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 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".
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.
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.
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.
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.
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.
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.
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.
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
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... ;))
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.
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.
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?
2:38 why these red LEDs are not glowing when you on switch electrons can flow easily so explain me and why it is glowing in reverse voltage why not in forward voltage tell 🙏
Very correct demo. While switching off collapsing field develops high emf which can damage semiconductor components connected to relay. Video appreciated.
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
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.
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.
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..
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).
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
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
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 ?
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.
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!! 😩
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.
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 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.
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.
Thanks for your video. Just want to correct when connecting diode it will not short circuited,it doesn't conduct. Because u connected diode in reverse direction. Hence it will not conduct.
COILS!!! All relays have coils! Coils like capacitors can hold a charge! However, Coils discharge by the degradation of their magnetic field. Note: this was a poor place to use a relay. As the switch should have been well withing the limits of the LED. With transistors, this circuit could have easily have been done! RELAYS, now are, mostly reserved for current isolation, as in using 5v to control 12v. And any coil will generate ringing into a system, unless measure are taken to have the correct dampener values. This is the part that makes new circuit building for a lot of people, quite hard. Also, you can put the diode between the switching device, and the relay. and a second diode, to allow a chirping or a clicking noise maker to activate. NOTE: The RED LED's are also diodes, (Light emitting diodes.) So the bounce, current can be used for making noise, and other fun such things. The important par is to make sure your bounce controls are rated to handle the pulse. One more thing, diodes coils and capacitors, can be used to make the beloved tank circuit. Just, do not zap ones self, and keep V ratings in mind at all times.
@ 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 ?
@@TechIdeasAG any specs or minimum amp or voltage I should use? Can I use a small diode like the one you use on the video? What's the spec of diode you use? At first I thought I should use a minimum of 6 amp diode or otherwise it wont work, but that 6 amp diode size is too big, I'm looking for the smallest diode I can use so it can save space on the PCB. Any suggestion my friend?
A pretty good way to show visually the "fly back" current caused by the decaying magnetic field of the relay. Just a note: the fly-back voltage will normally be negatively equal to the energized input voltage. 9 volt battery will produce a - 9 v reverse current in the coil when de-energized.
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.
@@TechIdeasAGI am planning to connect a front camera to my car using this relay and switch the feed on the screen using a latch push button switch. Do I still need a diode if I am adding front and back camera on this relay? Is it just safe to connect diode to + and - on the relay?
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.
If you want to learn the easy way(pain makes one learn their lessons real quick 😉) disconnect a coil touching the negative side. 🤣 Just so you know that phenomenon is what we use to create higher voltage to drive the leds on screen backlights.
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.
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 👍🏽👍🏽👍🏽
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.
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.
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".
Jhovanie Benliro ooooh
Thank u
You are nice man
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.
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.
EXCELLENT practical example to visually demonstrate the need and use of a "Flyback Diode".
It was also added to prevent what used to be called relay chatter. Had to do with voltage on the coil.
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.
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.
Thank you
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 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
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.
Thank you
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.
Hadn’t realised that at all. Very clever way to show it. Very clear. 👍🙂🏴
Thank you, it’s a very good graphical way to explain it.
+Javier Garcia Alvarez
Thank you
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.
Thank you
Clever! You did very nice, clearly demonstrativ job. No words at all, but worth than 1000000 words. Keep doing it.
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
Oh, i was working on access door system before and i found this diod and didn't know what it's for.
Thanks a lot.
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... ;))
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.
Interesting display. What is more amazing is that you manage to make a working solder joint with that cruddy iron tip.
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.
Should also talk and explain about the things he is doing. Thanks for hardworking
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?
2:38 why these red LEDs are not glowing when you on switch electrons can flow easily so explain me and why it is glowing in reverse voltage why not in forward voltage tell 🙏
Very correct demo. While switching off collapsing field develops high emf which can damage semiconductor components connected to relay. Video appreciated.
Thank you
Standard LED function - Long lead is positive. It's easier than finding the notch on the case.
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
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.
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.
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..
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).
@@leeackerson2579 halo is this thing diode same as snubber diode
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
Great video! I want to start experimenting with computer controlled circuits using relays, and this was very helpful.
Thank you
Excellent video. Well done.
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.
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
Clamps counter EMF from the relay coil when power is removed.
Very very very very cool way to explain. Thanks
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 ?
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.
Well explained thanks. As per comments below, a zener diode works better/faster to prevent the relay contacts being welded by the sparks ignited.
What is better (or reliable ) a Schottky or a Zener ?
I always wondered why a diode was required. What rated/type of diode would I need for a 5v and a 12v circuit?
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!! 😩
Well done. Exactly the info I needed.
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.
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 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.
Good information.
I know using diode relay with seris. but i don't know what this using. Thank you for your grate job.
We can avoid this problem through a diode(1N4001/7) ..it is called free wheel or fly back diode
¡Very useful, and very dydactic your video. Thankyou vm. Health for you!
simple!!! this is for voltage spike like -300v to -400v at tiny little amp which also kill you another sensitive electronic component
Now lets put a fuse in the mix lol, so it's ok to get any old relay from the junkyard, I've seen people do that, but which one?
