Energy / Electricity Usage Tracking

[WayneW]

I finally got a reply from energymonitor.com to my Feb 4th email requesting info on the EUM-2000.


----- Original Message -----
From: "Juan O. Gonzalez" <info@energymonitor.com>
Sent: Saturday, February 26, 2005 4:27 PM
Subject: Re: EUM-2000 & Smart Homes

> Thank you for your interest in the EUM-2000.
>
> The interface for interphase with a Smart Home or with a Home Automation
> System is for future development.
>
> If you are interested in obtaining (1) instantaneous KW , this is available
> by pressing the KW buttons and it appears in the display; and (2) cumulative
> KWh usage information is maintained in memory or by pressing the "kWh"
> button. All thi information is store in the EUM-2000, including daily
> usage.
>
> The EUM-2000 only tracks the total usage?
>
> The unit does not require a battery, because it has non-volatile memory and
> thus all unit is maintained during a power outage of any duration!!
>
> The unit has a 12 Volt input, but to measire power you need either 12 Volts
> or a 240 Volt input. We chose 12 Volt as being simpler.
>
> The maximum KWh that can be accumulated in any one month is 9,999. But the
> unit accumulates 62 days of individual daily usage.
>
> Yes, the only difference in the units is based on the number and type of
> current clips? Therefore you can
> save some money if I don't need the split core feature on a permanent
> installation?
>
> If any other question, please advise.
>
>
> Juan O. Gonzalez,
> Energy Monitoring Technologies, Inc.
> 7516 NW 55 Street
> Miami, FL 33166
> Tel: (305) 470-9716
> Fax: (305) 406-9469
> Toll Free: (800) 206-1230

[Mario from Spain]

Ready to go!

After some weeks waiting to have some time to work on this project I'm ready to start!

Now I have to work out the interface between the energy meter and the 1-wire counter.

The energy meter has an open collector output and needs 5V min to work. The 1-wire counter has an MCT6 optocoupler and needs an input of 1.5 Volts and beetwen 20 and 60 mA.

Whats the best way to connect them?

My setup:

DIN rail energy counter with pulse output + 1-wire dual counter TAI-8586 from AAG + DS9097 Interface + mcsTemperature plugin

The energy counter open collector output:

• optocoupler 30V/20 mA max. and 5 V min.
• impedance 100 ohms
• pulse duration 50 ms
• output pulses 1000i/kWh

The AAG's 1-wire dual counter TAI-8586:

There isn't information or datasheet about at in the AAG web site. So I'm just guesing...

According to the TAI-8586 schematics the input is via a MCT6 optocoupler with a 150 ohm resistor in series with the input.



What I guess I need is a resistor divider network that using a 9-12 VDC power supply gives 5-12 VDC (max 20 mA) to the energy meter and, taking into account the 150 ohms resistor already in series with it, give 1.2 V (20 mA) to the TAI-8586 dual counter.

Is this correct?

[jon00]

melkati,

This should work for you. VCC on the 470 ohm resistor should be 5-10V and not connected to the 3V battery supply.

[Mario from Spain]

Many thanks Jon.

The issue is, as I understand, the meter optocoupler needs min. 5 Volts to work.

I already tried with a 1.5 volts battery and the counter works when wired directly to the battery but not when wired across the meter's optocoupler.

[Mario from Spain]

May be what I need is more like this?

[jon00]

melkati,

My circuit should be sound. You need to make sure that the polarity of the opto isolator is correct, otherwise it will not work.

Basically when a the meter LED flashes, I suspect it is also connected to the LED in the meter's opto isolator. This causes the collector and emitter (the leg with an arrow) to become virtually short for the duration of the flash.

Now if you had a 10V supply at Vcc, this goes through the 470 ohm resistor then the 150 ohm resistor (to limit the current), through the AAG's opto isolator LED, and then finally through the meter optoisolator transistor switch. As you can see, when the meter flashes, the opto will short causing the LED in the AAG opto to fire.

If you have a meter that measures low resistance or better still diodes, you can put that accross the meter's opto output legs to see if it is firing or not. Again, make sure you try the meter probes in both ways across the opto.

