Can a homebrew FET circuit replace a NTC?

EvilLithiumMan

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Nov 20, 2003
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Chula Vista, CA
(As it turns out, I've just reinvented the wheel. But my wheel isn't nearly as round or a pretty as the others, so here it is)

My original goal was to see if I could a design cicuit using a power mosfet to replace the NTC (thermister) that many Mag623 users employ to prevent their bulbs from flashing. Although inexpensive and easy to hook up, NTC's run quite hot. They are also affected by ambient temperature, require a cool down period and generally have a tolerance +/-20%. I was hoping a simple FET circuit would be a practical solution. And by simple, I mean 6 or 7 discreet components, readily available, less than $20, and easy to tailor to one's specific needs.

This is my initial attempt, a FET with a basic RC network to control the gate:
Schematic.jpg


It has a terribly large time constant and for good cause: I don't have a scope. So I needed a prolonged, clearly visible soft start to verify functionality. Well, I got what I wanted. The startup delay was nearly 3 seconds, with a soft start period of approx. 1/2 second. Here is the breadboard: (It's nasty looking, you may want to wear safety googles)
Breadboard.jpg


Unfortunately, the turnoff cycle is nothing short of comical. 130 seconds to shutoff! From 0 to 110 seconds, the lamp appears fully lit. From 110 to 120 seconds, the lamp slowly turns off. But it takes another 10 seconds before there is no visible glow from the filament. I had to laugh.

My next step was to Google search CPF to determine how short of a soft start period would work. Perhaps if 20ms was adequate, a turnoff period of 10-12 seconds wouldn't seem so bad. But in my searching, I came across this thread (and others as well):
http://www.candlepowerforums.com/vb/showthread.php?t=100286&post#17 (post #17)

Clearly, this has all been done before. But they too had to confront the long shutoff period, which meant more components and complexity. To keep my circuit simple, I had to find a simple solution to discharging the gate capacitor. I then recalled I had an assorted collection of small signal PCB relays. Would a relay work? Would a relay fit? Turns out a relay works just fine! A Magnecraft W172DIP-7, plus a resistor, does the trick. And despite the length of the relay, there is no interference with the Mag's cammed reflector.
Breadboard1.jpg


I now wonder if I can fit all of this stuff into the ubiquitous KIU switch. To that end, I've ordered a couple of KIU kits and a couple of 64623 bulbs. I don't yet have a tri-bore host or battery pack to complete a 623, and I'm not sure I ever will. While I don't doubt for a second that it must be a Hell of a lot of fun to have an armor piercing 3000-5000 lumens in one's hand, I'm not convinced such a monster is for me. For now, I'm content with the challenge of replacing a NTC with a "simple" FET solution.
 
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Re: Can a "simple" FET circuit replace a NTC?

Can you provide a parts list for this thing? Looks like it'd solve my ROP's 20-30 clicks to "on" problem quite nicely.
 
Re: Can a "simple" FET circuit replace a NTC?

WOW, vantastic job. Looks really really promising. Please do the board view again with the relay in place. We have some reall good EE types here that as you have found have created solutions, but we can always use a new one and especially if it is a simple and straight forward as yours. I am sure they will chime in with comments as this comes along.

As long as it can handle around 10 amps max well, this should be a wonderful addition for most of us to try and build.

Bob E.
 
Re: Can a "simple" FET circuit replace a NTC?

You can get instant turn off if you switch the power going to the bulb also.

Connect the + side of the bulb to the same side of the switch as R1. That will also eliminate any potential issue with leakage current when the light is off. :whistle:

You also need to make sure the voltage on the gate is enough to cause the transistor to saturate and turn fully on, but not too much. Vgs is to saturate a FET is usually ~5V in a high current TO-220 power FET. You've got more than enough voltage on that pin. Increasing R1's value and decreasing R2's value will lengthen the start up time, and decrease the discharge time.
 
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Re: Can a "simple" FET circuit replace a NTC?

*bump* with my ideas for improving this circuit.
 
Re: Can a "simple" FET circuit replace a NTC?

I'm waiting for different parts to arrive. I feel I can fine tune the circuit somewhat without adding anything.

Stereodude - like other FET approaches (AW, AWR, JM-SST), I want the FET to handle both the large on and off currents. Your approach will work during the turn on period. But switching off the bulb with 8-10 amps through it means arcing and premature wear of the switch contacts. With the switch handling only a few milliamps, it should last a lifetime. The FET is already there and I want to take full advantage of it. I do appreciate your input and doubtless there are other approaches to consider. Thanks!
 
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Re: Can a "simple" FET circuit replace a NTC?

