A third installement about transistors in this Easy Electrons series.

So far, I’ve shown how to get more current out of an I/O pin from an ATmega, since this will probably be the most common reason to use transistors in combination with a micro-controller. But these circuits all act as switches, i.e. they turn current on and off (or in the case of the voltage regulator: adjusting current flow to a certain value).

What if we wanted to control one or two DC motors for a little robot? Lots of fun stuff to do in that area, especially with wireless communication. To do this, we also need to be able to reverse the voltage placed on the motor, so we can make it turn forward or backward under software control. And if we want to make it a bit fancier, it would be nice if we could control the speed of the motor as well.

First things first. Reversing the direction of a motor can be done with a double-pole double-throw (DPDT) relay:

This low-tech solution will switch the +12V and the -12V poles to make the motor run clockwise or counter-clockwise. And if we were to use a transistor for the -12V (i.e. GND) side, we could also turn it on and off.

But that’s clunky! – let’s see if we can do differently. What we need is a way to place either a high or a low voltage on either side of the motor. Here’s a first (flawed!) attempt:

Look what happens when we put the proper voltages on A, B, C, and D:

• with A high and B low, the left side of the motor is tied to “+”
• with D low and C high, the right side of the motor is tied to “-“
• it will start running

And now the other case:

• with A low, B high, the left side of the motor is tied to “-“
• with D high, C low, the right side of the motor is tied to “+”
• it will start running in the opposite direction

And of course, when A = B = C = D = low, the motor will stop.

What the two transistors “on top” of each other do, is create sort of a push-pull circuit, since you can tie the central connection to either the “+” or the “-” voltage rail. This type of circuit is called an H bridge, due to it’s shape.

(note that I’ve left out 4 protection diodes, i.e. one across each C-E junction – they do need to be added in a real-world setup with DC motors)

There are several serious problems with this particular design, though:

• to pull A or D high, we have to apply 12V, since 3.3V won’t be high enough to raise the base 0.7V above the emitter voltage level
• if we pull A and B high, then we’ve got ourselves a short-circuit, with huge currents through both transistors on the left!
• same for C and D…
• and lastly, this thing needs a whopping 4 I/O pins

Let’s tackle that last point first: we can halve the I/O pin count by tying A and C together, and by tying B and D together. Now three out of the possible combinations will get us just what we want: stop, turn clockwise, turn counter-clockwise. But with both signals high, we still get a short circuit. Not good – we don’t want a software error to be able to start a fire…

The bigger problem though, electrically speaking, is that the input voltages involved are no longer suitable for an ATmega. This can be solved by adding an extra NPN transistor on both sides, for a total of 6 transistors. Instead of explaining the whole setup in detail, let me point you to some articles I found on the web:

• this one describes the basic idea using relays
• this page uses 6 transistors (lots more interesting pages on that site)

As you can see, it takes quite a few components to drive one small motor. Fortunately there are lots of H-bridge driver IC’s with various voltage- and current ratings. Some of these are quite small – such as the TC4424A I used on the DC motor plug, which is why I was able to actually put 2 of them on a single plug.

The second task we’d like to be able to do is control the motor speed.

This turns out to be fairly easy. The trick is to use pulse-width modulation (PWM). This is just a fancy term for a simple concept: we generate a set of pulses, and we control the on-time vs. off-time ratio of these pulses. As it turns out, DC motors are far too slow to follow these pulse trains if you generate them at 100 Hz or more. Instead, they will tend to average out the 0/1 values sent to them. And sure enough, a pulse train which is 100% off will cause the motor to stop, and a pulse train which is 100% on will cause the motor to run at full speed. Everything in between will lead to a motor running at intermediate speeds – simple!

For completeness’ sake, let me mention that the on-off power control circuits I’ve been describing in these last posts often use MOSFETs nowadays, instead of the traditional BJT transistors. For simple experiments and small DC motors, BJT’s are fine though.

Now if you think transistors are so great… wait ’till you see what MOSFETs can do!

I’ll go into those next week. Enough electronics for now.

Today, a second example of a transistor circuit. As announced yesterday, this one is about boosting the output current further…

With a transistor which amplifies current say 100 times, there are still limits to what you can control from an ATmega I/O pin: 10 mA x 100 = 1A. What if you want to drive a DC motor with a stall current well above that?

Quite simple, really: use two transistors “in series”. But how?

Note that this is not the way to do it:

On first sight, it looks exactly like what we need: after all, when the inout voltage rises above 2x 0.7V, both transistors will start to conduct, right? And since each of them can amplify their input current by 100, that means we could control a current which is a whopping 10,000x as large as the input current! So with the proper high-power transistors, this ought to work, right?

There is a sneaky problem, though. Keep in mind that the base voltage is always capped to at most 0.7V above the emitter, since the B-E junction is like a diode. So when the first transistor starts conducting, it’ll drive the second base up. Up to 0.7V, in fact. And that’s where things go wrong: the first transistor will drive 100x its base current into the second base. With 10 mA, that might well end up being around 1A. But most transistors can’t handle that: they expect small base currents to drive large collector currents. Another way to explain the problem, is that the C-E junction of the first transistor will always be 12 – 0.7 = 11.3V, so with a large current, it’s bound to overheat.

