Hello, i'm looking for a simple delay off circuit. I have a switch which
controls the 120vac via a relay which turns on a computer power supply. I
am looking for something to be able to turn off the switch and then have the
120v be turned off about 3-5 minutes later (cooling purposes) the relay
would still have the 120 going to it, when switched off (aka no battery
system or anything). I was planning on paralleling a second relay with the
first, one controled by the switch and the second controlled by the timer
circuit powered from the power supply. any suggestoins are welcome, i'm
looking for a simple design that wouldn't be to cost prohibitive. Thanks
for any input.
-krem

On Sat, 18 Jun 2005 12:02:15 -0400, "izzi4"
wrote:

Hello, i'm looking for a simple delay off circuit. I have a switch which
controls the 120vac via a relay which turns on a computer power supply. I
am looking for something to be able to turn off the switch and then have the
120v be turned off about 3-5 minutes later (cooling purposes) the relay
would still have the 120 going to it, when switched off (aka no battery
system or anything). I was planning on paralleling a second relay with the
first, one controled by the switch and the second controlled by the timer
circuit powered from the power supply. any suggestoins are welcome, i'm
looking for a simple design that wouldn't be to cost prohibitive.

---

http://www.grainger.com/Grainger/pro...784362&ccitem=
--
John Fields
Professional Circuit Designer

On Sat, 18 Jun 2005 11:08:07 -0500, John Fields
wrote:

On Sat, 18 Jun 2005 12:02:15 -0400, "izzi4"
wrote:

Hello, i'm looking for a simple delay off circuit. I have a switch which
controls the 120vac via a relay which turns on a computer power supply. I
am looking for something to be able to turn off the switch and then have the
120v be turned off about 3-5 minutes later (cooling purposes) the relay
would still have the 120 going to it, when switched off (aka no battery
system or anything). I was planning on paralleling a second relay with the
first, one controled by the switch and the second controlled by the timer
circuit powered from the power supply. any suggestoins are welcome, i'm
looking for a simple design that wouldn't be to cost prohibitive.

---

http://www.grainger.com/Grainger/pro...784362&ccitem=

---
Oops...

Here's one with a 5 minute timeout:

http://www.grainger.com/Grainger/pro...mId=1611716878

--
John Fields
Professional Circuit Designer

"izzi4" wrote in message
news:1119110533.86be9c2fef43db175e83f860ac71cb7b@t eranews...
Hello, i'm looking for a simple delay off circuit. I have a switch
which
controls the 120vac via a relay which turns on a computer power
supply. I
am looking for something to be able to turn off the switch and then
have the
120v be turned off about 3-5 minutes later (cooling purposes) the
relay
would still have the 120 going to it, when switched off (aka no
battery
system or anything). I was planning on paralleling a second relay
with the
first, one controled by the switch and the second controlled by the
timer
circuit powered from the power supply. any suggestoins are welcome,
i'm
looking for a simple design that wouldn't be to cost prohibitive.
Thanks
for any input.
-krem

One really simple, really cheap way to do this is to get the mechanical
timer out of a scrapped microwave oven. This is the kind you have to
physically turn the knob to so many minutes, and it usually goes ding at
the end of the cycle. These should be easily obtainable from a thrift
store. But this has the disadvantage of having to be set manually.

I would also consider connecting power to the fans just by themselves.
To keep the external power out of the PS, you can put a diode in series
with the power lead from both the power supply and the external power
supply, probably a wall wart. This is what's called a diode OR.

Here's one simple time delay. You can make the delay much longer by
using another transistor as an emitter follower to the first one. With
two transistors I've had no problem getting 2 or more minutes delay.
http://ourworld.compuserve.com/homep....htm#relay_i.g
if

Hello, i've tried to get this circuit to work without much luck. I"m
using the one on the right. I was hoping to use the 12v from the power
supply which is hooked up to the relay to power this circuit but i'm
having some wierd issues. For some reason the circuit never turns off
unless i disconnect the connection to the base. i don't know if my
delay is for some reason ridiculously long and something else is
different (I'm using the same values as shown). two questions that i
had was about the first diode right after the "ignition switch" does it
matter what type of diode? also the diode around the relay, what
purpose does it serve and again does the value mean much. Thanks for
any hlep. I"m looking to get into electronics and actually looking for
a good place to start, anyone recomend a book or something else to get
my feet wet. I understand the basics of resistors capacitors voltage
and current sources but active components give me issues. Thanks

