Purpose of the Diode in Doorbell Circuits? Test circuit and Experiment!

[@brandonsaffell4100]

The day before you posted this I was working on how to make an electronic doorbell. I used your video as a guide for the circuit and everyone was spot on. I changed from a half wave bridge rectifier to a full wave, since I was concerned that the power drawn by the speaker would cause voltage dropout with only a half wave. This meant that I had to install a switch mode power supply between the two since the Uno didn't handle 21 volts well. In hindsight I could probably just have gotten an appropriate switch mode power supply in the first place, but the trigger portion of the circuit was on the same perf board as the full wave so I left it. Regardless, it works amazingly! Thanks for this video.

[@somegirl9877]

I know u were paying full attention to the homee improvement show. Glad you were. Excellent job on this video! I really appreciate it.

[@marhar2]

Really clever! And a great level of detail. The hand-drawn schematics are fine, perfectly legible, and it's actually better than if you had added the rest of the doorbell functionality, since we can concentrate on the new information.

[@ronpeterson2484]

Kevin, very well done. Thanks for taking the time to educate us all.

[@mitchellpurelife]

thanks a mil Kevin. I was just wondering about the diode in the doorbell yesterday. I hooked up a friend's new door chime received from China, and he put a diode in my hand. I said what the heck is this for? so knowing we've got AC going on I figured that somewhere it needed rectification. thanks to your video, I can answer this question now, and hook it up correctly. Awesome!

[@Chowmix12]

Love the whole back story to this video, great and very informative

[@maxximumb]

Very interesting. I like the method for detecting a negative peak as an input.

[@af1kradioham929]

Enjoyed the video and it confirms my suspicion on why it was needed, thanks! Now, riddle me this. I got a new "westminster chime" unit for my wife as a xmas gift since she suggested it (that was easy). But, I'd now like to take the old "ding-dong" unit and install it in my basement/mancave/hamshack/workshop. I thought it would a 10 minute job. But, with the diode there now in the push button, the half wave ac is more than enough to put the "ding dong" doorbell into a state of continuous humming. Luckily there is an unused third wire in the cable from transformer to new doorbell chime and then to front door push button. So, if I could find a double pole push button I could split the two chime units onto two separate circuits (one w/ diode & one w/o diode). But, I though I'd pass this dilemma by you for comment, ideas, or suggestions. Let me know what you think when you get a chance. Thanks & 73's de Rich AF1K

[@FredHerrman]

Awesome! Thanks for putting out the effort to explain this. It is much appreciated.

[@lasersbee]

Yeah the older non-diode/pushbutton circuits were used on simple solenoid ringers/bells and are still used today. The newer electronic annunciators require a power supply bypass through the button. Well explained.

[@jgrunder]

It might be useful to mention that many new digital door bells include a space to install batteries. Installing these batteries keeps the circuit active, and most times resolves the transformer/diode issue on its own.

[@ScrewThisGlueThat]

So my neighbor was installing a "Ring" door bell and he showed me the diode he was going to install.... and I went "WHAT" why would you have to install a diode? It has driven me crazy trying to understand why and then I come across this video. Thanks, now I can sleep again at night... ;) BTW... I was almost right in my thoughts about why it would be used, but I am not nearly as smart as Kevin

[@jayherde0]

This kind a video is well worth seeing. .. I don't know what they use for a light source in lighted buttons, but the only thing keeping that diode from being an LED is the low current that would pass through it.

[@Kevindarrah]

thanks! hmmm, sounds like a good idea for a part 2!

[@johnbuckley41]

Jay Herde qqq

[@ColinRichardson]

Nice. Makes a lot of sense. Now I just need to figure out how to get the cable to go everywhere in a pretty way.

[@sikamikan]

great video man, Please keep the circuit explanation videos they are inspiring

[@JanSeewald]

Great! Didn't know that and it's genius! Thanks for describing! :)

[@mikes9939]

Excellent video, I was wondering also what the diode did as I am installing a Ring doorbell Pro on a friends house and I saw the diode when I did some preliminary checking. I had just installed one on my house but it has the conventional mechanical doorbell and my friends house has a newer chime. Great idea on your part to use an opto-coupler as it isolates one side from the other. Thanks for this video.

[@oetken007]

Great video. You need the sevond diode for an other reason too. If you press the button without the second diode, the negative portion hits the positive side of the cap and discharges it with a relative high current.