What we need, is a way to limit the current into the second transistor. Here are two ways to do it:

Both should work. The former is like a double amplification stage, whereas the latter uses the voltage follower approach described yesterday. I don’t really know which one would be better, because I’ve never tried either one.

That’s because there is also another circuit called the Darlington transistor, named after the person who invented it, Sydney Darlington:

The difference is subtle but important: the collector of the first transistor is no longer tied to the positive power supply rail, but to the collector, i.e. the lower side of the load.

The practical advantage of this combination, is that it’s still a 3-pin device, so you can use it wherever a transistor is being used, and that’s indeed it’s main raison d’être. Roughly speaking, a Darlington transistor acts like a transistor with a much higher current gain than a single transistor.

Its main drawback is that the saturation voltage, i.e. the voltage over the (combined) C-E junction, is higher than with a regular transistor – i.e. more like 1.1V than 0.4V. This is the voltage you get when driving the transistor all out – and it affects the amount of power absorbed by the transistor (P = E * I). So although a Darlington can switch higher currents from an I/O pin, it also generates roughly three times as much heat.

The reason I’m going into Darlingtons, is that they are very convenient and widely used. There is a 18-pin chip with 8 Darlingtons, tied into a tiny package, the ULN2803:

This is what I used on the Output Plug, by the way. Not only does each pin drive 500 mA at up to 50V, you can actually tie them together to control larger currents. Each Darlington on this chip includes the necessary resistors, so it’s simply a matter of tying an I/O pin to its input, and it will “sink” up to 500 mA to ground.

An extra benefit of the ULN2803 is that it has a built-in reverse kickback protection diode, as needed when hooking up relays and motors. So that’s 8 more components saved, all by using this single chip.

Don’t get you’re hopes up too high, though. When driven from a 3.3V input, the output current of the ULN2803 might not go much higher than 300..350 mA on each pin. The components on the chip seem to have been designed for 5V. But it does work fine at 3.3V!

To summarize: the Darlington transistor pair is an easy way to get just a little bit more current (or more accurately: a higher gain) than from a single transistor.

So much for switching current on and off. Tomorrow, as last part of this mini-series, I’ll describe how to reverse the current as well, so we can also control the direction of a DC motor. And how we can control their speed. All with just a bunch of transistors.

With the basics and some real-world details out of the way, it’s time to start looking at some other circuits with transistors.

I’m going to describe three different types of circuits, to give you a feel for transistors – while noting that I’m just scratching the surface w.r.t. these amazing devices:

• regulated power supply
• more current with chained transistors
• push-pull drivers

There’s way too much to say about this. Let’s get’s going!

Regulated power supply

In the post about diodes, I described a circuit based on a zener diode. One problem with that circuit was that it only worked for a few milliamps of current, the other was that this current was being drawn even under no load.

With a transistor, we can fix both problems at once:

Well, we’re not changing the fixed current draw really. But we can use a higher resistor value to draw less current, since only a little bit will will be needed to feed the transistor base.

So what’s going on here? Well, remember that the B-E junction is essentially a diode, connected in the forward direction. So from base to emitter, we have a 0.7 volt drop. Since the zener keeps the voltage fixed, the output voltage will be 0.7V below the zener breakdown voltage: i.e. 4 – 0.7 = 3.3V, tada!

The neat thing is that this is basically all there is to it!

The output voltage doesn’t really depend on the load current. With an amplification factor of 100, with a very light 1 mA load, the transistor will need to draw 10 µA on its base to generate that amount of current. It may seem like it does this as if by magic, “seeking” the proper settings all by itself. Let me try to describe what happens anyway. What we’re doing is keeping the base voltage at a fixed level. Here’s what happens when things start up, i.e. when the emitter is still at zero volts, right after turning on the power:

• the base will then be at 0.7V, because it’s always 0.7V above the emitter level
• this is much lower than the zener voltage
• so all the current through R1 goes into the base
• the transistor will amplify this and start conducting
• that will “pull” the emitter voltage up, since the collector is tied to the input voltage
• this in turn means that the load will be supplied with power
• assuming the load is not a short circuit, the voltage on the emitter will rise
• so will the base voltage – by 0.7V more, as always
• at some point, the base reaches the zener voltage
• that’s when current will start going into the zener
• meaning: less current into the transistor base
• where does it end? simple: when the emitter is precisely at 0.7V under the zener voltage
• this adjustment process takes place continuously, so when the load current changes, the transistor will keep on adjusting its current to keep the base at the zener voltage
• similarly, this will keep the emitter voltage at a fixed level even when the input voltage changes, since the voltage at the base remains the same

Voilá, a regulated power supply with a constant voltage output, a.k.a. a linear voltage regulator.

