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Interfacing high current leads with a PCB

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Hi,

I now have my 12V thermosat schematic ready for board design. While thinking about the PCB I got to thinking about the track width. I expect to be switching about 16A through a Mosfet. How does one interface the high current carrying leads with the PCB. I was hoping to use PTR Connectors AK500/2 and /3.

What confuses me is the following;
- theortically I need about 6,5mm², (from wire tables) this translates to a track width of ~ 162mm?
- Why if I need 6.5mm² are the legs on the Mosfet barely 1mm².

Evidently I am missing something here, could someone point me in the right direction?

Cheers
Andrew
 

ericgibbs

Well-Known Member
Most Helpful Member
Hi,

I now have my 12V thermosat schematic ready for board design. While thinking about the PCB I got to thinking about the track width. I expect to be switching about 16A through a Mosfet. How does one interface the high current carrying leads with the PCB. I was hoping to use PTR Connectors AK500/2 and /3.

What confuses me is the following;
- theortically I need about 6,5mm², (from wire tables) this translates to a track width of ~ 162mm?
- Why if I need 6.5mm² are the legs on the Mosfet barely 1mm².

Evidently I am missing something here, could someone point me in the right direction?

Cheers
Andrew

hi Andrew,
Look at this pdf, pages 6,7

You can 'beef up' tracks by soldering onto them a heavy gauge tinned copper wire or copper braiding. [recovered from old coaxial cable screening]
 

Boncuk

New Member
Hi Andrew,

keep the traces as short as possible and use flat male connectors (females are crimp connectors as used in car electric)

Here is a picture of one to be soldered onto a PCB.

Boncuk
 

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Hi,

thanks for that, do you know if these connectors are in any Eagle library?

Here is the board, the two T220 packages will stand vertically and the 540 will be heatsinked. Is it OK, the screw connectors I am unhappy with and like the spade terminals.

Cheers
Andrew
 

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Boncuk

New Member
Hi Andrew,

as far as I know mounting tabs of that kind are not contained in Eagle.

Standard tabs are 6.3mm. I made a quick library for that one.

Try to get long ends to be bent opposite underneath the PCB for maximum mecanical strength. The females fit very tight and it requires quite some force to separate both parts.

The screw head terminal you used in your design doesn't look like it can stand high current. If neccessary use two mounting pads each for one connection and also use two wires in parallel (total of four)

Regards

Boncuk

P.S. One hint concerning PCB layout: Avoid right angles and route 45 degrees. At T-junctions add manual traces to meet the junction at a 45 degree angle. When done, check if all elements have the same net name. Click "show" and all traces must be highlighted. If not, check for different names and rename the unconnected trace accordingly. Check for air wires after renaming and reroute.
 

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Hi Boncuk,

thanks for the advice. Will redo the tracks.

What Zip program are you using, Winzip does not unzip the file.

Cheers
Andrew
 
Hi Boncuk,

is this new board getting closer? I have not put the spade terminals in as I am unable to open the Zip file yet.

Cheers
Andrew
 

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Boncuk

New Member
Hi Andrew,

looking much better! You might route the trace from the right bottom connector terminal to the heater connector around the screwhole of the IC.

I zipped the file using 7-zip, but it should open using winzip. Did a double click not unpack the file?

It does on my PC though.

You can download 7-zip free and retry.

Here is what I have done to your design.

The third image is the same design shorted by 10mm.

Regards

Boncuk
 

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shortbus=

Well-Known Member
Most Helpful Member
Hi,


What confuses me is the following;
- theortically I need about 6,5mm², (from wire tables) this translates to a track width of ~ 162mm?
- Why if I need 6.5mm² are the legs on the Mosfet barely 1mm².

Evidently I am missing something here, could someone point me in the right direction?

Cheers
Andrew


Hi Andrew, I'm new at this myself, but you might want to check this out;
FET Current Ratings -- Chuck's Robotics Notebook

It really opened my eyes to how the mosfet data sheets mislead you!

I see from your schematic that your using a IRF540. The data sheet says it's capable of 130W@25*C Now you want to put 12V x 16A = 192W through it!
And the 130W was at 25*C that would take a giant heatsink to keep it at 25*C. Let alone trying to put 192W through it.


It doesn't take too long to do the simple math in the link I posted to chose either the right mosfert or the right number of mosfets to do what you need.

I may be wrong in what I just said, if so I'm sure some one will correct me. If not now is the time to change your PCB not after it's made!

Cary
 
Hi Guys,

thanks Boncuk, a much cleaner and neater presentation. Will have another go at mine. I am always trying to make the components as compact as possible and when I see the printed board the are too small to work on sometimes.

