DC and AC Sharing A Wire

[big john]

I have a DC circuit that I'm building that has switch points, but I'd like to be able to supervise the continuity of the circuit.

My original thought was to to simply parallel a 60Hz power supply with my DC supply and jump the switch points with capacitors. The AC would still complete the circuit even as the DC was broken.

But then it occurred to me that I probably stand a good chance of damaging the DC supply by paralleling a transformer with it.

I need to filter out the AC component before it hits the DC supply. I thought of using inductors, but for 60Hz they'd have to be massive.

Is there a way to do this without trying to solder together a super-computer (I'm not much on soldering), or do I need a different idea?

-John

 

[gar]

110211-1927 EST

You have an interesting idea.

Can you describe your DC circuit a little more?

I am going to guess at one DC source, one DC load, and many switches, with all these items in one series circuit.

Add in series into this circuit a parallel resonant circuit at maybe 25 kHz close to the DC power supply. Choice of frequency will depend upon wire length and RFI problems. Now you can parallel your AC source after the resonant circuit. At resonance a parallel resonant circuit has a high impedance. The inductance you choose in the circuit needs to have a low DC resistance. The design requirement will be a function of your actual load.

Does this give you enough hint on how to proceed?

.

 

[big john]

110211-1927 EST
I am going to guess at one DC source, one DC load, and many switches, with all these items in one series circuit.

Spot on guess. It's a 24VDC supply that will be driving a time delay relay, and there's about a dozen switches in series between the two.

Add in series into this circuit a parallel resonant circuit at maybe 25 kHz close to the DC power supply. Choice of frequency will depend upon wire length and RFI problems. Now you can parallel your AC source after the resonant circuit. At resonance a parallel resonant circuit has a high impedance. The inductance you choose in the circuit needs to have a low DC resistance. The design requirement will be a function of your actual load.

Does this give you enough hint on how to proceed?

If I follow this correctly, the only way to really do it is to have the AC supply in the kilohertz range and then use a low-pass filter to knock it out (one also capable of passing the necessary DC current, of course.) My problems become two-fold at that those frequencies: One, I'd have to figure out how to get such a power supply. Two, I'd have to figure out how to drive a load with a power supply of that frequency.

Unfortunately, I have the feeling that if high frequency is my only solution, I'll have to abandon the attempt; I'm just not that much of an electronics whiz.

-John

 

[gar]

110211-2057 EST

You can do it at 60 Hz. A time delay relay probably implies a fairly low DC current. 60 Hz will require a larger inductance than higher frequencies, but at a low DC current level may not be impractical.

You may want a capacitor across the load as well as the switches. Then you want a current limited AC source and you measure the AC current to check your circuit. Instead of capacitors across the switches you could use diodes. These would be reversed biased for your DC path and would conduct on half of the AC cycle. You could also use a pulsed DC of reverse polarity with the diodes to check your circuit.

Another way to inject AC into the loop is to put the secondary of a transformer in series with the DC supply.

Still another method with the diodes is to test with a reversed DC voltage. See if you can figure out how.

Note: if you work with 60 Hz as the test frequency, then you will need larger shunt capacitors than at higher frequencies.

Study a little on electronic circuits and have some fun.

.

 

[zbang]

Will this work?

Use only an a/c supply.
Jump all the switch contacts with a diode pointed one way (call it 'up').
Put an indicator lamp with a 'up' series diode in the line.
The lamp should always be lit by up polarity half cycles. (Lamp voltage will depend on how many series diodes are in line, and closing all the switches -could- pop the lamp, depending on it's voltage rating. With a 24v supply and lamp, you won't have any problems.)

Across the indicator/diode, put a relay with a down polarity series diode.
Closing any switch will allow down half cycles in the line which will operate the relay.

** no, scratch that, won't work. next time I'll draw it out before writing. been a long day.

 

[gar]

110211-2121 EST

zbang:

You have a good idea and it will work.

The lamp will always see a half wave rectified current. This will vary some in going from all switches closed, to all switches open.

If there are 20 diodes and switches this might be about a 20 V or less drop from diodes, choice of diodes can reduce this drop. So maybe a 50 V AC source is used with about a nominal 28 V pilot lamp. Or use an LED. Also could use a current limiting device in series with the lamp. Current limiting does not mean a fuse.

A filter capacitor across the time delay relay will convert the half wave into a DC signal.

.