This voltage regulator will work fine for fairly large currents. After all, if we have a 10 mA current through R1, then a transistor with hFE 100 will be able to amplify this up to 1A.

But there is a catch: power consumption – again!

Note that the collector is held at the input voltage, and let’s assume this is fixed at 5V. And the emitter is held fixed too, at 3.3V in this example. So we have 1.7V on the C-E junction, with all the load current going through it. With a 1A current draw, that’s 1.7 watt of power. Not huge, but already well beyond what a little transistor can handle. This is why such circuits need to use power transistors and large heat sinks.

It gets worse with higher input voltage. Suppose we connect our setup to 9V instead of 5V. Now the C-E junction will have 5.7V over it, while still drawing 1A = that’s 5.7W of power, i.e. generated heat! So keep in mind, no matter how fancy your regulated supply is: if it works according to this “linear” principle, then it’ll turn all excess voltage into heat. See also an earlier post about this topic.

Our circuit also gives us a new way of looking at a transistor: while regulating, as shown above, it acts like a variable resistor, with the resistance being controlled by the amount of current flowing into the base.

So there you have it, the basic properties of a linear voltage regulator. I don’t want to stray too far from microcontrollers in this Easy Electrons series, but this really is a good example of how unusual transistors can be, compared to resistors and capacitors. Besides, voltage regulation is so common that it’s really useful to understand how it works and to be able to reason a bit about power consumption.

Tomorrow, we’ll go into “Darlingtons”! :)

Yesterday’s post started off a little mini-series about transistors. Today, I’ll go a bit more into the practical side of things. And in case you’re curious: tomorrow I’ll describe some interesting circuits with’em.

Let’s look at that NPN transistor symbol again:

First thing I want to point out is that there’s essentially a diode between the base and the emitter. The most important property of that, is that when there is current flowing from base to emitter, then there will be an ≈ 0.7V voltage drop across this “B-E junction”.

So yesterday’s calculation was a bit inaccurate. When putting 3.3V on the IN pin, current will flow through the resistor into the base, and from there to the emitter and to ground. Since the base is 0.7V above the emitter, and the emitter is tied to ground, there will be 3.3 – 0.7 = 2.6V over the resistor. Which means that the current through the base will be 2.6 mA, not 3.3 ma as reported yesterday.

With 0V on the IN pin, there is no current, so the voltage between base and emitter is irrelevant (0V in fact).

There was another inaccuracy in yesterday’s post (thx Reinhard), in that the current into the base also comes out the emitter. So with 2.6 mA into the base, and 3.3V current going into the collector, the total current coming out of the emitter is 2.6 + 33 = 35.6 mA (not 33 mA).

We’re not out of the woods yet. Transistors are not perfect conductors. Even when driven all out, there is a small residual voltage from collector to emitter. Usually it’s around 0.4V – this means we can now make much a more accurate calculation:

• with 3.3V on IN, and 0.7V on the base, there is 2.6 mA into the base
• with a current amplification factor (hFE) of 100, the collector-to-emittter current might reach 260 mA
• however, the R2 resistor will limit this
• with one side of R2 at 3.3V and the other at 0.4V, current through R2 is (3.3 – 0.4) / 100 = 29 mA
• currents will be: 2.6 mA into the base, 29 mA into the collector, and 2.6 + 29 = 31.6 mA out of the emitter

Now we’re ready for a calculation which is crucial: the amount of power dissipated in the transistor!

It’s really very simple (P = I * E): the base current corresponds to 2.6 mA x 0.7 V = 1.82 mW, and the collector current corresponds to 29 mA x 0.4 V = 11.6 mW, for a total of ≈ 13.4 mW.

How do we know this? Well, if there is a current flowing and a voltage drop, then that’s by definition a form of power consumption. And it can’t turn into anything but heat, really!

Fortunately, 13.4 mW is very little heat, so even the simplest small-signal transistor will be able to handle it. But with larger currents involved, this type of calculation can tell you whether the circuit will: 1) work, or 2) overheat and fail.

Here’s an example to illustrate the point of such calculations – suppose you want to turn a motor on or off, controlled from an ATmega I/O pin:

(I’ll explain the “kickback” protection diode some other time – it just blocks, mostly)

We know that an ATmega can’t supply more than a few milliamps. Let’s assume it can output 10 mA, and that when doing so the output pin voltage drops to 2.5V – what value do we need for the R1 resistor? And what type of transistor would we need?

Clearly it depends on the motor. Let’s assume it’s a fairly small motor driven from a 12V power supply. What we need to make sure is that the circuit won’t break down. The worst case is when the motor “stalls” and then starts drawing a large current – that’s when something prevents it from turning while power is being applied. Don’t be surprised to see stall currents of several amps, even for small motors which only draw a few hundred milliamps under light load conditions!

