Computing stuff tied to the physical world

Archive for April 2013

Meet the Manson NSP 3630

In Hardware on Apr 30, 2013 at 00:01

It’s always nice to explore more equipment, to see how it behaves in the lab. I already have a very nice dual-voltage lab power supply, but this one is interesting due to its relatively low cost (€90 + shipping), fairly high power, and very convenient small size:

DSC_4441

It’s the Manson NSP 3630, available from Reichelt in Germany by mail-order. At 27 x 15 x 7 cm, it really should easily fit in most home labs. With many thanks to David Menting for letting me play with it a bit and report some first impressions.

The display is quite simple, yet still nicely readable:

DSC_4442

Power consumption, without any load is 0.07W when turned off with the power switch on the back of the unit, and about 2.0W when turned on, without load.

There are two (optical encoder) rotary knobs to adjust the maximum voltage and current, respecively, over a range of 0.7 .. 36V and 0 .. 3A – i.e. plenty for most situations.

A drawback with this particular supply, is that it’s a bit of a hack to pre-set the current limit. You can turn it down during use to reduce the limit to the point where the voltage starts dropping, but if you want to set it up ahead of time, then the way to do it is to short out the power supply while adjusting the knob, to see the value it displays.

This is a switched-mode power supply, which explains why it is so small and requires only a small fan, but it does lead to some residual noise on the output. Hooking it up to the scope, still under no-load conditions, you can see that the output varies between about +10 mV and -20 mV of “ripple” above and under the preset value, respectively (5V in this case):

SCR02

To produce this image, the scope was set to “peak-detect” mode, which captures the high and low value at each point in time, and the trace is drawn in “envelope” mode, which is a variation of persistent display showing the most extreme values ever reached as long as the scope is kept on. The last trace is the one shown in the middle, while the top and bottom lines are the largest variations ever reached – I think I left it running for a minute or so.

It’s a pretty good result, actually, for a switching power supply. Variations of this kind should not cause any problems for most digital circuits, which usually can tolerate fairly large variations on the power supply lines.

Three laws – part 3

In Hardware on Apr 29, 2013 at 00:01

Yesterday’s post was about Kirchoff’s Current Law.

Now let’s look at the second version of that law, Kirchhoff’s Voltage law:

JC's Grid, page 72

This law says something about the voltages in a circuit: voltages cancel out, regardless of which way you follow a path in the circuit, i.e.:

V1 - V2 - V3 = 0    (hence V1 = V2 + v3)

But also:

V3 - V4 - V5 = 0    (hence V3 = V4 + V5)

And:

V1 - V2 - V4 - V5 = 0    (hence V1 = V2 + V4 + V5)

The signs are based on how the arrows are drawn: voltages are always relative, so it all depends on the direction of each of those arrows. Since V2 and V3 are opposite to the direction of “following” a single path through the circuit, they end up negative in the above equation. Same for V4 and V5.

Now this version of Kichhoff’s law is really useful. If each of the resistors has the same value (it doesn’t matter what the value is), then we can reason about the voltages in different parts of the circuit, even when only V1 is known at the start.

Let’s look at this circuit again in terms of voltages, with V1 set to 3V:

JC's Grid, page 72

Since V1 = V2 + V3, and all resistors have the same value, it’s a safe guess that the voltage over both will be the same. A more exact reasoning is to include the currents, and since they are the same for the two resistors on the left, we know that the voltages will be the same (Ohm’s law, regardless of the value of these resistors – as long as they are the same).

Perhaps a bit more surprising is that V4 is 0V. This is again a consequence of Ohm’s law: since the pins on the right are not connected, no current will flow. With no current, the voltage over a resistor is always 0V (Ohm’s law, again).

Since V4 is 0V, and V2 = V4 + V5 as we saw above, V5 must be 1.5V, the same as V2.

Note how a voltage over a resistor can be zero, and over two open pins can be non-zero.