[@tomasulicek]

Very good explanation, great work!

[@andrewp.9824]

+Kevin Darrah - This is good work, but I would be inclined to use a bridge rectifier and maybe a larger electrolytic capacitor in a real doorbell application, instead of the single 1N4007 rectifier that you showed in your test circuit schematic. In standby mode the regulator would still produce 4V, but when the sounder needs more power, i.e., the pushbutton is depressed, a full-wave rectifier would make twice as much power available to the doorbell. Since you have an optoisolator in your scheme, the loss of ground reference on the power supply wouldn't matter; it would still work.

[@cursoderobotica]

Interesting!! Good video! Thank you

[@jpalm32]

The prices for stupid doorbells at Home Depot are crazy. You could put Arduino at every door with power supply & add sound module and still have money left over. Good vid! PCB it and I'll buy it! Need 3 doors though

[@FishKungfu]

Very cool!

[@AlCatrraz]

you say 100 micro FARAD yet the scheme shows 100F or is it 10uF ??? I am not an electronic engineer... I have some understanding... So, I have this door-bell PROBLEM, it does not activate the WHOLE TUNE/RING unless the button is held DOWN.. I used the DIODE as recommended, tried both polarities and it does not work... The batteries, when installed have no affect either.. I was wondering if there is a way to add A SMALL CIRCUIT with a capacitor that will charge when the button id pushed, and will keep the CHIME GOING as it discharges? Is this here your solution? THANX

[@nathonbrown3596]

another great one. even though there wasn't a arduino in it.

[@Kevindarrah]

Thanks!

[@bobmist1010]

Great video I know you can help me.I have a Nutone Concerto LA61 Door Chime that does not play the programmed chime when front door button is used.Front & back door bell buttons and wiring to unit checked & are ok.Rear door chime is fine and all buttons for programming on unit are fine.The unit plays the chime when tested from unit.It appears the trigger circuit is bad for the front door.I wired the front doorbell to initiate the rear door trigger so I have a simple single chime for now.I am not sure where and what it would look like on the circuit board.Thank You Bob

[@danielbrothers2914]

how can i make the chime continue to ring when the button is held in...down. I want ring after ring when the power comes into the button, which is held in. Kinda like the opposite of a freezer lost power alert... I want the opposite...which is ... alert me when the AC power comes on...got any ideas??

[@joecardenas9681]

Great video, thanks. The reason I'm looking into this is that the bell doesn't work with the diode installed, but removed plays the first portion of the tune. Should I suspect a bad diode?

[@nathanbriggs74]

diode backwards?

[@joecardenas8651]

@@nathanbriggs74 I can't remember anymore, I'm sure I tried both ways. I quit that doorbell and went with a Ring. Thanks for the response.

[@marko6394]

Well done video, but if you already have a 1N4007 diode in your circuit prior to the LM317, then the diode at the button is redundant is it not?

[@DiffEQ]

No. If the diode is not across the pushbutton then the circuit is open and the chime is only powered during the button press. That said, this is not typically how the circuit is implemented as it required a specific orientation of the bypass diode. See my other comment for a full explanation.

[@BujingsTV]

hi there I've watched all over the video and looking what was the diode all about and I am understand that the diode can use a rectifier but I really don't get why the diode need to placed in the switches, In layman's term what will be the significant used of the diode there.Is it for the sound quality of a doorbell, for the delay sounds if you press it, or much longer sounds..why it should be there..Please elaborate Sir, only electronics enthusiast and electronics engr. understand the point of that experiment. , it was explained the flow of current and the output but as a viewer who wants to understand the real use of it in laymans term wasn't explained clearly to us.

[@DiffEQ]

The button is normally open so the power circuit to the chime is open. The diode allows half the power to the chime until the button is pressed and full power is sent to the chime. The chime doesn't need this power to operated, it just uses it to detect the button press. By the way, not everyone is writing to the audience that has ZERO understanding/background. Asking someone to teach you fundamentals is asking a lot. You have plenty of resources for learning electronics if you want to learn... which I think you don't. So, get used to not understanding deeper explanations.

[@BujingsTV]

Man thank you for your reply,Im an electronics engr though I practice in broadcasting industry that is why I am more on rf transmission...I am requesting for the explanaition for those who are not clearly understand the electronics and not for me

[@mitchelletzkin1321]

Still dont understand how the doorbell knows button is pressed. It looks to see when wave pattern is changed? Does doirbell have microprocessor?

