reed activated mosfet switch diagram

archer6817j

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Hi all, I've been playing around with reed activated MOSFET switches. It took a long time to learn and understand the information I needed since I'm not an electronics person.

I thought I'd post this diagram to help others that want to make this type of switch but don't necessarily have the know how. I also wanted to go with a "non technical" diagram so it can be more easily understood for those of us who aren't electrical engineer types.

Other types of MOSFET will work but I picked this one up at Fry's without having to mail order anything. Same with the resistor. This particular MOSFET package is fairly small (good for installing in a light) but I'm not sure what the technical "package size" is. The drawing is not to scale of course.

I have this circuit "working" in a light but I wanted to check with the experts here with respect to the location of the resistor. Do I have it right? Are there any other errors on the diagram? The MOSFET is intended to work with an 18650 battery (or similar) because the gate threshold is 3.0V If you wanted to use lower voltage batteries you might need a MOSFET with a lower threshold voltage. The threshold should be less than the lowest charge state of your desired battery configuration. Then all you need to know is the current capacity of the MOSFET. In this case 35A is more than enough :)

Thanks in advance and I hope someone finds this useful!

NOTE: the colors on the resistor are not accurate, I just made them up to illustrate the fact that it's a resistor.

reed-activated-mosfet-switch.jpg


EDIT: DO NOT USE THE NTE2389, IT WILL OVERHEAT
I pulled this link from a recommendation below for a "better" MOSFET for applications under 6 amps: http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=497-9093-5-ND
 
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uk_caver

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The NTE2389 seems to have a typical gate threshold of 3V, but with min and max of 2V1 and 4V.
Also, the threshold voltage is where the FET starts to turn on, not where it's completely on.

Depending on luck with the individual FET, it is possible that the circuit might not work.
 

DIWdiver

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That looks pretty good to me! A resistor value anywhere from 1K to 100K should be fine.

Gonna show us your build?

By the way, it's good that you picked a large FET. The current rating (35A in your case) is for use with massive heatsinks that don't get warm when you use them. If you leave it just hanging in the air, you need to use a lot less current, maybe by a factor of 10 or more! This is called derating, and is more important in the current and power ratings of FETs than almost any other piece of electronics. Interestingly though, you can count on the voltage ratings of FETs. Go figure.

For this particular FET, you're good for up to a few amps without a heatsink, depending on the battery voltage. You could go higher with a heatsink.

For most on/off applications, I like to run the gate voltage at least twice the threshold voltage. The NTE2389 is tested at a 10V gate drive (see the TEST CONDITIONS column for the Rds-on spec), and its 3V threshold voltage isn't actually low. Many FETs have threshold voltages of 1.4V or less, and would be better suited to single-cell (LiIon) lights.

Why so much over the threshold voltage? It's all about resistance. The higher the gate voltage, the lower the resistance of the FET, and resistance is bad in this case. At the threshold voltage, the FET is just beginning to turn on and you may or may not begin to see light. At twice the threshold voltage, the FET is firmly on but the resistance isn't as low as it can be. At several times the gate voltage the resistance will be substantially lower. So I wouldn't recommend the NTE2389 for use with a single LiIon cell, but it would be fine with 2 cells, and great with 3 or more cells.

D
 

archer6817j

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Hey thanks for the input so far! Right now I have this switch on a 2s and 3s setup. I'ts basically "what I could find" at Fry's that seemed in the right range. Could anyone suggest a part number for a 1 cell configuration? I can update the drawing to show what FET works best with what battery config.

This is really just a prototype to test some engineering concepts and isn't intended to be a usable light. For example, the ring holding the magnet is just held in place by a couple of o rings that I rolled onto the body. The battery is 3s 18650 with a 11.4v protection PCB. The LED is a luxeon rebel (does anyone remember those?) tri star that is rated at 540 lumens (18 degree optic); driven by a luxdrive 700ma buck.