If we use a multi-meter with a really high input resistance (any modern multi-meter will do, usually they have 10 MΩ or more), to make sure no current starts flowing, then we can indeed measure the voltage to be half of the 3V between top and bottom, i.e. 1.5V.

The main point to take away from this, is that the three laws are intimately inter-connected. Given known resistors, which is often the case, Ohm’s law dictates the relationship between voltage and current, and with the two variants of Kirchhoff’s law, it is usually possible to “reason your way into” the different points in a circuit, given an externally applied power supply.

One last example: what is the equivalent resistance of two resistors of 1000 Ω in parallel? Well, say we apply voltage V1, then the current through one resistor will be V1/1000. The same current will go through the other resistor, since it too has V1 on both pins. So the total current of the two in parallel will be 2 x V1 / 1000 = V1 / 500. What is the resistance when we apply V1 and see a current V1 / 500 flowing? Again, Ohm’s law: R = U / I, i.o.w. R = V1 / (V1 / 500) = 500. So the general answer is: the equivalent resistance of two resistors in parallel is half their resistance. A direct consequence of Ohm’s and Kirchhoff’s laws.

Get used to the three laws, and with a bit of practice, the rest should fall into place!

Three laws – part 2

In Hardware on Apr 28, 2013 at 00:01

After Mr. Ohm yesterday, let’s see what Mr. Kirchhoff has to say.

Kirchhoff’s First Law is about current. In a nutshell – what goes in, must come out:

JC's Grid, page 72

I’ve drawn resistors, but that’s really irrelevant here – each component could be anything.

The current I1 must be the same as the current I2, It does not matter what I3 is in this case (assuming the two pins next to it are not feeding or drawing any outside current).

Also, if I3 is zero, i.e. if the pins are not connected to anything at all, then the current through the leftmost resistors is identical. In this case, they are essentially in series, with the resistor on the right not doing anything at all (no current = no voltage drop = Ohm’s Law).

This makes it possible to reason about that point in the middle, where the three resistors meet. The currents at that point must cancel out: that’s what Kirchhoff’s Current Law says.

Suppose all three resistors are 1 kΩ, and the current I1 is 1 mA:

JC's Grid, page 72

If the two pins on the right are left open, no current will flow there. So the same current I1 (which is also the same as I2) will flow through both resistors on the left. Total voltage drop from top to bottom will be 1 mA x 1 kΩ = 1V on the top resistor and another 1V on the bottom one, for a total voltage of 2V across the left two pins.

Or to put it more practically: if you place a 2V supply across those left two pins, then 1 mA will flow. The voltage in the center point will be halfway, i.e. at 1V.

What will happen when we short the pins on the right?

JC's Grid, page 72

Again, there’s 1 mA flowing in from the top, so there will be 1 mA coming out the bottom. The bottom-left and right resistor will together see a current of 1 mA going through them. Since they are both the same 1 kΩ, it should not come as a surprise that each resistor will get half the current, i.e. 0.5 mA each. Total voltage from top to bottom will be 1.5V.

This isn’t such a great example in terms of practical use, since normally the reasoning goes the other way around: what current will flow when I apply voltage X to the entire circuit?

That’s the other version of this law, described tomorrow.

Three laws – part 1

In Hardware on Apr 27, 2013 at 00:01

It’s always a puzzle to predict just what will happen when you hook up some components.

But as mentioned in the what-if series, it’s really useful to be able to do so, to avoid surprises and damage. Also for questions such as: Why does a higher voltage cause a higher risk of damage in one case and less so in the other? Why do I need a resistor in series, and of what value? What if I don’t have the right resistor or a different voltage?

Lots of complex issues, but the simplest and most important case usually is static analysis and DC (direct current) voltages, i.e. when only one or two states are involved, and not so much the switching and AC (alternating current) behaviours.

You just need to get familiar with three “laws” of electricity:

  • Ohm’s Law – given two of: voltage, current, and resistance, we can predict the third.
  • Kirchhoff’s Current Law – current always adds up: what goes in, must come out.
  • Kirchhoff’s Voltage Law – how voltages “spread” across interconnected components.