 

[big john]

Thanks for the replies so far. Some really good ideas here, I gotta sit down and really think about how I'll implement them.

I've pretty much abandoned the idea of trying to parallel the AC supply because I foresee way too many possible problems (revere bias on the DC supply, over voltage of the DC supply, the transformer coil acting like a short circuit for the DC supply, etc.)

If I put the transformer in series with the DC, have I really changed anything? I guess I've put the impedance of the DC load between the DC supply and the transformer, but would that really help to protect the DC supply...?

And aren't I going to experience similar issues with even with two DC supplies? I need to put on my thinking cap for a while.

-John

 

[Smart $]

Spot on guess. It's a 24VDC supply that will be driving a time delay relay, and there's about a dozen switches in series between the two.

If I follow this correctly, the only way to really do it is to have the AC supply in the kilohertz range and then use a low-pass filter to knock it out (one also capable of passing the necessary DC current, of course.) My problems become two-fold at that those frequencies: One, I'd have to figure out how to get such a power supply. Two, I'd have to figure out how to drive a load with a power supply of that frequency.

Unfortunately, I have the feeling that if high frequency is my only solution, I'll have to abandon the attempt; I'm just not that much of an electronics whiz.

-John

If this is a one-, two-, or few-off design, consider using a micro-PLC. It could function as the TDR also, though for high current load switching you'd need a slave contactor. You'd only need one analog input to monitor the continuity and trigger the TDR-function, putting high impedance resistors across the switches and prorgamming the PLC to do various tasks based on input level.

 

[ronaldrc]

Jumper each switch with say a 1k ohm resistor and use a neon light lamp as indicator.

 

[hurk27]

You can couple AC onto a DC line through a cap but you can't couple DC onto an AC line, cable company's do it all the time to power their line amps, but they de-couple the 60 volts DC through a cap before it gets to the house.

The reason is if you directly coupled a DC voltage into an AC line the source would receive a reversed bias every half cycle, this would short out the AC source, no different then placing a diode across a transformer output, placing a diode in series with the AC supply will just result in a DC supply, now place a cap in series with the AC supply just before it connects with the DC circuit and you will couple the AC onto the DC circuit, then at the other end place another cap between the circuit and AC load and you will decouple the DC allowing only the AC to flow to the AC load without affecting the DC supply, but to power the line amp requires filtering out the AC component leaving the DC intact, building an RC low pass filter is what does this, but this tank would be quit large for 60hz to be of any usable current.

 

[gar]

110212-0759 EST

ronaldrc:

Your suggestion totally neglects the important details of how to make your idea work.

However, your idea does present a DC method for a reasonable number of switches. Consider your 1000 ohm resistors and a maximum of 10 switches and assume the time delay relay input resistance is less than 100 ohms or could be made less than 100 ohms with a shunt resistance. Might be able to work with a load somewhat more than 100 ohms, but not likely as high as 1000 ohms.

Maximum voltage across time delay relay is 24*100/1000 = 2.4 V when just one switch is open. So the time delay relay must not operate at 2.4 V input.

When all switches are open the total loop resistance is between 10000 and 10100 ohms. The current will be at least 24/10100 = 2.3 MA. Use a 1 MA threshold detector. Less than 1 MA and the circuit has an open.

But note that leakage resistance of 24,000 ohms might prevent the detector from sensing an open circuit further along the circuit path from where the leakage was present.

If the load itself was open and there was 10,000 ohms leakage across the load, then the system would fail to detect the open load.

.

 

[big john]

At this point I think the idea of using resistors to jump out the switch contacts is the most feasible.

Luckily, the load relay on the DC circuit draws 100mA, which means I can use a pretty low resistance without passing enough current to allow the relay to operate. I'm thinking of using a solid state relay in parallel with my time-delay: The solid state relay has a voltage range of 4-32 and an operating current of 10mA so if I put in a dozen 200 ohm relays in series (or maybe a lower value because of cumulative wire resistance) that should allow enough current to pass to keep the solid state relay closed, but not enough to pass to keep the TD relay closed.

-John

 

[gar]

110212-1009 EST

big john:

How many switches do you have in the series loop?

What are the characteristics of the time delay relay?
I calculate 240 ohms for its resistance from 24 V and 100 MA.
Is this a linear resistance? Is this an electronic time delay relay with an electro-mechanical output relay? This might have a changing resistance between relay energized and de-energized. Tell us more about the TD relay.

