# Can a power MOSFET for switching application be used as a linear amplifier?

Power MOSFETs nowadays are ubiquitous and fairly cheap also at retail. In most datasheet I saw power MOSFETs are rated for switching, without mentioning any kind of linear applications.

I'd like to know whether these kinds of MOSFETs can be used also as linear amplifier (i.e. in their saturation region).

Please note that I know the basic principles on which MOSFETs work and their basic models (AC and DC), so I know that a "generic" MOSFET can be used both as a switch and as an amplifier (with "generic" I mean the sort of semi-ideal device one uses for didactic purposes).

Here I'm interested in actual possible caveats for practical devices which might be skipped over in basic EE university textbooks.

Of course I suspect that using such parts will be suboptimal (noisier? less gain? worse linearity?), since they are optimized for switching, but are there subtle problems that can arise by using them as linear amplifiers that can compromise simple amplifier circuits (at low frequency) from the start?

To give more context: as a teacher in a high school I'm tempted to use such cheap parts to design very simple didactic amplifier circuits (e.g. class A audio amps - a couple of watts max) which can be breadboarded (and possibly built on matrix PCB by the best students). Some parts I have (or I could have) available cheaply, for example, include BUK9535-55A and BS170, but I don't need specific advice for those two, just a general answer about possible problems wrt what I said before.

I just want to avoid some sort of "Hey! Didn't you know that switching power mos could do this and this thing when used as linear amps?!?" situation standing in front of a dead (fried, oscillating, latched,... or whatever) circuit!

• Getting good behavior is probably going to require using an op amp that takes feedback from a point past the transistor, but also includes some circuitry to prevent oscillation. A class A amplifier may pose some difficulties because even turning the transistor totally off won't make the output rise very fast, and a class B amplifier may pose some difficulties if one wants to avoid nasty shoot-through currents. It's possible to get good results using power MOSFETs as you describe, but trying to get stuff to actually work well may be "educational". Of course, if that's the point... – supercat May 7 '15 at 20:05
• @supercat I'm not aiming at HiFi level distortion. Just some simple circuit that can show that a MOSFET can actually amplify signal (the same way you could do with jellybean BJTs like BC337 or similar in a 4 resistor CE circuit, just to draw an analogy). The audio band is nice for students since they could plug the output of their iPOD or iWhatever to the input and hear the sound in a little speaker (it is cooler than to see it on a scope - yep with the average student it work like this!). Yes, I know I'm describing a very low-tech context. – Lorenzo Donati May 7 '15 at 20:17
• @supercat BTW thanks for the other points, just the sort of things I needed to know. Just a question: what do you mean with the term "shoot-through currents"? Do you mean the inrush currents needed to charge the gate capacitance? – Lorenzo Donati May 7 '15 at 20:19
• In a class B amplifier, one transistor will have the job of driving the output high, and another will have the job of driving it low. Shoot-through currents are those which pass through both transistors. – supercat May 7 '15 at 20:22
• @supercat Ah! Ok, thanks! Perfectly clear now! I didn't know the English term for that. – Lorenzo Donati May 7 '15 at 20:24

I had a similar question. From reading application notes and presentation slides by companies like International Rectifier, Zetex, IXYS :

• The trick is in the heat transfer. In the linear region, a MOSFET will be dissipating more heat. The MOSFETs made for linear region are designed to have better heat transfer.
• MOSFET for a linear region could live with higher gate capacitance

IXYS app note IXAN0068 (magazine article version)
Fairchild app note AN-4161

• (+1) Fantastic! Thanks! Just the info I needed! I suspected that also university books (at least the ones I read) didn't tell the whole story! – Lorenzo Donati May 7 '15 at 20:32
• I was going to post more or less this. The Fairchild app note is a good source. – gsills May 7 '15 at 21:18
• @gsills Really interesting material, indeed! – Lorenzo Donati May 7 '15 at 21:21

The Spirito Effect, which is a thermal instability caused by the fact that threshold voltage $V_{TH}$ has a negative temperature coefficient, is usually more of a problem in new MOSFETs.