That and learning what the basic behaviour is of resistors, diodes, capacitors, etc, and you’ve got all the knowledge you need to “see” what a circuit does, before even building it and trying it out. And by this I really mean literally “visualising” what is going on – it only takes a little practice to turn this into a very intuitive skill.

You just have to grasp the essence of those three laws. So let’s get on with it, eh?

Ohm’s Law

This is by far the most important one. It says everything about resistance. The unit of resistance is – not surprisingly – the “Ohm”: a resistor of 1 Ohm will let 1 Ampère of current flow when you apply 1 Volt of electric potential difference over it. The formula is:

U = I x R

(U = voltage, I = current, R = resistance)

Same law, different uses, by simple algebraic manipulation: I = U / R, and R = U / I. If you know two of the units, you can calculate the 3rd.

  • What happens when I touch both poles of a car battery?
    My skin resistance will be some 100 kΩ, so I = U / R = 12 / 100,000 = 0.000,12 A = 120 µA. A tiny current, I wouldn’t sense a thing, so the answer is: “nothing happens”.

  • What happens when I place a metallic nail across that same 12V car battery?
    Let’s say the resistance of that nail is 0.1 Ω, i.e. almost a short, so I = U / R = 12 / 0.1 = 120 A, a huge amount of current. The nail will heat up like crazy, it might even melt!

Same battery, very different outcomes.

And it’s not just useful to predict such extreme cases. It also helps understand why some hookups are inherently safe: if my power supply delivers no more than 5V, and I am playing with resistors of 1 kΩ or more, then no matter how I hook things up (apart from shorting things out), there will never be more than I = U / R = 5 / 1,000 = 0.005 A = 5 mA of current through my circuit. A 1 kΩ resistor in series with just about any component will “limit” the current to 5 mA, which virtually prevents damage to just about any component.

Another example: suppose I am using a 12 V power supply, and want to turn on an LED with it. Most LED’s glow nicely at 20 mA. So if I put a R = U / I = 12 / 0.020 = 600 Ω resistor in series with the LED, I can be certain that the LED won’t get damaged. Better still, if I only have a 1 kΩ = 1000 Ω resistor, I can predict that it’ll probably work just fine in this same circuit as the current will be at most I = U / R = 12 / 1,000 = 12 mA. Using a 100 Ω resistor would be a bad idea (max current 120 mA), and using a 10 kΩ resistor probably also wouldn’t work (1.2 mA might be too little to make the LED light up).

These are all approximations, but they are extremely useful – even as such.

Some consequences of the simple ” U = I x R ” formula Georg Simon Ohm gave us:

  • twice the voltage => twice the current
  • twice the resistance => half the current
  • twice the current => twice the voltage drop

One thing to take away from all this, is that it’s not a bad idea to buy some 100 Ω, 1 kΩ, and 10 kΩ resistors. Having a bunch of spare resistors can often come in handy, as a way to limit the current (and hence avoid damage), and these three values are often all you need to try out a few things in circuits running at 1.5 .. 12V.

Tomorrow, we’ll give the stage to Gustav Kirchhoff!

What if I mix 3.3/5.0V – part 3

In Hardware on Apr 26, 2013 at 00:01

Welcome to the weekly What-If series, also available via the Café wiki.

There’s still much more to say about all this – as can be expected. One of the suggestions made in the comments was to use a few diodes in forward direction. Since these each have about 0.65V drop, three of them ought to bring down the voltage from 5.0V to 3.3V.

Time to hook up the signal generator and scope again. Be prepared for some surprises!

As I mentioned in my original post, the “official” way to handle this, is to use a “level converter” chip, which is based on some active circuitry, i.e. some transistors or MOSFETs. The resistor solutions described yesterday are less accurate, as you’ll see…

Here is a 1 MHz square wave @ 5V (yellow) feeding yesterday’s 4.7 kΩ + 10 kΩ resistive voltage divider to produce a 3.3V signal (purple):

SCR94

That’s quite a different signal coming out! A typical capacitive charge / low-pass effect.

