Go Down

Topic: Using Mosfets so I dont shock myself (Read 1 time) previous topic - next topic

tjones9163

Nov 18, 2019, 01:44 am Last Edit: Nov 18, 2019, 01:56 am by tjones9163
Hello, I have 2 pictures below,

One is a coil of magnet wire around an empty pencil cap hooked up to a push button and i hook this up to my benchtop power supply and i set the voltage to 12v and the max amps i can get is 5 amps, when i push the pushbutton to watch the effects where the electromagnet attracts an iron screw. But the problem is i get shocked when pushing the button.

The other picture is my hopeful solution, Where i press a push-button connecting it to a positive voltage and this activates a Mosfet, thus isolating me from the circuit.
Is my schematic correct as long as i am using a MOSFET that can handle more than 5 amps?


larryd

Add a kickback diode across the coil.


No technical PMs.
If you are asked a question, please respond with an answer.
If you are asked for more information, please supply it.
If you need clarification, ask for help.

raschemmel

Quote
Is my schematic correct ?
Electrically, yes (except for the missing FLYBACK diode across the coil Larry suggested.

However, FYI, it is customary to draw the load (the coil) directly above the mosfet , instead of on
the other side of the page. Obviously, you could put any component anywhere if they were wired
correctly but anyone looking at your schematic would know you've never drawn one before. If
you want to "appear" to know what you are doing , white out the coil and draw it on the left
just above the mosfet, and put the flyback diode across it.

PerryBebbington

I don't understand why you are getting a shock. Any decent push button should have enough insulation to ensure you don't get a shock when using it. Having said that, I would not expect the button on your photo to switch 5A into an inductive load for very long before it is damaged.

The thing that will stop you getting a shock, other than a better insulated button, is a flyback diode across the inductor.

The other thing that is bothering me is that I would not have thought an air cored inductor with the small number of turns I (think I can) see in the photo would produce much of a voltage when the current is interrupted. Maybe I am wrong, maybe it does produce a lot more voltage then I imagine, but I am concerned something else is going on. When exactly do you get the shock? You say:
Quote
I get shocked when pushing the button
So, do you mean you get a shock as you push it? All the time your finger is on it? As you take your finger off?

gilshultz

Somewhere along I think you forgot to install the MOSFET, this will give you lots of volts when the switch is opened.  Use an avalanche protected logic level MOSFET and you will not need a diode.
Good Luck & Have Fun!
Gil
This response is to help you get started in solving your problem, not solve it for you.
Good Luck & Have Fun!
Gil

MarkT

There is no such thing as an avalanche protected MOSFET.  Some MOSFETs have avalanche ratings, that's not protection, just that the avalanche robustness of the MOSFET has been _measured_ - each device has its own limits in terms of the one-off or repetitive avalanche energy it can survive.

Use a free-wheel diode across any inductive load (at DC) or an RC snubber or MOV (for AC).  You never want the high voltage spike causing interference in the rest of your circuit, so snub it somehow.
[ I DO NOT respond to personal messages, I WILL delete them unread, use the forum please ]

gilshultz


from; AN2344 Application Note
"Rating 'ruggedness' in a datasheet was very difficult because of the great confusion
regarding the meaning of this feature, as well as poor theoretical knowledge of it.
Nonetheless, all of the Power MOSFET manufacturers started to produce avalanche-rated
devices and propose datasheet ratings (although imperfect), to protect themselves and the
end users from this incomplete knowledge. " This was written in 2006 and is still true today.
https://www.st.com/content/ccc/resource/technical/document/application_note/05/13/69/ee/aa/87/49/b6/CD00100956.pdf/files/CD00100956.pdf/jcr:content/translations/en.CD00100956.pdf
Good Luck & Have Fun!
Gil






This response is to help you get started in solving your problem, not solve it for you.
Good Luck & Have Fun!
Gil

tjones9163

Add a kickback diode across the coil.



I have added the kickback diode to my circuit and it seems to be better. Do i still need one across the MOSFET? and if so what is the purpose of that?

larryd

No you do not need the diode across the MOSFET.

There is, however, and internal diode within the MOSFET which is there as a result of manufacturing.


You will often see the internal diode drawn in the schematic symbol as seen below.


