Yesterday was about MOSFETs and the heat they generate. Pretty impressive, those components which can control a 15A current with just a little bit of voltage.
Unfortunately, it’s not quite as simple as that. We have to check what happens with a lowly 3.3V applied to the gate. This can be found in the IRLZ34N datasheet:
This is an important graph for MOSFETs. Each line corresponds to a different voltage applied to the gate – this graph is in fact eight graphs in one.
You can see that they really do have some limits at lower voltages. At 3.3V, the IRLZ34N will not go much higher than 7A – quite a bit lower than the 30A maximum specs on the front page of the datasheet. Most MOSFETs are still at the end of their range when driven by a 3.3V microcontroller.
What this graph also shows, is the drain-to-source voltage at different current levels. This makes it very easy to estimate power consumption: at 3V and roughly 5.5A, that voltage will be 1V. In other words: 5.5W – quite manageable if the MOSFET is mounted on a suitable heat sink. But again: quite a bit lower current handling capacity than the 15A from yesterday. Note that we could drive it at a higher voltage by adding an extra stage in front, using either a BJT transistor or a MOSFET to get better current handling capacity.
As you can see, there is a very distinct switch-over point at each gate voltage level, where the MOSFET stops conducting more current. Under the switch-over point MOSFETs act essentially like a pure (low-value) resitor, but above that point they start acting more like a current limiter. This has major implications for power consumption, since the drain-to-source voltage will rise.
With this particular IRLZ34N, driving it at 3.3V, it is best not to push it beyond about 5..6A, and to use a heat sink when power consumption rises above say 1W (i.e. at 62°C/W, that makes the bare MOSFET rise 62° above ambient). Looking at the 3V line in the graph, I’d estimate that this MOSFET would work just fine without heat sink up to at least 2A, perhaps even 3A.
But what if we want to regulate the current? I.e. what if we want to use the MOSFET in an analog manner, and not just as on-off switch?
We could lower the gate voltage somehow, to force the current down. It won’t be a linear relationship, and I’m not even sure a MOSFET will conduct any current at all when the gate voltage is lower than 1 or 2V.
But in general, it’s a bad idea. The reason is (again!) power consumption. Say we use the following circuit:
(the resistor is a pull-down, explained in a future post)
That’s a 12W incandescent light bulb, powered by 12V, and controlled by a MOSFET, in other words: it draws 1A when full on. Let’s say we want to dim the light by halving the current through it. The lamp acts like a resistor (it just get pushed so hot that it starts glowing, that’s all). So half the current is what you get when you apply half the voltage to it. In this case 6V.
Suppose we figured out exactly what gate voltage to apply to get a drain-to-source voltage over the MOSFET of 6V. That would essentially accomplish our goal: the lamp would be dimmed. Another way to look at this, is that we’re tweaking the gate voltage until the MOSFET acts like a 12 Ω resistor between drain and source. Then the lamp and the MOSFET both get half the voltage.
What about power? Well, the MOSFET will have to have 6V across it, as we just saw. At 0.5A, this means that it will have to burn 6 (V) x 0.5 (A) = 3 Watts of power.
This is totally counter-intuitive: when we switch a lamp full on, the MOSFET will consume less than 0.1 W (as gleaned from the graph), but to dim it, that same MOSFET would need a heat sink!
There is some logic in this, though, if you think about it. The 12V supply voltage is a given. So if we want to apply less voltage to the lamp, we have no other choice but to waste the excess energy. Which is exactly what the MOSFET (or BJT transistor for that matter) will do. So although we’re reducing the total power consumption in this circuit by halving the current, we’re forced to do so by wasting power – as non-visible light, i.e. heat.
Can we do better? Yes, fortunately, we can.
This is where “pulse-width modulation” (PWM) comes in. Instead of eating up the excess power, we can take advantage of the fact that incandescent lights are fairly sluggish in their response. What we do is pulse the power on and of in very rapid succession. So the lamp will constantly heat up and cool down, and the result is that it won’t be burning at full brightness.
Why is PWM so incredibly useful? Several reasons:
- we don’t need a way to generate a regulated analog gate voltage, we can simply generate digital on-off pulses with no extra circuitry needed
- the MOSFET is again being used as pure on-off switch, and remains maximally efficient – so we probably won’t need a heat sink
- power is no longer wasted, it is now effectively throttled instead – in very short and rapid bursts
It turns out that PWM works in a lot more cases than just incandescent light bulbs. DC motors are also sluggish, so controlling their speed with PWM also works extremely well. And better still, even LEDs work well with PWM, even though they respond instantly – because it turns out that our own vision is sluggish too!
