# Constructing a "Power on time delay relay" with 5 sec. delay. max 6V

Discussion in 'Electronic Basics' started by [email protected], Jul 10, 2005.

1. ### Guest

I'm trying to build an autonomous "sumo"-robot for competition in
sumo-robot contests.
One of the rules state that the robot must be inactive for 5 sec from
the time it is switched on. Now I know that it probably would be much
easier to buy a basic-stamp processosor and program it, but my budget
is quite thin and I think it would be quite fun to construct a robot
without software.

I have found a circuit-drawing of a time delay relay, but it demands
12V and has a delay of roughly 7 sec.

http://ourworld.compuserve.com/homepages/Bill_Bowden/page2.htm#delay.gif

Since my motors run on 6V I would like the relay to do that as well.

QUESTIONS:
1. How do I calculate the time delay? In the text the author states
that the delay can be reduced by reducing the R or C values (R=47K,
C=100uF)but is there a formula that can be used?

2. Will this solution turn the robot on at "full power" after the delay
or will it start softly (when I push the button or when the time is
up?)?

3. How do I have to change the circuit to reduce the voltage to 6V?

4. Is there a better alternative that wouldnt cost me more than perhaps
20\$?

As you can see I have'nt got much experience of electronics, so pelase
be patient with me

2. ### Tom LeMenseGuest

[...]
The time delay will depend on the reverse breakdown voltage of the
base-emitter junction in the '3904 used as a Zener. You can figure that the
2N3904+2N3053 pair (in a "darlington" configuration) will be fully 'ON' when
the voltage at the base of the 2N3904 (of the pair) is 1.2V. So, as the
100uF cap charges through the 47K resistor, its voltage rises. This voltage
is reduced by (according to the web page's author) 6-9V due to the "Zener"
and then applied to the base of the "darlington".
If you use a relay to switch the robot power, it will be ON or OFF. I
wouldn't use this circuit to control the motor directly, as the motor
current will "ramp up" somewhat during turn-on. Stick with a relay to
switch the motor current ON or OFF.
a) Use a relay with a 6VDC coil, making sure that 'pull-in' votlage is no
more than 5V.

b) Replace the "zener" 2N3904 with something that will drop less voltage.
The 6-9V dropped by the '3904 clearly won't be compatible with a 6V supply
rail. Some people use an LED which will drop between approx 2V and 4V
depending on its color, others use several 1N4148 or 1N4001 diodes in series
to obtain the desired drop. If you were to use 3x 1N4148 diodes (very
common "switching" diodes) you'll end up with the darlington pair saturated
(ON) when the cap voltage is approximately 1.2V + 0.6 + 0.6 + 0.6 = 3V.

c) Adjust the 47K resistor to change the delay time, keep it above 1K ohm
to keep excessive current from flowing into the base of the darlington.
This shouldn't cost anywhere even close to \$20.

This circuit that you found will not behave so well over large temperature
extremes, or if the 6V supply voltage is going to vary quite a bit (the
threshold of the transistors doesn't track the supply votlage). Nor will it
be that repeatable from unit-to-unit -- so if you're going to build a whole
army of sumo robots, they'll all have slightly different turn on delay
times. For an army of one that only has to work indoors, however, it'll
probably work just fine.

I would recommend a means to help discharge the 100uF cap when the power is
interrupted - the web page author put a 10K from V+ to ground, but the 100uF
cap will have to discharge through the 47K resistor in SERIES with the 10K,
that's a pretty long discharge time. What this means is if the power were
applied briefly, then disconnected, then reapplied briefly, then
disconnected, &c, the cap will charge up a little bit each time, and
eventually the relay could turn ON before 5s elapsed when the power was
applied "for good" If you place a diode across the 47K resistor with the
anode at the 100uF + terminal, and the cathode at the +6V point, the
capacitor will "discharge" across the power supply rail when power is
removed.

A more precise turn on delay could be generated with the venerable LM555
timer circuit - a Google search will reveal lots of references to this very
common (and cheap) IC. While you might not find the circuit you need
exactly, it ought to give you lots of good ideas.
Have fun!

TJL

3. ### Guest

Thanks a bunch.
I can´t say that I grasp it all at first sight,
but hopefully I´ll figure it out .

/David

4. ### John FieldsGuest

---
See below.
---
---
Try this: (View in a non-proportional font like Courier.)

+6V>---> |
S1 |
O
|
+-----------------------+------+
| | |
| | |
| | |
+-------+-------+ | |
| | | +-------+ |
| +---+---+ [1M] |K | O COM
| |_ _| | [DIODE] [COIL]- -|
+--O|R D|O--+ | | O--> |<--O--->6VOUT
| |_ | | +-------+ NO NC
+--O|T TH|---+ |
| | | C
| OUT|---|--[100R]-----B
+-------+ |+ E
7555 [4.7µF] |
| |
GND--------------+---------------+
_____ _______
When you close S1, the 7555 will come out of RESET before the TRIGGER
input goes high, initiating an output pulse which will energize the
relay for about five seconds. This will pull the common terminal away
from the normally closed contact, disconnecting it from the switched
6V coming from S1. After about five seconds the 7555 will time out,
de-energizing the relay, and 6V will appear on the normally closed
has to supply current to the relay for five seconds, but the maybe bad
part is that S1 has to be able to supply all the current needed by the

Will that work for you, or would you like something else?

