I need help with my very simple relay circuit - Page 1
Forum section: electronics / beginners

author: kafor1
author_rank: Contributor
timestamp: November 28, 2018, 06:54:22 am
content: So i made a simple controller/light indicator so my problem is my relay buzzesComponents used555 timer6v relay10k ohm6.3v flashlight bulb idk the input current9v battery470 ohmPushbuttons3v led41 ohmProblemMy relay buzzes after connecting the 6.3v bulb to the NO contact relay (assuming the relay is already energized)Well it buzzes after energizing the relayThings tried:I tried connecting my 6.3v directly to 9v battery and it glows but light is slowly fadingI tried using many resistors lowest value is 41 ohm and highest is 10k ohm it doesnt glow at allTried directly connecting battery to relay, still buzzesChecked if the relay is broken, its not ,because after tryint it with a regular led the relay triggers, and it only buzzes the moment i connected.the 6.3v to the NO(+) and the (-) to the ground.Plz help me i really want to make this

author: vk6zgo
author_rank: Super Contributor
timestamp: November 28, 2018, 07:12:15 am
content: Quote from: kafor1 on November 28, 2018, 06:54:22 amThink about this---- is that telling you something?A 6.3v flashlight bulb is normally used with a lantern battery, what is different about your 9v battery?QuoteYour schematic is really hard to decipher------- what exactly are you trying to achieve?

author: Nerull
author_rank: Frequent Contributor
timestamp: November 28, 2018, 07:32:39 am
content: 9v batteries are hot garbage when it comes to internal resistance, your lamp is likely drawing enough current to significantly affect the voltage supplied to your circuit. You have zero bypass caps for power or pin 5 of the 555, so it's likely oscillating due to the messy supply voltage.I'm pretty sure this is the OPs circuit, it's should be working in bistable mode:It does function in the simulation world, at least.

author: spec
author_rank: Frequent Contributor
timestamp: November 28, 2018, 07:44:49 am
content: Quote from: Nerull on November 28, 2018, 07:32:39 amYeah, I am not a fan of 9V batteries either. Be nice if you could get LiIon versions. Not only is the basic structure difficult with 7 small cells connected in series, but the performance varies radically from manufacturer to manufacturer.IIRC the output resistance of a PP3 is around 60R which makes them unsuitable for supplying high currents. One dodge to improve the performance of a PP3 is to put a 1mF or lager capacitor across the terminals.Incandescent bulbs, if that is what is being used, are a nasty loads. Their filament resistance has a high positive temperature/resistance coefficient, so much so that their initial resistance can be around 10% of their hot operating resistance. This means that they take a huge gulp of current when first connected to a voltage source.

author: vk6zgo
author_rank: Super Contributor
timestamp: November 28, 2018, 07:53:55 am
content: OK, I think I can see what you are doing.The (unnamed) pin of the 555 goes low (In other words, goes close to zero volts, assuming the negative terminal of the 9v battery to be the zero reference)This, in turn, operates the relay, via the 41 ohm resistor, closing the NO contacts, & connecting the 6.3v globe across thesamebattery which powers the 555 and the relay.Looking back to my earlier answer, what did you conclude about the globe's glow "fading"when you connected it directly across the 9v battery?Two things to think about:-Try operating the globe from a totally separate battery.Do you really need the 41 ohm resistor?Seriously think about whether operating the relay directly from the 555s pin is really the best way to do things.

author: spec
author_rank: Frequent Contributor
timestamp: November 28, 2018, 07:57:38 am
content: Crossed postsQuote from: Nerull on November 28, 2018, 07:32:39 amNice work on the schematic.I would guess that the 9V supply line would collapse when you connect a 6.3V incandescent bulb across it. Of course, depending on the wattage of the bulb.

author: vk6zgo
author_rank: Super Contributor
timestamp: November 28, 2018, 08:11:17 am
content: Such a thing is pretty much obvious to those of us with real world experience, but it seems this is much less prevalent amongst beginners these days, many of which have never used, or even seen, a flashlight with an incandescent bulb.I was trying to nudge the OP in the right direction so he/she would work it out for themselves.