[@DiffEQ]

His reasoning is no longer applicable. Think of is this way: The diode prevents half the voltage from reaching the chime until the button is pressed and then the full voltage (diode shorted) is seen by the chime.

[@firesurfer]

Gaahh, I ordered a chime from amazon warehouse and the diode was missing. I guess the 1N4007 is the right one? It's a Westminster chime.

[@DiffEQ]

any small rectifier diode will work. 1N4001, 4002, etc. Nothing larger than a 4001 is necessary. The 4007 has a PIV of 1,000 Volts and that's completely unnecessary. the 1N4001 has a PIV of 50V and will do just fine.

[@oraszuletik]

Put a counter on the square signal and let the patient guests (=press the button long) get in.

[@docnez8562]

I bought a Ring doorbell and it won’t stay on for setup and video. Can I eliminate the old ding-dong bell and relay to provide continuous power to the doorbell cam? I don’t need the old bell since it came with a chime that is plugged into an outlet.

[@DiffEQ]

Depends on the model. Use Ring as your source as your problem is specific to that product. Really? I happen to know that older Ring doorbells required a limiting resistor if you bypass the original bell (which wouldn't work, but provides a current limiter for the Ring doorbell.) I think that requirement is now gone. But, again, ASK RING!!

[@victorperrin5891]

I have a Ring doorbell and a digital chime so there is a diode installed in the Ring. My problem is that I am getting ghost rings when no one is there. This happens at various times of the day and night. Do you think the diode could be breaking down and causing a ghost ring? I have been in contact with Ring and they are scratching their heads. They graciously sent me a replacement but after a week or so the ghost returned. Appreciate any ideas.

[@DiffEQ]

The problem is likely a cheap chime that is overly sensitive to voltage fluctuations/spikes or your power source (LV transformer) is too close to the voltage limits and then line voltage variations will cause the chime to see that as a button press. I've always wondered why people think when they call the manufacturer they think the person in the call center knows anything more than what's on their screens. Bottom line: it's clearly either an improper power issue or your chime is just hyper-sensitive. If you put another silicon diode in the circuit (same orientation, in SERIES, not parallel,) then you will know the voltage down about 0.7V and that may just be cure your issue. a 1N4001 will do just fine.

[@jp040759]

So based on your thought process the diode has to be wired in with the correct polarity for your circuit to work. Plus if it was installed backward you could destroy some components. Did the TV show specify anything about the installation of the diode in the correct direction??????

[@Meneltour27]

i dont think it will destroy anything of you have it backwards because the positive side is protected with another diode. the chip will be inactive offcource and some voltage will go to one of its pinouts but thats about it from what i can see

[@Kevindarrah]

yes, that's right, nothing bad would happen, just wouldn't work right. You wouldn't have the always on power, so the chime would only get power when you press the button

[@MrPaulLiu]

Since your opto is already a diode, couldn't you skip the diode in the front?

[@Kevindarrah]

thought about that and maybe, just need to check the reverse voltage rating of the opto.

[@over2there]

What happens when you reverse the diode in the switch? The system would not work. I would propose to use a full Bridge diode rec. and do the detection over the frequency. You can do this with eg with an ne555.

[@DiffEQ]

Detect frequency?? Overly complicated. The half-wave rectifier reduced the FW bridge output until the button is pressed. A simple voltage divider will suffice to detect this difference. (100% increase)

[@djfriz1]

Where are you getting the ground reference in the circuit from? It is not clear in your drawing.

[@Kevindarrah]

just the leg of that transformer (the N in the drawing)

[@i8BBQ4Lunch]

Nicely done. :-)

[@robertopuig7862]

I don’t have a doorbell, it just goes to my phone, do I need a diode?

[@mytruth2152]

Great Video, but I am having a different problem. When I connect my new video doorbell it does not stop ringing. Seems to be a constant power supply. Any ideas?

[@DiffEQ]

Seriously? You think this poor guy is now going to field all the doorbell questions on the Internet because he made a video? Really?

[@sidrausman308]

My door bell rings to push any other button of the home.what is the reason.

[@DiffEQ]

Try again... in English, this time.

[@AlecKristi]

@@DiffEQ what is the point of being rude to someone who commented 3 years ago and clearly doesn't speak good English? trolling?

[@tablatronix]

is this a zero crossing detector ?

[@Kevindarrah]

I suppose it could be. Only thing is that the input voltage will not be quite at zero when the opto kicks on, so you'd have to account for that.

