# 9VDC SPDT Relay Question

Discussion in 'General Electronics Chat' started by timothydog76, Oct 1, 2008.

1. ### timothydog76 Thread Starter New Member

Sep 30, 2008
9
0
Hello all,
Just found this site yesterday and it's very helpful. I'm quite a circuit building n00b and I have a project I'm developing. I used the search and couldn't quite find the answer I was looking for.

Basically, I'm building a circuit to switch the path of a relatively low voltage signal between two outputs. I figured that a SPDT relay is exactly what I needed. I plan on using a 9V battery for the relay coil circuit and it will be closed using a reed switch. This all makes quite good sense to me but I'm not exactly sure how much resistance I need to add to regulate the current to the coil.

The coil rating is 7.9VDC, 18mA, 500 Ohms (standard issue RS relay). Is 18mA the minimum amount if current needed for the relay switch? I'm trying to figure out what would be the best amount of resistance to add to the circuit to optimize battery life and still operate the switch properly.

Any help would be greatly appreciated!
Tim

2. ### scubasteve_911 Senior Member

Dec 27, 2007
1,202
1
Hi,

Welcome to AAC

You should know that a 9V battery doesn't actually give 9V, especially under decaying life and under load.

This can give you some insight to this:
http://www.powerstream.com/9V-Alkaline-tests.htm

If I were you, I would use the 9V directly without any extra resistance. You may not have enough overhead to activate the relay as the battery is drained.

What are you switching with the relay? You can probably simplify with a mosfet.

Steve

3. ### SgtWookie Expert

Jul 17, 2007
22,183
1,729
Like Steve said, standard 9v "transistor" batteries start off at around 8.6v.

Since I=E/R, 8.6v/500 Ohms = 0.0172 A = 17.2mA, well within the current spec of the relay.

9v alkaline batteries are typically rated for 500mAh, or more realistically, 50mA for 10 hours. If the relay is your only current draw, 500mAh/17.2mA = 29 hours of battery life (rounded off) while the coil is energized.

MOSFETs would certainly be a better way to go as far as battery life, but MOSFETs are vulnerable to static electricity.

It would help a great deal if you could post a schematic of what you're considering doing, as well as explaining it as much as possible.

4. ### timothydog76 Thread Starter New Member

Sep 30, 2008
9
0
Thanks for the quick responses!

I will try to explain my project a little better. I don't have an easy way to draw up the circuit at the moment. Is there an easy-to-use program you guys can recommended for this? I tried Microsoft Visio and was a bit overwhelmed.

Basically, I have single electronic drum pad with a piezo sensor inside and a foot pedal with a reed switch (normally open) inside. When this pedal is depressed it closes the circuit. I want to direct the signal from the piezo between two different output paths depending on whether or not the foot pedal is up or down (reed switch open or closed).

Would a MOSFET still be a simpler solution for me considering all this?

Also, would maybe using a lower voltage relay be better considering the overhead? (i.e. a 5VDC)

Last edited: Oct 1, 2008
5. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Could the foot pedal operate a switch, and so eliminate the relay?

6. ### timothydog76 Thread Starter New Member

Sep 30, 2008
9
0
I would rather not modify the pedal in any way and leave it the way it comes from the factory. Thus, doing all the work within my circuit.

7. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
So you don't need a constant current through the relay coil, a latching relay might work. Look up a Panasonic ASX2104H. It's \$6.64 from Digi-Key, part #255-1583-5.

It need the current reversed to latch the other way, but the relay is DPDT. Use one set of contacts for the control function, and steer current with the other.

Last edited: Oct 1, 2008
8. ### timothydog76 Thread Starter New Member

Sep 30, 2008
9
0
Hm, I'm not sure a latching relay is what I want. If I'm understanding what I read correctly, the latching relay will keep its switch position after voltage is taken from the coil. I want to keep the two paths dependent on whether the pedal is up or depressed. I don't want it to be a momentary thing. In other words, one path only when the pedal is down and switch to the other path only when the pedal is up.

