at your thought train, but the optoisolator is to somehow protect the timing circuit from the current on the mains? Also a cap would be placed in line with motor 2 to enable it to stay running long enough for the timing circuit to actuate? I would love to see what you come up with. I can see a rough sketch in my head to some degree :)
 
Last edited:

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
OK, I've gone for a circuit that's permanently powered from 110VAC, doesn't use a transformer (uses a "capacitor fed" power supply), and gets its timing from the mains.

It uses two ICs, but it's pretty accurate and doesn't need any adjustment. I hope you don't think it's too complicated to build.

268425.001.GIF

A constant 110VAC comes in on wires W1 and W2.

D1~4 are a bridge rectifier (not drawn the usual way, but that's what it is) that is fed from C1 in a circuit called a "capacitor-fed power supply". C1 has a characteristic called capacitive reactance, which allows a limited amount of current to flow from the mains input into the bridge rectifier; most of the voltage is lost in C1, and no actual power is dissipated in C1, due to the magic of reactance.

R1 limits the surge current when power is applied, limits the current due to noise spikes on the mains supply, and acts as a fast-acting fuse if C1 fails short.

R2 discharges C1 if mains power is removed, so the circuit does not store a charge that could bite you if you touched the wires.

The incoming 110VAC also feeds further along into other parts of the circuit.

The bridge rectifier output is clamped at about 50V by D5 and smoothed by C2. When power is initially applied, because of the limited current available, C2 does not reach 50V immediately. It takes about a second to get within a few volts of the target voltage. This is normal.

Because the circuit is powered directly from the mains via a bridge rectifier, the 0V rail should be considered potentially live at all times. The whole circuit (everything shown on the diagram) must be properly insulated - for example, enclosed in a fully plastic box, with only the insulated wires emerging, and with markings on the inside of the box lid, and on the circuit board, saying "All circuitry is LIVE" or similar.

Q1 is a Darlington transistor - a combination of two transistors that has a high current gain. It is used as an emitter follower (aka common collector amplifier) to buffer the 12V reference voltage set by D6 and provide a (crudely) regulated 11V supply to the rest of the circuit.

The 220VAC supply from the other motor is fed into wires W5 and W6. It goes through a circuit similar to the capacitor-fed power supply, which provides a low-current AC signal into U1, a 4N33 optocoupler. The optocoupler provides electrical isolation and propagates the mains-frequency signal through to its output on pin 4. This signal is pulled down to 0V by R8 when there is no AC on W5/W6, but when that AC is present, pin 4 is high for every positive half-cycle of the AC waveform.

This signal is filtered by R9 and C6 to filter out any disturbances due to noise or interference on the 220VAC feed, and drives the RESET input of U2.

U2 is a CD4040BE CMOS "12-bit ripple carry binary counter". This means it counts transitions at its CLOCK input (pin 10) and reports the count on twelve binary outputs, Q1~Q12. When RESET is high, the count is held reset, and all the Q outputs are low (0V). When RESET is low, each falling edge (transition from high to low) on the CLOCK input causes the count to advance by 1.

The CLOCK input is driven with a signal derived from the 110VAC mains frequency via R11 and one gate in U3 which I will explain later. Its frequency is equal to the 110VAC mains frequency, which in this case is 60 Hz.

So 60 times per second, the count in U2 increments by 1. After one second, the count will be 60, which is 111100 binary, where those six digits correspond to Q6~Q1 in that order.

But if AC is present on W5/W6, U1 will reset U2 on every mains cycle, so the count in U2 will never advance very far. So although the count might advance as far as 1 or 2, all of the higher Q outputs will remain low while AC is present on W5/W6. Once the other motor is turned OFF and the AC disappears, U2's RESET input will remain low, and U2 will start to count upwards.

The Q9, Q10 and Q11 outputs of U2 are combined using one gate in U3 and Q2. This circuit detects when Q9, Q10 and Q11 are all high simultaneously, and when this occurs, brings the signal marked "RUN" low. (RUN is normally high.) This corresponds to a count of 11100000000 in U2; in decimal, that number is 1792. Dividing 1792 by 60 (the number of CLOCK transitions per second) gives 29.8667. So this count will be reached, and RUN will go low, about 30 seconds after the other motor is switched OFF.

