Building a high power mosfet switch with NTC

Charon

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Just looking to see if there is any feed back on this schematic.

Based on the fet switch in this thread. But higher amp components in an attempt to add a bit of a softer start to try to get more volts to the bulb.

Strongly based on the schematic by JimmyM.

Any thoughts ? Or can you see more than the expected :poof:
NTC should handle 36A
MOSFET is rated
Drain to Source Voltage (Vdss)55V
Current - Continuous Drain (Id) @ 25° C240A Power - Max480W

high_power_mosfet_switch.JPG
 
i comment but i am so far from a real EE its rediculous , so take it with a grain of salt, or a whole sea of salt :)

just things i learnt the hard way
as long as mosfets (in general) go from full open, to full closed they handle a lot of power, its when they are Partly closed or open that they become resistive items heat up and could die immediatly in this shown ammount of power.
if you were really ever going to run 400full watts switched through it, then 480W possible mosfet isnt giving you much leeway, EVEN if it is snapping quickly from full on to full off. if you expect better longevity and handling of switch burping (manuel or switch flaws) you should have a handling capacity of more than 2x that , or at least some good heat removal should a burst hit the mosfet.

The Upper Resister:
to trigger the mosfet gate , you need very tiny ammounts of "current" hence the use of the resisters on the gate, to avoid frying the gate.
to FULLY trigger the gate it will need the full trigger voltage of the gate (at very low currents) if it does not get the full voltage of the gate trigger, then it will be part open (see above). Note: full voltage FOR the gate (prob about 4v), not full voltage of 36v battery.

The lower resister:
once the gate thing closes the mosfet, via hitting it with the correct voltage, it (sorta) stays locked closed, even when your switch here is off, it takes Very little to open it again, hence the second resister, the other resister pulls the gate back down (in voltage) and stops the flow through the mosfet.

the resister RATIO will set the voltage of the "voltage divider" that the pair of resisters are, setting the voltage to the gate.
got it?

with such (conciderably) small pull-up pull-down resistances as shown you should get a very fast SNAP to gate triggering, both up and down. potentially a lighter version of that curcuit would use much higher ammount of resistances for the pulldown resistance, as only the smallest ammount of Drain off the gate is actually needed.
This very fast snap back down, should assist in keeping the mosfet fully opened or fully closed, and not in-between which would be bad.
plus you could overvoltage the gate (i think around 24v), so the ratio would be important.
sooo
Although your going to get some power consumption (resistive short) between the poles of the battery, and maby more than is nessisary, you will get a good fast switching, which you need to keep the mosfet from handling resistive power.
Most of the power consumption your worried about would be minimal (comparative to total on load), and only when it was switched on.

As far as the current limiting resister to keep bulb pop from occuring, somehow it seems like a waste to me. although it can "bend" a bit when the filiment is cold, and reduce total current flow to achieve a starting, its going to be the biggest waste of power to the whole curcuit, and will only provide a bit of soft starting, and a lot of wasting.

higher level of experts would trigger the mosfet in sharp (fast) pulses (PWM) for the soft starting, once the mosfet was an integral part of the curcuit, of course complicating the whole thing by a lot.

another thing, is a FUSE, a fuse will have some resistance itself as they must to heat up and break, so some of your inrush current control resistance could come from a fuse, which could also assist in the case of other issues, sort of kill 2 birds with one stone with a fuse there instead, or Also.
You may need to selectivly add resistance in ANYWAY just to get to the desired power to keep from popping your bulb over time,but myself i would more likly to drop out a cell of the battery, instead of adding in resistive inneficency.
to calculate (easily) your losses through a "inrush" resistance put in, just check the voltage drop there relative to the total voltage, it will give you a quick and dirty idea of its losses.

so all you need to do is "test" what you end up with, by using a smaller wattage of 36V bulb thing (or 3x12v) then switch your switch on, and test the heat hitting the mosfet, and/or checking the voltage drop from source to drain (on the mosfet). Checking the voltage drop on the mosfets power switching area BEFORE putting the full amp load on will tell you if the triggering of the gate is sufficient both on and off. that would be a good test prior to putting on the full load, using the same voltage but less amps for testing.
if you observed to much drop through the mosfet still, or to much heat hitting the mosfet while on or off, then you would adjust the pull-up pull-down resistances, prior to putting your full load on and blowing the mosfet.
so with that, you can test with 3x12V festoon (dome) bulbs or something, and check for yourself and adjust if nessisary.