Stereodude - like other FET approaches (AW, AWR, JM-SST), I want the FET to handle both the large on and off currents. Your approach will work during the turn on period. But switching off the bulb with 8-10 amps through it means arcing and premature wear of the switch contacts. With the switch handling only a few milliamps, it should last a lifetime. The FET is already there and I want to take full advantage of it. I do appreciate your input and doubtless there are other approaches to consider. Thanks!
That's where my comments about changing R1 and R2 can come into play. Vgs at saturation is probably under 5V for your TO-220 FET, so you don't need 13V on the gate of it. The soft start and soft stop occur when Vgs is somewhere between 0V and the saturation point. That's usually a very small voltage range. For example, on a Vishay SUP90N03-03 that soft start and stop occurs between 2 and 3 volts. In the case of the SUP90N03-03 if you set Vgs much above 3V for full on it's just extra time you have to wait for it to turn off after opening the switch. In your FET is similar you have to wait for C1 to discharge through R2 down to 3V from ~13V before you get a soft stop. That's why you have such a long delay for turn off. :sssh:
 
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Re: Can a "simple" FET circuit replace a NTC?

That's where my comments about changing R1 and R2 can come into play. Vgs at saturation is probably under 5V for your TO-220 FET, so you don't need 13V on the gate of it. The soft start and soft stop occur when Vgs is somewhere between 0V and the saturation point. That's usually a very small voltage range. For example, on a Vishay SUP90N03-03 that soft start and stop occurs between 2 and 3 volts. In the case of the SUP90N03-03 if you set Vgs much above 3V for full on it's just extra time you have to wait for it to turn off after opening the switch. In your FET is similar you have to wait for C1 to discharge through R2 down to 3V from ~13V before you get a soft stop. That's why you have such a long delay for turn off. :sssh:

As I stated in the OP, I need to use a horribly large tc (time constant) to visibly verify the circuit's operation - I don't have a scope. I have no doubt the R1/R2 ratio will change quite a bit once I'm satisfied the design is feasible.

As for the FET, the IRF530 used in the original breadboard is gone. It was all I had available at the time. I'm using an Infineon IPB050N06L now, but waiting for some IPB048N06L, the latter having nearly 1/2 the gate threshold voltage. These devices are rated 60V, 100A and 300W and very low on resistance. I do not claim these to be the best FETs for the task. I just want something rugged, available and in a TO263 package so as to fit inside the KIU switch.

The Mouser catalog alone appears to have nearly a dozen different nFETs that will work. That will be one of the benefits of using discreet components - use the FET you want! I'm also planning on trying out the IPB048N06L with some 4V, 3.5A axial bulbs I have from Fivemega. If this circuit can operate from 4 volts to 20 volts, it will be that much more versatile. Again, thanks for your suggestions.
 
Re: Can a "simple" FET circuit replace a NTC?

How warm is the FET getting during the ramp up/down period and what is this timing at the moment? (when the FET is being operated in its linear mode).
 
Re: Can a "simple" FET circuit replace a NTC?

How warm is the FET getting during the ramp up/down period and what is this timing at the moment? (when the FET is being operated in its linear mode).
The FET will get hotter when the light is fully on than it will when it's in linear operation area because the current through the FET is higher. A good TO-220 FET has an Rds of < .01 ohms, so even with something like 10A going through it, there is less than 1W of power lost in the FET being turned into heat. The IRF530 that EvilLithiumMan first tested with is an older FET with a Rds of .11 ohms, so with 10A going through it, it would have to dissipate 11W of power. That's not gonna work without extensive heatsinking.

Vishay Siliconix makes some of the lowest Rds FET's around. One of their best N-channel TO-220 FETs has a Rds of .0033 when Vgs is 4.5V. That means you could push 10A through it with it only losing .33W to heat.
 
Re: Can a "simple" FET circuit replace a NTC?

The FET will get hotter when the light is fully on than it will when it's in linear operation area because the current through the FET is higher.

True but only because the time spent in the fully ON region is much much longer than during the linear region.

Power dissipated by the FET is much higher in the linear region but the period of time spent in this region is so short that the case doesn't rise much. This is why I asked him how hot it was getting.
Keep the FET in the linear region for a few seconds (10's of seconds if you're cruel) and you can easily exceed the max operational temp of the part, even fry it, though your current through the part is less than 10A.

Dave
 
Re: Can a "simple" FET circuit replace a NTC?

True but only because the time spent in the fully ON region is much much longer than during the linear region.

Power dissipated by the FET is much higher in the linear region but the period of time spent in this region is so short that the case doesn't rise much. This is why I asked him how hot it was getting.
Keep the FET in the linear region for a few seconds (10's of seconds if you're cruel) and you can easily exceed the max operational temp of the part, even fry it, though your current through the part is less than 10A.