U3 is a "quad 2-input NAND gate with Schmitt trigger inputs". It contains four independent gates. Each gate has two inputs and one output, and performs a logical NAND function - it drives its output to the logical opposite of the AND of its inputs, which is true only if input A AND input B are high. So its output will be high unless both inputs are high, in which case its output will be low.

The gate connected to pins 1~3 creates a logical NAND of the Q11 and Q10 outputs from U2. Normally, its output (U3 pin 3) is high; it only goes low when Q11 and Q10 are both high. Also, transistor Q2 is driven from the Q9 output, and it pulls its collector high unless Q9 is also high. R11 combines the two signals, producing a combined signal called RUN that goes low when U2's count reaches 11100000000, i.e. after 30 seconds with the other motor switched off.

When RUN goes low, the clock input to U2 is disabled via the second NAND gate in U3, because when the gate input on U3 pin 5 goes low, the output on pin 4 is forced high regardless of the signal on pin 6 (the mains frequency feed that provides the clock source for U2).

This stops U2 from counting, so it remains frozen at that count, and can only be un-frozen by an active signal at RESET, which occurs the next time the other motor is turned ON.

The signal at RUN is also fed through the remaining gates in U3, which are cascaded so they behave like a buffer. The final output on U3 pin 10 drives Q3 via R12. While RUN is high, Q3 is turned ON and keeps relay K1 energised. When RUN goes low, Q3 turns OFF and K1's contacts open, removing power to the 110V motor that is powered from wires W3 and W4. D8 protects Q3 from the back EMF from K1's coil when current through it is interrupted by Q3 turning OFF. Without D8, when Q3 turned OFF, K1's coil would generate a large "flyback" voltage that would pull Q3's collector voltage well above its rated maximum collector-emitter voltage and damage it.

When the circuit is powered up, U2's output states are undetermined. If there is no AC present at K5/K6, the circuit may initially start up with the motor enabled, and will switch it off after an undermined amount of time (less than 30 seconds). Once the other motor has been run, U2 is forced into a known state by its RESET input, and the circuit will then give the specified run-on period.

R13 and C9 are an "RC snubber" network that protects K1's contacts from arcing due to back EMF generated by the motor on W3/W4 when power to it is interrupted. Because of its inductive nature, the motor can generate a high voltage, which can cause arcing in the relay contacts. The RC snubber suppresses this voltage and protects the relay contacts.

The components connected to U3 pin 6 provide a clock source that's derived from the 110VAC supply frequency. From the point of view of the circuit, i.e. relative to the 0V rail, the voltage at the bottom end of R11 swings between about 0V and about 50V at mains frequency. R11 has a high resistance so it limits the current available, to prevent damage to U3. C7 provides a small amount of smoothing on the signal to prevent possible bogus clocking due to noise on the mains supply.

U3's inputs all have a "Schmitt trigger" feature which gives them a very clean switching behaviour. Instead of having a single voltage threshold, like normal logic ICs, they have two thresholds. When the input voltage is low, it must rise above the higher threshold before U3 will recognise it as high; once it has been recognised as high, it must drop below the lower threshold before U3 will recognise it as low. This characteristic is called hysteresis, and it creates a deadband between the low and high thresholds, where input variations will be ignored. This feature is only needed on this one input; all of the other signals going into inputs on U3 are well-defined logic levels.

C5 and C8 are decoupling (aka bypass or reservoir) capacitors for U2 and U3 respectively. Logic ICs, especially counters (U2), draw brief spikes of significant current through their supply pins, but for reliable operation, they need a clean supply voltage. These capacitors provide a tightly coupled local reservoir to ensure reliable operation. They must be connected as directly as possible (with leads as short as possible) between the VDD and VSS pins of their respective ICs - pins 16 and 8 for U2, and pins 14 and 7 for U3.

For operation from 50 Hz mains, the count detection logic needs to be changed, so RUN goes low at a lower count (as U2 is clocked more slowly). This can be done by deleting R10 and Q2, and replacing R11 with a link. This causes RUN to go low at a count of 11000000000 binary, which is 1536 decimal. Dividing 1536 by 50 gives the timeout period, 30.72 seconds.