Hello, i've tried to get this circuit to work without much luck. I"m
using the one on the right. I was hoping to use the 12v from the power
supply which is hooked up to the relay to power this circuit but i'm
having some wierd issues. For some reason the circuit never turns off
unless i disconnect the connection to the base. i don't know if my
delay is for some reason ridiculously long and something else is
different (I'm using the same values as shown). two questions that i
had was about the first diode right after the "ignition switch" does it
matter what type of diode? also the diode around the relay, what
purpose does it serve and again does the value mean much. Thanks for
any hlep. I"m looking to get into electronics and actually looking for
a good place to start, anyone recomend a book or something else to get
my feet wet. I understand the basics of resistors capacitors voltage
and current sources but active components give me issues. Thanks

wrote:
Hello, i've tried to get this circuit to work without much luck. I"m
using the one on the right. I was hoping to use the 12v from the power
supply which is hooked up to the relay to power this circuit but i'm
having some wierd issues. For some reason the circuit never turns off
unless i disconnect the connection to the base. i don't know if my
delay is for some reason ridiculously long and something else is
different (I'm using the same values as shown). two questions that i
had was about the first diode right after the "ignition switch" does it
matter what type of diode? also the diode around the relay, what
purpose does it serve and again does the value mean much. Thanks for
any hlep. I"m looking to get into electronics and actually looking for
a good place to start, anyone recomend a book or something else to get
my feet wet. I understand the basics of resistors capacitors voltage
and current sources but active components give me issues. Thanks

it sounds like you maybe using too high of a value cap at the base.

you can calculate the Time constant using the T = RC which means
(R) Resistor feeding or draining the cap, (C) = Cap value in farads!

(T) would be the time it would take to reach a 63.2 % charge etc.

so lets assume for now your relay creates 50 ma at 12 volts DC on the
coil. this would be 240 Ohm coil.
now since your using a transistor (bipolar type btw), you need to
use the Bata/hfe spec on the transistor to increase the simulated
effect of the Resistor value since the transistor is going to aid in
generating most of the current to the coil for you.
lets assume your transistor is a 150 HFE/beta.
HFE*COIL-R = 150*240=36,000 ohms;

now this is not the exact math to use but should be close enough to
get you in the ball park.
now lets assume your using a 100 uf cap..
T = 36000*0.000100 = 3.6 seconds to get to a 63% charge
etc..

the diode on the input to the base is to isolate the circuit from the
electronics in the auto to prevent it from influencing your cap at the
base from other devices on your the ignition line.
this the Cathode must be pointing to the cap and base and the anode
from the ignition line. the diode going acrossed the relay coil is to
protect your transistor since the release of voltage from the coil can
generate a reverse polarity of potential voltage.
if you have the time constant working, you may not even need it
because a slow decay of the charge in the cold will keep the HV
potential down/

now this only a rough note. you will need to experiment a bit.

wrote:
Hello, i've tried to get this circuit to work without much luck. I"m
using the one on the right. I was hoping to use the 12v from the power
supply which is hooked up to the relay to power this circuit but i'm
having some wierd issues. For some reason the circuit never turns off
unless i disconnect the connection to the base.

If we are talking about the same circuit, the relay will
stay energized (turned on) as long as 12 volts is connected
to the base, so your circuit may be working properly.
Connect your circuit to +12, with nothing connected to the
left side (in the diagram) of the 1N4002 diode. Then, with
a jumper wire, temporarily connect +12 volts to the left
side of that diode. The relay should energize when you
connect the jumper. Once it does, remove the jumper.
The relay should de-energize in roughly 15 seconds. At
this point, you don't care exactly how long it takes,
just make sure it eventually does de-energize. You can
work on setting the delay to 15 seconds later.

Just to make sure we're talking about the same thing:

+12
|
+---------+
| |
----- (||
D2 / \ (||Relay
--- (||Coil
| |
+---------+
D1 |
1N4002 R1 / -- (collector)
A o---------|-----+------/\/\/\-----| 2N2222
| + \ -- (emitter)
--- |
1000uF --- C1 |
| |
Gnd Gnd



It is supposed to stay on as long as +12 is applied to the base,
from point A through D1 and R1. You must disconnect 12 volts
from point A for the relay to de-energize.


i don't know if my
delay is for some reason ridiculously long and something else is
different (I'm using the same values as shown).

