Time Dealy Circuit

Richard Harris wrote:

You can do this with an RC delay. One way would be to use a switch transistor, as follows:

(look at this with a fixed-width font, or it won't make sense)

On-Off

/ / s d .--o o----o---------+^+-------------. | | ||| | | | === | | | |g | | | | | | | | | | | | | | --- | |+ | --- | .------------------. | 9V | | | | --- | | | | - | | | | | o---------' | Your | | | | Circuit | | | | | | .-. | | | | | | | | | | | | | '-' | | | | '------------------' | | |- | | | '-------------o-------------------------' (created by AACircuit v1.28 beta 10/06/04

The idea is that when the switch is closed, the capacitor will charge up to 9V, and slowly decay down to 0. The P-MOFSET will turn on when it's gate at some point as the voltage decays. The voltage at which it turns on will be determined by the cap and resistor values.

The problem with this circuit, through, is that the transistor will turn on slowly. Because of the fact that you want a long delay, the transistor will stay in it's 'linear region' for a long time, causing the voltage across your circuit to come up quite slowly.

A better way is to use an opamp (or comparator) to sense the delay and turn on a drive transistor with. Here is a rough sketch

On-Off / P-MOSFET / .-o o---o-------o------------o---------+^+-----------. | | | | ||| | | | |+ | === | | | --- 47uF | | | | | --- | | | | | | | | |+ | | | | | .-----o-----. | | | | | | | | | | | | | | | | 1MEG | | | | Your | | 9V .-. | |\| | | CIrcuit | --- | |--------o | | | '-' o-|___|-o -|+/ | | | | | | 1MEG | |/| ___ | | | | | | '--------|___|-' | | | | | | 10MEG '-----o-----' | | .-. | |- | | | | 1.8MEG | | | | | | | | | | '-' | | | | | | | | | | | | '---------o-------o------------o-----------------------' (created by AACircuit v1.28 beta 10/06/04

When you throw the ON switch, the 47uF cap will be dragged up to 9V, and will gradually decay down to 0V. The 1MEG potentiometer is used to adjust when the P-MOSFET comes on. Depending on the leakage of the 47uF cap and the mosfet, it's possible that the thing will decay more quickly than expected, so the potentiometer can be used to adjust the time delay. You can get LM324 chips from Radio shack (it's an opamp). There are 4 opamps on it, so if you use that chip, make sure you disable the other opamps by tying the - leads to 9V, and the + leads to 0. Otherwise, it might cause some kind of oscillations. If you can get a comparator instead, you may need what's called a 'pull up' resistor on the output. Look at the package. If so, a 1k resistor from the output to

The P-MOFSET should have an off voltage of something like 3V, since the LM324 will only get up to about 7V, and it should be completely off. The ON voltage of the P-MOFSET must be less than 9V. My guess is that all of the P-MOFSETs you can get at radioshack will fall into this category.

The voltage at the - input of the opamp will be near

V = 9 * exp(-t/RC).

Where t is the time after the switch is closed.

You can probably get the parts for a few dollars.

You could easily adapt one of these three delay circuits I posted a few years ago for a 'Delayed On Motor':

IRFD9120 on this page

should work.

No, initially the MOSFET will be off. It's only after a minute or so that the MOSFET will turn on. It turns on when the voltage between the 'source' lead and the 'gate' lead gets to be about 8V. It's off when the voltage is less than about 3V.

It won't turn off again until you turn off the entire circuit.

Could you sugest a p-mosfet for this task,

Thanks.

be

Thanks for that, I have another question, On your circuit diagram the capacitor will charge up and then the mosfet will switch and the circuit it will power will come alive. What happens when the capacitor discharges, will it then stop the circuit being powered then be cut off. Will this form of delay then be more like a flasher. Charging and discharging?

Thanks for your time and patience.

for

The formula to use to compute the voltage at the mosfet gate after a bit is

V = 9 * e^(-t/RC)

where t is time, R is resistance, and C is capacitance.

You haven't said which circuit you are going to use. If you are going to use the simple circuit with only the P-MOSFET, then the IRFD9120 starts to turn on when the gate somewhere between 2V and 4V below the source lead, which is at 9V. So, given a 1M resistor and 220uF capacitor (electrolytic is fine), then if it's 2V, you have

t = -ln(2/9)*1e6*220e-6 t = 17 seconds.

