Low-power AC presence detection?

[BreakingOhmsLaw]

Hi everybody and a happy new year.
I'm currently designing a controller for my two drainage pumps that keep my basement turning into an unwanted swimming pool.
Both pumps are on their own circuit breaker, and I want to detect if that breaker has triggered. So, I need to detect whether AC is present.

Now, there is the classic way of taking a big resistor, or capacitive dropper, and triggering an optocoupler. But that will constantly piss away a couple of mA. Something rubs me the wrong way to waste a couple of kWh per year just to detect if voltage is present. There must be a better way.

I do have a current sensor in place that monitors the power consumption of the pumps and checks them against spec to detect a failing or blocked pump. But that only is useful if the pump is running, and that's not always the case. It runs 2-3 times per hour for a couple of minutes at most.

So, what's a better and efficient way that doesn't use more than a couple of mW?

[BreakingOhmsLaw]

Putting something across the breaker probably violates a hundred rules. You can buy breakers with a mechanical interface to a reporting switch like the ABB S2C-H11L, but rebuilding the switchboard is not an option.
I have three phases (breaker on each) going into my controller. L1 feeds the controller, L2 feeds Pump 1, L3 feeds pump 2.

[ajb]

How low is low enough power for you?  I think it will be tricky to get much better than a sensitive optocoupler.  Maybe use a relay to only enable the ac-referenced measurement circuitry periodically to check if AC is present?  Or do similar with a PV-output optocoupler to power some low power/high sensitivity detection circuitry on the AC-referenced side? 

[Benta]

Current transformer. No loss (except a bit in the burden resistor).
But what you have on the secondary side is up to you.

[BreakingOhmsLaw]

Maybe use a relay to only enable the ac-referenced measurement circuitry periodically to check if AC is present?

Not a bad idea actually. Will take that into consideration.

[PCB.Wiz]

So, what's a better and efficient way that doesn't use more than a couple of mW?

You can add some active parts to pulse that optocoupler, which slashes that average current.
It gives one pulse per cycle, for simplicity.  Feed to your MCU, or a monostable or a slow RC as needed

Here is an example I did a while back, mains side rms of just 68uA and gives ~3mA peak into the opto.

[bdunham7]

What is your detection system connected to?  Does it need to be an optocoupler?  IOW, what circuit or device needs to be informed that there is AC power and what sort of signal does it require?

[BreakingOhmsLaw]

What is your detection system connected to?  Does it need to be an optocoupler?  IOW, what circuit or device needs to be informed that there is AC power and what sort of signal does it require?

An STM32 microcontroller, regular GPIO.

[BreakingOhmsLaw]

You can add some active parts to pulse that optocoupler, which slashes that average current. [...]

I gave that a thought. Might be even simpler by just using a 300V+ zener diode so it just conducts during a brief moment  when the sine is at peak. Going to simulate that.

[BreakingOhmsLaw]

Current transformer. No loss (except a bit in the burden resistor).
But what you have on the secondary side is up to you.

I actually considered that, but I couldn't find one that has a sufficient isolation rating to ensure electrical safety.

[xvr]

You can use oscillating circuit to reduce power consumption. Main idea of this circuit is to charge capacitor from rectified AC line with very very small current (via resistor). Next part of circuit is a dinistor (Shockley diode or analog) + optocoupler. When voltage across capacitor raised above dinistor threshold it open and discharge capacitor to LED part of optocoupler.
Carefully selecting capacitor value and duty cycle of optocoupler signaling you can achieve almost any desired input (AC) current.

[BreakingOhmsLaw]

You can use oscillating circuit to reduce power consumption. Main idea of this circuit is to charge capacitor from rectified AC line with very very small current (via resistor). Next part of circuit is a dinistor (Shockley diode or analog) + optocoupler. When voltage across capacitor raised above dinistor threshold it open and discharge capacitor to LED part of optocoupler.
Carefully selecting capacitor value and duty cycle of optocoupler signaling you can achieve almost any desired input (AC) current.

Ah yes. Haven't used a DIAC in anger for many, many years. Completely forgot about them.
Something along the lines of this ought to give a pulse once per second, going to spice this.

[Benta]

[ Specified attachment is not available ]

Current transformer. No loss (except a bit in the burden resistor).
But what you have on the secondary side is up to you.

