Computing stuff tied to the physical world

Easy Electrons – Transistors #2

In Hardware on Jan 20, 2011 at 00:01

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:

Screen Shot 2011 01 19 at 23.45.31

(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!