So i made a simple controller/light indicator so my problem is my relay buzzes
Components used
555 timer
6v relay
10k ohm
6.3v flashlight bulb idk the input current
9v battery
470 ohm
Pushbuttons
3v led
41 ohm
Problem
My 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 relay
Things tried:
I tried connecting my 6.3v directly to 9v battery and it glows but light is slowly fading.
Think 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?
I tried using many resistors lowest value is 41 ohm and highest is 10k ohm it doesnt glow at all
Tried directly connecting battery to relay, still buzzes
Checked 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.p
Plz help me i really want to make this
Your schematic is really hard to decipher------- what exactly are you trying to achieve?
so 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?
As
spec pointed 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 copying
Nerull's drawing.
The 9v battery won't "burn" the relay coil without the 41

resistor, 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"?