For 4 volts, you have

t = -ln(4/9)*220 = 98 seconds

Thus it will start turning on between 17 and 100 seconds. You can't tell beforehand any more accurately. These are the limits given in the datasheet:

It gets worse. The time it takes to turn on the thing completely depends on your load. If you are powering something that draws 9mA (like a 1k resistor) it will turn on in something like 9 seconds. That means the voltage will start at 0, and will increase over a period of 9 seconds to

However, if you are powering a load that takes 10x as much current (a

The second circuit I posted, the opamp circuit, doesn't have this problem. You can set the timeout experimentally using the 1MEG pot to be exactly 60 seconds. Once the timeout is complete, the MOSFET snaps on, because the comparator turns on completely as soon as the voltage drops to equal the voltage at the pot's trimmer.

Regards Bob Monsen

What type of capacitor should I use, electrolytic? And will the capacitor and resistor in the circuit have and influence to the output voltage of the circuit? i.e I can use a high ohm resistor 1-2M and a small capacitor like a 220uf and all will be fine.

Thanks alot, sorry Im very new to electr> > Thanks for that, I have another question, On your circuit diagram the

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of

Look for yourself. Here is the datasheet:

Thanks, Ill have to try your op amp circuit out then, could you give me the pin information for your circuit design. I found the LM324.

the

and a

is

datasheet:

Does this look good.

On-Off / P-MOSFET / s d o-o o---o-------o-------------o---------+^+-----------o | | | | ||| | | | |+ | === | | | --- 47uF | |g | | | --- | | | | | | | | |+ | | | | | .-----o-----. | | | | | | | | | | | | | | | | 1MEG | | LM324 | | Your | | 9V .-. | |\4 | | CIrcuit | --- | |1-------o | | | '-' o-|___|-o--3|+/ | | | | | | 1MEG | |/11 | | | | | | | | ___ | | | | | | o---------|___|-o | | | | | | 10MEG '-----o-----' | | .-. | |- | | | | 1.8MEG | | | | | | | | | | '-' | | | | | | | | | | | | o---------o-------o-------------o-----------------------o

PS Ill send ya some free op amps if this works.

"Robert Monsen" wrote in message news: snipped-for-privacy@comcast.com...

the

Yes, I think that will work.

Ok, let me know.

I've got enough LM324s, but thanks.

The issue with the initial power surge is that the LM324 isn't getting power until after the P-MOSFET, causing the mosfet to be on briefly.

Put a 1k resistor between the output of the LM324 and the gate of the mosfet. Then, put a cap, like a 0.1uF ceramic cap, between the gate and the source of the mosfet. When the power comes on, it should drag the gate up, keeping it from turning on until the LM324 wakes up. If it still has the brief blip, add more resistance, maybe 10k.

Thank you so much, your a genious, works almost perfect.

If ya send me your address ill post ya some LM324's, are you in the UK?

I have one more requirement for this circuit, if you can help. The problem is when power is conected the circuit being powered by the time delay circuit gets power for a fraction of a second. I need it to only get power when the time delay has passed. No current can flow until the time delay has passed, this is critical, is there away to stop this breif power surge.

Thank you agian.

time

..-----o-----.

'-----o-----'

problem is the "unknown" behaviour of the opamp during power-on, thus a typical error in design overlooked by most in this circuit you can only handle it by slowing down the fet method would be to use a series resistor towards the gate, and a cap from gate to ground this again would slow down your output a bit, but might be reasonable compared to thhe surge now

I am not understanding the issue. Why do you want to drain the capacitor? What is the problem you are solving?

That solved the problem perfectly. I have discovered a new problem theo, I needed a way to drain the 47uF capacitor when the circuit is switched off, so I used a switch that when off connects the positive battery terminal to the negative capacitor pin. This drains the capacitor ok but for some reason there is a breif serg to the output circuit. Can this be solved are am i stuck.

Thanks again.

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get

power

me

Yes, I'm in the US, near Oakland, CA.

-- Regards, Robert Monsen

"Your Highness, I have no need of this hypothesis." - Pierre Laplace (1749-1827), to Napoleon, on why his works on celestial mechanics make no mention of God.

no problem, I have fixed. Thanks again, by the way are you in the US?

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