Selecting proper MOSFET

[x2dz]

It's really hard to tel. I thought that heatsink was going to be MUCH larger.

The dimensions of this heatsink are 30X55X20mm... That's 33cm^3, right? So how much degC/W is that? I'm really mad that the manufacturer doesn't give that info for the product...

(also, measuring the voltage drop across the mosfet would be useful too. You can then determine the actual heat being dissipated and compare this to the results you got from the datasheet.

To get the voltage drop I would have to measure the D-S resistance, right? Wouldn't it change when I start putting current through the element - e.g. I'd have to measure the resistance "on-the-fly" as I do the test run. Meaning I'd have to attach a multimeter across the terminals with current going through them, which as far as I know is a no-no... How would you go about doing this measurement?

If I'm able to get my hands on an IR thermometer I'd get a much clearer picture of the situation, but after seeing the results I don't think its a must since the FET doesn't even get warm...

As for the efficiency of MOSFETs vs regular Bipolar transistors - I've taken a few shots of the PSU's radiator and the mounted transistors:

Comparing this sink to the one, fitted to the FET - I'd say... WOAH

I know that the Bipolar transistors actually do have more power through them, but the efficiency difference is undeniable and its most definitely the FET's ball game

Ironically, one of the assistants at the University I spoke to regarding my project told me I'd be better off using Bipolar transistors He really couldn't explain why but to me it was just a matter of him not having enough knowledge regarding FETs. To me its as clear as day that FETs are the way to go in this application - they're more cost-effective, easier to implement and more electrically efficient. Enough said

I was rather curious as to how much current I'd be able to push through the FET without it getting hot, but as I already said I won't be able to get much past the 4A marker with my electronic PSU. I do have another one - an old pre-transistor age supply which I'm currently using to supply the logical voltage...

This beast weighs about 15KG and is rated at 5Amps (although I'm sure it will be fine up to at least 7A)... Ah how I love vintage electronics - it really is something different when you flip the switch and hear that distinctive transformer brrr...

The only thing keeping me from switching the supplies around and trying to see how much current I can pump through the FET is that the transformer-based supply doesn't have current limiting (other than the fuses )... Meaning than I really can't use the coax cable for testing as I did up until now because of its ultra-low resistance.

I'll put this test on the back burner for now (and probably try it later on with a car battery or something as the source).

As per the heatsink/FET insulation - I'll be getting the fittings and the insulating material come Monday I'll then be able to use regular old steel bolts and nuts to tighten it since I wasn't able to find plastic ones anywhere... Go figure The only plastic ones I managed to find were for toilet fastening

What remains now is for me to complete the heatsink mounting and print out a PCB mock-up so that I can verify all the components' dimensions - whether everything lines up okay so that I can go ahead and start making the PCB...

Do you have any experience in this field - because I'm basically thinking of printing it out on a sheet of paper and putting the components on top to see if the legs line up... Any advice would be helpful

Thanks in advance!

 

[(*steve*)]

The dimensions of this heatsink are 30X55X20mm... That's 33cm^3, right? So how much degC/W is that? I'm really mad that the manufacturer doesn't give that info for the product...

The volume occupied by the heatsink is fairly meaningless. What counts is the material it is made of (copper is very good, aluminium is pretty good), the total surface area (so more fins mean more surface area), The ability for heat to get to those fins (which means long thin fins are poor because the heat will have escaped to the atmosphere before it gets to the end of the fins -- rendering them useless), the ability for air to get to the fins (more widely spaced fins will allow cooler air to get to them Very narrow spacings may mean the air will all heat up, rendering the fins useless), the surface coating (some are more emissive than others -- black surfaces are generality more efficient than silver ones), and probably more that I can't think of.

If the manufacturer doesn't tell you, you *can* calculate it after doing some tests. The figures you get will be specific to the way you're using the heatsink (orientation and placement of the device does matter).