5. ### Guest

That sounds great.
If I have understood it correctly that is .
Correct me if I´m wrong: When power is switched on the the current
will flow through the 7555 (and therefore through the base of the
transistor) and the coil. The coil will act as an electromagnet
"shutting of" the rest of the circuitry by pulling away the common
terminal (COM?).
When the 7555 times out power will stop flowing through the coil and
start flowing through the rest of the circuit!

Sounds like just the thing ive been looking for.
A couple of questions arise:
1. How do i set the 7555 to time out after 5 sec?
2. The closest thing i found att my components shop is something they
call "ICM 7555" would that be what your thinking of? (do you know what
ICM stands for?)
3. What is the diode doing in the circuit?
4. Is the "[4.7µF]" a capacitator?
5. What does NO and NC to the right of the circuit stand for?

Phew, I think I´m running out of questions for the moment .
Thanks a lot (and again for your patience, if this project works I sure
will have learned a lot from it .

/David

6. ### Guest

OK, I found out what NO and NC stands for (Normally Open and Normally
Closed), but then another question arose... Can a relay be both NO and
NC?
Do you buy a specific NO or NC relay or is it just a question of how
you turn it when hooking it up to the rest of the circuit?

/David

7. ### Rheilly PhoullGuest

One day got dressed and committed to text
It's usually the norm to have a relay with 2 contacts and a moving 'arm'
that will move between the contacts and connect to 1 of the contacts when
energised and to the other when de-energised. Thus it can be connected as NO
by using the open contacts when de-energised or NC by using the NC contacts
when de-energised.

----0\ 0-------
|
|

Hope that comes out but no doubt you will get many alternative explanations
to consider !

8. ### Randy DayGuest

Yes. Look for a DPDT (double pole, double throw) relay.
That way you'll have two sets of contacts which can be
used either as NO or NC. Assume the relay below is at
rest (not powered). Contacts a and c are NC, e and f
are NO.

o- b
a -o--__
| o- c
|
|
|
| o- f
e -o--__
o- g
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)
You can buy SPDT, DPDT, 4PDT, etc. relays. It's in how
you hook them up...

HTH

9. ### John FieldsGuest

---
It's already set. The 1 megohm resistor and the 4.7µF capacitor set
the pulse time. Get the data sheet at:

http://pdfserv.maxim-ic.com/en/ds/ICM7555-ICM7556.pdf

to find out all about it.
---
---
Yes, that's the part. "ICM" is MAXIM semi's prefix
---
---
It's clamping the high voltage pulse generated by the relay coil when
it turns off to about a volt over the supply, keeping it from
destroying the transistor.
---

10. ### ehsjrGuest

I'll post a simple circuit - but first, there is something
*critical* that no one has mentioned. The relay you use must
be able to handle the current that the sumo draws, and that
is likely to be fairly large. I have specified a relay that
should work. Ok, now to the circuit:

+------+--------+-----+
| k| | |
| [Diode] [Coil] o COM
| | | |
+6V----+--------+ +--------+ | <--o--->6VOUT
| | | NO
| [R3] e|
[R1] | /
| +-----[R4]-----b Q1
/ k| \
P \ ----- c|
O /<-----r/ \ TL431 |
T \ --- |
| a| |
+---+ | |
| |+ | |
[R2] [C1] | |
| | | |
Gnd----+---+----+---------------+

C2 is a 100 uF 16V electrolytic capacitor
Diode is a 1N4001 diode
POT is a 20 K linear potentiometer
Q1 is a 2N5401 transistor
R1 is a 56K 1/4 watt resistor
R2 is a 100K 1/4 watt resistor
R3 is a 330 ohm 1/4 watt resistor
R4 is a 100 ohm 1/4 watt resistor
R5 is a 47K 1/4 watt resistor
TL431 is part number 497-3035-5-ND at Digikey. It is
a programmable (by resistors R1, R2 and the pot) shunt.

Relay: Part # 255-1109-ND from Digikey is a 6 volt single pole
normally open relay with a 68 ohm coil. The contacts can
switch and carry 16 amps.

The potentiometer (POT) allows you to adjust the timing
to exactly 5 seconds.

k
Diode: ----|<---- (Banded end at left)

TL431 2N5401
----------- -----------
| | | |
| Flat Side | | Flat Side |
| | | |
| | | |
| | | |
----------- -----------
| | | | | |
r k a e b c

See the datasheets at:
http://www.fairchildsemi.com/ds/TL/TL431A.pdf
http://www.fairchildsemi.com/ds/2N/2N5401.pdf

It will be great to hear how you make out with