Note that the signal generator has a ≈ 10 nS rise-time, i.e. the edges are not completely vertical, and that the Hameg scope probe has a 14 pF loading capacitance, according to the specs. So some of these effects are artefacts of this measurement setup.

Let’s raise the frequency to 10 MHz (the horizontal scale is now a very fast 10 ns/div):

SCR95

Hardly a square wave, and no well-defined 0-to-3.3V transitions either, as you can see.

Now let’s try this circuit:

JC's Grid, page 71

The reasoning being that the diodes will “drop” the voltage from its high 5V level to 3.3V:

SCR96

Quite an asymmetric effect, although lowering the resistor to 1 kΩ ought to improve it.

The above is again a 1 MHz square wave input, and here’s the same at 10 MHz:

SCR97

This final setup is so far off the desired voltage levels that it probably won’t work.

Given these outcomes, I’m inclined to stick with a single 10 kΩ resistor in series. Or perhaps drop it to 1 kΩ to get better rise and fall times. Active MOSFET-based level-shifting circuits are starting to make a lot more sense now – they exist for a reason!

What if I mix 3.3/5.0V – part 2

In Hardware on Apr 25, 2013 at 00:01

Welcome to the new weekly What-If series, also available via the Café wiki.

Yesterday’s post was about mixing 3.3V logic levels with 5V logic levels.

The real trouble, apart from things not working, is damaging some component, of course. And in electronics, almost all damage comes from overheating. At some point (well over 100 °C, usually), chips really do get “fried” and irrevocably damaged occasionally.

The trick is to avoid overheating. This in often easy to do by limiting the current that can flow. How do you limit the current? Simple: make sure there are some resistors in series!

But first, let’s examine what’s really going on when we connect a 5V logic “1” to an input which handles only 3.3V logic levels:

JC's Grid, page 71

On the left, the typical circuit inside a digital chip at each I/O pin. The essential ingredients are two ESD protection diodes to VCC and GND, respectively, and the MOSFET input gate.

The diodes are placed in the “blocking” direction, i.e. they normally do nothing but block the current. The MOSFET gate is a very high impedance input, essentially it acts just like a capacitor, in fact. The rest of the input circuitry can be ignored for our purposes.

So normally, input signals on such a pin just “float” and follow whatever voltage is applied. Until the voltage is too high, or too low, that is.

On the right, I’ve redrawn the exact same circuit, but “lifting” the input well above VCC level. To make this clear, I’ve drawn the input pin above the VCC pin, as an intuitive way to represent voltage as height in the diagram.

Now you can see what happens above VCC + 0.65 or so: the top diode will start to conduct in this situation. So if the input is set to 5V, and VCC is 3.3V, then the diode will become a conductor and try to pull either 5V down or 3.3V up, to reach equilibrium again.

This is a danger sign. The amount of current flowing will rise as long as these voltages are more than about 0.65 V apart. If both power supplies are very strong, there could be several amps of current – the 5V supply could even start powering the 3.3V circuit!

And that’s where the damage-through-heat comes in: a 1 A current over a diode with a 0.65 V difference leads to 0.65 W of heat being dissipated. Far more than these tiny on-chip diodes can handle. In fact, they really are only designed to carry up to a milliamp or so. The result: a “blown” on-chip diode, and since the voltage difference continues to exist, the damage will probably propagate to other diodes and components on the chip.

There are a few ways to prevent this. One is to use special “level-converter” chips, designed to take one voltage on the input, and then generate a different voltage on the output.

But there are much simpler ways to deal with such small differences, especially when the pins are all close together and on the same board:

JC's Grid, page 71

On the left, a traditional 4.7 kΩ + 10 kΩ resistive “voltage divider”, which takes one voltage and divides it down to a lower voltage. This works, but in this case we can even get by with a single resistor, as shown on the right.

