Inductive Kick and Snubber help

[Njguy]

I believe a 1N4148 would be blown by that amount of current it's trying to clamp.

The problem with using a 1N4001 type is that they are slow and don't clamp quickly, so the leading edge of the inductive spike goes by unclamped and damages things. A Schottky diode is much better because it's a lot faster. What you want is the class of diode known as "ultra fast recovery" for clamping inductive transients.

A series R-C snubber is faster yet.

Are these the Schottky diodes you were referring to? If so which one exactly cause I'm not very competent when it comes to diodes.
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[kinarfi]

Here's a more protective design

 

[Njguy]

Here's a more protective design

I added in a battery and removed the words negative and positive to avoid confusion. Are the diodes still in the correct position? I thought the other end had to be facing against the negative side.

 

[crutschow]

Whatever. Considering how cheap fast diodes are, I just don't see the wisdom of using junk diodes for clamps. You know that transistor the diode is protecting? Well, if the diode is a shade slow for the peak of the kick, then the transistor gets beat on with narrow duration voltage spikes that gradually punch it through. If you are betting the life of the transistor on it ALWAYS being so much slower than the diode you know the di-dt rate, you better specify the transistor. Me, I use fast diodes for snubbers and junk diodes for rectifiers. makes life simple and means the transistors live their natural life span. I am thrifty, but snubbers ain't where I try to save three cents. I have reworked too many power converters that smoked because my boss pinched pennies on the snubber.

And BTW, turn on times for junk diodes are never a guaranteed (or tested) parameter and I never pin the survival of the main part on an untested parameter. I have seen semiconductors from Taiwan and China that were garb age all wearing the industry standard JEDEC number because they meet the TESTED limits... and the rest of the specs, who knows?

Whatever.

Fast diodes of any nature don't normally specify turn-on time, only turn-off. The semiconductor parameters that affect turn-off time aren't the same that affect turn-on time.

If there is concern about high frequency spikes, these are better suppressed by a small series capacitor-resistor directly across the transistor. That also absorbs any spikes from wiring inductance.

The flyback diode in a switching regulator converter is a different story. These require fast recovery for best efficiency and to minimize current spikes.

 

[giftiger_wunsch]

I added in a battery and removed the words negative and positive to avoid confusion. Are the diodes still in the correct position? I thought the other end had to be facing against the negative side.

Note that a voltmeter should be connected in parallel. It has high impedance and you have it connected in series, which will prevent anything but a very small current flowing through.

 

[Njguy]

Note that a voltmeter should be connected in parallel. It has high impedance and you have it connected in series, which will prevent anything but a very small current flowing through.

There I think I fixed it, also I rotated the diodes, I'm pretty sure they are correct now.

 

[giftiger_wunsch]

There I think I fixed it, also I rotated the diodes, I'm pretty sure they are correct now.

No, the diodes were correct before. As explained previously, they should be in antiparallel with the inductive load. Turn them back around The voltmeter is correctly connected now.

 

[Njguy]

No, the diodes were correct before. As explained previously, they should be in antiparallel with the inductive load. Turn them back around The voltmeter is correctly connected now.

How?? I thought they were supposed to give a clear open path for the excess voltage to go back to the negative terminal. If I put them back, power will flow from the negative terminal through the diodes and cancel out the relay. The relay will never break the circuit.

 

[giftiger_wunsch]

Well for a start, the way you have the diodes connected in that picture, there is a clear conductive path across the battery terminals, with only the two diodes and the (low-impedance) restricting the current. The majority of the current would flow through this path and not through the electromagnets or the relay.

From your statement, you don't seem to understand the polarity of the diodes; the start of the "arrow" is the anode, which is the positive end; the other side is the cathode, the negative end. The diodes need to be given a reverse bias to be used as protection diodes for the inductive loads; that is, they need to be connected "the wrong way round".

 

[Njguy]

Ok fair enough. How about this instead, one diode for each magnet, that way its in a nice self contained package.

 

[Speakerguy]

Whatever. Considering how cheap fast diodes are, I just don't see the wisdom of using junk diodes for clamps.