Ok, now let’s figure out how much of a motor current we could handle with the circuit shown above:

• to get max current output, we’ll feed as much current into the base as we can, i.e. 10 mA
• so we’re feeding 2.5V, and the base is 0.7V, meaning that R1 needs to be (2.5V – 0.7V) / 10 mA = 180 Ω
• let’s assume we get a transistor with an amplification factor of 100
• IOW, it can only produce 100 x 10 mA = 1 A current

If we have a motor with a stall current larger than 1 amp, then something nasty will happen: the voltage between collector and emitter will rise above the 0.4V saturation voltage that you get when driving a transistor all out.

So what’s the problem? Heat!

We’ve got 12V on the one side of the motor. Say it has a 2A stall current. The transistor only amplifies its base current to let 1A through. What happens, is that the collector voltage will rise – my hunch is that it will reach about 8V (4V over the motor and 8V over the transistor, matching the 2:1 current ratio).

Now we’re in big trouble. The transistor is conducting 1A with a 8V voltage drop. That’s eight watt of power consumption, i.e. heat. If you take a simple little transistor which isn’t designed for that sort of heat dissipation, then it’ll heat up and go up in smoke. Fireworks!

Note how the lack of output current is what caused this. If the transistor had been able to deliver those 2A, then it’s power consumption would be “only” 2A x 0.4V = 0.8 watt – one tenth of what happened here.

As you might expect, there are several solutions around this. I won’t go into them here, though – I just wanted to go through these calculations to show you that you can’t drive everything directly from an ATmega I/O pin.

Tomorrow, some real circuits!

Another installment of the Easy Electrons series. This time I’m going to follow up on the diode post, and go into transistors.

Transistors. Phew. Biiig topic!

There are several types of transistors. The most common one is the BJT. There are two variants, which are schematically drawn like this (from that same Wikipedia page):

A transistor has three pins, called Base, Collector, and Emitter. Using the convention that current “flows” from “+” to “-“, it’s easy to remember which in an NPN transistor (the most common variant):

• the collector, eh, “collects” current – IOW, this is where current flows into the transistor
• the emitter is where the current flows out again, to ground at some point
• the base is the main control which determines what the transistor does

So what does it do?

One way to look at it, is as a current amplifier: put 1 mA into the base, and the current from C to E will be a multiple of that – anywhere between 50 and 500 times depending on the particular model, i.e. 50 .. 500 mA!

Note that I’m not suggesting that a transistor generates current. Only that when a voltage is applied between C and E, then the current flowing will be a multiple of the “control current” fed into the base.

Things are not quite that simple, though. How do you “put a current” into a component? You can’t just put a voltage on all the pins… before you know it, the current might easily exceed some allowed maximum rating! Transistors have pretty rigid limitations. When you exceed them, they can easily overheat and get damaged.

We don’t really have an easy way to control current, but what we can do is turn ATmega I/O pins on and off, i.e. control the output voltage on those pins. It’s relatively easy to turn that into current though – Ohm’s law: use some resistors!

Here’s a simple circuit to let us explore this behavior:

The input pin can be fed a voltage, and the R1 resistor will make sure that the current flowing into the base of the transistor will be limited. Likewise, R2 makes sure that no matter how much current the transistor wants to pass from its collector to its emitter, there will be a limit.

How much are these limits? Depends on the voltages and resistor values, of course. Let’s assume the following:

• the “+” voltage is 3.3V (and GND is 0V, as always)
• the input pin can have voltages between 0 and 3.3V applied to it
• resistor R1 is 1000 Ω
• resistor R2 is 100 Ω

The maximum current flowing into the base will never be more than when IN = 3.3V, i.e. 3.3V over 1000 Ω = 3.3 mA (using “I = E / R”).

The maximum current flowing from collector to emitter is what you’d get if the transistor were a complete short circuit between C and E, i.e. 3.3V over 100 Ω = 33 mA.

Note how I’ve made all sorts of worst-case assumptions about the transistor. In fact, I’ve calculated these values as if B, C, and E were all shorted together, and hence tied to ground (since E is tied to ground).

A real transistor behaves somewhat differently. But no matter how it actually behaves, I know that no more than 3.3 mA will be flowing into the base, and not more than 33 mA will be flowing into the collector and out of the emitter. Just about any NPN transistor can handle that.

Here’s what happens when IN is 0V (i.e. GND):

• no current will flow into the base
• since the transistor amplifies the current, no current will flow from C to E
• if no current flows from C to E, then it’s basically like an open (unconnected) circuit
• if unconnected, there will be no current flowing through R2
• with Ohm’s law, the voltage across R2 is (E = I x R) = 0 x 100 = 0
• so the voltage on the OUT pin will be 3.3V, the same as on the other side of R2

Now, when IN is 3.3V, this is what happens:

• current will flow through R1
• lots of current will flow from C to E
• that’s the same as saying that C and E are more or less shorted
• that means the OUT pin will now be (close to) 0V

We’ve created a signal inverter: 0V in = 3.3V out and 3.3V in = 0V out!

As I said, big topic. I haven’t even touched on some important details of a “real” transistor.

Stay tuned…

It’s time for some Easy Electrons again.

