# 12 VDC Relays

Discussion in 'Electronic Basics' started by Conan Kelly, Nov 27, 2006.

1. ### Conan KellyGuest

Hello all,

My background is not in electronics. It is in car stereo & security. So please dumb-it-down for me.

What is the minimum voltage required to energize the coil of a standard 12 VDC relay? How can I wire 2 relays with resistors so
that the output line will be switched off when the voltage drops below 9.6 Volts?

I think that I might have a diagram on how to wire 2 relays and a momentary push-button switch (and maybe diodes) so that each push
of the button toggles the output line between OFF and ON. I was thinking that there is probably some similar way of wiring 2 relays
with resistors to accomplish what I'm trying to do.

The reason that I'm trying to do this is because I have a Makita 12V cordless drill. I gutted an old Makita battery charger and
wired it so I could drain the battery or power small 12V devices off of the drill battery. I read somewhere that the ideal
discharge for NiCad batteries is 1.2 V per cell before recharging. (I'm guessing that there are 8 cells in a Makita 12V
battery--12V / 1.5V per cell = 8 cells--8 cells * 1.2V per cell = 9.6 V ideal discharge) Right now, it is wired directly to the
battery so the thing can be completely discharged if a cell phone cigarette lighter charger is left on it too long. I was hoping to
find a way to use relays and resistors and what-not to get my contraption to shut off when it reaches the target voltage.

2. ### John FieldsGuest

---
Usually 75% of the rated coil voltage at 25C, but the best bet would
be to get the data sheet for the relay because it'll vary with
temperature. And manufacturer.
---
---
The short answer is, "You can't." because the relay release voltage
varies all over the place. What you need is a comparator, a
reference, and a voltage divider to determine when the voltage goes
to 9.6V and then when that happens, open up the relay.
---
---
Since a relay needs current to keep it closed it shortens the amount
of time the battery can supply current to the load, so a better
solution would be to use a P Channel MOSFET for the switch since it
needs no current to keep it turned on. If you like, I can post a
schematic for you which will do what you want, and I'm sure we'll
all be happy to help you get it up and running if you're willing to
get the parts and do the wiring and assembly. Interested?

3. ### Conan KellyGuest

John,

Thanks for the feed-back.

Thanks again for all of your help,

Conan Kelly

4. ### kellGuest

John,

A very similar question, and you even posted a spice simulation.

7. ### John FieldsGuest

---
You're welcome, but just to clear up a couple of things, a NiCd
battery, when fully charged, starts off at about 1.4V. Then, when a
load is placed across it slowly falls to 1.2V, then very quickly
falls to 1.0V, which is considered its 'cutoff' voltage.

So, your 12V battery pack is probably made up of 10 cells, with an
expected cutoff voltage of 10V, which is when you'd want to

This should do it for you:

View in Courier

IRF4905L
| | | G
| [150k] | |
| | +---|---[1M]--+
[412k] | | | |
| +-[1k]--+--|+\U1A |
| | | >-------+
+----------|----------|-/LT1017
| | |
| K|2.5V | +--|+\
[137k] [LM385] | | | >
| | | +--|-/U1B
| | | |
GND>---+----------+-----------+---+----------->GND

8. ### ehsjrGuest

Some general information not discussed in other replies:
Nominal voltage for a NiCd is ~1.2V per cell.
When fully charged, and immediately after being
removed from the charger, voltage from the cell
is ~1.43V. NiCd packs should not be discharged
below ~.9V per cell. The "standard" 14 hour charge
rate is ~C/10 where C is the Ampere-hour rating
of the cell.

When you have the specific recommendations from
the manufacturer, they should be followed. Otherwise
you can use the general information above.

Ed

9. ### Conan KellyGuest

John Fields,

Thanks for the feedback.

You where right, my drill's battery pack has 10 cells.

Remember that I do not have a background in electronics, so I do have a couple of questions about your scematic.

I'm not going to post them here because I have pix to go with them. Go to http://home.att.net/~ctbarbarin/schematic_questions.htm
to see the pix/questions.