Notes: if this had not already been demonstrated as working, i would try putting much higher resistances on the gate, just seems to low more like 50K is what i would have attempted, specially with such a high inital voltage on the battery. i was just pointing out the positive ramifications of doing that.
 
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Thanks for the info.

I am just getting ready to start testing this using a regulated power supply.

The NTC has a 1 Ω resistance @ 25C and 0.009Ω resistance at current and 0.028Ω resistance at 50% current (approx 18A). Once going it has very low resistance, but it will be relative to the resistance in the rest of the circuit. May or may not work as well as I want, but I am curious to test it out.

At the moment I expect to see 14 A. But I will be better able to confirm once I hook up the power supply and measure voltage and current.

I would like to learn more, given time and money I would take an EE program for fun. But I had spent my time in school for programming, one or the other.......

So I am probably on your path, trying to learn via the internet, books, advice and trying things out.
 
Hi, if you do decide that you need some power resistors, let me know. You should probably know that mine are surface mount, 1206 size, but of substantial power capability. (rated 50 watts with proper thermal path)

Make sure you have soldered 1206 size parts before trying these. These were custom produced for me, so slightly higer value to me than commodity resistors. :eek:
 
OK. The "7.4K" short is actually a voltage divider to knock down the Vbat of 40V to something the FET gate is more likely to ba able to handle. Using the values Charon has proposed, the max Vgs will be ~15V. A little on the high side but should be OK. I'd switch the 4.7K to a 5.1K.

The resistors are not there to protect the gate from too much current. You want as mush current available to the gate as possible. A voltage divider is a cheap easy way to do things. Ideally, you'd have a voltage divider charging a capacitor and have the switch between the Cap and the gate. This way you could have the Cap dump into the gate for a really fast turn on. Then let the 2.7K resistor pull it down to shut it off. This would take a little more figuring, and would be a constant drain on the batteries, even when off. There are other ways of doing this, but the arraangement you have is probably the simplest and easiest.

The 480W rating of the FET is the amount of power it can safely dissipate with proper heat sinking. This FET is rated at 3.6 mOhm so the amount of heat dissipated in the "ON" condition will be very low. 100 Amps would cause 36W of dissipation. The bulb will probably draw about 12-14 Amps. That's only 0.6 Watts. Unless the light is flashed repeatedly, the switching event shouldn't dissipate much heat either.

One thing though. I'd move your NTC to between the bulb and the FET. You want the FET Source leg as close to ground as possible.

Other than that, it should work just fine.
 
I would ditch that NTC (it only wastes power) and USE that MOSFET in the linear mode! Just a small heatsink on it will do, as the state of "not-fully-on-and-not-fully-off" is only needed for a few tenths of a second, in order to save the lamp.

DON'T hard drive the gate, and limit Ugs-voltage to about 20V. Instead, add a small capacitor between the gate and the source (which is GND also).

What is important, is a low Rds(on). Look for a MOSFET that has a resistance under 10milli-Ohm.
A 480W, 36V lamp will draw about 13.3 Amps! Power in the MOSFET will be
13.33 * 13.33 * 0.01 = 1.77W so only a small heatsink is needed for continuous running.

RC time should be about 0,3 seconds I assume.

With fully charged batteries, I assume we will have 40V of power supply voltage. Divide this with 2 pieces of 10k resistors, and Ugs will be 20V.
The gate is driven with a "voltage source" of 20V and an internal resistance of 5k (2X 10k in PARALLEL!)
R * C = 0,3
with R=5k
C = 0,3/R
C = 0,3/5000 = 0.00006 [Farad] = 60µF.

Take any value between 47µF and 68µF.