DL-V1.jpg


This is a shot of the two FET's I bought and the reed switch. For scale, the pegboard holes are about 1/4" in diameter.

FET-and-reed.jpg
 

VidPro

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Just a note:
i hate the NEEDED pulldown resister on a mosfet, because when the device is on, notice the PATH created? frilling power consumption through the pulldown.

On some items i got far superior efficency by using a relay instead (of all things)
because:
A) there was way less ohmies lost through the relay contact points.
B) Unseen power losses through the pulldown didnt make up for the seen losses through the relay (old school) coil.

This of course depends on how fast you needed to pull the gate back down to get a clean fast shutoff.
I have spent hours TUNING a pulldown to get it to close the way i want, at all the possible battery voltages i would have, then insure that the mosfet isnt getting too hot via its resistive opening , and have the least power consumption too, i was unimpressed :) when compared to old school.
SO I also tuned the pullup connection with the second resister (to the switch) in higher voltage application to reduce this resistive path as much as possible, but still get a full closure at its lowest possible battery voltage.
I am not an EE at all, only do hands on stuff measuring and guessing and all.

But that isnt what i was getting at.

you shouldnt NEED a pulldown, if you have a SPDT, and they make REEDs in SPDT, so being a proto type, you might test such things for the power savings. depending on how it is done, the power savings can be significant TO ME at least :confused: stop the resistive path, even if you put resistance on the second position of the switch to reduce the load on the reed for the pulldown.

when using way higher voltages , then sending them to the gate to do the pullup, make sure you measure that current too. you need to get full gate closure at the ~4V, at the lowest battery voltage, but you dont need to be wasting any unnessisary power there either. not that i saw any wasted there, but it wouldnt hurt to check when hitting it with the higher voltages, and reduce the intital load on the reed switch again.
 
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Steve K

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This is really just a prototype to test some engineering concepts and isn't intended to be a usable light. For example, the ring holding the magnet is just held in place by a couple of o rings that I rolled onto the body.

For a prototype, it's pretty nice!

I think the concept is good, but could be tweaked and improved a bit. For instance.. I've spent a lot of time working with hall effect or MR sensors, and one big issue is always magnet retention. After seeing magnets come loose in various designs, we've adopted the policy of either completely surrounding the magnet with other solid objects, or using two retention techniques (i.e. perhaps using an adhesive as well as swaging the magnet in place).

In a production environment, it seems that the process for one of the two retention techniques will eventually go bad. It's only by using two techniques that you can be very sure that the magnet won't get away!

In your design, I'm assuming that the little circle on the ring is a rare-earth magnet inserted into a hole in an aluminum ring. You could drill a blind hole in the inner diameter of the ring and insert the magnet there, knowing that there's no way the magnet could fall out once the ring is installed. Of course, drilling a blind hole from the I.D. would be rather tricky! Maybe get a long drill bit and drill from the other side of the ring? You'd end up with a hole in the ring that didn't have any real function, although you could use it as an indicator in the same way that the magnet now functions.

An alternative would be to peen the hole at the outer diameter of the ring. That would reduce the hole diameter and be pretty simple.

Electrically... try to find a mosfet that is fully turned on with a gate voltage of 2.5v or so. A logic-level mosfet is a good start. Flipping through the Digi-key site, the Zetex ZXMN4A06G has a resistance of about 0.1 ohm at Vgs = 2.5v. It's in a sot-223 package, which is smaller than what you have now, but would benefit from being mounted on a circuit board to dissipate heat. An alternative would be to find a high gain BJT, or cascade some BJT's, but that has its own issues.

regards,
Steve K.
 
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MikeAusC

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Just a note:
i hate the NEEDED pulldown resister on a mosfet, because when the device is on, notice the PATH created? frilling power consumption through the pulldown.

On some items i got far superior efficency by using a relay instead (of all things)
because:
A) there was way less ohmies lost through the relay contact points.
B) Unseen power losses through the pulldown didnt make up for the seen losses through the relay (old school) coil.