No technical PMs.
If you are asked a question, please respond with an answer.
If you are asked for more information, please supply it.
If you need clarification, ask for help.

wvmarle

Hello, I have 2 pictures below, the max amps i can get is 5 amps
I don't think so. One or more of the below will happen if you actually have 5A available, and the resistance of that coil is indeed no more than 2.4Ω.
1) the breadboard won't support that load so it won't ever get near to the 5A you hope.
2) the button won't support that load.
3) the brown wires will start melting.
4) your coil will start glowing.

You're probably not getting more than about 0.5A in that contraption. At this current your coil contains enough energy to parduce some serious voltage when the circuit is broken. A flyback diode will take care of that. A MOSFET is also a good idea, as you don't have all that current flow through the button.
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

krupski

Add a kickback diode across the coil.



Correct advice, but if the OP doesn't know why he's getting a shock then do you expect him to know what a "kickback diode" is or how to connect it?
Gentlemen may prefer Blondes, but Real Men prefer Redheads!

raschemmel

#11
Nov 20, 2019, 05:21 pm Last Edit: Nov 20, 2019, 05:21 pm by raschemmel
"kickback diode": diode to prevent kickbacks?
(just a wild guess...)

larryd

Hopefully people will say, hummm new phrase, don't know what they are talking about, I will give it to Google to see what 'she' says.

:)



No technical PMs.
If you are asked a question, please respond with an answer.
If you are asked for more information, please supply it.
If you need clarification, ask for help.

krupski

"kickback diode": diode to prevent kickbacks?
(just a wild guess...)
Kick-back, fly-back, back emf clamp, "no shocky diode thingy"... whatever. I knew what he meant. :)
Gentlemen may prefer Blondes, but Real Men prefer Redheads!

DrAzzy

#14
Nov 23, 2019, 06:42 am Last Edit: Nov 23, 2019, 06:51 am by DrAzzy
"kickback diode" is not the standard phrase for it; flyback diode is. Band goes towards the side of the coil that's getting the positive voltage (so it normally doesn't conduct). Diode should be sized to carry the maximum current that will be going through the coil.

I am impressed that you're managing to get a shock from a 12v supply - IME it takes like 40v or so to feel a shock on your fingers (though you can feel much lower voltages on your tongue or sensitive parts of your body)

There are only two ways I can think of that you could be getting a shock from it; either:

1. when the coil is disconnected (including via contact bounce - when you flip a switch or press/release a button, the contacts "bounce" briefly (on the scale of milliseconds)), and the coil generates a reverse voltage pulse (an inductor, such as a coil, acts against a change in current - when you disconnect it, it will try to keep current flowing, generating a brief spike in voltage opposite that of the voltage you'd applied to it - if there's nowhere for that current to go (because you disconnected one of the wires) this voltage spike can be many times the magnitude of the voltage you had applied to it*) - this can provide a high enough voltage to give a shock. Without the diode (which gives a path for that current to flow through while the magnetic field dissipates) if you're switching using a MOSFET, it will trash that (and if using a relay, it's good at welding the contacts) -

2. you are using an old non/poorly isolated supply, and it's coupled to the power line. With the supply on, but not connected to any other components, you should be able to touch either or both of the wires without getting a shock. If this gives you a shock, the power supply is not isolated and is not suitable for use with hobby electronics (not only will it shock you, it'll also be liable to trash devices you connect to it).

Also, I am amazed that the poor button is still working - those buttons are meant for at most a couple hundred mA. Are you sure that what you think is a shock isn't actually just heat from the button heating up from the massive current you're putting through it (relative to what it's designed for)? honestly, I'd expect the *wires* to be getting hot at that current - the copper in those dupont jumpers is pretty thin.

* This phenomenon is widely used - it's how DC-DC converters (including the output stage on modern switchmode power supplies) work, and how the spark coil in a car generates the high voltage for the spark plugs from 12v.
ATTinyCore and megaTinyCore for all ATtiny, DxCore for DA/DB-series! github.com/SpenceKonde
http://drazzy.com/package_drazzy.com_index.json
ATtiny breakouts, mosfets, awesome prototyping board in my store http://tindie.com/stores/DrAzzy

Go Up