At high overdrive voltages (overdrive $V_{OV}=V_{GS}-V_{TH}$), MOSFETs have no thermal instabilities because their channel resistance has a positive temperature coefficient. This causes good current sharing between devices. At low overdrives however, current sharing is poor because threshold voltage $V_{TH}$ has a negative tempco. Under the right circumstances, this leads to thermal instability.

New MOSFETs (generally optimized for switching, because that's where the market is) have much higher subthreshold currents -- in other words, at low overdrive voltages, they carry more current and dissipate more heat. Another way of saying this is: at currents that are practical for linear amplifiers, even despite running amps of current, newer MOSFETs need very little overdrive (a regime that exhibits thermal instability), unlike their ancestors which needed lots of overdrive (a regime with great thermal stability).

Thus, even if the newer MOSFETs were placed in the same packages with the same heat removal capacity, they would still have smaller SOAs (Safe Operating Areas). Further complicating the matter, as sort of a general rule, most transistors' datasheets don't have accurate SOA curves.

When using newer MOSFETs, design with wide margins (e.g., a MOSFET that sees 200V might be spec'd for 400V) and don't expect them to hold up to their datasheet SOA curves unless you test them to.

• Would you care providing some links or additional info about "subthreshold currents" and "spirito effect"? I never heard those terms. Whereas I can guess what the former refer to, I'm completely clueless about the latter. – Lorenzo Donati May 7 '15 at 21:19
• Yes, probably few will know what the Sprito Effect is, at least by name. But see the app note an4161 – gsills May 7 '15 at 21:28
• The Spirito Effect is explained here. The Fairchild app-note in Nick's answer mentions it as a "thermal instability limit." As for subthreshold currents, that's just another way of saying that the current at low overdrives (overdrive $V_{OV}=V_{GS}-V_{TH}$) the MOSFET carries a lot of current. This (combined with $V_{TH}$'s negative tempco) causes the thermal instability described by the Spirito Effect. – Zulu May 7 '15 at 21:32
• Ok, thanks for the explanations! I just skimmed over those documents linked by Nick. – Lorenzo Donati May 7 '15 at 22:12
• Extremly interesting reading the article you linked to in your comment about the spirito effect. This quote is remarkable (emphasis mine): JPL looked into this destruction, talked to the manufacturer, and discovered the auto industry had found the problem in 1997. JPL then reverted to “older parts,” and trusted the manufacturer to advertise the problem; however, this never occurred. Would you care editing your answer to include what you said in the comment? It would be a useful improvement. – Lorenzo Donati May 9 '15 at 12:06

Yes, you can use power MOSFETs intended for switching applications in their linear region, but this is not what I recommend for your purpose.

Stick to BJTs for demonstration amplifiers. The reason is that their bias requirements are more predictable in voltage, and it is therefore easier to create circuits to bias them usefully.

MOSFETs have significant part to part variation in the gate threshold voltage, which is the gate voltage at which a small dV causes the largest output change. With FETs intended for switching, it is desirable to minimize this transition region, but for linear operation you would like it to be spread out. Put another way, you want some "forgiveness" in the gate voltage. Switching FETs may give you less. The design for biasing such FETs in their linear region ends up being very pessimistic, usually with larger source resistors than you'd otherwise use, just to get some predictability.

It can be done, but the extra circuitry to set the bias point, probably with additional deliberate DC feedback, will detract from the other concepts of the amplifier design, unless of course that's what you want to teach. However, it sounds like any amplifier is already a stretch for the students, so adding this complication may make the whole thing impenetrable to them.

• (+1) Thanks for the useful insights! Unfortunately I'm not teaching any kind of EE design this year. It is just an "umbrella" course about electronics for future maintenance technicians in the thermotechnical field. I just aim at making them understand that some components exist, what are their main applications and why these applications are feasible using the littlest amount of math possible (Ohm's Law, KCL, KVL and empiric characteristic curves). After covering diodes, I went on to teach MOSFETs because they are a bit easier to explain to my audience. ... – Lorenzo Donati May 7 '15 at 21:09
• ... The lab part is not really about design but to aid familiarizing with the components and the measurements instruments. For those students it's not so important to understand the finer details, but rather to see in practice that all my waffling about load lines was not just hand-waving or BS. In other words, it's me who will design the circuits, they will only mount them and verify that they work as explained. – Lorenzo Donati May 7 '15 at 21:12

