A very well done demo of why a diode is required to direct a collapsing magnetic field to prevent circuit damage. The use of LEDs to make your point was 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. :)
In the (g)olden days before transistors, relays and switches controlled everything, and "low voltage" control was often 24VAC, sometimes rectified to 12V with a bridge. With nothing installed across the coils of the relays, switches contacts would become welded shut from the back EMF. Initially caps were used to shut the back current, but when the price of diodes came down they became the standard
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.
This is the basis of boost converters, having the inductor increase the voltage. The down side is a diode slows down the relay opening by two to three times allowing more arcing of the relays contacts. Since you only need to protect the switching device, a resistor can be placed in series with said diode to increase the clamping voltage or a zener. RC networks can also be effective.
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.
The diode provides a path for back EMF current to flow delaying release of the relay. For high speed relays, sometimes a resistor of the same resistance of the coil results in decay times = to the attack rate. On release the emf voltage is the same as the applied voltage was, so no damaging spike. Sometimes a zener diode is used instead of the resistor. This is sometimes used in protection circuits for fast response. The string of LED's in the video dissipate the coil energy quicker than the shunt diode as the higher reverse voltage changes the current quicker. This can be seen on a scope. For most applications just the diode for protection is required.
This is known as flywheel diode. It provides and easy path to the current which is produced due to back emf. Without this diode the current due to back emf can easily destroy the switch activating the relay.
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.
The main purpose of diode in this connection is to protect additional circuits/power source. We connect diode in parallel connection on this relay coil (Anode to -volts ; Kathode to +volts) in this set up the diode is inactive, {it doesn't affected by the inputvolts/12v}. But right after you cut the power in the coil, this coil generate Back EMF volts in opposite direction > "Anode to (+) ; Kathode to (-)" this voltage polarity activates the diode and in this circuit the diode acts like a wire to short circuit the back EMF. But if you connect diode (Anode to +volts ; Kathode to -volts) in parallel correspond to the load at first, this diode will initially be activated {and is affected by 12v} this diode will short circuit the connection and current will not be able to flow in the load/coil. # We can't use LED/resistor as circuit protector, (it won't fully absorb the back emf) and it can't fully protect the circuit . # The major part of a relay is a Coil, This Coil acts like an inductor (inductor can generate very high voltage or "Back EMF" [current x resistance = Back EMF volts]) that relay coil has the ability to generate thousand of volts and it can cause a considerable damage in your circuit. "this is why we must provide a path for the back EMF/current to flow".
A snubber is a device used to suppress "snub" the “inductive flybackback” from motors, solenoids, relay coils, or any electromagnetic device energized with an inductive load is abruptly interrupted (switched off). When the flow of current is suddenly eliminated, A sharp rise in voltage across the switching contacts polls can create a high voltage electrical spike within the device. If this voltage spike is higher than what the internal electronic components can tolerate, (max voltage rating) permanent damage may result. However, that's not all... The sudden power interruption can also create electromagnetic interference (EMI) pulses that can affect other electronic devices within close proximity. These transients (noise) sometimes produced clicks that can be heard through radios, TVs, amp speakers, etc. Snubber circuits are use to minimize these situations by providing an alternative current discharged path away from the coil or switch that can produce these phenomena.
very useful, thanks. how do we know what diode do we need there? since, for example, for a fan we need very much amps? what kind of diode should we use if there's a 500A draw?
I've explained back EMF to many people but showing in this manner is a great idea. I will help greatly seeing with their own eyes then just trying to imagine it in their heads.
All our DC relays have flyback diodes. That’s what we call them. The term normally applies to cathode ray tubes (CRTs) but we use the term for this application, too.
Hi I 've used solid state relay for my car dome light trigger, as we know car dome light is negative trigger. so I connect pin 3 to continuous +12v and pin 4 to switched ground negative. When the pin 4 swithed to ground the relay work as expected it joint/closed pin 1 and 2, however once the pin 4 ground is switched off the relay is still closed. Any explanation or solution to this issue ?
I usually prefer videos with actual voices explaining what is going on. But your visuals with the LEDs works even better for a novice like me. Now I REALLY understand. Thank you.
@ Tech Ideas Can you suggest 5 volt ac supply relay for activating the relay coil and common pin should be able to handled 12 to 30 DC voltage with around 5 to 10 amperes ?
Since the led and the diode are connected in parallel so the reverse voltage choose to be short by the diode. I would assume it would flow through both but half and half?
I can't fathom why this has over 500+ thumbs down. You couldn't describe the issue any better. Thanks for the technical help, this really clears up a lot for me.
