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Electonic timer

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kawakiman

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I am looking to make a circuit board that when I initially provide 12v to it, it provides 12v to an LED for around 3 seconds and then stops. The 12v power supply will remain powering the circuit but I need the LED to remain off after 3 seconds and only restart once the supply ends and is then reapplied.

I have found a photo of a circuit and would buy it but I am in the UK & the supplier wants $12 to ship it. Can anyone please provide the circuit and the value of the items needed to make a similar circuit?View attachment 62733
 
hi,
This is a simple Timer, the output goes High for ~3secs when the 555 is powered ON.

If you make R1 a 10K fixed resistor and a 500K variable in series, you will be able to adjust the ON period.
 
hi,
This is a simple Timer, the output goes High for ~3secs when the 555 is powered ON.

If you make R1 a 10K fixed resistor and a 500K variable in series, you will be able to adjust the ON period.

Hi sir e,

I am just curious. I could not find this kind of connection in the datasheet. I don't know if this is astable or monostable. What confuses me is the connection between pin 2 and 6. For that, it should be astable. But the simulation looks like it is a monostable circuit. What is R2 doing there? Can't the circuit be changed to monostable? I think it will meet the requirement of the problem posted by the OP. And sir, what are the two outputs for?

Kindly help me understand the circuit sir.

Thank you. :D

m8
 
Hi sir e,

I am just curious. I could not find this kind of connection in the datasheet. I don't know if this is astable or monostable. What confuses me is the connection between pin 2 and 6. For that, it should be astable. But the simulation looks like it is a monostable circuit. What is R2 doing there? Can't the circuit be changed to monostable? I think it will meet the requirement of the problem posted by the OP. And sir, what are the two outputs for?

Kindly help me understand the circuit sir.

Thank you. :D

m8

hi m08,
If you look at the 555 , you will see there is no discharge path for the timing capacitor, so its not an Astable, so it acts as a Monostable.

R2 connected to the Discharge pin, is just an alternative output.

As its open collector it requires a pull up resistor, IF the output is used.
 
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hi m08,
If you look at the 555 , you will see there is no discharge path for the timing capacitor, so its not an Astable, so it acts as a Monostable.

R2 connected to the Discharge pin, is just an alternative output.

As its open collector it requires a pull up resistor, IF the output is used.

Hi,

If there is no discharge path for the timing cap, then it is monostable.

I'm ok now with R2. :D

Another question sir, what happens when pin 2 and 6 are connected? When the cap is charged up to 1/3 VCC, the on time starts? I'm not sure with this. Please verify. :D
 
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Many thanks for the diagram, can you please confirm V1 is my 12v battery and the triangle symbols indicates a connection to ground? I assume that the 'Out' pin on the 555 (01) is my 3 second power output?

I really appreciate your help.



hi,
This is a simple Timer, the output goes High for ~3secs when the 555 is powered ON.

If you make R1 a 10K fixed resistor and a 500K variable in series, you will be able to adjust the ON period.
 
Many thanks for the diagram, can you please confirm V1 is my 12v battery and the triangle symbols indicates a connection to ground? I assume that the 'Out' pin on the 555 (01) is my 3 second power output?

I really appreciate your help.

hi,
The V1 is your 12Volt supply and the triangle is 0V [ the minus terminal of your battery]
Use pin3 as the 555 output pin, it goes high for 3 seconds when the V1 is connected.

If you connect an LED to pin 3 be sure to include a series resistor or ideally a transistor to drive the LED or relay..
 
Here's all you need:
**broken link removed**
When this circuit is connected to a supply (from 3v to 12v), the LED turns on and gradually fades after about 3 seconds. The circuit takes no current when the LED is off.
 
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hi me08.
As I explained on Chat removing those components will stop the circuit from working.

Compare these the LED ON times for the two circuits.

An LED on flash of only 0.1 sec.
 
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Hi sir e,

C2 and R4 determines the on time of the transistor of the 2nd schematic. When the cap is fully charged, no current flows to the base so the transistor will eventually turn off which will also turn the LED off. What principle am I missing here? The simulation proves my idea to be wrong. :p
 
Hi sir e,

C2 and R4 determines the on time of the transistor of the 2nd schematic. When the cap is fully charged, no current flows to the base so the transistor will eventually turn off which will also turn the LED off. What principle am I missing here? The simulation proves my idea to be wrong. :p

hi m08,

As I explained the first transistor has a very input impedance so the 47uF charges via the 1meg.

In your version the 47u charges via the Base , Emitter of Q2, look at this plot
 
Hi sir e,

Thank you for the reply. I think I'm understanding this now. On the first schematic, the cap charges through the 1meg resistor. The input impedance of the left transistor is Beta*(10k). Granting that Beta is equal to 100 like your example yesterday, Zin would be 1meg also. Is it correct to say that the cap will already charge through a 500k resistor, the parallel combination of the two 1 megaohms?
 
Hi sir e,

Thank you for the reply. I think I'm understanding this now. On the first schematic, the cap charges through the 1meg resistor. The input impedance of the left transistor is Beta*(10k). Granting that Beta is equal to 100 like your example yesterday, Zin would be 1meg also. Is it correct to say that the cap will already charge through a 500k resistor, the parallel combination of the two 1 megaohms?

hi m08
I ran these simulations so that you can see the charge current values, when comparing the two circuits.
Note: the Beta could be much greater than 100.

E

EDIT:
Look at the calculations, compare to Post #10
 
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