The reason the single resistor works, is that current will start to flow through the diode as before, but now also through the 10 kΩ resistor. Since the circuit seeks equilibrium, in this case, there will be about 5.0 – 3.3 = 1.7 V across resistor + diode, i.e. ≈ 1 V across the resistor. With a 10 kΩ resistor, it only takes 0.1 mA of current to generate a corresponding voltage drop, so that’s when equilibrium will be reached. These current levels are completely harmless and can easily be sustained by the on-chip diode.

So there you have it: we’ve explained why direct connections might lead to overheating, and how a 1..10 kΩ resistor in series can prevent it, while still allowing the circuit to work.

All this required, was some theory and a basic understanding of the internal circuitry.

PS. These resistor solutions are sensitive to noise and capacitive loading (they act as low-pass filters), so this only works well when signal lines are short, a few cm or so. For reliable high-speed signaling over longer distances, a level-shifter chip would be a better way to go.

What if I mix 3.3V and 5V?

In Hardware on Apr 24, 2013 at 00:01

Welcome to the new weekly What-If series, also available via the Café wiki.

As you probably know, mixing 3.3V logic with 5V logic is usually a bad idea, but why and what if you need to do it anyway?

This topic is too complex for a single post, but let’s just start and see where it leads to:

  • VCC in these examples is the supply voltage, i.e. either 3.3V or 5.0V, depending on which chip we’re looking at. The goal being to connect a mix of these.

  • Digital I/O pins work by treating everything below a certain voltage as “0” and everything above another voltage as “1”. In between, the levels are undefined and could be interpreted either way, depending on temperature, stray capacitances, moon phases, who knows…

  • For the ATmega328, for example, everything between -0.5V and 0.3 x VCC is treated as “0” and everything between 0.6 x VCC and VCC+0.5V as “1”.

  • For 5V signals, that translates to: under 1.5V is “0” and over 3.0V is “1”, respectively.

  • Also relevant, is that unloaded output pins tend to be very close to the 0V and VCC ground and power supply levels, respectively.

So mixing is in fact not a problem at all for the following scenario: 3.3V levels for output signals, tied to 5V levels for the input signals. If all you need, is to read logic levels on say a 5V Arduino Uno from a 3.3V JeeNode, then just tie the signals together and you’re all set.

Connecting a 3.3V level output pin to a 5V level input pin works fine.

The problem occurs in the other direction: a “1” output on a 5V logic level is about 5V, whereas the maximum allowed input level for a “1” on a chip powered by 3.3V is 3.3+0.5, i.e. at most 3.8V.

Hooking these together without taking care of the problem will lead to problems, such as overheating chips and even smoke or fire (although this is almost impossible with simple battery- or USB-powered hookups).

Tomorrow I’ll describe the cause of these problems, along with some simple solutions.

New series – What If?

In Hardware, News, Software on Apr 23, 2013 at 00:01

Questions are very useful: “what would happen if…” is the foundation of science, after all.

Conjectures and Refutations is a famous book by the late philosopher Sir Karl Popper. I could not possibly summarise it (heck, I haven’t even read it), but what I take away from what I’ve read and heard about it, is that theories can be judged on their predictive value. A theory in itself is no more than an intellectual exercise, but its real value lies in being able to apply it to what-if questions. The stronger a theory, the better it should predict outcomes. The way to “refute” a theory, is to come up with an example where it fails. Rinse and repeat, and you’ve captured the essence of science.

Want to predict what will happen when you place a 100 Ω resistor across a 9V battery? That’s easy, given the proper theory: take Ohm’s Law (i.e. a theory which has stood the test of time), and apply it – a current of ≈ 11 90 mA will flow. Actually a bit less due to the internal resistance of the battery, which goes to show how strong theories can be refined further, leading to even more accurate predictions.

The what-if question is a great way to experiment, especially in electronics and electro-mechanics, because it lets you be prepared and avoid silly (and sometimes catastrophic) outcomes, such as a damaged component, a harmful burn, or even an explosion.