This of course depends on how fast you needed to pull the gate back down to get a clean fast shutoff.
I have spent hours TUNING a pulldown to get it to close the way i want, at all the possible battery voltages i would have, then insure that the mosfet isnt getting too hot via its resistive opening , and have the least power consumption too, i was unimpressed :) when compared to old school.
SO I also tuned the pullup connection with the second resister (to the switch) in higher voltage application to reduce this resistive path as much as possible, but still get a full closure at its lowest possible battery voltage.
I am not an EE at all, only do hands on stuff measuring and guessing and all.

The LED is drawing an Amp from the battery - why are you worried about one-thousandth of an amp being drawn by the pull-down resistor ?

The value of the resistor will affect the on-state resistance of the MOSFET as you always have the full battery voltage across the Gate.
 

DIWdiver

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For a prototype, it's pretty nice!

I think the concept is good, but could be tweaked and improved a bit. For instance.. I've spent a lot of time working with hall effect or MR sensors, and one big issue is always magnet retention. After seeing magnets come loose in various designs, we've adopted the policy of either completely surrounding the magnet with other solid objects, or using two retention techniques (i.e. perhaps using an adhesive as well as swaging the magnet in place).

In a production environment, it seems that the process for one of the two retention techniques will eventually go bad. It's only by using two techniques that you can be very sure that the magnet won't get away!

In your design, I'm assuming that the little circle on the ring is a rare-earth magnet inserted into a hole in an aluminum ring. You could drill a blind hole in the inner diameter of the ring and insert the magnet there, knowing that there's no way the magnet could fall out once the ring is installed. Of course, drilling a blind hole from the I.D. would be rather tricky! Maybe get a long drill bit and drill from the other side of the ring? You'd end up with a hole in the ring that didn't have any real function, although you could use it as an indicator in the same way that the magnet now functions.

An alternative would be to peen the hole at the outer diameter of the ring. That would reduce the hole diameter and be pretty simple.

Electrically... try to find a mosfet that is fully turned on with a gate voltage of 2.5v or so. A logic-level mosfet is a good start. Flipping through the Digi-key site, the Zetex ZXMN4A06G has a resistance of about 0.1 ohm at Vgs = 2.5v. It's in a sot-223 package, which is smaller than what you have now, but would benefit from being mounted on a circuit board to dissipate heat. An alternative would be to find a high gain BJT, or cascade some BJT's, but that has its own issues.

regards,
Steve K.

That's a pretty cool looking light, Archer!

Steve's points are very sound.

Magnet retention is critical. On my light I cut the ring in half, so I could easily drill from the inside. I then used an o-ring to hold the two halves onto the light. This accomplishes several things: it allows you to mount the magnets inside the ring where they cannot escape; you can mount the ring in a groove cut into the body of the light; you can control the friction of the ring on the body; it allows you to remove the ring for cleaning or other maintenance (this is important in my light, which is a diving light, and the magnets collect iron particles from the shipwrecks that I dive on).

If you cut the ID of the ring slightly larger than the OD of the body, then the saw cuts away part of the ring when you cut it in half, the result is that the o-ring can hold the ring halves against the body of the light, and the tension in the o-ring determines the friction of the ring against the body. Friction, glue, and peening are all good secondary (maybe primary?) methods to retain the magnets.

The suggested ZXMN4A06G is great for applications where space is limited and LED current isn't much more than 2A. For up to 10A I've had good luck with the FDP8860, but it's much larger. There are literally thousands of FETs to choose from depending on the specifics or your application. The important characteristics are: max voltage, max current, efficiency vs cost tradoff. If you post more specifics of your application, we can better help you select the appropriate FET.

As far as the pull-down resistor being worthy of hatred, I think that's best applied to other kinds of circuits than what you propose. In switching regulators, for example, where the switching frequency often exceeds 10^5 Hz, pull-down resistors have to be low values, thus they cause high power losses. In your case you can use 10K or even 100K resistor for pull-down, which results in trivial power loss.