Why is the negative connected to 2 pins.? It should only be one.?
Good information through simple explanation.
Great demo!
Thank you
so it makes a short circuit , but how can the current go to the relay when the switch is on , i mean the short circuit we made is still there right ?
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.
@@abbas38073 and where does to current flow in the end
Hi, how would you know what diode to use in a circuit?
You should give explanation in the place of background music...
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.
12 volt relay me kitne number ka diod lgega Maine 4007 lgaya tha Jo blast ho gya.
You can use any
I understand that when a diode is present the reverse voltage goes directly back to the battery. Doesn't that damage the battery though?
The diode short-circuit the voltage, so there is no reverse voltage going to the battery.
@@TechIdeasAG does that mean that the current goes in a loop between the inductor and the diode?
The reverse supported voltage of a single LED is not much more than 5V, so, I would not have connected it to the 9V battery.
Hii
What is the Difference between AC Wires and DC wires ?
Thanks for your video.
Just want to correct when connecting diode it will not short circuited,it doesn't conduct. Because u connected diode in reverse direction.
Hence it will not conduct.
I usually hate techno music. But it actually complimented your explanation. Strange, but satisfying.
+atdlusdriver
Thank you
Is it required for AC currents?
COILS!!!
All relays have coils!
Coils like capacitors can hold a charge!
However, Coils discharge by the degradation of their magnetic field.
Note: this was a poor place to use a relay.
As the switch should have been well withing the limits of the LED.
With transistors, this circuit could have easily have been done!
RELAYS, now are, mostly reserved for current isolation, as in using 5v to control 12v.
And any coil will generate ringing into a system, unless measure are taken to have the correct dampener values. This is the part that makes new circuit building for a lot of people, quite hard.
Also, you can put the diode between the switching device, and the relay. and a second diode, to allow a chirping or a clicking noise maker to activate.
NOTE: The RED LED's are also diodes, (Light emitting diodes.)
So the bounce, current can be used for making noise, and other fun such things.
The important par is to make sure your bounce controls are rated to handle the pulse.
One more thing, diodes coils and capacitors, can be used to make the beloved tank circuit. Just, do not zap ones self, and keep V ratings in mind at all times.
I'm so glad I found this video thanks for sharing 👍👍👍👍👍
Really good video
Lekin switch ko stand mod pe kaisea rakhna??!ex.0n rahe ya off
Very well explanation bro thank you!
Not only for Relay it also mandatory of all kind of inductive load like dc motor
Nice tutorial, No need of bla bla blaa..speech .Every thing was well defined in this video.Any way ,really good demonstration..Thank you.
Thank you
please think and make and idea to use this near 10v return voltage for any project,
thanks good explain.
@ 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 ?
Music a little distracting but very nice job on presentation. Well done sir!!
Thank you
@@TechIdeasAG you are welcome, thank you!
Thanks
nice video, keep making video like this one
Thanks, you saved my day
What diode size should I use for motorcycle relay, I want to create a relay board for my bike.
Any doide will work fine
@@TechIdeasAG any specs or minimum amp or voltage I should use? Can I use a small diode like the one you use on the video? What's the spec of diode you use? At first I thought I should use a minimum of 6 amp diode or otherwise it wont work, but that 6 amp diode size is too big, I'm looking for the smallest diode I can use so it can save space on the PCB. Any suggestion my friend?
I was thinking if you can use a whistle as a switch and im searching for any ideas related to that
Crude but effective.
Larry
Very good knowledge booster
A pretty good way to show visually the "fly back" current caused by the decaying magnetic field of the relay. Just a note: the fly-back voltage will normally be negatively equal to the energized input voltage. 9 volt battery will produce a - 9 v reverse current in the coil when de-energized.
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.
Put leds in series with diode...that will tell you if it's actually stopping it...
All issues clear nice vedio 👌
Hello master excuseme in this vedio which model relay you use 5V 9V or12V relay thank you for answer
Here I've used 12v relay, but you can use any
@@TechIdeasAG Excuseme master i have a question 9V battery can control 12V relay? it can normal working? thanks for answer
@@TechIdeasAG Thank you master
thanks for the very nice explanation sir😍
is it possible if i use this method in Solenoid??
Where the diode connect to the battery positive parallel or series or connect to the circuit in parallel please explain sir.
Perfectly explained bro
+Shankar Kumar
Thank you
What diode did you use? Please share details
You can use any diode, like 1n4007, 1n4148 etc.
@@TechIdeasAGI am planning to connect a front camera to my car using this relay and switch the feed on the screen using a latch push button switch. Do I still need a diode if I am adding front and back camera on this relay? Is it just safe to connect diode to + and - on the relay?
Self induction current??? It's because of the coil in relay
Awesome project. Very helpful.
Many thanks for.making great video .....
We understand now why
Now I know why I did not understand electronics.
Can I use smd diode? Thank you
Very useful video, thanks for sharing
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.
as relay is both side open, so diode, to flow current in one direction
Not even close..
If you want to learn the easy way(pain makes one learn their lessons real quick 😉) disconnect a coil touching the negative side. 🤣 Just so you know that phenomenon is what we use to create higher voltage to drive the leds on screen backlights.