[@DiffEQ]

So, your circuit idea does work but this is not, typically, how it's actually implemented, commercially. The problem with your "solution" is that it requires that the push-button diode to be installed in a specific direction. If you look at instructions for Hampton Bay, for example, the diode direction does not matter. It's also why there is no mention of having to reverse the polarity of the diode if the doorbell isn't working correctly. (I believe older Ring doorbells did only work in one direction, but this is the problem with the Internet as information becomes dated or is only specifically correct.) The actual implementation for this is not polarity sensitive and works no matter the orientation of the bypass diode. Easily accomplished since the output of a full-wave bridge rectifier will only be half of what it could be if there is no half-wave rectifier (the bypass diode) on the AC input. Now, when the button is pressed the voltage of the FW bridge is twice what it was and THAT is what is detected. The voltage regulator takes care of the temporary voltage increase (ringing a doorbell is a momentary operation, anyway) and the FW bridge ensures that no matter which half of the AC wave form you randomly detected the circuit works as intended. This is actually no more difficult to construct then what you suggest and the quiescent state of the system limits the power drop across even an inexpensive linear voltage regulator.

[@AlecKristi]

you do not get "half voltage" from a half bridge rectifier in any normal operation. you will only get "half voltage" in a low pass filter. there are plenty of detection circuits out there, but the first generation of these doorbells was in fact polarity sensitive, I have dealt with this a lot back in the early 2k's.

[@DiffEQ]

@@AlecKristi Incorrect. The RMS voltage (what is used in AC measurement) is the equivalent DC working voltage. With only half the wave present the resulting DC voltage averages to half of what it could be since HALF THE WAVEFORM is missing. Period. This, in essence, is a form of pulse-width modulation which is commonly used to dim LEDs. Not sure why you chose to argue a subject you don't really know (electronics) but your only anecdotal "argument" is that you "dealt with this a lot" TWENTY YEARS AGO. Ugh. BTW, I did say this video was dated. I'm pretty sure people from 2002 aren't here reading this for information to take back to 2002. Specifically because the first video doorbell was available in 2012 (DoorBot) and Ring didn't have one until 2014! SMH Stay in your lane. There is no such thing as a "half bridge" rectifier, genius. It's either a Full Wave rectifier (which requires a CT transformer and is exactly half of the secondary voltage winding) or a Full Wave Bridge rectifier, which is WHAT IS IN MY COMMENT.

[@AlecKristi]

@@DiffEQ sounds like you have great knowledge of theory, but zero understanding of reality. also, you're confusing energy with "voltage" and "rms measurement". half wave rectified DC isn't "half voltage". especially, since in case of electronic doorbells, the very next component after the rectifier is a capacitor, and unless the circuit draws all the available power that is supplied (hint hint - it doesn't), then the voltage of the power supply will be higher than the "rms voltage" prior to recertification. don't believe me? set up a transformer, a diode, and a capacitor on your desk, and get out your meter. also, I'm very familiar with PWM, I have been manufacturing PWM dimming controllers for over a decade...

[@DiffEQ]

@@AlecKristi Alec, you really shouldn't argue with a 40+ year retired electrical engineer about basic electronics. You don't even use correct nomenclature, so it's really not something you should be arguing. I made no mention of energy and neither should you as the point is one about voltage, not power or energy. I clearly said that this is NOT how it's done in the present and did not suggest FW modification of his circuit. Clearly. That said, my original statement is correct and stands that the output OF THE FULL-WAVE BRIDGE will be half of what it is when half of the input is clipped. Now, if you want to put filtering/storage caps on the output of the rectifier to allow it to charge to Pk or Pk-Pk values, then you're arguing something different than what I stated. My FW bridge output will definitely detect when the input is changed from HW clipped sine wave to a full sine wave. Period. If you add circuitry that hides this detector function, then that's on you. Don't chide me about understanding or pretend that I'm some academic individual with no real world experience. Undoubtedly, you or someone you know is using a device that I helped bring to fruition. How's that for reality?