To get the voltage drop I would have to measure the D-S resistance, right? Wouldn't it change when I start putting current through the element - e.g. I'd have to measure the resistance "on-the-fly" as I do the test run. Meaning I'd have to attach a multimeter across the terminals with current going through them, which as far as I know is a no-no... How would you go about doing this measurement?

Yes, you are measuring the resistance, but it changes as the current changes (it is non-linear). What you need to measure is the voltage across the D-S and, knowing the current (you're measuring it too) you can use ohms law to calculate the effective resistance. More easily, you can just multiply the current by the voltage dropped and get the power dissipated.

If I'm able to get my hands on an IR thermometer I'd get a much clearer picture of the situation, but after seeing the results I don't think its a must since the FET doesn't even get warm...

It will get a little warm. an IR thermometer my be the only way of getting an accurate indication. You can also then measure the temperature across the heatsink and you will probably find that the temperature varies so that it is warmaet near the device, and coolest near the ends of the outer fins -- all because the heatsink is doing its job.

If you had a copper heatsink, you would find that the temperature differential acorss the heatsink was less because copper is a better conductor of heat. You will find that some heatsinks (think of those used for processors) have parts made of copper to get the heat out across the rest of the heatsink more efficiently.

I know that the Bipolar transistors actually do have more power through them, but the efficiency difference is undeniable and its most definitely the FET's ball game

Well, they are dropping a higher voltage -- and that's as per the design because they are regulating the voltage coming from the transformer/rectifier/filter.

The point is, that out of all the power going to the motor, so very little actually is dissipated in the mosfet. Another device may get somewhat warmer -- it's actually a fairly poor comparison because we're comparing apples and oranges.

Ironically, one of the assistants at the University I spoke to regarding my project told me I'd be better off using Bipolar transistors He really couldn't explain why but to me it was just a matter of him not having enough knowledge regarding FETs. To me its as clear as day that FETs are the way to go in this application - they're more cost-effective, easier to implement and more electrically efficient. Enough said

Well, mosfets are easier to damage, and they have a D-S resistance, so the voltage across them rises with current. Bipolar transistors have a voltage drop across C-E which (for simplicity) stays relatively constant with current. So it is certainly possible that a bipolar transistor could dissipate less energy than a mosfet in switching the same load.

However, there are issues with driving a transistor into saturation, mostly these are that a large base current is required, and if you need more current carrying capacity you can't just connect bipolar transistors in parallel.

In contrast, although you may need a high gate current to quickly switch a mosfet, you just need to maintain a voltage to keep it turned on, and for higher current you can much more easily connect them in parallel.

The low gate current (almost zero) required to keep them on or off means that for low switching speeds, the driving circuit can be very simple.

I was rather curious as to how much current I'd be able to push through the FET without it getting hot, but as I already said I won't be able to get much past the 4A marker with my electronic PSU. I do have another one - an old pre-transistor age supply which I'm currently using to supply the logical voltage...

The datasheet, and some measurements should allow you to predict this without too much trouble.

If you don't want to kill the device, you could estimate the current what would achieve a tJ of about 15C lower than Tj(max) for your current test conditions.

As long as that is less than maximum drain current for the device, you could probably achieve it in practice.

I'll put this test on the back burner for now (and probably try it later on with a car battery or something as the source).

No trouble with current from there!

As per the heatsink/FET insulation - I'll be getting the fittings and the insulating material come Monday I'll then be able to use regular old steel bolts and nuts to tighten it since I wasn't able to find plastic ones anywhere... Go figure The only plastic ones I managed to find were for toilet fastening

You must ensure that the bolts are suited to the mounting material. metal bolts can be used, but you generally have an insulating washer which protrudes into the hole so that the bolt remains free of any metal, be it the heatsink or the tab of the device.

Another option is a device with a plastic package or an insulated tab. These are already electrically isolated and mean that no special mounting is required.