Using a reed relay to improve the efficiency is not going to work in your case. Reed relays require 5 mA or more to operate, while a 100K pull-down resistor at 12V would only require 0.12 mA.

Given that you have a switch that is controlled by your hand, having a switching speed in the micro seconds is unnecessary. Tuning the pullup to optimize switching speed is completely unnecessary in this application. Using a second resistor may be necessary if your battery voltage is greater than 20V, but even that's not difficult. I am an EE and don't have to resort to guessing about these things.

VidPro's point about not needing a pull-down resistor is true. If you use an SPDT reed switch, you will not need a pull-down resistor at all, and the FET will switch much faster. Eliminating the pulldown resistor will increase the efficiency of your circuit. However, SPDT reed switches are less common and more expensive, and in a hand-operated application a well-chosen pull-down resistor can cause less than 0.1% efficiency loss (for a 100K pull-down, 7.6V supply, 3W LED it's 0.02%), so the improvement may be trivial.

D
 

VidPro

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The LED is drawing an Amp from the battery - why are you worried about one-thousandth of an amp being drawn by the pull-down resistor ?

The value of the resistor will affect the on-state resistance of the MOSFET as you always have the full battery voltage across the Gate.

your right why would you worry about 1.2ma using a 3S of li-ion? and if the led uses 1 amp with one battery, and the OP suggest that he is using 3, then mabey the wattage consumed total, instead of the Ma should be concidered. because if using 3xli-ion a single led should use more like .33 amps, not 1A. so already were 3times off what the actual is vrses the math :)

when you have a low modes , that run for 100 times as long the ammount is not 1/1000th it it is 1/10th and for me it was WAY higher as measured not calculated. plus if the led is running off the 3x li-ion as suggested, that makes it 1/3rd that, and puts us at 1/333rd already, when in a low mode 1/3rd of the total power . Dontcha just love math, that is why i do it hands on and actually measure it.
if the pulldown is to slow (because of to high resistance), especially when using tiny logic level mosfets, that dont switch FULLY at 2.5v anyway (measure it) , the switching rate is to slow with too high of resistance, and the mosfet heats up. But then i have seen mosfets on stuff made by real EEs char into oblivion, so i guess they know thier electronics, and me i just fix what they screwed up :)

what is the problem? are you assuming that there is some difficulty in adding a 2 way switch instead of a 1 way switch? is it the extra 12cents? if it is mabey i can make a donation.
 
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SemiMan

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Don't use that Zetex part suggested. The specs are really not that good and the package is not good for what you are trying to do. This part would be more in order:

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=IRLB8743PBF-ND

And it is reasonably priced. I would still prefer a lower threshold voltage, but this is a start. It will likely have about 20 - 50 times lower resistance in your application versus the NTE part. That pulldown resistor should draw almost no current. It looks like you are using 1K resistors. That is way smaller in value than you need. 10K or even 100K would work fine though would need to look at gate leakage but likely to not be an issue at 100K at all.

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=497-9093-5-ND

This part is about as ideal as you are going to get. It is a bit more money, but much better for your application.

Good luck with your project. Note that there will be some current flowing through that FET even when off... microamps.

Semiman
 

DIWdiver

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Okay, if you want to do the math, let's do it carefully or what's the point? I'll push everything toward the worst case to make the pull-down look as bad as possible.

3 Luxeon Rebels with Vf of 3.4V (from the data sheet) adds up to 10.2V. Running at 700 mA, that's 7.14W. Assuming 95% efficiency from the buck, the load on the battery is 7.14/ 0.95 = 7.52 W.

With 3 18650's at near full charge, we'd have 12V supply. So the load is 7.52W / 12V = 0.626A.