Maybe a time of 0.3 seconds is a little bit too long, then take a 22µF or 33µF capacitor, and you're done :wave:

When you only switch the upper resistor to the plus, you'll end up with a switch on AND a switch off delay. This is undesirable, so take a SPDT switch, the "mother" goes to the resistor, one of the daughters to the plus and the other to GND. In order to quickly get rid of the charge of the capacitor, add a diode across the upper resistor, cathode towards the top of the resistor (going to the plus). In order to save the switch and the capacitor, also add a small series-resistor to the diode (about 100 Ohms).

I can't add a schematic, so I hope you understand it this way ;)


Timmo.
 
You'll cook that FET. I promise you.
Look at the "safe operating range" graphs in the FET spec sheet.
About half way through the start, the Vdiff across the FET will be 18V at half voltage it will be drawing half the amps. So 18*(13/2) = 117 Watts.
 
You'll cook that FET. I promise you.
Look at the "safe operating range" graphs in the FET spec sheet.
About half way through the start, the Vdiff across the FET will be 18V at half voltage it will be drawing half the amps. So 18*(13/2) = 117 Watts.

That is 117 watts instantaneously, not continuously. It will only occur for microseconds at best. Not an issue.

Semiman
 
Yes, but during turn on with a cold filament, it will pull 60+ amps.

Give it a try. That's the problem AWR faced with his HotDriver.
Keep your RC constant lower than 0.3s to give that FET a fighting chance.
 
just jumping in here.....

Adding a cap at the gate is the better way to slow the turn-on. The NTC could work, but then you get into other issues. I've seen designs with plain resettable fuses that had problems when tested over a somewhat wide temperature range (-40C to 85C). The cap will be more consistent than the NTC will be.

Add a diode across the upper resistor in the divider network used at the gate. The anode should be at the gate, providing a slow turn-on, but a faster turn-off. No point in dissipating a lot of power at turn-off. Or.. put a small resistor in series with the diode. Slow the turn-off just enough to keep the wiring inductance from producing a large spike at the mosfet drain. Or add a snubber circuit across the mosfet.

I think I'd start conservative when developing the circuit. Put a big heatsink on the mosfet, and see what the heating is like with the desired slow turn-on. Try smaller heatsinks, and when the mosfet feels pretty warm, that's the smallest heatsink to use. This approach takes a little longer, but kills fewer mosfets.

regards,
Steve K.
 
I am just saying (still without knowing)

i now know WHY i would use higher resistance on the gate resisters :)
because i am a sloppy ON-HANDS type of assembler, and if you notice , when the cliplead (sloppy) temporarily connected to the resister comes undone, the gate will get hit with the 36V (through 4.7K), and that is where higher resistance would keep the gate from being destroyed :confused:?

obviously if your running a hotwire bulb your probably not using 10amp clipleads and having the things stretched out across your workbench trying different things :sick2:
but transister gates always got the bad end of being sloppy on the bench :huh: for me.

if its a clean voltage divider, well connected then thats different. but when slopping stuff around on the bench hitting a gate too hard, makes gate not work anymore :thumbsdow.

also how the heck do they drive these gates with the tiny microcontrollers , is only ever with leetel currents?
so sure fast current hits to get to the voltage fast, but if to much current AT to much voltage hits it, it will die (like my other projects) that can happen IF the other side comes loose.
 
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How much current you need really depends on your required switching speed. My PWM regulator uses a 150 Ohm resistor in series with the FET gate (46 nC). At it's rather low switching speed, 248Hz, that is plenty of current to turn it on or off nicely. Much faster PWM 100s kHz -> MHz requires a brief spike of current a couple of volts above the FET gate enhancement volotage (typically 5 or 10 volts). An FET driver is used for that. It's like a relay that can pass amps of current in little spikes to drive the FET gate.

But you want just the opposite. A slow gate. I still think you're going to kill FETs. But you learn quickly when you're letting the smoke out of $7 FETs. Many have tried the slow gate approach and failed. That's whay I developed the JM-SST (PWM based soft starter) and myself, Alan, and wquilles have built PWM based regulators.
 

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