Shortbus= that was an interesting read, the heat factor was a problem to me and before finalising the board was goind to ask a couple of questions on the heatsink I should use.

Cheers
Andrew
 

mneary

New Member
The 130W rating is not how much power you can put through the component.

The power the transistor is dissipating is the current, times the voltage across the transistor at the same time (i.e. when it is on.)

When the IRF540 is "on", its resistance is nominally 44mΩ. The power when it is carrying 16A is I^2*R, which is 11.3 watts. (16A * 44mΩ = 0.7V). Yes, 11W will need a lot of heat sink!
 
Hi,

thanks for that clarification Mr Neary.

I was going to fit a small fan for forced air cooling, something like a 25mm fan. For that matter could even have a 90mm fan cooling the entire enclosure.

Is it better to parallel a couple of devices? They are reasonably cheap in the scheme of things.

As the board is not made I would you mind offering a recommendation as to how you would overcome the problem, it is becoming a little vexing for me.

Thanks
Andrew
 
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Hi

Oh and that 0.7V drop across the device also explains why my calculated current draw was not as per the Spice model. I put three in parallel and 2A more reaches the heater.

Cheers
 

shortbus=

Well-Known Member
Most Helpful Member
The 130W rating is not how much power you can put through the component.

The power the transistor is dissipating is the current, times the voltage across the transistor at the same time (i.e. when it is on.)

When the IRF540 is "on", its resistance is nominally 44mΩ. The power when it is carrying 16A is I^2*R, which is 11.3 watts. (16A * 44mΩ = 0.7V). Yes, 11W will need a lot of heat sink!

Hi Mneary, yeah its been a while since I read that link and your right about it being I^2*R, sorry my bad.

But doesn't the Rds(on) have to be figured from the graph (fig.4 on the data sheet)? Using 160*C for the junction temperature, that would make the Rds(on) about 2.6Ω instead of the .44mΩ @25*C.

Cary
 

mneary

New Member
True, Rdson has to be figured at the operating temperature. Hopefully that isn't 160*C. Figure 4 is a little confusing; it's 'normalized'. At 160*C Rdson is (2.6 * 44mΩ) or 114mΩ. That's the figure to use if you actually allow 160*C.

I agree the cheapest way to fx this with these old transistors is to put a few in parallel. Every transistor decreases the Rdson and at the same time increases the number of devices which share the dissipated power.

Three in parallel gives Rdson of (44mΩ / 3) at 25*C, or if you get it to 160*C it's 38mΩ total. The power shared by the 3 transistors is 9.7 watts, 3.2 each. Too much to run bare, but with forced air it might go.

If you were running PWM, the extra gate capacitance would kill the switching speed, but that's not a problem here..
 

Boncuk

New Member
Where is the problem?

Hi Andrew,

here is a suggestion with two parallel IRF540 (not the best choice).

It uses an outdated Pentium IV heatsink with fan. If necessary you might place three transistors on that heatsink.

3 resistors are added: R19 is a gate resistor for the second transistor and R20 and R21 are zero Ohm resistors instead of wire jumps.

Boncuk
 

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Hi guys,

this is all great info thanks.

Boncuk,

a) you said this is not a good choice, I assume you mean the choice of Mosfets rather than the fact that they are paralleled? I happen to have a P4 heatsink and fan combo waiting for a job. Any suggestions for the Mosfets?

b) I don't mind whacking 4 Mosfets in, there is lots of space where this is going. If I used 4 would probably take them offboard and arrange around the heatsink in a North, E, W and South arrangement with a ringmain connection

c) what drawing program are you using?

d) on the gate resistors, based on your comment one obviously needs one per Mosfet? I have been trying to teach myself electronics with LTSpice, playing around with values etc. The one 390R resistor feeding 4 Mosfets gives the same output trace, what am I missing here?

e) Would the Mosfets in the Spice model be better?

f) What are the connections for the P4 fan, G, B, Y


Cheers
Andrew
 

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Hi,

Okay I arbitarily selected IRFP240 Mosfets rated at Id = 20A. Applying the formulae from the spreadsheet was interesting. I calculated the "free air" parameters i.e. what the limit of the devices would be without cooling.

Max Operating Temp 25 to 125°C
Ambient 40°C (yes this is real)
Rds ON = .0.14Ω
Rθ(ja) = 30°C/W

Maximum allowable power dissipated from the device
P = ((125-40)/30)
P = 2.8W

Maximum switching current
P = I²R
√(2.8/.14) = I
I = 4.5A

So 4 of these devices in parallel should be OK. Would anyone care to check if I am correct please.

Cheers
Andrew
 
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