This approach lends itself to all sorts of practical questions:

  • What if I short out a 3x AA battery pack?
  • What if I connect my chip the wrong way around?
  • What if I have to use a 12V power supply instead of 5V?

But also issues as varied as:

  • What if I omit a certain component from my circuit?
  • What if I unplug the Raspberry Pi without shutting it down?
  • What if I wanted to use HouseMon in combination with MySQL?

Properly phrased, what-if questions are essential for practical experiments, and – by extension – also the key to building useful circuits and automated installations.

A useful variation of the what-if question is to help predict “bad” outcomes and estimate the risk of an experiment, such as: can shorting out my power supply cause real damage?

Starting tomorrow, I’m launching a new series on this weblog, titled “What-If Wednesday”. As far as I’m concerned, it can run as long as there are interesting questions I can answer, so please feel free to suggest lots of topics in the comments below. These weekly posts will be tagged What-If, and I’m also setting up a new wiki page to collect them all.

Sooo… please help me out folks, and send in some nice what-if questions!

Winding down

In Musings, News on Apr 22, 2013 at 00:01

The JeeDay 2013-04 event is over.

I would like to warmly thank the 40 or so people who attended on Friday and Saturday. It is clear to me from the kind follow-up emails that the event was appreciated by many of you and I really hope that everyone got something useful and stimulating out of this.

Allow me to also thank the “anonymous sponsor” at this point for funding the venue, the coffee and drinks, and Saturday’s lunch. I’ve passed on your and my appreciation, and it has gratefully been accepted. As several people have pointed out, this whole concept of an anonymous sponsor is really a contradiction in terms, so let’s all just cherish the fact that philanthropy (and mystery) still exists, even in today’s western societies.

This is probably the point where I’m expected to write sentences full of superlatives, self-congratulatory remarks, let’s-conquer-the-world type of pep-talk, congratulations for the speakers and their choice of interesting topics, all sorts of grandiose plans, and where I’d also describe how stimulating all the discussions on the side turned out to be.

I could, and it’d be true. But I won’t…

Instead, I’d like to give this a somewhat different (personal / philosophical) twist.

We’re focused on success. We crave rewards. We seek recognition. So when something good (for some definition of “good”) happens, we want to take it further.

Again. Better. More.

Yet to me, that’s not what JeeDay was about. Sure, we could do it again. In fact, I’d love to and I’ve even sort-of committed to organising another JeeDay a year from now. We’ll see.

But to me, JeeDay is not about the next step or some future trend. It’s about this event we just had. Some 10 talks from people describing what they like to do in their free time. That’s quite a special situation, when you stop and think about it: here we all are, a few dozen geeks with a common techie interest, and this what we choose to spend our time, our creative energies, and our money on. We could do anything, yet this is what we want to do. In. Our. Free. Time.

Now of course, everyone’s reasons will differ. But to me, it’s pretty amazing: there’s rarely a financial reward (heck, it usually costs money!). There’s often not much recognition. These are not TED talks, we’re not working on some high-visibility successful project and showing the world. We just tinker in private, we come up with stuff, we learn, and we like doing it.

In my view, this is about the top two tiers of Maslow’s hierarchy of needs:

Maslow's_Hierarchy_of_Needs.svg

The basic idea being that you can’t really get to focus on the levels above before the levels underneath have more or less been covered.

This is – again, in my perception – not about success, and probably not even about peer recognition, but about the intrinsic fun of discovery, invention, creation, and problem-solving. And about finding out how others deal with this. It’s no accident that most of it happens as open source, either: open source (hardware + software) and sharing is what floats to the top when the intrinsic puzzles and their solutions dominate.

In a world where so much is about ownership, money, and time, I think that’s precious.

I hope JeeDay has helped you find and follow your passion. Everything else is secondary.

PS. The mystery topic in my presentation was JeeBoot – more to follow soon.

