mosfet? transistor? hexfet?

Generally, you want the lowest on resistance and the fastest switching times. So, look at the spec sheets and pick the best part within your price range.

It's more complicated than that, but it is a start. Maximum ratings and drive requirements are also critical.

Another part of the answer is to use what you know better about and can design. Getting something working reasonably well is better, even if it dissipates more, than having something not dissipate as much but also not work quite right.

Jon

First off, I should say that I design and use discrete BJTs into circuits, but don't use Mosfets that much. Partly, because I just find BJTs more interesting and I enjoy learning and partly because I don't design anything for others to use. Plus, I suppose, I get BJTs in groups that work out to about 1-3 cents each, so I don't mind freely giving away some to interested students without worrying whether or not they will actually ever get around to using them. I'm decidedly NOT a designer of motor controls or anything else. Just an occasional hobbyist. That said....

Mosfet's source-drain connection act resistive (which is why their on-resistance is commonly specified) and BJTs don't act quite that way, which is why several of their saturation Vce's are given in a table, often along with a chart. (There is _some_ resistance there, though. So it is a mixture of effects.)

Neither STMicro nor OnSemi seems to provide a Vce vs Ib graph for various Ic on the TIP36C, like I usually find for the 2N3055 or MJ2955, for example. But the TIP36C table shows 1.8V at 15A for Vcesat (and 4.0V at 25A.) At 5A, it will be an even better experience than what 15A will show, but at 15A this works out to 1.8V/15A or 0.12 ohms. Probably, you will get a little better than the extrapolated

0.6V Vcesat, though, at 5A. You will also be driving current into the base, though, and this will also dissipate. The 2N3055/MJ2955, by comparison, shows slightly more than 0.5V at 5A, so it seems in the same ballpark. I'd estimate .1 ohms equivalent. Then take into account the drive currents working against the Vbe, which may be around 1.5V itself at say .5A, so not insignificant. Maybe 2.5W + 0.75W, for 3.25W total. Just rough numbers for comparison.

The IRLZ34N shows an on resistance of 0.06 ohms at Id=16A with a Vgs of 4V. And it will be a little better than that at 5A. So it will probably dissipate 1.5W at 5A. I don't think the rough 1nF input capacitance will cause you trouble in switching it fast enough for your needs and it probably won't dissipate that much during that transition time. So 1.5W is probably good.

From that, you can probably see that the hexfet looks better in terms of dissipation. About twice as good. Driving either case is probably not too hard, but the hexfet drive will probably be a little cheaper for which to buy parts (supplying .5A will use somewhat larger parts than supplying say 1mA for a short time to the mosfet requires.)

There are other considerations. Runaway thermal problems are possible with BJTs, but I seem to recall some discussion about similar effects with IRF's hexfets, in particular, on sed. I don't know about it, though.

Why not try both and see what you think about it? That's probably what I'd try doing in a case like this.

Jon

Thank you for the very informative reply.

I *think* I've figured out the difference between MOSFETs and bipolar transistors: the gates of MOSFETs are opened by application of a

*voltage*; the base of a bipolar transistor is activated by application of *current*. (Right? Wrong?)

So here's a spec sheet of a 2N3055 (what a beauty):

So, is the 5 mA for I_ebo the minimum current needed to "activate" the base (fully open the "gate" from collector to emitter)?

So, would applying 1 mA to 5 mA activate the base linearly (for example, 1 mA at the base = 20% collector-emitter, 2 mA base = 40% collector-emitter?)

I guess I need help figuring out how to read these spec sheets.

Thanks,

Michael

This page explains the relationships between Ice, Vbe and beta for BJTs and Id and Vgs for FETS:

No, not fully. The parameter you really want to look at is the hFE (or "beta") which is the DC current gain. From the data sheet you can only count on switching 20x more current than you supply for turn-on. So if you have a 5A motor you'd need to supply 250mA or more to the base to ensure the transitor was saturated. And you need to do it *fast* to avoid having the transistor live in the linear region where it is dissipating the power (and heat!).

The FET presents a different problem: The gate appears to you as a capacitance and you have to drive Vgs up to the saturation point quickly (which will be a higher voltage than the BJT). Once you *get* there, though, you don't have to supply any more current. The bigger the device, the higher the Ciss. Then, to turn it off, you have to suck that charge back out.

--
Ben Jackson AD7GD

http://www.ben.com/

Also, don't forget Cds which works against your efforts here. It may appear small but there is a multiplier effect. The higher the drain voltage when open the greater the detrimental effect on gate turn-on.

When looking at Vgs turn-on levels it is prudent not to pay too much attention to what the front page says. That's often marketing spin. What matters are the guaranteed values in the tables. There needs to be a specified maximum Rdson at a given Vgs. Often you'll find 2-3 values there. Anything else can't really be relied upon IMHO.