Semiconductors are at the heart of today’s electronics designs: diodes, transistors, and the huge advances made possible by Integrated Circuits (i.e. IC’s) is what makes all those electronic devices around us possible. And it all happened in the time scale of just a few decades…

I can’t possibly cover everything in this series, so I’ll cover the two most common ones for now: the diode today, and the “normal” (BJT) transistor in the next installment.

Very roughly speaking, a diode is a one-way conductor. This is also indicated by the schematic symbol used for diodes (image from Wikipedia):

The arrow-like symbol points in the direction of the current flow, if you stick with the convention that current flows from “+” to “—”. If you keep forgetting which side is called the Anode and which the Cathode, this trick might help: in the alphabet, we (eh, I mean currents) go from A (to B) to C.

Diodes are useful to protect a circuit against connecting a power source the wrong way around (we’ve all been there, right?):

With this diode, the circuit is protected. Hooked up the wrong way, the diode will block, so no current will flow through the circuit.

But diodes aren’t perfect. There’s a voltage drop when conducting current, usually around 0.7V. So with the above circuit, if the power supply is 5.0V, then the circuit will only get about 4.3V. With low-voltage components, and especially battery-powered devices, this sort of voltage loss is awkward – and often unacceptable.

Which is why you won’t see this reverse voltage protection very often in circuits operating at 5V, 3.3V, or less. There is another type of protection, however:

This one is a bit nasty. It doesn’t really prevent the circuit from getting a reversed voltage at all. Instead, it will act (almost) like a short circuit when the voltage is applied the wrong way around. The idea being that this will cause the power supply to shut down (or the battery to drain very quickly). The RBBB uses this type of protection to overcome the voltage drop problem. The (cheap) diode will protect the (much more valuable) ATmega, as well as all other components hooked up to it.

This sort of protection is tricky. If you were to connect a LiPo battery, for example, then the short circuit can cause HUGE currents to flow, since many LiPo’s are able to supply them. Think many Amps… and now something else may happen: even if the diode can handle the current, the rest of the power lines might well become overloaded. Especially thin copper traces on a PCB are likely to act like a fuse and simply… evaporate!

There are other ways to deal with the voltage drop and still end up with diode protection. One of them is to minimize the voltage drop – this is where Schottky diodes can be useful. They usually have only half the voltage drop of normal diodes, i.e. around 0.3V. That might just be low enough for your particular setup.

Another option is to build an “ideal diode”. This might sound like an impossible task, given the properties of diodes, but there is actually a way to do this using a MOSFET. I won’t go into MOSFETs here, but basically they can switch current while having almost no resistance and (Ohm’s law!) therefore also almost no voltage drop. Trouble is: MOSFETs don’t know which way the current is flowing, so you need considerable extra circuitry to tell them when to turn on and off, based on comparing voltages on both ends. And although it is not a simple or cheap solution, this datasheet of the LTC4413 chip shows that it is indeed possible to beat the diode characteristics with some clever engineering. Electronics is often like that: people have come up with the most amazing tricks to overcome certain drawbacks, for all sorts of electronic circuit tasks. That’s why it’s so much fun just exploring and discovering it all, IMO :)

The graph of what a diode does is very characteristic: in reverse mode it blocks, and in forward mode (i.e. above around 0.7V) it conducts, albeit not perfectly. For some good example graphs see this page on Wikipedia (just skip the math formulas and look at the pretty pictures).

Now, assuming the voltage drop is no problem, because you’ve got some extra volts from the power supply anyway, then diodes can be extremely useful. The bridge rectifier for example, can be used to get a properly polarized voltage out, regardless of how the power supply is hooked up. This is particularly useful with alternating current, as present on the AC mains lines and on the coils of a transformer (a lot more Easy Electrons articles will be needed to present all this stuff!).

Another interesting diode is the Zener diode. It’s like a regular diode, but one which can’t support a very high reverse voltage. With Zeners, this voltage is called the “breakdown voltage”, and it ranges from about 2..200V. The value is fixed for any particular model.

Zener diodes make very simple (low-current) regulated power supplies:

Note how we’re putting the Zener in reverse mode, and counting on it to break down. As it does, current will start to flow. Until enough current is flowing across the resistor (Ohm’s law!) to take up all the “remaining” voltage.

So with a resistor of 100 Ω, and a Zener of 5.0V, we could power it with say 6..9V. At 6V, the current would be (6 – 5) / 100 = 10 mA. At 9V, the current would stabilize at (9 – 5) / 100 = 40 mA.

The reason this can be used as a regulated power supply, is that we can connect our circuit in parallel with the Zener, and it would always get 5V. The only drawback is that we can’t draw more than 10 mA from it:

• at 6V, the resistor needs to “eat” 1V, so that the Zener ends up with 5V
• if the circuit draws 10 mA, then 0 mA will go through the Zener
• if the circuit draws 5 mA, then 5 mA will go through the Zener
• if the circuit draws 0 mA, then 10 mA will go through the Zener
• at 9V, the current will increase to 40 mA (to get 4V over the resistor)
• in all cases, the circuit will see a 5V input voltage

Cool, so now we have built ourselves a simple regulated power supply!

