Thanks

Nice demo of the relay coil's inductance, and the importance of a diode across the coil to dampen the jolt.. Great job.

you kidding dude ? genius

@William Custinne That's a good idea for teaching students, Once they get that shock, they will never ever forget to put that diode in this relay circuit in future. :)

Loved this 👍🏽👍🏽👍🏽

There is a trend in higher current relays with DC loads to use a zener diode in series (opposite polarity) with this diode to collapse the field. It still protects the driver but it collapses the field faster by having a higher clamp voltage. Still safe, maybe between 10 and 40 volts, with a driver rated higher than this. This arrangement opens up the contact faster and offers longer contact life and a lower chance of contact welding which can be a problem in higher current loads and relays, especially with DC relays and loads like the ones used to run car cooling fans. With a simple diode clamp, usually AC loads are fine but a DC load with inductance can arc, welding the contact on if the contacts don’t move away from each other fast enough.

I used to see buckets of relays welded shut that my dad collected from HVAC work. It is cools to now understand what was happening. I also remember these big capacitors we would add to struggling systems to help with compressor and fan motor load fluctuations. The were really simple back then.

Yes - I think the zener is the best solution as, correctly chosen, it gives the fastest relay response for a given permitted inductive spike voltage.

Yeah. Some years ago I bought a 12V mini-UPS device, and noticed it had a standard 3.7V Li-ion battery inside, yet was able to output the required 12V. I obviously needed to learn how such a thing is achieved, and thus learned about the fascinating theory behind the step-up converters.

@Michael David Robinson the pilot If you select the correct diode and resistor the failure rate is infinitesimal, therefore no need for a duplicate. The use of the highest voltage diode, as you mention 4007 that is ok and cheap, try a 1k resistor and then go to a 250ohm, to notice the difference. The problem is each type of relay is different, number of turns on the coli, the amount of flux stored in the core, etc. Each needs to be tuned.

Jhovanie Benliro ooooh

Thank u You are nice man

Thank you

The Engineer back emf ???

you mean flyback

Yes Back EMF

The current is kept flowing by the collapsing magnetic field,if there is no path for that current to flow in then the voltage across the coil will peak at about 7X.

Not Bemf, but self induction

+Javier Garcia Alvarez Thank you

Thank you

Thank you

Thank you

@Y B , you are 100% correct. However, I have seen the term "snubber" used to describe circuits that utilize resistor-capacitor (RC), as you say, in addition to diode only and resistor-capacitor-diode (RCD) circuit designs.

Thank you

simply, the magnetic field in the coil collapses when current is interrupted. That generates a current in the same direction as the original current. When you put a diode in reverse polarity across the coil of the relay it allows the current to flow from the output of the coil back to the input with very little voltage out of the coil(that voltage which forward biases the diode). When you put the string of LEDs as the conduction path from the output of the coil back to the input no current will flow until the collapsing field generates a high enough voltage to forward bias the diodes. So, the collapsing field will rapidly rise to the voltage needed to maintain the current which was flowing before the switch was interrupted, until all the coil's energy is dissipated. The most interesting thing about the coil(almost like there was some intelligence built into the coil) is that inductors are volt-second device, that is the output volts X time is equal to the input volts X time. So if you put 9 volts for 1 second into the coil you get 9 volt-seconds out of the coil. If you put a bunch of diodes in the feedback path(or flyback as it is usually called) voltage will have to rise higher to make the coil conduct. For example, if the output voltage has to rise to 27 volts to conduct, then the conduction time will be 1/3 of a second(27v X 1/3 second = 9 volt-seconds).

@@leeackerson2579 halo is this thing diode same as snubber diode

What is better (or reliable ) a Schottky or a Zener ?

lol you dont understand electricity

@@legobuildingsrewiew7538 I'm glad you do

@@vladstrulev you think voltage and current are disconnected from each other bruh

@@legobuildingsrewiew7538 no. I don't. There is a linear relationship of V and I. Voltage is potential and current is the actual measurable electron flow. In context of a resistor, you must choose a resistor to limit CURRENT flow trough the led. You may use the same led with 5v an 100v if you limit current flow trough the led. For 5v resistance will be lower than for 1000v. If disagree, search for any led resistor calculator and you will see. But in your electronics world it is different, I guess

Thank you

Thank you

Thank you

@@TechIdeasAG you are welcome, thank you!

The instructor probably does not speak English

He did - in text titles - I had no problem following it and thought it was a excellent demonstration of the back voltage effect.

The instantaneous voltage across an inductor is equal to the derivative of the current multiplied by the inductance. (Derivative being the instantaneous rate of change in current.) So, to drop -9 volts across a 1 Henry inductor, one possible derivate would be: 1H * (-d0.9 A / d0.1 s) = -9V That is, a drop in current of 900mA per 100 ms multiplied by 1 H (for simplicity). Now, try the same equation but use 1s, 100ms, etc. Considering all that, I don't think it's correct to state that the voltage across an inductor is normally the opposite of the source when it's de-energized.

Generally speaking (intuitively speaking, lets say) The amount of magnetic flux that forms around the coil is determined by the electrical energy flowing through the wires of the coil. When the electrical flow is cut off, the magnetic flux collapses and generates a reverse electrical flow in the wire determined by the energy of the collapsing magnetic field - how dense it was to start and how quickly it collapses. Without the math to calculate it accurately, the energy that went into the system to create it is proportional, in this case in the opposite direction, to the energy that's released when it falls apart, minus a bit of loss in the form of heat. For anyone who hobbys in electronics and designs circuits using the "hack, hunt-n-peck" method it's a good rule of thumb to assume the reverse fly back voltage will be nearly the same as the input voltage. The speed of the collapse has a large influence over the generated voltage and a polarized metal core can either slow it down or speed it up depending on the direction of polarization relative to the direction of the magnetic flux. Not too many manufacturers use permanent magnet cores in their relays.

I think I understand where you are coming from now. When I don't have the instruments necessary to accurately determine the derivative, it is generally a good assumption that the voltage drop will be approximately the opposite of the source voltage. Thank you for clarifying.

necroseraph7, Holy Molly, that was a hard one to read, let alone understand.

The video demonstrated a fly back voltage in excess of 10V, when using a 9V battery, so it does not appear to be equal.

+Sandeep naik Thank you very much

+Shankar Kumar Thank you

+atdlusdriver Thank you

Farsi

Not even close..

Quick response

Lol!!!

@@TechIdeasAG 😂

+vivek ravi Thank you