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:

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:

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):

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.

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

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

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:

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!

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:

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:

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?

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.

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!

Remember this screen shot?

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:

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.