Capacitor-fed power supplies cannot supply much current, and the current budget for this circuit is around 15 mA, most of which is used by the relay coil. Therefore, not all 48V-coil relays are suitable. See below.

Notes on components

I have tried to use a limited number of different types of components to make the bill of materials a bit simpler. For example, D7 and D8 don't need to be rated for 1A but it's simpler to keep them the same as D1~4 instead of adding a new, different component.

The three 100 ohm fusible resistors should be rated at 0.5W. I suggest http://www.digikey.com/product-detail/en/FRM-50JR-52-100R/100DTCT-ND/2813208.
If you decide to fully encapsulate the circuit, you should ensure an air gap around the fusible resistors; if they are fully enclosed, they may not be able to fuse properly (there will be nowhere for the magic smoke to go). I'm not sure how to do that, but I have an idea, so let me know if you want to encapsulate it.

D1~4 and D7 and D8 can be any member of the 1N400x family rated for at least 100V. That is, 1N4002~4007 are all suitable.

C1, C4 and C9 must all be "X2 rated" or better. X1, Y2 and Y1 ratings are all better than X2. These ratings mean that the capacitor is designed to have mains-frequency AC voltage across it permanently, without failing. These capacitors have a special construction. It is NOT safe to use capacitors that are not thus rated, or are rated for lower AC voltages.

C1: 0.33 µF 150VAC: http://www.digikey.com/product-detail/en/B32922C3334M/495-2321-ND/778983
C4: 22 nF 275VAC: http://www.digikey.com/product-detail/en/PHE850EB5220MB06R17/399-5420-ND/1927365
C9: 0.1 µF 150VAC: http://www.digikey.com/product-detail/en/ECQ-UAAF104M/P14779-ND/2674011

Suitable relay types for K1 are listed. These all have 16A contact ratings, with coils rated for 48VDC with coil currents less than 12 mA. That corresponds to a coil resistance of 4000 ohms or higher. Don't use a relay with a coil resistance less than 4000 ohms, because the capacitor-fed power supply cannot supply a lot of current.

These relays are all available from http://www.digikey.com and are listed in ascending price order.

Panasonic ALE1Px48 ('x'is either B or F for the temperature class: B is up to 85°C; F is up to 105°C)
http://www.digikey.com/product-detail/en/ALE1PB48/255-2373-ND/1680241

Panasonic ALZxyz48 ('x' is either 1 for SPDT contact or 5 for SPST contact (preferred); 'y' is either 1 for flux-resistant or 2 for fully sealed (not important); 'z' is either B or F for temperature class - see above)
http://www.digikey.com/product-detail/en/ALZ51B48/255-3369-ND/2224606

Omron G2RL-1(A)(4)E DC48 ('A' present means SPST contact (preferred); '4' present means fully sealed (not important))
http://www.digikey.com/product-detail/en/G2R-1A-E-DC48/Z2309-ND/368728

TE Connectivity / Schrack RTxyz048 ('x' is either D for "wash tight" (sealed) or 3 for "flux proof" (not fully sealed) (not important); 'y' is either 1 for SPDT contact or 3 for SPST contact (preferred); 'x' is either 4 for standard or 5 for gold plated contact (not important))
http://www.digikey.com/product-detail/en/RT314048/RT314048-ND/1128613

TE Connectivity / Schrack RPxy0048 ('x' is either 3 for "flux proof" or 7 for "wash tight" (not important); 'y' is either 1 for SPDT contact or 3 for SPST contact (preferred)).
http://www.digikey.com/product-detail/en/3-1393230-3/PB1674-ND/2397842

This circuit can be constructed on stripboard. Keep U2 and U3 away from the high-voltage sections of the circuit, and cut the tracks that connect to them, close to the ICs, to reduce the possibility of signal coupling or shorts between those tracks and other higher-voltage parts of the circuit.

I recommend using IC sockets for U1~3. Decoupling capacitors for U2 and U3 can be soldered on the underside of the board (use sleeving on the capacitor leads to prevent shorts) and can be soldered directly between the corner pins of the IC socket.

Heh! Just as well I can type fast, eh :)
 
Last edited:

chopnhack

Apr 28, 2014
1,576
Joined
Apr 28, 2014
Messages
1,576
That is a lot to absorb!! I really appreciate the care in design and the explanation :) I love how the flow is altered by the logic of the circuit, very cool to understand an IC package, I might learn electronics yet!!!