[@AlecKristi]

@@DiffEQ ok, you win a semantics argument. English isn't my first language, and I didn't learn physics in English, so my nomenclature may not be as exact as you'd like. however, if your skip the semantics, and stop counting "technical points" by taking the power supply circuit components out of the real world, and assuming that you'll be using a "spherical bridge rectifier in a vacuum" instead of a real power supply circuit, your may start to realize that you are in fact not correct, despite being a retired electrical engineer. while it would be possible to "detect the voltage drop" the way you describe it, it would be just as cumbersome in the real world application as the OPs solution in the video, if not more so. you'd either have to introduce another diode before the filtering in the power supply, and another load resistance, or have a separate rectifier for the doorbell circuit from the detection circuit. also, please keep in mind that I'm not arguing the concepts with you, but a real world application in a real world circuit, that has to do with half-wave trigger detection, specifically in an electronic doorbell. also, it looks funny when you like your own posts ;)

[@ezraburton6152]

The push of the button opens the circuit not closes it. My doorbell is lit...closed circuit. The solenoid is energized and hammer retracted on chime. When doorbell is pressed, circuit opens, light goes off, and spring loaded hammer is released and strikes bell.

[@DiffEQ]

I, with 40 years of electronic and engineering experience, have never seen a normally-closed bell loop. In fact, with the old mechanical chimes (all were) a normally-closed loop would NOT HAVE WORKED and what you described is a mechanical chime! The spring keeps the plunger retracted and energizing the solenoid is what opposes that spring. You couldn't be more wrong if you tried. So, keep your chiding to yourself because are not just wrong, you are VERY wrong. The button is illuminated, genius, BECAUSE THE CIRCUIT IS OPEN and the applied voltage is measured across an open. Ugh. Seriously? Did you notice the light goes out when you press the button? That's because the voltage across the light goes to ZERO when the button shorts across it. Some people.

[@AlecKristi]

a capacitive dropper with an optioisolator would be a whole lot more efficient, and the polarity of the supply would not matter.

[@DiffEQ]

Makes no sense. Which means you either don't know what you're talking about or you just don't have any communication skills. Full-wave bridge solves the polarity issue and no opto-isolator (huh???) necessary for any of this. Button open: half voltage. Button pressed: full voltage. Simple voltage comparator done with passive components.

[@AlecKristi]

@@DiffEQ let me enlighten you, since it makes no sense to you... where he's showing the button detection circuit that relies on the optoisolator and the diode at about 9 minutes into the video, you can replace that by a simple capacitive dropper and an optoisolator. at half wave, regardless of polarity, the opto will stay off, because the capacitor is charged in the single direction of the current. at full wave, the current will pass, and the opto will trigger. that way, it will not matter which direction the diode inside the button is pointing, and the rest of the doorbell can use a full bridge rectifier for its power supply. now, this was 5 years ago... now I would use a detection circuit that's even simpler ))) also, to answer the second half of your reply: no, you will not get "half voltage" with a diode... which tells me that you have no understanding of what you're talking about...

[@pyroslavx7922]

Why you need constant power to MCU, those "singing greeting cards" play in miliseconds after activated, why a doorbell "player" can't "boot" in relatively no time?

[@DiffEQ]

You miss the point. The chime would quite playing as soon as you released the button. It's not so the chime will work when the button is pressed, it's so it will work when it's no longer pressed. THINK! BTW, that is a design issue and would have nothing to do with this explanation, anyway.

[@thoughtyness]

So if your doorbell goes ding doooooong then it won't have the diode.

[@Kevindarrah]

yea, I guess. Unless that is the electronic sound the chime was programmed to simulate

[@rj44319]

na bra

[@slyspy9819]

This is your idea of a simple doorbell circuit ? Who is your target audience because the average guy is not understanding your electronic terminology and components . The question was why the "Diode" ? You went a little crazy !

[@DiffEQ]

It's about as simple as it gets. If simple AC rectification is hard for you to understand, then you are just out of your league. It doesn't get much simpler than this. Sorry you think he should also have given you an electronic fundamentals primer while explaining what's going on. The information is only specifically correct, anyway, and is rather dated, now. See my full explanation if you care... but it's not written for people who don't even know what a diode is or why it has to be used in the power circuit when using an electronic chime. Hint: electronics need to be powered to work.

[@slyspy9819]

@@DiffEQ Hi Richard I hear ya but I was just wondering if this video is directed to average homeowner or was this tech advice ? At first I thought the former but as he went on it got more and more technical and I said wait who is he talking to for a homeowner would be lost . i did go to Electronics school back in the 80's and worked as a tech for 6 years so i do have some experience . I decided after 6 years I was looking for a change so i went back again to school for Electricity . I've been a Licensed Electrician for 34 years now .