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Energize a relay with a short delay

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k7elp60

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I needed to turn on a relay a short time after the power was applied. Since I wanted the relay to remain on I figured that a 555 would not
work for the timer. After a little thought I bread boarded the attached circuit. Initially it was tried with out D2. Then I realized that if the
power was removed and I had to start the circuit again it takes time for the capacitor C1 to discharge. Then I installed D2 and the discharge
time of C1 was almost instant. What I haven't been able to use some math to figure the time delay and how the discharge path for the capacitor
when the power is removed.
Any help would be great.
RELAY TIMER.jpg
 
Put a resistor in series with the diode? Then you have a simple RC circuit and should be able to compute its time constant pretty easily, no?
 
As the power is removed, Q1 is already turned on. Current from the top of C1 flows into the source of Q1, out the Drain, through the relay coil and back up D2 to discharge C1. The peak discharge current is limited by the resistance of the relay coil.

What is the purpose of R2?

What is the max allowed source-to-gate voltage on your PFET?

R1 can be up to 1megΩ, which makes C1 smaller for a given time delay.
 
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R2, helps prevent parasitic VHF oscillations.

As the power is removed, Q1 is already turned on. Current from the top of C1 flows into the source of Q1, out the Drain, through the relay coil and back up D2 to discharge C1. The peak discharge current is limited by the resistance of the relay coil.
Until the FET turns off. A residual voltage will be help across C1 unless the +12v switches to 0v.
 
R2, helps prevent parasitic VHF oscillations.
With a bypass cap (otherwise known as a timing cap) from Gate to Source?

Until the FET turns off. A residual voltage will be help across C1 unless the +12v switches to 0v.

The residual gate-to-source voltage can be further discharged by putting a 10KΩ resistor from the source of the PFet to bottom of the relay coil. That still leaves one diode drop across the capacitor. Using a older style, non-logic gate, higher threshold voltage PFet would minimize the effect of that.
 
As the power is removed, Q1 is already turned on. Current from the top of C1 flows into the source of Q1, out the Drain, through the relay coil and back up D2 to discharge C1. The peak discharge current is limited by the resistance of the relay coil.

What is the purpose of R2?

What is the max allowed source-to-gate voltage on your PFET?

R1 can be up to 1megΩ, which makes C1 smaller for a given time delay.
The max source to gate voltage is +- 20V, and R2 helps supress VHF oscillations.
 
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And why would a 555 timer not work for this???
A 555 timer has two states timed bi-stable and mono-stable. In the mono- stable mode it would supply the delay, but after the delay time it would return to the idle mode.
And why would a 555 timer not work for this???
 
A 555 timer has two states timed bi-stable and mono-stable. In the mono- stable mode it would supply the delay, but after the delay time it would return to the idle mode.
You can trigger it on power up, isn't that what you want?
 
.... and if you connect the relay coil between pin 3 and V+ the relay will be off until the timed period ends. It will then switch on and stay on until power is removed.
 
.... and if you connect the relay coil between pin 3 and V+ the relay will be off until the timed period ends. It will then switch on and stay on until power is removed.
Yes you are 100% correct. 4pyro you are also correct. I am sorry for not telling the whole story, in that the negative lead of the relay was all ready connected to ground and I couldn't disconnect it from ground. Thinking further, I could still have used the 555 and used a transistor or mosfet with the control signal from pin 3 of the 555. However I achieved the same result with fewer components.
Thanks to all for your input.
 
Put a resistor in series with the diode? Then you have a simple RC circuit and should be able to compute its time constant pretty easily, no?
Actually no as when the power is first applied the capacitor voltage is 0V
As the capacitor charges to the turn on voltage of the pmosfet (about 4V),
is the time constant I was trying to calculate.
 
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