Is this right, or do I have latching relays wrong?

9. ### timothydog76 Thread Starter New Member

Sep 30, 2008
9
0
OK,

Forgive me if the diagram is wrong. I'm still new at this... but I believe this is what I want.

File size:
7.8 KB
Views:
70
10. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
That is correct. I suggested it as a battery life extender. 9 volt cells are pretty weenie for current.

11. ### Bernard AAC Fanatic!

Aug 7, 2008
4,248
424
Might try a CMOS IC 4066. Shelf life for battery. It has 4 NO switches, callthem A_A', B_B' C_C', D not used. Controls GA, GB GC [G for gate ]. Connecta 27k resistor from batt.+ to GA & C', Cto gnd. Switched 9 V toGB & GC Input sig. to A & B, out A'& B'. Sw A NC , When reed sw closes B closes GA is shorted to gnd. opening swA. Resistor value not critical 10k range. I have used this IC for low level analog signals. Sorry can't find its pin out.

Nov 29, 2005
781
58
13. ### timothydog76 Thread Starter New Member

Sep 30, 2008
9
0
Thanks for the suggestion. I looked up the chip. Could you provide a wiring diagram for how you would wire it? I didn't quite follow everything you said.
Thanks.

Apr 20, 2004
15,815
283

15. ### timothydog76 Thread Starter New Member

Sep 30, 2008
9
0
I downloaded the data sheet but I didn't quite follow Bernard's instructions on how to wire it for my application.

16. ### Bernard AAC Fanatic!

Aug 7, 2008
4,248
424
I found my 1977 CMOS data book;but printer/scanner still off line. So if you still want the 4066 version I'll try. Equate letters to pin numbers as:
A-1 , A'- 2 ; B-3 ,B' -4 ; C-9 , C'-8 ; GA-13 , GB-5 , GC-6 , GD-ground [,was eroneously listed GA onearlier submission ].For connections :Ground _9V, +9V to 27kΩ and reed switch, other end of reed sw to pin 5 and pin 6 , free end of resistor to pin 13 and pin 8 , ground pin 9 and 7, +9V to pin 14. Signal input to pins 1 & 3 , sig. out from pins 2 or 4 . Note pins 1& 3 are tied together as are pins 5 & 6 . Sw. D is not used. A schematic would be easier.

17. ### timothydog76 Thread Starter New Member

Sep 30, 2008
9
0
Thanks, that explains it a little better.

18. ### KMoffett AAC Fanatic!

Dec 19, 2007
2,586
234
Migel,

Is there something I'm missing? Wow, what an awful web site. I thought I'd purchase some of those switches. The only link on their Web page http://linearparts.com/ is an email link. ??? Is this a link from another link?

19. ### Externet AAC Fanatic!

Nov 29, 2005
781
58
Hello KMoffett.

I have no clue of that web site other than it had a picture of the exact type of latching switch I would use instead of any elaborated solution for the asked circuit. Ignore the site, focus on the item displayed.

That type of switch is in most old PC towers front panel, used for 'turbo' , available usually free, canibalizable from defunct junk.
One push routes the signal to one place; next push to the other, next to the one place, next to the other... SPDT or DPDT.

Put in a box and wired to 1 (piezo) source and 2 inputs. It is a latching switch, requieres no relay, no battery, no circuit, no nothing and will route the signal the way Tim wants.

Miguel

Last edited: Oct 4, 2008
20. ### timothydog76 Thread Starter New Member

Sep 30, 2008
9
0
Miguel,
Thanks for the suggestion. Unfortunately adding this switch is not and option for me for a couple reasons. First, this button would have to be added to my pedal which I do not want to change. I want to make the circuit run without changing any of the stock parts. Second, I am hesitant to use any kind of physical push button switch for this function as it will be depressed many times and will undoubtedly break more quickly than the reed switch that is in it currently.