I need to ask a question about W5 and W6 to clarify. Motor 1 (220v) is a unit that can be plugged into the mains directly. In order for this circuit to work as drawn, W5 and W6 would need to receive the switched 220 - i.e. after the motor turns on to be of value in this schematic, correct?

If so that would make it very hard to implement in my case. I was envisioning a unit that would "sense" when the 220v was being drawn and thus trigger off of that. In a previous mains circuit that I constructed, I needed to switch a circuit on a different breaker by means of a light switch. (X-mas decorations too high of a load for the light switch, so I used the switch to throw a relay which switched a separate breaker "on" for a series of outlets outside). By triggering the 110V side of the circuit on when needed there will be no waste - as the circuit won't need to be on all the time. Is there a way to construct this so that the circuit will only go live when the 220 is sensed?

Follow up question to that - Is there a way to add a level of logic to the turning on of the 110v motor such that the motor wouldn't turn on accidentally when the registers are cleared and reseting?

Thanks for your time and efforts!!
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Yes, W5/W6 need to have 220VAC across them when the triggering motor is running. When that happens, the controlled motor (connected to W3/W4) will start. When the 220VAC on W5/W6 disappears, the circuit will keep the controlled motor running for another 30 seconds, then it will shut off too.

The circuit doesn't "sense when the 220V is being drawn". You need to connect W5 and W6 across the triggering motor. When the switch (or whatever) that controls that motor is turned ON, 220VAC is switched onto the motor, to make it run. That is the voltage that you need to connect to W5/W6.

As designed, the circuit takes its power from the 110VAC supply. From what I understood, this is necessary if the circuit is going to continue to run after the triggering motor has been turned OFF. If it took its power from the 220VAC feed from the triggering motor, then when that motor was turned OFF it would lose power. It could use an electrolytic capacitor to store charge for 30 seconds so it could flip a latching relay to the OFF state, but assuming the 110VAC supply will always be present, it's simpler to use that, and use a standard relay.

The motor run will only happen when 110VAC is initially applied to the circuit. If it's always powered, there won't be a problem. Yes I could add a kludge to avoid it but I would rather not, if it's not necessary.
 

chopnhack

Apr 28, 2014
1,576
Joined
Apr 28, 2014
Messages
1,576
Ok, I will need to think on this some more to make sure I understand everything. I am having some difficulty visualizing the w5/w6 connection, thanks!
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Just straight to the two terminals on the 220VAC motor. When the motor runs, it runs because it has 220VAC across it. This is the voltage that is sensed by the optocoupler.
 

chopnhack

Apr 28, 2014
1,576
Joined
Apr 28, 2014
Messages
1,576
Is there a way for me to safely wire my 220v feed, w5 and w6 and the motor in series? That way the circuit will be between the power and the motor's switch - with the switch creating the open in the loop, when latched w5 and w6 will be powered. (The switch on the motor is a magnetic switch if that matters.)
Won't the inrush current of the motor affect the circuit?
Can this be mitigated with higher cap. and res. values?