As I mentioned, this circuit is not very powerful. If we draw more than 10 mA, then the voltage drop over the resistor may increase, leaving less than 5V for our circuit.

There is another drawback with the above regulated supply: it’s grossly inefficient. The reason is that it will always draw 10 mA, whether our circuit needs it or not. And that’s at 6V – at 9V it will always draw 40 mA!

I’ll show you how a transistor can easily increase the current and improve the regulating efficiency in a future installment. Exploring these simple electronic circuits can be great fun, and most of the time you can reason your way through without even having to build them!

Next time: transistors – incredibly useful devices, with tons of ways to use ’em!

PS. Does anyone have tips on how to improve these diagrams? I really want to continue drawing them by hand, but the texts don’t come out very nice, no matter what I try!

Yesterday’s post tried to create a new data point for measuring the forward drop voltage over a red LED at a slightly higher current. To summarize:

• the voltage over the LED was read out as 2.229 V
• the LED current was calculated to be 10.71 mA

But not so fast. What about component tolerances? What if the 100 Ω resistor isn’t exactly 100 Ω? And what if the supply voltage isn’t exactly 3.3 V?

It turns out that there’s quite a bit more going on here. Time to get the multimeter out and measure these values. If nothing else, this will be useful as cross-check against everything we did so far.

Here’s what I found (expected vs. actually measured):

• 3.3V power supply = 3.298 V
• “1” on VIN = 3.036 V (!)
• “0” on VLOW = 0.213 V (!)
• voltage over LED = 2.011 V
• 100 Ω resistance = 99.7 Ω
• current out of VIN = 7.85 mA
• current into VLOW = 7.72 mA

Whoa – very different outcomes!

First off, this is a Voltcraft VC940 multimeter with 5-digit readout and a specified accuracy on DC voltage and current ranges (substantially) below 1%. For up to 4V, that means it shouldn’t be off by more than 20..30 mV, worst case.

First the good news: the power supply voltage on the 3.3V rail is very close to the desired voltage (3298 mV).

But check out those “1” and “0” digital I/O voltages: they are 3036 and 213 mV, respectively, i.e. around 10% off!

There’s a very simple explanation for that, and as it turns out, it’s in fact fully according to specs. Here is the output voltage you get when powering an ATmega at 3V, while drawing some current from one I/O pin:

As you can see, there’s just about a 0.3V drop at 25°C when drawing 10 mA. These I/O pins were not meant to be used as power supplies for other circuits, let alone as reference for analog circuitry.

Another way to put it, is that an I/O pin on an ATmega has an internal resistance of around 30 Ω. It’s easy to see that when we start drawing several milliamps through a 100 Ω resistor, then those “perfect” I/O signal lines are in fact a lot less like a real switched power supply.

The same holds for the “0” pin voltage levels. Once you start dumping current into such an I/O line, it will start to look more and more like a resistor tied to ground, and not just a direct connection to ground.

So the first conclusion is that the LED isn’t being fed as much current as we had expected, because the voltage levels are not quite 3.3V and 0V.

Let’s recaclulate our results with this new insight:

• VIN = 3.036 V
• VHIGH = 2.229 V (same as before, via the ADC)
• VLOW = 0.213 V
• voltage over the LED = 2.229 – 0.213 = 2.016 V
• 99.7 Ω resistor voltage = 3.036 – 2.229 = 0.807 V
• current through resistor = 0.807 / 99.7 = 8.1 mA

Ah, that’s a lot better: 2.016 V calculated vs. 2.011 V measured, and 8.1 mA calculated vs. 7.85 mA measured (about 3% off, oh well).

One last puzzle: why is the current measured out of VIN not the same as the current into VLOW?

You’ll need to check the schematic again to see what’s going on. Since the VLOW voltage is not 0V as would be the case with a perfect I/O output, there is now also some current flowing from VLOW to GND, through that other 1 kΩ resistor.

How much? Well, Ohm’s law points to the answer, as always: 0.213 V over 1 kΩ means there is 0.21 mA flowing through that resistor. Since the only source of power is VIN, and since all current must flow either into VLOW or GND, this means we can add that 0.21 mA to 7.72 mA to get 7.93 mA total.

Ok, it’s not conclusive in that some unexplained differences still remain, but this is nevertheless a lot closer to the real values in the circuit than the original 2.2V @ 11 mA we arrived at earlier.

Note that we don’t really need a multimeter to be able to make these corrections. I could have tied VIN to another analog input, to be able to use the digital I/O as (imperfect) power supply for the circuit, while at the same time measuring the voltage drop due to the higher current. I’ll leave that as an exercise for the reader.

There’s a lot more to be said about this sort of design and measurement work. Electronics is a very large field, and all we did so far, was hook up 4 simple components!

One very interesting option, is to simulate the circuit instead of building it on a breadboard. There is a well-known application for that called SPICE which takes a circuit description and computes, time step by time step, how the circuit behaves when subjected to certain input signals and power conditions.