What remains now is for me to complete the heatsink mounting and print out a PCB mock-up so that I can verify all the components' dimensions - whether everything lines up okay so that I can go ahead and start making the PCB...

yep

Do you have any experience in this field - because I'm basically thinking of printing it out on a sheet of paper and putting the components on top to see if the legs line up... Any advice would be helpful

Most components will line up is the spacing is 1/10 inch between pins (2.54mm).

You can draw it freehand if you wish, then transfer this by hand (using a special pen) to a printed circuit board.

Or you can get some software to help you design it.

Another alternative is veroboard. This removes the issue of having to drill holes and would be quite suitable for a project of this complexity (i.e. not very complex). I believe there are also programs out there for assisting with veroboard (also called strip board) designs.

The most important thing to remember is that the pattern you draw will be on the opposite side from the components, so you either need to imagine the devices upside-down, or draw the design as a mirror image.

 

[x2dz]

Thanks for your response!

I calculated the volume, because I'd found a chart which roughly showed cm^3 vs deg C/W :

By my accounts 33cm^3 is about 7-6 deg C/W - as you predicted!

Just for fun, I did the calculations to see the max Amps I could pump through the FET as it is now with the current heatsink. I got some interesting numbers.

- I assume 10 degC/W for T(j-a) for the heatsink
- 1.5 thermal resistance for washers heatsink mounting and other losses
- 150 maximum junction temp
- 50 ambient temp

This gives the following equation:

100/X - 1.5 = 10

Where X is Power(Watts).

When solved this gives - 8.7Watts of power for maximum conditions with a 10 degC/W heatsink.

We got to this equation to calculate the Amps of current:
- R ds(on) = 0.1 conservative value

x^2*0.1 = 8.7
x^2=87
x= 9.33Amps of current

About what I thought I guessed something about 10Amps With this being a conservative value I guess you could get 2-3 Amps above the 10Amp mark without overheating the component...

p.s. 13.27Amps to be precise (175 max operating, 25 ambient, 7 degC/W heatsink)

Yes, you are measuring the resistance, but it changes as the current changes (it is non-linear). What you need to measure is the voltage across the D-S and, knowing the current (you're measuring it too) you can use ohms law to calculate the effective resistance. More easily, you can just multiply the current by the voltage dropped and get the power dissipated.

I'll do that test later today to see the actual Watts and whether they're close to the calculated ones

However, there are issues with driving a transistor into saturation, mostly these are that a large base current is required, and if you need more current carrying capacity you can't just connect bipolar transistors in parallel.

Yeah, even more so when you're interfacing with a mC... With a 50mA max rating for the outputs I don't think I'd be able to get away with using any old Bipolar transistor - I'd probably have to go with a Darlington and I think it would be more expensive than the FET I'm using now. And being able to parallel together the FET is really a nice trait for me because the loads controlled can be enormous.

You must ensure that the bolts are suited to the mounting material. metal bolts can be used, but you generally have an insulating washer which protrudes into the hole so that the bolt remains free of any metal, be it the heatsink or the tab of the device.

If I were a little sloppier I'd probably bolt it on and not even worry about what it (the heatsink, respectively the D channel of the FET) could come into contact with... But for me that's just bad design and even if it were relatively safe I wouldn't do it. I have half-assed solutions.

Another option is a device with a plastic package or an insulated tab. These are already electrically isolated and mean that no special mounting is required.

Oh, how I wish my little FET's tab was insulated... That'd have saved me so much trouble in finding the insulation materials... But alas it isn't insulated. By the way do the manufacturers point that out in the sheets since I think I haven't seen it written before in a datasheet?

You can draw it freehand if you wish, then transfer this by hand (using a special pen) to a printed circuit board.

Or you can get some software to help you design it.

Thanks for the advice! However, I already have the PCB designed - I just have to see whether all the component dimensions I've used while designing are on mark.

Thanks again for your help!