What are you using for R1? (Should be ~ 4.7 K). Is your
relay coil rated at 120 ohms?

two questions that i
had was about the first diode right after the "ignition switch" does it
matter what type of diode?

It is not critical. Use a diode from the 1N400x family
(1N4001 through 1N4007) for both diodes.

also the diode around the relay, what
purpose does it serve and again does the value mean much.

The diode around the relay eliminates a "spike" that
results when power to a relay is disconnected. It is
good practice (and often necessary) to put a diode
across the relay coil in DC circuits. In this particular
circuit, it is not needed if the circuit is working properly.
But don't omit it. When you first start out with electronics,
do exactly what you are doing: ask about things! But follow
the plans until you fully understand why changing them is
a good idea.

By the way, the diode is marked with a band at one end.
The banded end of the diode corresponds to end of the diode
with the vertical line in the schematic symbol: ---|---
so the banded end of the D1 diode connects to R1 and C1 in
the circuit.

Ed

Thanks for
any hlep. I"m looking to get into electronics and actually looking for
a good place to start, anyone recomend a book or something else to get
my feet wet. I understand the basics of resistors capacitors voltage
and current sources but active components give me issues. Thanks

wrote in message
oups.com...
Hello, i've tried to get this circuit to work without much luck.

What circuit??

Just to clairify that i am using the circuit on the right at
http://ourworld.compuserve.com/homep...tm#relay_i.gif
(may be messy due to word wrap) Thanks for all the input. Through
messing around with this circuit before posting I did hook things up
incorrectly. I do have a multimeter and was wondering how to check the
transistor to see if it is still viable. I have replacements but i'd
like to know for future applications too.

Watson A.Name - "Watt Sun, the Dark Remover" wrote:
wrote in message
oups.com...
Hello, i've tried to get this circuit to work without much luck.

What circuit??

On 23 Jun 2005 09:11:49 -0700, "krem" wrote:

Just to clairify that i am using the circuit on the right at
http://ourworld.compuserve.com/homep...tm#relay_i.gif
(may be messy due to word wrap) Thanks for all the input. Through
messing around with this circuit before posting I did hook things up
incorrectly. I do have a multimeter and was wondering how to check the
transistor to see if it is still viable. I have replacements but i'd
like to know for future applications too.

---
Place your multimeter in "diode test" mode and connect it across a
known good diode. With the diode conducting, the lead connected to
the anode will be positive and the other negative.

With the multimeter still in diode test mode, connect the positive
lead to the base of the transistor and the negative lead, first to the
emitter and then to the collector. The multimeter should indicate a
conducting diode in both cases. Now, connect the negative lead to the
base and connect the positive lead, first to the emitter and then to
the collector. The meter should indicate an open circuit in both
cases. Finally, connect one of the leads to the emitter and the other
to the collector and then reverse the connections. The meter should
indicate an open circuit in both cases.

If the transistor passes all the tests it is most likely good. The
possibility exists, however, that it has been damaged and its gain is
not what it should be. In order to test that you can do something
like this: (View in a non-proportional font like Courier)


C---[MILLIAMMETER]--+
+---[10K]---B |
+| E |+
[1.5V] | [9V]
| | |
+-------------+-------------------+


Assuming a Vbe of about 0.7V for the transistor, there'll be
about 80µA of base current in the transistor. If the transistor is
good, the current indicated by the milliammeter (the collector
current), divided by the base current, should lie within the range of
beta specified for the device at that collector current.

--
John Fields
Professional Circuit Designer

On Sat, 25 Jun 2005 11:45:47 -0500, John Fields
wrote:

On 23 Jun 2005 09:11:49 -0700, "krem" wrote:

Just to clairify that i am using the circuit on the right at
http://ourworld.compuserve.com/homep...tm#relay_i.gif
(may be messy due to word wrap) Thanks for all the input. Through
messing around with this circuit before posting I did hook things up
incorrectly. I do have a multimeter and was wondering how to check the
transistor to see if it is still viable. I have replacements but i'd
like to know for future applications too.

---
Place your multimeter in "diode test" mode and connect it across a
known good diode. With the diode conducting, the lead connected to
the anode will be positive and the other negative.

With the multimeter still in diode test mode, connect the positive
lead to the base of the transistor and the negative lead, first to the
emitter and then to the collector. The multimeter should indicate a
conducting diode in both cases. Now, connect the negative lead to the
base and connect the positive lead, first to the emitter and then to
the collector. The meter should indicate an open circuit in both
cases. Finally, connect one of the leads to the emitter and the other
to the collector and then reverse the connections. The meter should
indicate an open circuit in both cases.