Cheap power analysis

In Hardware on Apr 21, 2013 at 00:01

Remember this screen shot?

annotated-room-packet.png

It was a carefully captured analysis of the power consumption of a JeeNode, running the RoomNode sketch, and sending / receiving wireless RFM12B packets. There’s a fantastic amount of info in there, to help understand which part of the code and which activity is drawing the most power. It was a great help at the time to reduce power consumption, allowing these nodes to run well over a year on a couple of AA batteries.

Trouble with this, is that you need an expensive piece of equipment, called an oscilloscope. Long-time readers might remember that I’ve written extensively about this. These things cost anywhere from a few hundred Euro, to thousands, or even tens of thousands for high-end units. I ended up settling for a Hameg HMO2024, which is a great instrument, but with a pretty hefty price of well over €1000.

So how would you go about analysing the power consumption of your sketch without plunking down this sort of cash? Well, there really are not that many alternatives, you have to see the current-vs-time graph to be able to understand what’s going on.

Luckily, there is a fairly capable little unit from Gabotronics, called the Xminilab. It pushes an ATXmega (note the “X”) to its limits, allowing it to capture quite a bit of information, just like its bigger brothers. It even includes things like FFT analysis, an 8-channel Logic Analyser, and an AWG Signal Generator! Last but not least, the software is open source.

Interested in how capable this $69 device is? Well, check this out:

DSC_4438

Do you recognise the waveform? The Xminilab has captured a packet transmission, a bit like the one shown at the start of this post (it’s a different sketch, i.e. radioBlip2, hence a different pattern). It may not look like much, but it should be sufficient to see the effect of changes in the code and to optimise power consumption with it.

So, do you need a scope? IMO, anyone wishing to explore electronics should have one. Whether second-hand or the above-mentioned Xminilab, it really helps to be able to see things in a way where our human senses fall short – such as these brief events. It’s the most versatile instrument in the lab, if you ask me – even with a 128×64 pixel LCD screen.

PS. I don’t recommend the even lower-cost $49 Xprotolab (which I also have). It has the same functionality, but with its tiny OLED display it really is too hard to read, IMO.

Automated + manual 230V

In Hardware on Apr 20, 2013 at 00:01

(This is based on an idea by Ard Jonker, who planted a seed and watched the coin drop, if you excuse the mixed metaphors…)

The recent experiment with direct relay switching suggests that it is possible to switch a latching relay with just two I/O pins tied together plus a 100 µF cap.

The thing about that circuit, is that it draws no current in either ON or OFF state – it only needs a little energy to change the state.

That means it could in principle be powered by a very low-power source, such as this other fun projects I had to shelve a while back. The reasons for this remain as valid as ever: I can’t realistically turn this into a safe kit, given the direct connection to AC mains. So while the thought of having 80 of these sprinkled around the house and consuming under 1 Watt total is a tempting thought, it just isn’t going to happen.

Which doesn’t prevent ME from using it anyway, of course…

Ok, now let’s bring a couple of components together:

  • a JeeNode Micro
  • a 12 mW AC mains supply
  • a directly-powered latching relay
  • a toggle switch
  • power cabling

Here they are, with a nice plastic case (whoops, forgot to include the JNµ, oh well):

DSC_4433

The toggle switch is the small but essential ingredient here. Let me explain:

JC's Grid, page 70

This is a switch which can be operated manually and remotely. Flipping the switch or the relay has the same effect: toggling power, regardless of the state of the other component!

This means it can be operated even when the automated system is off or disconnected, or has crashed. And likewise, the power remains under remote control regardless of the state of the manual toggle switch. This solves a key problem with all those cheap remote power switches out there: the necessity to find the remote, because there is no local switch anymore. And the fact that it breaks down when the home automation system fails.

Given the relay used, I doubt that this solution will be able to control more than 30..50 W, but there are plenty of such devices around the house these days, even LED lighting.

I think I’m gonna have to start messing with AC mains again… with caution, of course.

See you in Houten!