In addition to being called a flyback diode circuit, it is also known by many other names as well: *Snubber circuit, *Clamping circuit, *Suppression circuit, *Protection circuit
@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.
When I wired this up to test it I noticed that the LED pulses faintly sometimes when the switch is closed as well as when it opens again. I suppose this could be due to the contacts in the switch occasionally bouncing when it is pressed.
Why don't the LEDs light up when still connected in parallel with the diode. I understand the diode stops current from negative to positive but why doesnt it still flow through the LEDs?
Very correct demo. While switching off collapsing field develops high emf which can damage semiconductor components connected to relay. Video appreciated.
The direct answer to the question in the title of course can be very brief ....... but you made it very visible to those that think it's not so bad to leave it out
Most of the designers use for home use diode on the relay coil.it is better to add a resistor instead of relay or to add a resistor with a series zenner diode. Have a look on a relay datasheet especially for higher grade application (automotive,military and aviation) The reason for not using diode is that by using it you short the coil induced voltage to a 0.6v and this will be translated in slower operation of the relay and will decrease the lifecycle guaranteed by producer . By using resistor you need to calculate more the max voltage that is allowed on the driver transistor.
i don’t understand . can i use a and 2v Led red light instead of a 0.7V diode? besides, why a 0.7v diode can withstand a reverse voltage which is more than 10 v.
What about the use of a double-pole switch, which also breaks the circuit between the sensitive load (your five LEDs) to prevent any back-emf from damaging a switching transistor or similar?? In this way, there would not be a voltage drop across the relay.
Thanks, I've been reading about this, helps to see a definitive real world example. Wondering if there is still enough to power the original single LED
You should do a follow-up video to show why diode protection aswell as diode protected relays and starter solenoids everywhere important in automobiles as this is why most vehicles on the road are failing with faulty switch panels broken climate control plus even burnt out sensors and engine computers plus transmission changes/overhauls. Even fuel injectors should be manufactured with built-in flyback protection..
How can I connect led lights in my car to come on when I unlock the car with my dome light and also be able to turn on with the fog lights as well?? ..... HELP!! 😩
I have 0 knowledge of electronics and I have to say: your explanation is very clear. Since this seems to be a well known issue, shouldn't all relays come with a diode to prevent reverse current from damaging the relay or the circuit? (I have a wine fridge PCB to trouble shoot and I think I have relay issues... ;))
Makes the point quite well, thank you. One thing I don't get though is how the diode helps relative to its direction. So when the backfeeding happens current flows essentially from "negative to positive leads of the circuit", which is why the LED lights up (anode is on the negative side, cathode on the positive side). But the regular diode is wired in parallel to that with the same anode/cathode arrangement. So I don't get how it helps - isn't it allowing current to flow the same direction as-is powering the LED's? Both are allowing current to flow towards the positive wire towards the switch, no? I'm missing something.
Hello, great video 😄 I'm a beginner and I have a question : Is it absolutely mandatory to put a protection diod in parallel with a relay? Thank you in advance 😉. Cordially.
Thank you. Have you considering a remake of this video with a theoretical explanation of the field in inductor. It would make it one of the best videos on this topic!
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).
Can you please show how to use the diode when working with 5v 4 channel relay module and operating electrical appliances. I just damaged 2 relays and I think the reason is reverse charge.
I'm using 12v relay in my project....I used diode 1N4007,...With 230v ac power supply... When relay gets off my 16*2 LCD display shows unknown characters.. How to solve this problem... Plzzz
The diode only acts as a low resistance path if it is forward biased, meaning plus to anode and minus to cathode (the side with the gray ring). When the relay is "on", the diode is reverse biased so it acts as an open circuit.
@@abbas38073 what i ment to ask is that before current reaches the relay .. it has the diode pathe -short cicuit to ground- so so no current should go to the relay . thanks brother but still confused .
@@aka0989 when the relay is on, the diode will be reverse biased so it is not a short circuit. That means that the current comes from and goes to the relay. But if you open the switch, the overvoltage has the opposite polarity. This means that the diode is now forward biased and acts as the path for the current to flow.
Sir I making blinking ckt with 555 ic timer &one take chargeable china torch light open the torch & saprate connect the blinking ckt parallel to inside the toch light , the the lights blinking after one hour not blinking & not change battery. what will be mistake ,where the connect diode.
1. Flyback diode protects other circuit elements from voltage transients caused by collapsing of the electromagnetic field, created by the relay coil when it energized/deenergized. It also keeps the noise at minimum. The voltage spike may easily be ~100V. So flyback diode is a must. 2. Each LED has to have a current limiting resistor in series. LEDs don't care about voltage flowing through them, but the current.
@@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