First, let's get the terminology straight. A switching transistor ideally is either always in cut off or saturation, whether it is bipolar or a FET. As a practical matter, transitions must pass through the linear region. FETs have an added complexity: the resistive region for small values of drain-source voltage. Moreover, the raw transfer characteristic of a FET is quadratic, not linear. When switched, a FET will quickly saturate, and if the external circuit is designed correctly, the drain-source voltage will equally quickly slide down to nominally one volt. At that point, it will be in the resistive region, but it will also, more importantly, be saturated. So, for example, if you are dumping 5 amps, the power dissipated in the FET will be about 5 watts.

You want to use the transistor in a circuit that is biased in the linear region. To be clear, this is all about the external circuit. A gain block is a gain block. It does not matter one whit whether it is a BJT, an FET, a MOSFET, or an op amp. The only thing you lose by using a switching transistor is manufacturer specifications for gain and phase shift with respect to frequency. For a switch, you don't care, so they make it easy for you by processing the data into a switching time parameter instead of frequency parameters.

If you were trying to manufacture amplifiers, you would care, but you are just demonstrating to a bunch of green kids, so you too do not care about frequency response. A switching transistor makes a perfectly good gain block, especially for your stated few watts of output - you can drive a small speaker with a common op amp for goodness sake!

You really don't need to worry about biasing: couple your input signal with a small capacitor. Your basic class A small signal amplifier with say a 30 volt rail would be:

1. A voltage divider setting bias, say 200K rail to gate and 100k gate to ground. This gives you a quiescent 10 volts at your gate node.
2. Couple the input to the gate node with a capacitor.
3. Place a resistor from source to ground - this controls your drain current bias. Use, say .5k to give a quiescent drain current of 20mA - easily endured by any power transistor.
4. Place a 100ohm resistor in series with your nominally 8ohm speaker coil - remember, a speaker responds to changes in current, not voltage - its coil creates a varying magnetic field in a bias field.
5. The transistor will pick up whatever power dissipation that is not carried by these other loads - at most 400 mW.
6. Your small signal transfer characteristic will be:

$$V_\text{drain} = 30-\frac{v*G*108}{500} = 30-\frac{v*G}{5}$$

where v is your peak to peak signal voltage, G is the transistor's transconductance, and the other values are the rail voltage and load resistances. If you want to get fancy, work in the inductance of the speaker coil and you'll see a circle instead of a load line on the I-V diagram.

Vary the external components at your pleasure. Simple, and no nonsense. Be sure to emphasize to your kids the irrelevant nature of the gain block. Specs only matter for production quality control, but for a one off hack, anything works.

• This doesn't really answer the question, although I appreciate the effort of providing useful informations. BTW, they are not kids, but teenagers learning to become technicians. As for terminology ("... let's get the terminology straight."), you got it wrong, sorry. See my reply to a comment to another answer here in this thread. Moreover compare the output characteristics of BJTs and MOSFETs. – Lorenzo Donati May 9 '15 at 11:36
• The etymology of the "saturation" term for BJTs and MOSFETs is not related to the shape and position of the output characteristics, but to the phenomena that happens inside the semiconductor. Thus, while a BJTs to be fully ON must be driven into saturation, for a MOSFET you must drive it into its ohmic region. The saturation region for a MOSFET is analogous to the active region of a BJT. – Lorenzo Donati May 9 '15 at 11:39
• "...the raw transfer characteristic of a FET is quadratic, not linear" This is true for ordinary FETs, not power MOSFETs, which are different technology. If you look at the datasheet links I provided in the question you'll notice that the transfer characteristic is fairly linear, after an initial knee. – Lorenzo Donati May 9 '15 at 11:45
• "... the drain-source voltage will equally quickly slide down to nominally one volt. At that point, it will be in the resistive region...". The Vds value which separates the ohmic (resistive) region from the saturation ("active") region is not fixed, it depends on the overdrive voltage, i.e. the difference between Vgs and threshold voltage. So it could be 1V, 4V, 0.2V or whatever (depending on Vgs level and the specific FET model). – Lorenzo Donati May 9 '15 at 11:51