Dave
Are you sure about that? The Rds will be higher, but the current will be much lower. P = i^2*R. The FET will limit the current going to the bulb, but the power difference between full operation and the linear region shouldn't be burned up in the FET because the the battery isn't being heavily loaded. I will admit though that I'm a little fuzzy on how a FET behaves in it's linear region, so my memory could be wrong. :thinking:
 
Re: Can a "simple" FET circuit replace a NTC?

It's not a FET thing, just resistance through the FET itself.

Here's 2 different ways to think about it.
- in your example of the older FET vs the newer one, the older having a higher Rds ON resistance, the older one would dissipate more power and hence heat. Take it to an extreme - say one FET had 1 ohm of resistance and the other had 0 resistance, the one with 1 ohm will dissipate more power and heat then the 0 ohm one at 10A.

Now consider that the 0 ohm and the 1 ohm Rds ON FET are actually the same FET. It's 0 ohm when fully on, and 1 ohm at a certain point in its linear region.

Light bulbs when off have very low resistance so they're pretty much a dead short. When turning on, even with the voltage drop across the FET being so high because it's now a 1 ohm resistor, a decent amount of power is going to be dissipated through the FET.

The other example is a toaster plugged into a worn wall socket, where the wall socket/plug is the 'FET'. Too much contact resistance there and the plug/socket heat up, enough to cause a fire, the toaster itself doesn't get as hot because it sees a voltage drop. New socket and plug in intimate contact - this part stays cool and the toaster heats up just fine.


If the FET were REALLY high in resistance, like 10K ohms, the bulb doesn't light up, the FET doesn't get hardly warm at all, things are good (except for your batteries going dead over time).
 
Re: Can a "simple" FET circuit replace a NTC?

Probedude is correct in the operation of a FET in it's linear range. Another illustration could be with a 10V, 100W bulb, which results in 10A of current. It also means the filament, when fully warmed up, is 1 ohm. We'll even use Stereodude's FET to drive it. And, as Stereodude calculated, the FET is disapating .33 Watts. Now let's take the FET into it's linear range. (For the sake of simplicity, we'll assume the bulb's filament stays at 1 ohm)

As we take the FET into the linear range, the current drops. Let's say we get to the point where the current is 5A, vs. the full on 10A. 10 volts, 5 amps, means our load (bulb and FET) is 2 ohms. We know the bulb is 1 ohm, therefore the other 1 ohm is the resistance of the FET. But now look at the power the FET must handle : 5x5x1 (I squared R) = 25 watts! Far greater than the .33 watts when the FET is fully on.

The technical term for a FET's linear range characteristics is SOA, Safe Operating Area. As far as I know, every power FET's data sheet gives a graph of the SOA. Here's the one for the FET I'm using:

SOA.jpg


As you can see, it's a 300 watt device. (3 volts @ 100 amps, 10 volts @ 30 amps, you can see it in the graph) But don't forget, that 300 watts is with the device properly heatsinked; temperature at 25C. Without a heatsink, far less power must be used. That's why I'm trying this 300W device, I'm hoping that without a heatsink, it will still tolerate 20-30W for a few seconds.
 
The good news is, by adding a 200K pot and wiring it as detailed in this link:http://www.candlepowerforums.com/vb/showthread.php?t=100286&post#17 , I can "dial out" the startup delay caused by the 1M/10uf RC network and have the light turn on immediately with the soft start period. The bad news is, this clearly is no longer a "simple" circuit. And a relay is still required to shunt the FET's gate at turnoff.

I built another breadboard, modding the Mag switch to specifically work with the 4V, 3.5A axial filament bulb from fivemega in his latest 1D-4AA cell creation:
Breadboard2.jpg

Breadboard3.jpg

It's working so well, I feel it is pretty much done. The only exception is to try to replace the relay with a pch FET. (As discussed in the link given) I am waiting for a KIU kit to arrive. It will be a challenge to build this circuit into the KIU's limited space. If it works, it will give the Mag superbulb (100-150W) modder a cooler running alternative to a NTC.
 
As for not having a scope, you might be able to see the brightness increase during a short turn-on period by taking a video clip with a digital camera. Your camera might be able to make a video at (say) 640x480 or 320x240 pixels at 60 fps or 30 fps. Press the shutter, turn on the switch, capture a few seconds of video. Then view it frame by frame in a simple video editing application.
 
Nice job!

For your next project, learn assembly and program an 8 pin (or even 6 pin) microcontroller and have it PWM the FET for your slow startup! :p You could then do modes, but the best part is when you turn off the switch, the FET turns off (parts counts likely lower and no fiddling with values).
 
Try it with a high power bulb. The FET will get so hot it will either fail or desolder the wire on the tab. FETs are not designed to spend a lot of time in their linear range. The power dissipated during turn off and ESPECIALLY turn on, is immense. This was the reason I went with PWM for my soft starter. It will work with low power bulbs, but beasts like the the 623 and 458s will cook the FET.
 
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