--
Regards, Joerg

http://www.analogconsultants.com

Kind of. Actually, the purist will point to the basic Shockley diode equation (it's a part of the Eber-Moll model for BJTs) and point out that a voltage determines BJT collector current, as well. But the voltage determines collector current in a very non-linear way (small increases leading to large Ic changes, collector load permitting) and to achieve that voltage, you have to drive current through a diode sufficiently to create that voltage, and this base-emitter voltage itself increases only slightly to large increases in that base current, Ib. Together, these two non-linear behaviors conspire together to be nearly close enough to each other that it will roughly appear that the Ic will vary nearly linearly with Ib. Hence, the idea of 'beta' arises and from that, the idea that the BJT's collector current is a function of base current (for some range of various values.)

In the case of switching, rather than amplification, your interest is mostly in just driving the Vce to a low value. As low as is possible, consistent without having to go to ridiculous extremes ramming in base current. Here, the usual beta figure isn't all that meaningful, so a rule of thumb is to estimate the base current as 1/10th of the collector current to drive Vce "suffiently down" that you've reached diminishing returns by then. My experience for smaller BJTs is that about 1/30th is where diminishing returns are met. But with 5A devices, like you are talking about, I'd definitely go with the 1/10th figure.

No curves to speak of. I think you'll find this better:

Take a look at Figure 6. To read it, just look at the curve for "4.0 A". This is your collector current curve. Now, note the y-axis is your Vce you can expect to get. Lower is better, here. Then look at the x-axis. This is your base current. You can see it knee-over around 200mA (1/20th of collector current) at around Vce=0.55V and that if you push this to 400mA (1/10th of the collector current) you get about 0.50V. As you can see, cramming in more base current pushes the Vce down a little bit (50mV), but that you have to double the base current to do it. I think that curve will help you think about this.

By the way, take note of the fact that the minimum guaranteed beta for this device is 20, maximum is 70. For switching purposes you need to be using an estimate that is substantially less than the minimum figure. So your base current should be well more than 1/20th of the collector current. At least 1/10th and if you are being really conservative about getting the worst of a batch of these parts, probably even 1/5th (which isn't so beautiful.) Look at Figure 4, though, and look up from the Ic=5A line. You can see that we are talking about a typical beta of 30, or so. As a hobbyist, I'd go with

1/10th (beta=10 for switching) for an estimate, though, and risk it to see how it works out.

No, not at all. I frankly don't recall ever thinking about that parameter, though. Perhaps someone here can explain it more fully. I could guess, but I'd rather here from someone about this. But keep in mind that this is a maximum specification (not a minimum, as you seem to suggest) and it isn't the parameter to look at for turning on the BJT. Basically, you need to go from about 0mA to about 500mA into the base. During part of this time, the BJT will be in its linear region and the load (motor) will experience a rising current that is kind of proportional to your ramping-up base drive current (consistent with what the effective motor inductance and voltage rail permits.) While this is taking place, your Vce will be much larger and the BJT will dissipate more for part of this short time.

No. In saturation, the Ic isn't a linear function of Ib. Only out of saturation might that be roughly true. Look closely at that Figure 6. Imagine you have a supply voltage of 10V and we can treat (I don't want to get into unloaded and loaded motor behaviors, which I can probably not do well at explaining) your motor as resistive and exhibits 2.5 ohms. This gets is to about the 4A curve in Figure 6. So let's say you are driving 400mA into the base. You will see Vce=0.5V. This means 9.5V across your motor. That works out to a little less than 4A (9.5V/2.5ohms). Now, if you take the curve literally and you look at reducing the base drive by half, to 200mA, then your Vce=0.55V. This means that your resistance (motor) experiences 9.45V. The current is still about the same, now 9.45A/2.5ohms. Not much change in current. So the 4A curve probably still applies, so far.

Cutting down to 150mA leads to a Vce=0.6V. Again, perhaps not so much change. But enough that you can imagine that the Ic=4A curve may be beggining to not apply so well, anymore. As you drop to Ib=100mA, you can see that the 4A curve goes almost straight up. This tells you that the 4A curve isn't applying anymore, at all. You are getting into the more linear region where the beta=30 will start to dominate (30*100mA is 3A, which you might use as a rough guess here), but keep in mind that as Ic continues to decline the beta rises, too. So it's not perfectly linear here, either. For example, in the case of 100mA for your Ib, you might guess that beta=30 because you've gone into the linear (non-saturated) region, because beta=27 at 25C at Ic=5A (Figure

4.) This would suggest Ic=3A. But if you look at Figure 4 closely, at Ic=3A, you see that beta=40. So that would suggest, at Ib=100mA, that Ic=4A. But at Ic=4A on Figure 4, beta=30, not 40, which again suggests Ic=3A. The actual Ic will be something between 3A and 4A, trusting these curves, with beta that is between 30 and 40. Where exactly, you won't really know without testing. At Ib=50mA, you might guess a beta=40 and compute Ic=2A, but at Ic=2A the beta is listed at more than 50. So this suggests a little higher than Ic=2A at Ib=50mA. So you can see it is 'kind of' linear with Ib, but not strictly so as we get into lower currents.

What you can consider as your minimum base current is roughly what value it is, at which your motor won't run at all even with you turning it slightly with your fingers, divided by about 100 or maybe even 150. If you get below that, the motor won't have much drive current.