The reason I ask is I need this device to be portable, able to be used for more than one 220v device. This way, I can move it from device to device as needed. I envision being able to plug in 110 and 220 mains, the trigger motor and the load motor into the unit, 2 lines in and 2 out.
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Is there a way for me to safely wire my 220v feed, w5 and w6 and the motor in series? That way the circuit will be between the power and the motor's switch - with the switch creating the open in the loop, when latched w5 and w6 will be powered. (The switch on the motor is a magnetic switch if that matters.)
Yes, you could monitor the current flow by connecting a circuit in series with it, but there's no good reason to do that. It would cause some problems too: the wires going from the load circuit to the box would need to be quite thick, because they would be carrying the full operating current of the motor; most circuits that detect current flow in series with a load will drop some voltage, so the load won't receive quite as much voltage as it should; there's a higher chance of damage to the monitoring circuit due to current surges; and the monitoring circuit is a bit more complicated and expensive. All those disadvantages and no advantages!
Won't the inrush current of the motor affect the circuit? Can this be mitigated with higher cap. and res. values?
The inrush current of which motor? The 220VAC motor, no. The 110VAC motor, maybe. It will presumably draw a higher current as it comes up to speed. I don't know how much more. It might turn out that you need an interposing relay with a 110VAC coil driven from W3 and W4, to switch the 110VAC to the 110VAC motor. For example:
http://www.digikey.com/product-detail/en/G7L-1A-TUB-J-CB-AC100/120/Z814-ND/127305 (30A)
http://www.digikey.com/product-detail/en/7-1393132-0/PB1335-ND/1127404 (40A)
The reason I ask is I need this device to be portable, able to be used for more than one 220v device. This way, I can move it from device to device as needed. I envision being able to plug in 110 and 220 mains, the trigger motor and the load motor into the unit, 2 lines in and 2 out.
That's another reason to monitor the voltage across the motor instead of the current in series with it. If you insert the circuit in series with the motor, then you would have to break the circuit and connect the socket (for the unit to plug into) across the break. Then when you unplug the unit, you would have to plug a shorting plug into the socket, to provide a path for the current. Or have a bypass switch. If you connect the socket across the motor, then the presence or absense of a voltage monitoring circuit plugged into it doesn't affect anything.

If you want to be able to plug the unit into different 110VAC motors, you're going to have to get some pretty grunty sockets. The alternative would be to fit a relay (like the ones linked above) to each 110V motor that you want to use the circuit with, so the circuit only needs to pass enough current to activate the coil of the relay. You could then have a small bypass switch that would feed 110V directly to the relay coil if the unit was not plugged in.
 

chopnhack

Apr 28, 2014
1,576
Joined
Apr 28, 2014
Messages
1,576
I sort of understand. I will make another diagram to explain better what I am working with - essentially my limitations to why I need the device to be portable between machines.
------------------------------------------------------------------------------------------------------------
As shown below, the entire circuit needs to be able to be put into a box and made portable to be moved between machines, i.e. multiple 220v machines, where the 110v will be the same motor at about 11A draw. The 220v motors are all induction motors, but with varying draws. I would need to have extension cords permanently made up to attach to the box (red lines in) with a plug on one side. The receptacles are where the motors would plug into.

Is there a way to use your circuit to satisfy the constraint of being able to plug the 220v motor into the receptacle? This was the area I was having a hard time understanding how it all comes together.

box.jpg
 
Last edited:

chopnhack

Apr 28, 2014
1,576
Joined
Apr 28, 2014
Messages
1,576
Since I can't tap into the switched side of the 220v from the motor and still keep this unit portable could an inductive sensor be used to detect the current flow in the 220 line to act as the "on" signal? Would this type of sensor shield the circuit and eliminate the need for the optocoupler? I think that this would allow me to simply plug in the triggering device. What do you think?
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Ah, I think I see what you're getting at now.

In post #15, the red box with the sockets on it represents the thing you want to build, right?

It plugs into the 220VAC and 110VAC supplies using two plugs on power cables at the left.

And the two motors plug into the two sockets on the front.

And the ON/OFF switch for the 220VAC motor is actually downstream of the plug that plugs into the unit, so the only way to sense whether the 220VAC motor is ON or OFF is by detecting the current flow in that part of the circuit, since you don't have access to the switched 220VAC that feeds the motor.

Is that right?

If so, yes, we can monitor the current flowing in the 220VAC line to detect whether the motor is running or not.

We can use an optocoupler. It will cause a small voltage drop but nothing to worry about.

Alternatively, devices are available that can sense the current without introducing any voltage drop, such as the Allegro ACS712ELCTR-20A-T: http://www.digikey.com/product-detail/en/ACS712ELCTR-20A-T/620-1190-1-ND/1284607 at USD 5 from Digikey.

They are used like this:

acs712.gif
There's a thick internal connection between pins 1 and 2, and pins 3 and 4. The 220VAC path is broken and the two sides of the break are connected to those pairs of pins. Current flowing in that part of the device is sensed by a Hall effect sensor that generates an output voltage that follows the AC current. They have 2 kV isolation. They come in an 8-pin SOIC package. Would you like to use one of these instead of an optocoupler, to avoid the voltage loss?