What is the minimum voltage where the TD relay starts to time? Is this an on delay relay? If this was a thermal time delay relay, Amprite, then any moderate residual current thru the relay would alter its time delay time.

After all the switches are closed is the timer started, and then at the end of the time delay does the TD output relay change state?

Assume the TD relay looks like a constant 240 ohms, then a 200 ohm resistor across a switch will produce 240/(200+240) = 0.545 times the source voltage across the TD relay when the TD relay is supposed to be NOT timing.

You have not specified the input resistance of your solid-state relay, and it is not linear. Nor is it a constant load current vs voltage. A P&B ODC5 is energized by 3 V at least. Its V-I characteristic on a sample of one is
03 V 06.9 MA = 435 ohms
04 V 10.0 MA = 400 ohms
10 V 32.0 MA = 312 ohms
The on threshold point of a solid-state relay will not be constant with temperature, life, and from device to device. In other words you can not depend upon a constant known on threshold point, and it is somewhat soft. Soft meaning a not sharp threshold.

Let's ball park 435 ohms in parallel with 240 ohms. This is 155 ohms. 12 200 ohm resistors in series = 2400 ohms. With all switches open the voltage across the relay pair is about 24*155/(155+2400) = 24*0.061 = 1.5 V. Not sufficient to guarantee that any solid-state relay of the model you choose will be energized.

At the other extreme condition to consider is with only one switch is open. Now the load relays look like something somewhat less than 312 and 240 in parallel, or about 135 ohms. The voltage across the relays is about 24*135/(135+200) = 24*0.4 = 9.7 V. How does the TD relay respond with 9.7 V applied to it?

Now consider zbang's diode approach. Use a 17 V AC source. Peak is about 24 V. At 10 MA and room temperature the voltage drop is 0.62 V on a sample of 1 of a 1N5060 diode. 12 of these would be 7.5 V. However the drop will be somewhat higher used to produce 10 MA from a half wave capacitor input filter. But also I do not need 10 MA input to make a successful optical coupling, the input to a solid-state relay.

Assume we have 12 V at 10 MA for our input, then the resistance in series with the optical coupler is about 1200 ohms, use somewhat less because of the diode drop in the coupler. The monitoring circuit has between 12 and 0 diodes in series for the switch section of the circuit. The monitoring portion has two diodes, a series resistor of about 1000 ohms, and a 30 mfd capacitor. One diode is in the coupler, and the other diode is for the rectifier and current routing. This circuit has lots of latitude for variations.

The TD relay will have one routing diode, oppositely phase to the monitoring diode, and possibly a 100 mfd filter capacitor. No problem from the shunting devices, diodes across the switches, because their reverse leakage is very small. Thus, the monitoring circuit contributes virtually nothing to residual voltage across the TD relay when one or more switches are open. Again very good margins.

I do not have time to check what I wrote, and have to leave now. But the essence of the method is here.

.

 

[dbuckley]

So if I understand this correctly - there are a bunch of (presumably push-button) switches you want to wire in series, and that series loop is connected to a timer, so you press any button, and the timer is triggered, and the light comes on or whatever. And you want the loop of switches supervised so you can tell if there is a loop failure.

Ok, use normally closed switches, all in series, so the loop is normally complete. Press any switch (or several switches!), the loop opens, so that triggers the lighting timer to start. Use another timer that notices when the series loop has been open for more than 30 seconds. If that timer completes then the loop is broken 'cos no-one holds down the button that long.

With two timers and a bit of logic, that looks like a candidate for a micro-PLC.

 

[gar]

110212-1600 EST

My interpretation is that big john has a time delay relay and this implies a time delay to activation after the application of a sustained input. And generally this type of timer would automatically de-energize on loss of input, and automatically reset the delay time to its beginning whether timed out or not.

This would be classified as an on-delay with instantaneous reset. Whether the output contact is NO or NC is immaterial relative to the above described logic.

See http://relays.tycoelectronics.com/pnb.asp Potter & Brumfield types CL and CU.

Momentary inputs without a sealing circuit would not work.

Since a series circuit of switches was specified and the relay was described as a time delay, then this implies that timing does not start until all switches are closed, and these all must continuously remain closed until after the time delay time for there to be any output from the timer.

.