Finally, it is possible that you have destroyed the opto on the meter by in-correct wiring and putting a voltage across it without limiting the current and destroying the opto transistor within.

[rocco]

Mario:
Jon's circuit should work. He is using the open-collector output of the meter's opto as a low-side switch.

Your circuit should also work, if you remove the resistor that is across the input of the counter. You are using the open-collector output of the meter's opto as a high-side switch.

But the problem may be the values of the resistor. If you take Jon's circuit, remove his 470 ohm resistor and supply it with 5 volts, you will get at least 23 milliamps through the LED of the MCT6. That is just over the minimum of 20 milliamps.

But with the 470 ohm resistor, even with 10 volts, you will only get 14 ma, which is well below the minimum. The 150 ohm resistor on the counter is ideal for driving a MCT6 input with 5 volts. So my recommendation is to use Jon's circuit with no added resistor and 5 volts.

Here is the rub:
The meter spec states that in can output UP TO 20 milliamps. The counter spec states that it's input needs AT LEAST 20 milliamps. So you need to get as close to 20 as you can get, and hope there is enough overlap in the margins.

[jon00]

Hi Rocco,

I took the cautious route as the spec of the Opto as posted by melkati is only 20mA. Seems very low and would indicate LED current in my books but this cannot be in this application??

Anyway, with only a 150 ohm resistor at 5V thats 33mA if Ic is only 20mA then......

[rocco]

Jon, the current would be 33 milliamps if it weren't for the voltage drop across the diode. The MCT6's diode has a maximum forward voltage of 1.5 volts, thus leaving 3.5 volts across the resistor. The MCT6 does not spec a minimum forward voltage, but does spec a typical forward voltage of 1.25 volts. That would give you 25 milliamps. If I was going to use a resistor at all, I would use a 39 ohm, which would get you 19.8 milliamps typical. Of course, with tolerances being what they are, you could be quite a ways away from that target.

The surest approach would be to replace the resistor with a 50 ohm potentiometer and adjust for 20 milliamps.

[jon00]

yes, you are right.

The meter opto transistor and the AAG opto LED drop will equate to around 2.1V. So at 5V using just the 150 ohm resistor gives just under 20mA.

Personally, working with such low voltages makes the choosing of resistor values far too critical. I much prefer using 12V supplies for this type of application.

The word is tolerences :-)

[rocco]

Oops . . .

I completely forgot about the drop across the output transistor.
Thanks for catching that, Jon.

I totally aggree with the 12 volt statement. When the voltages are higher, the variations in diode drop become less significant. But I made the assumption that one of the boards needed 5 volts, so that Mario had that supply already.

(my spelling checker changes "tolerences" to "tolerances". I, myself, have no idea which is right. If I didn't have a spelling checker, I couldn't communicate at all. Maybe it's another English/US thing, like "grey" and "gray").

[Mario from Spain]

I totally aggree with the 12 volt statement. When the voltages are higher, the variations in diode drop become less significant. But I made the assumption that one of the boards needed 5 volts, so that Mario had that supply already.

I also prefer to do it with a 12 volts supply as I have two very close to this place (one for the Ocelot and another for the alarm system).

If I need the 5 Volts I should use a 7805 IC for this (or another 5 volts wall wart).

[jon00]

(my spelling checker changes "tolerences" to "tolerances". I, myself, have no idea which is right. If I didn't have a spelling checker, I couldn't communicate at all. Maybe it's another English/US thing, like "grey" and "gray").

No I can't spell!


Melkati - change the 470 ohm to 330 ohm and run at 12V - should be fine.

[Mario from Spain]

The surest approach would be to replace the resistor with a 50 ohm potentiometer and adjust for 20 milliamps.

The problem with this is how to measure the 20 mA as I guess a regular digital multimeter will show less current as it's pulsed.

[Mario from Spain]

Melkati - change the 470 ohm to 330 ohm and run at 12V - should be fine.

I'll give it a try today!

Thanks.

Mario.-