I actually considered that, but I couldn't find one that has a sufficient isolation rating to ensure electrical safety.

What?
They are available with isolation into the kV ranges. It's just an isolated blob whit a hole in the middle.
Example:

[inse]

Why not monitor the water level instead?
That would report any failure in the whole system.

[bdunham7]

An STM32 microcontroller, regular GPIO.

An LDA111 Darlington optocoupler has a typical control current of less than 100µA, so 12mW or 105Wh per year just using resistive dropping and a lot less with a capacitor.  I suspect just your GPIO pin will use as much.  The only other isolated method I can think would be to use an optotriac switched by the STM32 to supply the feed to the LDA111.  This would use more power when activated but would not use any (other than leakage) when the STM32 wasn't polling the power supply.  A really clever method might be able to do all that and still only use one GPIO pin...

Still, I'd calculate out how much power a 100µ capacitor dropper with series resistor circuit would really use.  3mW or less and still having sufficient series resistance to protect against a shorted capacitor?

[BreakingOhmsLaw]

What?
They are available with isolation into the kV ranges. It's just an isolated blob whit a hole in the middle.
[...]

I see, I was thinking of something else.
However, don't these usually transform the current down?

[BreakingOhmsLaw]

Why not monitor the water level instead?
That would report any failure in the whole system.

I'm doing that too with a 190 Ohm level sensor. The values get sent to a Syslogd server and I can monitor the amount of water going into the drainage.
However, when the water level is too high, both pumps have failed. A bit late for a warning 

[BreakingOhmsLaw]

The only other isolated method I can think would be to use an optotriac switched by the STM32 to supply the feed to the LDA111.  This would use more power when activated but would not use any (less leakage) when the STM32 wasn't polling the power supply.  A really clever method might be able to do all that and still only use one GPIO pin...

Not a bad idea actually. It certainly would have the advantage that the "alive" signal isn't coming when the STM32 is sleeping. It could poll the status when its active.

[Zero999]

Current transformer. No loss (except a bit in the burden resistor).
But what you have on the secondary side is up to you.

I actually considered that, but I couldn't find one that has a sufficient isolation rating to ensure electrical safety.

It should be easy. Nearly all current transformers I've seen have sufficient isolation to ensure electrical safety. Indeed, there are some where I work on 3kV bus bars and are connected to a device powered by 24VDC.

[TimFox]

What?
They are available with isolation into the kV ranges. It's just an isolated blob whit a hole in the middle.
[...]

I see, I was thinking of something else.
However, don't these usually transform the current down?

Of course, but they are designed to work with a reasonable burden resistor to develop a measurable voltage without a large power loss in the sensed circuit.  The resistance is usually specified for the part.  The primary can be one turn (single wire) while the secondary has many terms.

For example, a 100:1 transformer with 100 termination would insert only 0.01 into the wire, and give 1 V rms at the resistor for 1 A rms current.

[bdunham7]

Current transformer. No loss (except a bit in the burden resistor).
But what you have on the secondary side is up to you.

How would that detect voltage at times when there is no current (pumps are off)?

[ejeffrey]

I'd say go with the resistor/capacitor dropper + rectifier + optocoupler.  It's simple, safe, and reliable.  A standby current of a fraction a milliamp is just not that much.  Whatever circuit you have observing it on the secondary side is probably going to to have a few milliamps of mains current draw for it's power supply too.  If you use a capacitor dropper it's going to be mostly reactive load anyway and a bit of extra capacitive reactance on the power lines is probably a net bonus.

[Benta]

Current transformer. No loss (except a bit in the burden resistor).
But what you have on the secondary side is up to you.

How would that detect voltage at times when there is no current (pumps are off)?

Why should it? The point seems to be monitoring if the pumps are running or not. Detecting the current will do that.

[bdunham7]

Why should it? The point seems to be monitoring if the pumps are running or not. Detecting the current will do that.

His first post states that he wants to detect whether the breakers are tripped.  He also says that he already has current sensors on the pumps themselves.

[Zero999]

Presumably this doesn't have to be fast? If a turn off time of a minute or so is acceptable, then how about a small mains powered switched mode power supply i.e. a wall wart? They typically have a very low standby power consumption. The downside is the smoothing capacitor will take a long time to discharge. Adding a load isn't an option if you want low standby consumption.