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basic capacitor question


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Hi Mike,
Welcome to this community.
So far, you have gathered most of the information that you need:
1) You have a power supply with a full description and its ratings.
2) You described a purpose: a certain time delay before energizing the load.
3) You know that a capacitor takes time to charge.
But the capacitor needs to be part of a timer circuit, which activates a transistor switch. Please look in our projects section for a timer and I'll look too.

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Try my timer it works well on 6vdc too!
Use a smaller capacitor like 1 uf to 47uf and experiment with it.
It is a delayed off timer. When you push the switch it turns on the load and turns it off after the timming cycle. off. If the switch is pushed again durring the timing cycle it resets the time limit.

3_min_timer.jpg

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Hotwaterwizard,
Your "delayed on relay" is a good circuit that will do the job, but with only a 6V supply, the input divider should be changed.
If you use 220K for both divider resistors, and a 100 microfarad capacitor, I figure that the delay will be about 6 seconds.
Use another 2N3904 transistor to replace the obsolete MPSU10, and use a small relay.
The power must be turned-off for at least 30 seconds, or else the next delay time will be shortened.

Mike,
Do you understand how to build this circuit?

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  • 3 weeks later...

Yevgenip,
The formula for charging a capacitor through a resistor from a fixed voltage source is:
R X C. Where R is in Ohms and C is in Farads. The result of the multiplication is in seconds, when the capacitor will have reached a voltage that is about 0. 63 of the fixed voltage. To calculate the time for it to charge to another voltage level, I refer to a graph in a book.
For example, A 1M resistor charges a 10 microfarad capacitor in 10 seconds.
The capacitor will almost reach its full charge (almost to the fixed voltage level) in a time that is 5 times as long.
These calculations depend on having a measuring device or sensing circuit that has an input resistance that is much higher than the resistor doing the charging.

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