Please confirm or correct my assumptions.

Are these components that can be purchased at Radio Shack? ...Fry's Electronics?

Thanks again for all of your help,

Conan Kelly

10. ### kellGuest

You where right, my drill's battery pack has 10 cells.

Conan,
I don't want to steal John's fire, but I'd like to jump in here
Yes, the numbered components with no other information are resistors.
K means 1000, M means 1,000,000.
Yes, the + means connected and the other pictorial that looks like
under-over means no connection. The positions of the components on the
actual circuit boards might (probably will) be different in the actual
execution than in that schematic, and the "wires" will actually be
copper traces, if you fab an actual pc board for the circuit. Or you
could do point-to-point wiring where you use wires soldered to the
component leads. This you can accomplish in different ways, for
example
1 wire wrap (I don't know how many people do it any more)
3 or you can insert the components in perf board and run wires from

similar to a zener diode. You can look up datasheets on websites like
digikey.com and mouser.com, which are also good places to buy from.
http://www.digikey.com/scripts/DkSearch/dksus.dll?Filter

The U1A and U1B are both contained in a single chip, the LT1017 dual
comparator. You only need one comparator, so you were right that U1B
does not function in the circuit.

As for how to change the voltage setpoint, look at the two resistors on
the comparator's negative input, 412k and 137k. They take the battery
voltage and divide it by a certain ratio for the comparator to look at.
They form a voltage divider. You could make the circuit adjustable by
replacing them with a potentiometer.

11. ### John FieldsGuest

---
I don't know, but they're surely available from Digi-Key and
probably also Mouser; check their web sites.
---
---
---

I’m assuming anything followed by a K or M is a resistor (412k,
137k, 150, 1k, 1M)

---
Correct. 'k' means that the value shown, in ohms, is multiplied by
1000, while 'M' means that the multiplier is 1 000 000. Also, the
resistors in the voltage divider feeding U1A- have a tolerance of
+/- 1%
---

connection where wires/components need to be soldered togther. And
the other (-|-) is where wires/components cross over/under one
another (I need to make sure that they are insulated and don't touch
one another).

---
Correct.
---

I don't know what this component is.

---
It's an LM385-2.5, a voltage reference.
---

I don’t know what these 3 components are. It looks like the lower 2
are similar. Is "LT1017" part of "U1A" or is it a separate component
all together.

---
They're both parts of an LT1017, a dual voltage comparator
manufactured by Linear Technology. U1A is one of the comparators in
the package and U1B is the other one.
---

I know I’m probably wrong, but it doesn’t seem like the lowest one
(U1B) serves much purpose. Is it necessary? What is it’s function?

---
It's a spare and serves no function. It's not necessary, but it
comes in the package anyway, so we spare it.

The IRF4905L is a P channel enhancement mode MOSFET being used as a
switch which allows the battery to be connected to the load when the
battery voltage is greater than 10 volts, but disconnects the
battery when its voltage falls to less than 10 volts.
---

What changes would need to be made if I wanted to get this to cut
off at 0.9 v per cell? You mentioned 1.0 v per cell as a cutoff
voltage, someone else who responded to my post mentioned 0.9 v per
cell, and I think I read somewhere ( Battery University or WikiediA
) that the cutoff voltage is 1.2 v per cell, so I’m not sure which
one to use. In case I decide to go the 0.9-v-per-cell route, what
changes would I need to make?

---
Change the 137k resistor to 147k.

looking at the circuit again, the LT1017 is probably a little pricey
for your application, so for about 10% of its cost you could use an
LM393, but the circuit would change somewhat and you'd have to add a
pullup resistor:

IRF4905L
| | | | G
| | | [10k] |
| [150k] | | |
| | +---|--[1M]--+-+----+
[412k] | | | |
| +-[1k]--+--|+\U1A |
| | | >------+
+----------|----------|-/LM393
| | |
| K|2.5V | +--|+\
[137k] [LM385] | | | >---+
| | | +--|-/U1B |
| | | | |
GND>--+----------+-----------+---+---------+------>GND