See also an earlier post about PWM on this daily weblog.
So there you go. MOSFETs are the workhorses of power control, due to their incredible properties, and PWM is the technique of choice when it comes to throttling power devices (lights, heaters, motors, and more).
nice explanation of some basics.
about lamps: keep in mind the “cold” (non glowing) resistance usually differs by a factor of 10 from the “hot” (normal on-state) resistance, here assumed to be 12 Ohm.
PWM will give you anything between say less then 1 and the 12 ohm (average) and a momentary current flowing can be as high as 10A In this case it will not do any harm though.
pls keep up the good work :)
kees
“I’m not even sure a MOSFET will conduct any current at all when the gate voltage is lower than 1 or 2V”
If Vgs < Vth the MOSFET won’t conduct any current.
In the case of the IRLZ34N the min Vth is 1V then if Vgs < 1V the MOSFET won’t conduct any current at all
You actually see a lot of examples of PWM on Top Gear. Most of the flash cars they test have LED rear lights, and guess how they control the different brightness of the red lamps between normal night running (5watt filament equiv output) and brake lights (21watt filament output), yes, you got it, PWM… We might not be able to see the flicker, but a TV camera can. So now as you watch the latest Ferrari zooming off down the straight you know why the rear lights appear to be flickering and doing odd things.
I don’t need Top Gear for that. I DO see the flicker in LED rear lights all the time, with my own eyes, and it’s quite annoying really. I don’t remember if I saw a Ferrari, but lot of Volkswagens and Volvos and Audis, with these damn disco strobo rear lights.
I wish the manufacturers used higher frequency with that PWM.
Here’s a small FET which can do over 20A even at 3.3V gate drive. On Semi NTMFS4897NF. (Not so easy to hand-solder in that package, though.) http://www.onsemi.com/pub_link/Collateral/NTMFS4897NF-D.PDF
Amazing specs!
In 3rd paragraph you wrote “What this graph also shows, is the drain-to-gate voltage at different current levels.” It must be obvious, but I can’t see how you read this information from the graph :(
I’ve been recently interested in MOSFETs because I want to control the speed of a slot-track car with PWM. I bought a FQP70N10 because it had a low Rds(on) and I hoped that below 1A I wouldn’t need a heatsink (the power source I currently use for running the car indicates that about 600mA are used at most). I however found out later that this Rds(on) value can be reached only when Vgs is 10V. Is there a way to estimate the Rds(on) (and thus the power disspated by the FET) when Vgs is only 5V? I can of course use a BJT to drive the FET, but the circuit would be simpler without it.
Moreover, after a bit more reading, I found out that the Arduino does’t like to be directly connected to a FET, because the FET, acting as a kind of capacitor, can have a very low impedance just when it’s switched on, and you therefore need to put a protection resistor between the output of the microcontroller and the gate to limit the current drawn from the microcontroller. Your circuit however doesn’t appear to have such a resistor. Is this resistor really needed in practice? At least the datasheet mentions such a resistor in the wiring diagrams.
And finally, I also read that the FET dissipates heat when switching (meaning used for PWM, I guess), depending on the switching frequency. Could you tell us how to estimate how much heat the FET will produce in this case, and whether it’s negligible?
Lowest line is the behavior with 2.5V on the gate. Next line up for 3.0V. Then 4.0V, 6.0V, etc.
As for your other Q’s. Yes, a series resistor will limit the current, but also switch (slightly) more slowly, and that will increase power losses in the MOSFET, especially when switching at high frequencies. At the 50..300 Hz I had in mind, I don’t expect that to make much of a difference. But I haven’t verified this.
What I’d do: design a good setup using just these basic rules, and then build it and measure it to see whether it behaves more or less as expected, especially when pushed to the limits it needs to go to (w.r.t. voltage, current, PWM rate, and such).
Please keep in mind that I’m only scratching the surface with this “Easy Electrons” series…
I suppose you meant
source-to-gate voltage'' (i.e., the Vgs indicated by the curves) instead of
drain-to-gate voltage” ? Or has the drain-to-gate voltage any particular signification in a FET?Each curve is for a fixed gate-to-source voltage. The plot axes are currents / voltages of drain-to-source.
I fixed a mistake in the post (it; said drain-to-gate i.s.o. drain-to-source). Sorry for all the confusion.