JGJ Veken (Joop on the forum & wiki) was very kind to help out with a simulation of the LED discharge circuit I’ve been describing in these weblog posts. Here’s what came out:

The voltage levels are slightly different, because the LED charateristics are different, but you can clearly see more or less the same graph as shown in yesterday’s post. Just to drive tht point home: yesterday’s results were measured in a real circuit, whereas the graph above is a simulation performed 100% inside a computer.

To find out more about this exciting way to try out circuits and explore different parameter settings, see these excellent wiki pages by JGJ Veken: 5Spice and LTSpiceIV. These are two different systems (both with common ancestry), the former commercial but free for personal use, and the latter available from Linear Technology as freeware (I don’t think either of them is open source).

So there you have it. This concludes my mini-series about experimenting with simple components, and using JeeNodes and Arduino’s to investigate the circuit and create a simple test-jig setups for it. Combine that with the SPICE simulation options available out there, and you’ve got a very simple and low-cost way to exlore electronics for yourself. Transistors? Filters? Op-amps? Go for it – there’s a world out there to explore and tinker with!

Following earlier posts about the LED discharge circuit (here, here, and here), let’s do a few more measurements. Here is the schematic again:

I’m going to apply a 1-second charge pulse and then disconnect for 5 seconds, as before. This time, I’m going to disable the pull-up during those 5 seconds, and improve the timing of the reporting so the time to send out characters over the serial port does not affect the whole measurement cycle:

The reason the timing is so much better in this version, is that it is now based on MilliTimer, which continues to count while the serial port deals with its outgoing data. Note that 20 chars @ 57600 baud takes a few milliseconds, which can quickly add up when reporting values 10x per second.

The values are sent out in a format which can be used directly in a spreadsheet – here are the first few lines of output:

The values are:

• voltage measured over the capacitor, i.e. LED + 1 kΩ resistor (Volts)
• voltage measured over the 1 kΩ resistor (Volts)
• difference, i.e. voltage measured over the LED (Volts)

Here’s a graph of the resulting measurements:

Due to Ohm’s law (E = I x R, i.e. I = E / R), that yellow line is also the amount of current going through the resistor in milliamps. Since there is no other path for the current than through the LED, by deduction this is also the current through the LED.

What you can see (blue curve), is that the capacitor charges up within a fraction of a second, and then discharges to under 2V in the next 5 seconds.

What you can also see (yellow curve) is that the current through the LED is proportional to the voltage on the capacitor. As the capacitor charges up, the current increases, and from the shape of the curves is should be clear that the relationship is essentially linear.

This is not surprising, since linear voltage-vs-current curves are inherent to resistive loads, i.e. the 1 kΩ resistor in this case. What is special is the LED placed in series with that resistor. What it does is “eat up” a fixed voltage, i.e. 1.8..1.9 V of the voltage present on the capacitor “stays” in the LED. The rest is passed on to the resistor, which does its usual linear current-to-heat conversion thing (only milliwats of heat in this case – we’re dealing with very small currents and very low voltages here).

But LEDs are not perfect devices. If they were acting like a perfect diode, they would have a fixed voltage drop, regardless of how much current goes through them. You can see from the green line that there is a slight change in voltage drop.

Does that mean a LED has something like a “hidden resistor” inside it? Not really. Here’s a plot over that same data, as current versus voltage:

That’s not very linear. Remember that resistors just do what Ohm’s law tells them to (or was it the other way around?): twice as much voltage = twice as much current, etc. Which in a graph would show up as a straight line.

In this case, the graph appears to be tapering off. In other words, as more and more current goes through the LED, its forward voltage seems to stabilize around a certain value. Probably around 1.9V, but this will be different for each type of LED.

Note that I placed the current on the horizontal axis, because that’s the variable we can control. The LED “decides” what voltage drop it wants to add.

What seems to be happening in this circuit, is that between 1.75 and 1.80 V there is a small amount of leakage. This is consistent with the video, showing that the LED continues to light up very dimly as the charge leaks away, even though the current is less than 0.25 mA, i.e. 250 µA, at these voltage levels!

Ok. Can we find out more about this forward voltage behavior? Sure. Let’s put 3.3V on VIN and 0V on VLOW, and then measure the voltage on VHIGH:

(Isn’t it fun to do this stuff under computer control?)

Sample output:

Whoa, looks like I need to revise my conclusions a bit. The LED forward voltage does go up quite a bit when more current is applied!

Let’s try to figure out how much current is flowing:

• the cap is fully charged after a few seconds, so we can ignore it
• therefore the cap voltage will be 2.229 V, as measured by the ATmega
• and the voltage over the 100 Ω resistor will be 3.3 – 2.229 = 1.071 V
• the resistor is 100 Ω so with “I = E / R”, we get 1.071 / 100 = 10.7 mA

IOW, at roughly 11 mA, the foward voltage drop of this particular LED is about 2.23V, quite a bit more than the 1.87 V we measured at 1 mA. But still not anywhere near a linear relationship.

