Power off time delay circuit revisited

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DonMuncy

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About 10 years ago, with the generous help of you guys, I built a 14 v timer. It works fine. Now I need another one, and assembled one again. When the power is applied and then removed, it will operate the relay but shuts off after about 1/2 second, no matter what resistor I use to bleed the capacitor. Since I (at least tried to) made it exactly the same, I can not understand what I have done wrong. I have assembled two of them, with the same result. I used a different source for the capacitor, but it is a 25v 470mf. Any ideas on how to trouble shoot the problem? Bear in mind, I am a rank amateur and am fumbling my way through.
Don Muncy
 
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Whjat is the hold in current rating of the relay ?

The bleed R is the 47.5K into the base of 2N222 ?


Regards, Dana.
 
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rjenkinsgb said:
Check what voltage the capacitor is charging to, while the switch is closed?
It should be within 1V of the supply voltage.
Click to expand...
The voltage across the leads of the capacitor reads 5.9v while the switch is closed. I have the supply voltage set at 12.9v. But doesn't that just indicate the drop across the 100 ohm resistor?
 
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DonMuncy said:
Please pardon my ignorance. It says it has current rating of 10A at 24v. Does that help?
Yes, that is the resistor I spoke of as the bleed resistor.
Click to expand...

No, that is the ratings of the contacts.

Holdin current is the minimum current needed, after the coil activates the contacts,
to keep the contacts in that state. The latter is much less than the former.

Regards, Dana.
 
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DonMuncy said:
The voltage across the leads of the capacitor reads 5.9v while the switch is closed. I have the supply voltage set at 12.9v. But doesn't that just indicate the drop across the 100 ohm resistor?
Click to expand...

There should be almost no drop across the 100 ohm resistor, if you're dropping 7 volts something is seriously wrong. Is the capacitor the wrong way round?, are the resistors the correct values, or have you wired it wrong. Post a picture for us to check.
 
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DonMuncy said:
The voltage across the leads of the capacitor reads 5.9v while the switch is closed. I have the supply voltage set at 12.9v. But doesn't that just indicate the drop across the 100 ohm resistor?
Click to expand...
Yes, but calculate the current that gives a 7V drop across that resistor.
Sounds like the 47.5kΩ resistor is less than a 100Ω.
 
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DonMuncy said:
The voltage across the leads of the capacitor reads 5.9v while the switch is closed. I have the supply voltage set at 12.9v. But doesn't that just indicate the drop across the 100 ohm resistor?
Click to expand...
Once the capacitor is charged to a stable voltage, you have a simple resistive divider.

If the resistor values were correct and the capacitor OK, there would be 1/476th of the supply voltage across the 100 Ohms: Around 25mV.

(100 Ohms is 1/475th of 47.5K, so a total of 476 lots of 100 Ohms with the separate 100 Ohm added in the chain).
 
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Is there a right and wrong way to install a capacitor? I was unaware of that. You see, my ignorance is profound.
Although I am ashamed to have anyone see how sloppy my soldering is, I will try to get a photo.
 
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I now see that is the problem. I can't believe I have been so lucky to have installed them correctly before now; not that I have that much experience.
Thanks so much for everyone's your help.
 
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DonMuncy said:
I now see that is the problem. I can't believe I have been so lucky to have installed them correctly before now; not that I have that much experience.
Thanks so much for everyone's your help.
Click to expand...
Just take it out, and measure the voltage without the capacitor in circuit - confirm it's pretty close to 12.9V - you might need to replace the capacitor, you may have killed it.
 
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