author: spec
author_rank: Frequent Contributor
timestamp: November 28, 2018, 08:37:23 am
content: I see

author: kafor1
author_rank: Contributor
timestamp: November 28, 2018, 04:32:04 pm
content: Quote from: vk6zgo on November 28, 2018, 07:12:15 amso my mistake is that i used 9v battery instead of a lantern battery?and yes my schematic is really hard to decipher can u recommend me a software to draw schematic?

author: kafor1
author_rank: Contributor
timestamp: November 28, 2018, 04:41:46 pm
content: Quote from: vk6zgo on November 28, 2018, 07:53:55 amill try using a different battery, ill probably use 4 AA battery to attain 6vand yes i need 41 ohm resistor, if i dont use 41 ohm resistor my 6v relay's coil will probably burn due to the output of the 555's 9v

author: kafor1
author_rank: Contributor
timestamp: November 28, 2018, 04:43:01 pm
content: hey man what software do u use to draw that schematic

author: kafor1
author_rank: Contributor
timestamp: November 28, 2018, 04:48:14 pm
content: Quote from: spec on November 28, 2018, 07:44:49 ami have no idea what ur saying, the only thing i understood is replace my battery with li ion bat, and put a capacitor across the terminals, but which terminals exactly?

author: Efe_114
author_rank: Contributor
timestamp: November 28, 2018, 05:42:43 pm
content: You need to put capacitors between the positive and negative rail of your IC. Also you can try easyeda for schematic designing

author: Nerull
author_rank: Frequent Contributor
timestamp: November 28, 2018, 06:32:51 pm
content: I used Proteus to draw the schematic, mostly because i've been using it for classes, but I don't actually like it all that much. It's nice for real time digital simulation, though. The inability to rotate a part while placing drives me nuts.There isn't a huge difference in the results you get from any schematic capture software, it's mostly just a matter of how well you know/like the UI. I could have done the same thing in kicad, ltspice, eagle, everycircuit, etc.

author: vk6zgo
author_rank: Super Contributor
timestamp: November 29, 2018, 02:19:04 am
content: Quote from: kafor1 on November 28, 2018, 04:32:04 pmAsspecpointed out, the 9v battery voltage will fall due the the large current drawn by the 6.3v bulb.When that happens, two things may happen.(1) The 9v battery can no longer produce enough current to hold the relay, so it will release.This removes the load of the 6.3v bulb, so the 9v battery can again produce the current to operate the relay, which pulls in, putting the bulb back in circuit, so the sequence continues, causing the relay to "buzz".(2) The 9v battery output voltage drops below the operating threshold of the 555, which causes its output operating the relay to no longer be at zero volts, releasing it.Or both may happen.A quick explanation of why battery voltages drop under conditions of excessive current draw:-The cells that make up batteries produce electricity by chemical reactions.These reactions are not perfect, so lose energy in operation.As there are no symbols to show these chemical losses on a schematic, we can "pretend" they are a resistor.They add up in series, just like real resistors, so we can draw an equivalent circuit of a perfect cell (or group of cells in a "battery"), in series with an "internal resistance".Drawing current through this "internal resistance" causes the voltage at the output terminals of the cell or battery to be less than that of the same device not producing current, due to the voltage drop across the internal resistance.Your 9v battery is intended to supply small current devices, & is designed to be compact in size.The downside of this, is it has a higher internal resistance.Put a 6v flashlight bulb across it, & its terminal voltage will drop radically.The lantern battery will be able to supply enough current for your circuit to work & the lower voltage probably won't bother the 555.Or, you could keep the 9v for everything else, & just switch the 6.3v from the lantern battery with the relay contacts.You don't really need software to draw schematics, although they are useful for posting on forums like this one.Try to get a bit proficient at drawing schematics on paper----just grab pencil & paper & try copyingNerull'sdrawing.The 9v battery won't "burn" the relay coil without the 41resistor, because of the battery internal resistance, & because the 555 output doesn't really look like a direct connection to zero volts.Do you have a multimeter, or are you "flying blind"?