Tomorrow, I’ll examine these results a bit more closely … because they aren’t quite right!

Welcome to another installment of the Easy Electrons series.

There have been a few posts about What is “power”?, but that was about the electrical concepts of power and energy, and how it relates to voltage and current.

In this post, I’d like to go into a different topic: how to control more powerful electrical devices. As you’ve seen in the LED discharge circuit post, the I/O pins of an ATmega can be used as weak power supply, sort of. This is because an output pin set to “1” can be used as a 3.3V (5V on Arduino) power source, as long as you don’t draw more than a few milliamperes of current.

Here are the specifications of what a single pin can do, taken from the ATmega datasheet:

A few milliamps is not much. Barely enough to light a LED, and far too little to drive even the simplest electric motor. Let alone a light, a heater, or any AC mains appliance.

The first thing to note, is that there are a lot of different ways to control power – and each approach has fairly rigid limitations. You could use a power transistor to drive a bright 12V halogen lamp, but that same transistor would fail (spectacularly and dangerously so) if you tried to use it to control even the smallest AC mains lightbulb. Controlling power is a tricky business. As voltages and currents increase, so do the risks and potential effects of failure.

Let’s start simple, with an LED strip. These strips often need 12V to light up, and use about 300..900 mA per meter. There are several ways to control an LED strip of up to a few meters – using a relay, transistor, or MOSFET, respectively:

The relay is is the oldest type of power switch, and easiest to understand: an electromagnet closes a switch while powered up. It has been around for centuries. And although I included it as option, in practice even the smallest relays draw more current than what an ATmega I/O pin can supply (apart from very low power reed switches).

The transistor was invented only a few decades ago, and is at the heart of every electronic circuit these days. It’s usually depicted as follows in schematic diagrams (image from Wikipedia):

There’s a lot more to say about this fascinating invention. For now, let me just say that you can feed it a small current and it’ll control a much larger current (and voltage) on another pin.

And then there’s the MOSFET, which is even more recent than the standard bipolar-junction transistor. MOSFETs are pretty incredible: feed it a few microamps, and it can switch a huge amount of current (dozens of amps!). Perfect for direct hook-up to a microcontroller.

So how do we control an LED strip?

Easy – with a MOSFET, for example:

Note that the LED needs 12V at a fairly high current, whereas something like a JeeNode tolerates 12V – and needs only a few milliamps. In this case, we can get away with connecting both power supply inputs to the same 12V. The built-in voltage regulator will immediately reduce that 12V input to the 3.3V used in the rest of the circuit. So the voltage on the MOSFET “gate” will only be 0V or 3.3V – fortunately, that’s enough for many common MOSFETs these days.

It’s important to note that all these voltages and currents are fairly independent. The following would work just as well, using two separate power supplies:

I’ve drawn some extra arrows to indicate how the different current paths work. The crucial point is that current always needs to go through a closed loop. If the current “came from” the power supply on the left, then it will also always “return” to that same left-side supply. Think of it as “tagged electricity” if you like: what goes out of a battery on one en must come back on the other end of that same battery. It doesn’t really matter whether you think of electricity as “flowing” from + to –, or electrons flowing from – to +.

Note the connection between “–” on the left and “–” on the right. Without it nothing will work. The reason for this, is that voltages are always relative to some other part of the circuit. You can’t have a voltage (or a current, for that matter) over a single wire. You always need a return path. So in the case of the MOSFET, to drive it and put a voltage on it, you need to connect two wires – even if the current is small, as in this case. That’s why “–” is usually called ground, i.e. the reference against which we measure everything else.

Does this matter? You bet. If the power circuit uses a lot of current, then all its conductors must be able to carry that current: it needs thick wires! – but at the same time, the rest of the circuit can still get by with thin copper traces, as usual on boards like the JeeNode. This means that the ground connections (“–”) may need to be thick on one side, but not on the other – even though all ground connections have to be electrically tied together!

Current flow and proper design of all the wiring and connections is essential in power circuits. It’s a lot more involved than with purely digital circuits, or circuits which carry just milliamps.

How about controlling a lightbulb connected to 220V (or 110V)?

One way is with a relay (designed for such high voltages). But a relay can’t be driven directly from an output pin – so we add a transistor in between:

The extra resistor and diode are required (for reasons I won’t go into right now).

Note that there is no connection between this circuit and the switch shown at the top which is used to control the lightbulb. That relay does something which is crucial in the case of AC mains: it adds an isolation barrier between the low-voltage (safe!) side of the circuit and the AC mains (potentially lethal!) side of the circuit.

The second common way to control such a lightbulb is a Solid state relay, which uses a built-in opto-isolator to provide the isolation barrier. The advantage of an SSR is that it has no moving parts – hence no mechanical wear.

I’ve only touched briefly on how you can hook up stuff which needs more power than a feeble ATmega output pin can provide. Follow the links above if you want to find out more.

And marvel at the amount of knowledge and understanding Wikipedia brings us, freely available to anyone with an internet connection, anywhere in the world!