In News on Apr 19, 2013 at 00:01

Psst, did you know? We’re getting together…

Here’s the location, right next to the railway station:

Theater-JeeDay

If you prefer to come by car, set your driving directions for Kamillehof in Houten to find a parking spot (or use the parking garage on the other side of the railway). Note that there is essentially only one way to get there:

map-kamillehof

Any attempt to drive differently will probably lead you to a bicycle path or one-way street…

JeeDay flier

Bring your pet projects (not the pets, please), bring your curiosity, and bring your energy. Let’s have some techie fun, eh?

(9)50 days and counting

In AVR, Hardware, Software on Apr 18, 2013 at 00:01

The new JeeNode Micro has joined the ranks of the test nodes running here to determine battery life in real time. The oldest one has been running over 2 and a half years now:

dsc_1822

That LiPo battery has a capacity of 1300 mAh, and since it’s still running, this implies that the average current draw must be under 1300/24/950 = 57 µA, including self-discharge.

The other two battery tests now running here are based on the new JeeNode Micro v3, i.e. this one – using a boost converter:

DSC_4401

… and another one using a coin cell. Here’s a picture of a slightly older version:

DSC_2858.jpg

The boost version is running off a rechargeable AA battery, of type Eneloop, which I use all over the house now. These batteries hold 1900 mAh, but there’s a substantial penalty for running off one AA cell with a boost converter:

  • energy is not free, i.e. drawing 10 µA at 3.3V means we will need to draw 30 µA at 1.1V, even if we could use a 100% efficient boost converter
  • real-world efficiency is more like 70..80% at these levels, so make that 40..45 µA
  • the boost converter has some idle current consumption, probably 10..20 µA, so this means we’ll draw 50..65 µA even if the JeeNode Micro only uses up 10 µA at 3.3V (actually, it’s 3.0V in the latest JNµ)

This would translate to 1900/.065/24/365, i.e. ≈ 3-year life. Or perhaps 2, if we account for the Eneloop’s 85% per year self-discharge.

The coin cell option runs off a CR2032 battery, which is rated at about 225 mAh, i.e. considerably less than the above options. Still, since there are no boost converter losses, this translates to 225/0.010/24/365, i.e. ≈ 2.5 years of life if the JNµ draws only 10 µA.

These figures look excellent, but keep in mind that 10 µA average power consumption is a very very optimistic value, particularly when there are sensors hooked up and packets are being sent out. I’d be quite happy with a 6-month battery life for a single AA or a coin cell.

For reference, here is an earlier post about all these power calculations.

Here are the current reports I’m getting via HouseMon:

Screen Shot 2013-04-17 at 11.12.08

That’s just about 50 days off a coin cell. Let’s see how it holds up. The nice bit in these tests is that the new nodes now report several different voltage level measurements as well (this also consumes some energy!). They haven’t dropped much yet, but when they do, I hope that we’ll be able to use the drop as a predictor.

Onwards!

Electro:camp 13.04

In News on Apr 17, 2013 at 00:01

As if one event is not enough – here’s a reminder for another one coming up soon:

elektro.camp.2013.04

This the next one in a series of bi-annual meetings, from people in Germany, Belgium, the Netherlands, the UK, and beyond – this time in Kaiserslautern @ the Fraunhofer institute.

See the wiki for further details. Don’t forget to register if you wish to attend.

As I see it, Electro:camp is more focused on electricity metering and monitoring, whereas JeeDay – which has yet to define itself, clearly – will be more of a “maker/hacker” style event with focus on DIY home projects, low-power wireless, and electronics as a hobby.

JeeDay next Fri+Sat

In News on Apr 16, 2013 at 00:01

(whoops, this went out one day earlier than planned… before it was ready)

Just as reminder, here’s the announcement (as you can see, I’m no graphics designer):

JeeDay flier

And this is the draft time schedule for the entire event:

JeeDay 2013-04

For more information, see the JeeDay wiki page.

PS – The is no pre-registration, everyone is welcome. The venue is large enough to accommodate some 25 people on Friday and about 35 on Saturday, which matches more or less the interest on the original list.