If the transistor passes all the tests it is most likely good. The
possibility exists, however, that it has been damaged and its gain is
not what it should be. In order to test that you can do something
like this: (View in a non-proportional font like Courier)


C---[MILLIAMMETER]--+
+---[10K]---B |
+| E |+
[1.5V] | [9V]
| | |
+-------------+-------------------+


Assuming a Vbe of about 0.7V for the transistor, there'll be
about 80µA of base current in the transistor. If the transistor is
good, the current indicated by the milliammeter (the collector
_____ ^ _______
/quotient of the\

current), divided by the base current, should lie within the range of
beta specified for the device at that collector current.

--
John Fields
Professional Circuit Designer

thanks for all of the input, i've managed to get it working. Not to
mention picking up a bit more of how things interact with one another.
Again thanks

Perhaps one of the turbo timer circuits would be suitable - the gizmos that
keep the car engine running for a short time after the ignition is switched off



David

izzi4 wrote:

Hello, i'm looking for a simple delay off circuit. I have a switch which
controls the 120vac via a relay which turns on a computer power supply. I
am looking for something to be able to turn off the switch and then have the
120v be turned off about 3-5 minutes later (cooling purposes) the relay
would still have the 120 going to it, when switched off (aka no battery
system or anything). I was planning on paralleling a second relay with the
first, one controled by the switch and the second controlled by the timer
circuit powered from the power supply. any suggestoins are welcome, i'm
looking for a simple design that wouldn't be to cost prohibitive. Thanks
for any input.
-krem

"quietguy" wrote in
message
...
Perhaps one of the turbo timer circuits would be suitable - the gizmos
that
keep the car engine running for a short time after the ignition is
switched off

.... pre-ignition?

These guys have all kinds of timers with differnt time settings. Perhaps
it's worth checking into


http://www.weisd.com/store2/45201.html




"quietguy" wrote in
message
...
Perhaps one of the turbo timer circuits would be suitable - the gizmos
that
keep the car engine running for a short time after the ignition is
switched off



David

izzi4 wrote:

Hello, i'm looking for a simple delay off circuit. I have a switch which
controls the 120vac via a relay which turns on a computer power supply.
I
am looking for something to be able to turn off the switch and then have
the
120v be turned off about 3-5 minutes later (cooling purposes) the relay
would still have the 120 going to it, when switched off (aka no battery
system or anything). I was planning on paralleling a second relay with
the
first, one controled by the switch and the second controlled by the timer
circuit powered from the power supply. any suggestoins are welcome, i'm
looking for a simple design that wouldn't be to cost prohibitive.
Thanks
for any input.
-krem

Here are some timers from National Controls

http://www.weisd.com/store2/45201.html





"quietguy" wrote in
message
...
Perhaps one of the turbo timer circuits would be suitable - the gizmos
that
keep the car engine running for a short time after the ignition is
switched off



David

izzi4 wrote:

Hello, i'm looking for a simple delay off circuit. I have a switch which
controls the 120vac via a relay which turns on a computer power supply.
I
am looking for something to be able to turn off the switch and then have
the
120v be turned off about 3-5 minutes later (cooling purposes) the relay
would still have the 120 going to it, when switched off (aka no battery
system or anything). I was planning on paralleling a second relay with
the
first, one controled by the switch and the second controlled by the timer
circuit powered from the power supply. any suggestoins are welcome, i'm
looking for a simple design that wouldn't be to cost prohibitive.
Thanks
for any input.
-krem

I don't think that can happen with a diesel, but....

David

Dingus wrote:

"quietguy" wrote in
message
...
Perhaps one of the turbo timer circuits would be suitable - the gizmos
that
keep the car engine running for a short time after the ignition is
switched off

... pre-ignition?

Perhaps one of the turbo timer circuits would be suitable - the gizmos
that keep the car engine running for a short time after the ignition is
switched off

... pre-ignition?


In article , quietguy wrote:
I don't think that can happen with a diesel, but....


I don't think a diesel engine can run without it (or something similar) happening.

Turbo timers run the engine at idle for a few minutes to allow the
turbocharger to cool, this allows the rotor to cool from orange-hot
in the relatively cool idle exhaust gasses rather than cooling by
conducttion up the shaft and cooking the bearings.

Bye.
Jasen