 

[big john]

Dbuckly,

The switches are part of an interlock system, the timer is for resetting a process to allow it to be reinitiated. The switches are actually magnetic switches on access doors: There's no time limit for how long one should be open: It's open as long as a person needs it open. My concern is having people defeat the interlock circuits, which is why I want it monitored: While the switches may open periodically, the loop should never open unless it's broken or been tampered with. If it does, the circuit needs to lock the process out until the problem can be addressed.

I appreciate the replies so far. I'm gonna give the dual relay approach a shot instead of superimposing AC on the circuit.

-John

 

[gar]

110212-1700 EST

big john:

If you have magnetic interlock switches, then how do you prevent people from simply using a magnetic(s) to defeat the system? A successful solution to this would seem to be the basic starting point.

Some sort of anti-tie-down method might be a start. Anything you do will necessitate that the personnel you assume will try to bypass the system can not get access to critical parts of this lockout system.

How does monitoring the continuity of your loop of switches in any way prevent bypassing the interlock. If all switches are of the normally open type and closed with a door magnetic when the door closes, then to bypass the interlock it is only necessary to jumper the loop close to the main panel, or use magnets.

How do you create a door sensor switch that is hard to bypass when the door is open?

I will suggest a coded thing with a unique address at each door. For example a Dallas button key. This would need to be attached to the door in such a way it could not be removed with any ease. No other button has an identical address. At each door is a reader of the button that also includes its own address.

At the main panel is a "master box" that interrogates via a three wire bus all of the door switches. Inability to address a door button means the door is open. Inability to address a reader means either a cable problem or a door reader problem.

Then the only easy way for your personnel to defeat the interlock is to do something with the "master box" output in the main panel.

.

 

[gar]

110212-1737 EST

big john:

You started the thread with a specific question about how to mix AC and DC signals on the same wire loop. But that was not your problem. Then the statement of the problem has evolved into something quite different. You have a system problem that needs adequate definition of the system parameters. These definitions still need to be determined.

Depending upon your requirements a simple system using your series magnetic reed switches and the closed loop might consist of the following:

At the far end of the loop a special impedance, basically a code, maybe done with a Dallas coded device.

At the main panel an electrically held relay, resettable only inside the main panel. Thus, requiring an electrician, with maybe a key, to reset it.

A special box monitors the loop. If the loop opens, then the electrically held relay drops out.

Until all doors are closed and the "special box" indicates the doors are closed the relay can not be reset. Shorting or opening the loop is the same as any door open because the special impedance can not be detected.

So long as you have an honest electrician that follows the rules set forth for this machine it can not be reset. One of the rules is:

Visual inspection that all doors are closed before reset.

.

 

[weressl]

I have a DC circuit that I'm building that has switch points, but I'd like to be able to supervise the continuity of the circuit.

My original thought was to to simply parallel a 60Hz power supply with my DC supply and jump the switch points with capacitors. The AC would still complete the circuit even as the DC was broken.

But then it occurred to me that I probably stand a good chance of damaging the DC supply by paralleling a transformer with it.

I need to filter out the AC component before it hits the DC supply. I thought of using inductors, but for 60Hz they'd have to be massive.

Is there a way to do this without trying to solder together a super-computer (I'm not much on soldering), or do I need a different idea?

-John

One of the simplest supervisory circuits with discrete signals is putting a high value resistor - high in comparison to the circuit resistance - in parallel with the contact and measuring circuit resistance. When it goes from high to low, you know that the switch is closed and when it goes from high to much higher, you know that you have an open circuit.

If you monitor multiple contacts you need to work out different resistance values for each, so no two, three or any number of variants of the switch closures would give the same result and that way you can determine which switches are closed. Simplest commercially available tool for this would be a PLC with analog inputs.

 

[dbuckley]

One of the simplest supervisory circuits with discrete signals is putting a high value resistor - high in comparison to the circuit resistance - in parallel with the contact and measuring circuit resistance. When it goes from high to low, you know that the switch is closed and when it goes from high to much higher, you know that you have an open circuit.

Now that I understand the problem, this is what I too would recommend,

This mechanism is a standard method in the security alarm industry to monitor tampering on alarm sensors, so the easiest realisation might be an off-the-shelf alarm panel. Pretty much all panels have auxilliary relay outputs that can be operated when things occur. Set one relay to the "tamper" status, and the other to "monitor" a zone. You dont even need to arm the panel to make zone monitor work.

Then you have two relay outputs as per spec.