Of course, keep in mind I'm not experienced in these things.

I hope that helps some.

Jon

... hear ...

Oh, well.

Jon

...

Actually, by "what a beauty" I'd meant the pic of the TO-3. ;-)

Ok. I found Figure 6, and I was a little puzzled by it. Let's say I have a light bulb (or, for argument's sake, simply a fuse!) in series between +12V and the collector, and the emitter is grounded.

Won't V_ce be 12V? Or is this a different V_ce?

Ok, so the name of the game is to minimize V_ce (as much as is practical; the knee-point is probably good enough.)

(snip lots of great stuff that will require more time for me to study.)

Thank you,

Michael

When would one use a transistor? mosfet? hexfet?

** Forget bi-polar devices completely for this app - they have high heat loses and take high power to drive.

Go with MOSFETS, the lower the on resistance the better.

You can simply parallel two or more if needed to reduce heat or increase current.

Test and find the "stall current" of the motor - it is WAAAAAY higher than the normal running current. This IS the current that will flow if power is suddenly applied or the motor is jammed and cannot turn.

Your drive circuit must be able to withstand the "stall current" of the motor or it will soon fail.

Switching frequency can be from 200HZ to 5kHz - what is best it depends on the motor and what you are doing with it.

Don't forget to put a fast diode across the motor.

....... Phil

I was kind of thinking that.

That depends. If you are using a dead short to +12, then probably the dead short will win out and then, yes, the Vce will be near 12V, for a short time, until it blows up. However, that isn't the usual case.

Your BJT has to be big enough to handle the current needed to allow its Vce to be low when turned on in the way I was discussing earlier.

One way to explain what I'm saying is this: You have some kind of load attached between the collector and some supply rail. Mentally assume for a moment that this load has some fixed resistance to it that we can use for talking purposes. That figure can be 100 ohms or

1000 ohms or 1 ohm or .1 ohm. Doesn't matter. Just pick a number and stick with it for a moment. Then select a voltage for your supply rail. 12V is fine. Or pick something else. We'll assume that the emitter is connected to the ground reference, for talking purposes. Whatever you pick for the supply voltage and the load resistance, you can now compute a current from them, I=Vsupply/Rload, that tells you the maximum current possible if the BJT were a perfect switch and the Vce across it was exactly 0V. If your transistor isn't able to stand this current, then you need a bigger one. If your transistor _can_ stand that level of current, then when you drive enough base current into it to equal about 1/10th or 1/5th of that level, you can be fairly sure that the transistor has a Vce that is small -- say .4V, or .6V, or .8V, or even .03V. But with that much base current flooding through, it will be able to pull that collector down pretty far.

All this assumes that the BJT can stand it.

So now, think about a fuse. Call it 0.01 ohms. If you had a 12V rail, then the maximum current would be 12/.01=1200A. Now, assuming you had a fuse that was rated for 1500A, let's say, and a BJT that was rated for a collector current of 2000A, and assuming you supplied some

1/5th of that, or 400A of base current, and assuming you had a power supply that could meet these requirements and more and that everything could easily dissipate 12*1200A = 14,400 watts without going up in smoke (this, I need to see), then your BJT might just be able to pull its collector down, less than 1V.

Of course, none of that is very likely except that the fuse might be .01 ohms and your voltage rail could be 12V. If you supplied a MJ2955 with a base current of 1A (a not unreasonable possibility), then you could expect that the MJ2955's collector would be pulled down below 1V or so if the load resistance were such that if 10A flowed through it then the voltage drop across it would similarly allow the collector to go below 1V. But if you were using a fuse, the fuse would probably just blow, since it's load resistance probably isn't an ohm or more.

So suppose you had a 10 ohm resistor as the collector load, here? Then if the collector current was 1.15A, which it could easily be if you were sinking in 1A of base current, then the load resistance would be dropping 11.5V, which means the collector would be .5V. The actual case here, looking over Figure 6 on the Ic=1A curve (close to 1.15A) and trying to follow it all the way over to where Ib=1A, would probably be more like a Vce of .1V. So actually, you'd get a collector current of (12-.1)/10 = 1.19A with a base current of 1A. And yes, the Vce would be very low. But if you knew your load was 10 ohms and you could estimate that the current should be about 1.1A to 1.2A, then you can also see that a base current of 15mA-20mA would suffice to get the Vce low enough for your needs. Just to be sure, you could push it to 50mA. As you can see, this is about 1/50th to 1/20th of the collector current. Which is why 1/10th is an extremely safe bet in most circumstances.

Yes and no. I hope the above explanation clears this up.

When you are thinking of the transistor as a switch, then you want the voltage dropped by the switch itself to be as low as possible so that it mimics a switch as well as it can. Another reason for getting the Vce low is that then the transistor itself reduces its dissipation (so long as you aren't going nuts in over-driving the base.) The total dissipation in the BJT switch is roughly Vce*Ic+Vbe*Ib. That is what you often want to minimize in your switch. In saturation (BJT used as a switch), Vce