Let me know if I've figured out what you want to do, and I'll upload an updated schematic diagram.
 

chopnhack

Apr 28, 2014
1,576
Joined
Apr 28, 2014
Messages
1,576
Ah, I think I see what you're getting at now.

In post #15, the red box with the sockets on it represents the thing you want to build, right?
Correct
It plugs into the 220VAC and 110VAC supplies using two plugs on power cables at the left.

And the two motors plug into the two sockets on the front.

And the ON/OFF switch for the 220VAC motor is actually downstream of the plug that plugs into the unit, so the only way to sense whether the 220VAC motor is ON or OFF is by detecting the current flow in that part of the circuit, since you don't have access to the switched 220VAC that feeds the motor.

Is that right?
Yes, that is why I was not understanding how the circuit would work with motor 1 :)
If so, yes, we can monitor the current flowing in the 220VAC line to detect whether the motor is running or not.

We can use an optocoupler. It will cause a small voltage drop but nothing to worry about.
Can you explain how the optocoupler can be used in series without frying it? In parallel, it would not be effective in gauging current.

Alternatively, devices are available that can sense the current without introducing any voltage drop, such as the Allegro ACS712ELCTR-20A-T: http://www.digikey.com/product-detail/en/ACS712ELCTR-20A-T/620-1190-1-ND/1284607 at USD 5 from Digikey.

They are used like this:

View attachment 12719
There's a thick internal connection between pins 1 and 2, and pins 3 and 4. The 220VAC path is broken and the two sides of the break are connected to those pairs of pins. Current flowing in that part of the device is sensed by a Hall effect sensor that generates an output voltage that follows the AC current. They have 2 kV isolation. They come in an 8-pin SOIC package.
That is a really cool chip! Can a SOIC be converted easily for thru pin mounting?

Would you like to use one of these instead of an optocoupler, to avoid the voltage loss?
I honestly don't know enough to make that decision. I think to determine this I would need to know the potential voltage drop - being an induction motor, the startup draws quite a bit of current, so every volt counts, no? If the V drop is less than a volt, I don't think it should matter, do you?


Let me know if I've figured out what you want to do, and I'll upload an updated schematic diagram.
Yes, I believe you have it very well, I just didn't know we were on different pages with respect to our input triggering method. Thanks for clarifying everything and for the explanations - I am still trying to workout how everything functions in your first schematic! :D (expand the quote to see more)
 
Last edited by a moderator:

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Right. The optocoupler option can use some high-current diodes to provide a fixed voltage drop. A typical diode drops around 0.8~2.0V depending on the current and the diode's characteristics. So you break one of the lines to the motor and insert a circuit consisting of two high-current diodes in series in each direction. One pair of diodes conducts on each half-cycle of the AC waveform, and this limits the voltage across the inserted circuitry to about 2V each way. The voltage across the diodes can drive the LED in the optocoupler, which needs about 1V across it.

Here's an updated schematic that shows the circuit.

268425.002.GIF

The diodes I specified as D7~10 are 1N1184 (http://www.digikey.com/product-detail/en/1N1184/1242-1013-ND/3594392) and 1N1184R (http://www.digikey.com/product-detail/en/1N1184R/1242-1095-ND/3594362). They are rated for 35A (maximum peak instantaneous current at any moment). All four of these can be mounted onto a small heatsink, as they will get warm while the 220VAC motor is running. This heatsink must of course be fully insulated from EVERYTHING!

Yes that's a clever chip alright. You can mount it on an adapter board: http://www.google.com/search?q=SO-8+adapter&tbm=isch

The voltage across the diode-based monitoring circuit above should be less than 2V DC maximum at the highest parts of the 220VAC sinewave. Voltage drop with the ACS712 will be practically zero, of course, and there will be no detectable heating.
 

chopnhack

Apr 28, 2014
1,576
Joined
Apr 28, 2014
Messages
1,576
I see! So you are recommending to go with the hall effect chip for practical reasons. That makes sense. Avoids a lot of problems too! I would appreciate it greatly if you could make that change to the schematic.

Just one question on the above schematic - if the 220v lines are broken and the circuit inserted in series, how would the motor get 220v? I don't follow how the circuit would reduce the voltage to 2v but the motor still get 220v.

Thanks again for your time and help, its been both educational and inspirational :)
 
Top