A 100K pulldown draws 0.12mA from the battery (not 1.2mA). This represents 0.00012/0.626. This is 0.000192 or 0.0192%. I'd say that qualifies as trivial, but that's a judgement on my part and subject to dispute. If you'd rather talk about power, the numbers are 0.00012*12/0.626*12 = 0.000192 (same result).

The OP never said anything about low modes, and his diagram makes no provision for modes, so we should stop there. But let's say someone goes to the trouble to make a low mode, 1/100 current. The Rebel data sheet shows no Vf for this low current, but we can estimate something like 2.6V. Total LED power is 2.6V * 0.007A * 3 = 0.055W.

Load on the battery is 0.055W/0.95 efficiency = 0.058W.

The quiescent current of the driver is 0.5mA, so it represents 12V * 0.0005A = .006W load.

Total load on the battery is 0.064W.

The pulldown adds 12V * 0.00012A = 0.0014W. This represents an increase of 0.0014/0.064 = 0.0225, or 2.25%. In the case where the light is going to burn for days, this is probably trivial, but again that's my judgement, and subject to dispute.

Now let's think about overheating the FET during turn-off. Gate charge isn't specified for the FET the OP mentions, but similar parts have gate charge in the range of 20-30 nC. Let's assume 50 nC worst case. Discharging via a 100K resistor would take worst case 50 nC/(2.1V/100K) = 2.4 mS. In reality it would take less than that. The maximum possible power dissipation in the FET is 7.52W, which is the max load we calculated earlier. In reality, it will be much less. But using these numbers the maximum energy of turn-off is 0.0024S * 7.52W = 18mJ. Again the datasheet for the OP's part doesn't specify a max for this part, but similar parts have specs in the hundreds of mJ. Thus it is very unlikely we will have a problem. I realize that 'very unlikely' may not be very comforting to all readers, but safety factors greater than 10 are considered excessive, even in bridges, buildings, and airplanes, where failures make national headlines.

D
 

MikeAusC

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what is the problem? are you assuming that there is some difficulty in adding a 2 way switch instead of a 1 way switch? is it the extra 12cents? if it is mabey i can make a donation.

Very sad. People try to help you based on 40 years of doing electronics design, and you think an appropriate response is to get sarcastic - DON"T EXPECT ANY MORE HELP HERE.

This is my last bit of advice - listen to DIWdiver's similar advice "Given that you have a switch that is controlled by your hand, having a switching speed in the micro seconds is unnecessary." If your FET is overheating, it has nothing to do with the resistor's value.
 

VidPro

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Very sad. People try to help you based on 40 years of doing electronics design, and you think an appropriate response is to get sarcastic - DON"T EXPECT ANY MORE HELP HERE.

.

i was not asking for help, i was just mentioning something that i have actually observed. and an alternate method.
if your offended that after 40 years you never thought about it LOL, then have at it.
I wasnt even begining to get sarcastic. now i will.
 

SemiMan

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Very sad. People try to help you based on 40 years of doing electronics design, and you think an appropriate response is to get sarcastic - DON"T EXPECT ANY MORE HELP HERE.

This is my last bit of advice - listen to DIWdiver's similar advice "Given that you have a switch that is controlled by your hand, having a switching speed in the micro seconds is unnecessary." If your FET is overheating, it has nothing to do with the resistor's value.

Note that VidPro was not the one asking for assistance, he was just the one providing less than valuable assistance (not to mention not accurate) and was most certainly being sarcastic. VidPro you come off as a hack with a chip on your shoulder who seems to have something against those with education and training perhaps because you do not have any? I almost wonder if you knew what a reed switch was when you made that 12 cent comment.

Your limited experience and knowledge shows in your comments. While most FETs will not fully turn on at 2.5V there are some that do, and odds are you do not have the equipment to measure whether they do or not as they are sub 10 milliohm unless quite warm.
 