Calling them 'junk' doesn't advance your argument. They cost less money, which is a good thing. They also work just as well, 100% as well, exactly as well, as 'fast' diodes in this application. 'Fast' diodes just cost more. The inherent forward recovery time in a standard diode is really, really fast. Faster than you can switch a MOSFET.

**broken link removed**

The turn-on time is about 4 ns, which represents the 2246's bandwidth (100 MHz) and the rise time of the 8012B pulse generator (less than 5 ns). The 1N4001's turn-on time is thus less than the test equipment used in my test setup.

How is single-digit nanosecond, or maybe even sub-nanosecond, switching time not fast enough? Why pay more for the more expensive 'fast' diode? Wasted money.

I never pin the survival of the main part on an untested parameter.

Then you've likely never used a 'fast' diode as a snubber, since they don't spec the parameter either?

 

[giftiger_wunsch]

Ok fair enough. How about this instead, one diode for each magnet, that way its in a nice self contained package.

All you've actually done there (besides removing the relay's protection diode) is move the protection diodes on the schematic to a less clear position. Hopefully you understand that other than the addition of the battery and the removal of the relay's protection diode, your schematic is the same as the one provided by kinafi.

Edit: Sorry, just noticed you also added a conductive path on the far left so that the electromagnets are also no longer in series with the relay. Not entirely sure why you've made that modification though...



A question to whoever may be able to answer it: why the two protection diodes for the two electromagnets, since both protection diodes are in parallel with both electromagnets anyway? Is there a specific reason why this would be required? Would the electromagnets not be able to share a single protection diode?

 

[giftiger_wunsch]

Here's my interpretation of what you're trying to do:

That should be sufficient to protect whatever other devices you add to the schematic which are sensitive to the voltage spike produced by the inductive loads from the electromagnets and relay.

 

[Njguy]

Here's my interpretation of what you're trying to do:

That should be sufficient to protect whatever other devices you add to the schematic which are sensitive to the voltage spike produced by the inductive loads from the electromagnets and relay.

Yours looks the best so far because it doesn't bypass the relay which would keep the mags on at all times. At least that was my interpretation. However with yours, even with the relay off, isn't there still a current flowing through the diode attached to the volt meter to the next diode and then the battery? Is the objective of the diodes, to reroute the kickback to the positive terminal?

 

[giftiger_wunsch]

The diode is connected with reverse bias so almost no current is able to pass through it; I believe the reverse current for most small signal diodes is in the order of nA (best to check the datasheet if this is important). The voltmeter will most likely cause more leakage current than the diode would though.

 

[Njguy]

The diode is connected with reverse bias so almost no current is able to pass through it; I believe the reverse current for most small signal diodes is in the order of nA (best to check the datasheet if this is important). The voltmeter will most likely cause more leakage current than the diode would though.

I feel like I'm gonna get a lot of voltage drop with all these diodes. Is there a cardinal rule, or a guide to this? I feel like I'm treading water and missing something simple. I can't seem to get anyone to answer as to what the diodes are doing. I can't picture in my head how diodes in parallel come into play.

 

[giftiger_wunsch]

I feel like I'm gonna get a lot of voltage drop with all these diodes.

They're in parallel, not series. They don't cause a voltage drop.

I'm not entirely clear on how the parallel diodes work myself, nowhere I've read seems to explain it particularly, I just know that it's the accepted method for protecting components from the high-voltage spike due to inductive loads. I believe the diode short-circuits the inductor, allowing the magnetic field to drive a small current back through the inductor, which slows the rate at which the magnetic field collapses and prevents this spike reaching sensitive components elsewhere.

If someone could correct / expand on that it'd be helpful

 

[kinarfi]

Basics, Electron theory, Current flows from - to + outside the source,
Current flows through a diode against the arrow, cathode to anode,
Charge up the electro magnet and when the power is removed and field collapses, It becomes a source of the opposite polarity trying to keep the current flowing in the same direction and you want the diodes to conduct.

 

[giftiger_wunsch]

While it is true that electrons move from negative to positive, it is convention that current flows from positive to negative. You're just confusing the issue by showing the opposite.

 

[Njguy]

While it is true that electrons move from negative to positive, it is convention that current flows from positive to negative. You're just confusing the issue by showing the opposite.

I was actually more confused when you said that. I figured power on a battery flows from negative terminal to positive. What were you saying