Mr Happy

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One thing puzzles me slightly: given that it is apparently possible to produce FETs having a very low on resistance and a low gate turn on voltage at a reasonable price, why are there so many FETs that have much higher resistances and higher turn on voltages? There must be some (even most) applications where these other FETs are useful, such as perhaps analog circuits or high voltage circuits?
 

VidPro

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, and odds are you do not have the equipment to measure whether they do or not as they are sub 10 milliohm unless quite warm.

assumes facts not in evidence.
also assumes that to measure the losses i am getting through a mosfet, that i would need to even measure the ohms. (whos skill were we faulting?)

there probably are mosfets that close fully at 2.5, dont doubt that, but even the spec sheets and curves and graphs show that a silicoln gate has certian properties, that are not as easily overcome as just picking one that claims it will.

Limited , yes freely admitted limited knowlege, based on HANDS ON work with the items, and have discovered that the math doesnt always add up like it should transferring from paper to reality. Completly my fault for having looked.
 
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VidPro

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I almost wonder if you knew what a reed switch was when you made that 12 cent comment.

.

Hilarious. :wave: i dont know what a reed switch is, and dont have any , but that is probably because i dont have a Magnet :sssh: either. :thumbsup:
They are too sofisiticated for my tiny brain to comprehend.

and just when i thought they can be bought on E-bay , what is wrong with me http://cghttp://cgi.ebay.com/Reed-G...136?pt=LH_DefaultDomain_0&hash=item563d944ae8

lets keep this secret alive, reed switches are only for professionals :)
 
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SemiMan

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One thing puzzles me slightly: given that it is apparently possible to produce FETs having a very low on resistance and a low gate turn on voltage at a reasonable price, why are there so many FETs that have much higher resistances and higher turn on voltages? There must be some (even most) applications where these other FETs are useful, such as perhaps analog circuits or high voltage circuits?


It is all about trade-offs .. on resistance, gate charge, on/off times, cost, voltage, etc.

In general, every new generation lowers one critical specification while keeping others the same. Hence some of the range of available parts is just that parts from several manufacturing generations are currently available.

However, in general, most parts are optimized for a particular gate voltage range. The most common range is for full enhancement at 10V (or so) which covers most most supplies whether AC\DC or higher voltage (>12V) DC\DC. It all depends how the FET is to be used which will cover whether it is optimized for low on resistance, switching speed, etc. Add to that the max drain source voltage for the part.

Most logic level/low voltage gate turn on FETs have very high gate charge compared to standard FETS. If you used them in a higher voltage drive design, the power dissipation from just switching would be high. However if you only switch at 5V (or less) then the dissipation is much lower.

Semiman
 

SemiMan

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Hilarious. :wave: i dont know what a reed switch is, and dont have any , but that is probably because i dont have a Magnet :sssh: either. :thumbsup:
They are too sofisiticated for my tiny brain to comprehend.

and just when i thought they can be bought on E-bay , what is wrong with me http://cghttp://cgi.ebay.com/Reed-G...136?pt=LH_DefaultDomain_0&hash=item563d944ae8

lets keep this secret alive, reed switches are only for professionals :)

From the add you quoted on EBAY "Please note that 3-way reed switches or reed switches with Normally closed contact are quite rare and more difficult to find."

That said they really are not hard to get (Digikey has lots). At issue was your sarcastic 12 cent comment which was not helpful or accurate .... more like showing off which you are not in a position to be doing.

"and have discovered that the math doesnt always add up like it should transferring from paper to reality. " ... actually I find out it pretty much always adds up properly when you do the math right and understand one designs for worst case, not typical values.

Semiman
 

MikeAusC

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Note that VidPro was not the one asking for assistance, he was just the one providing less than valuable assistance (not to mention not accurate) and was most certainly being sarcastic. VidPro you come off as a hack with a chip on your shoulder who seems to have something against those with education and training perhaps because you do not have any? .

Some people, when told publicly that they're wrong, ask questions to learn more - some people just lose it.

The latter certainly shouldn't post information -and VidPro certainly fits in that second category.
 
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