Multiple Transistor switches to control one relay

Thread Starter

Rickm

Joined Jun 24, 2009
20
I have several transistor switch circuits (all on 5v with resistor and led) that I want it be able to denergize a relay if only one of the transistor switches is off, I guess I should be asking how an if transistors as a switch can have the outputs placed in series. if not any other thoughts
 

mik3

Joined Feb 4, 2008
4,843
Do you have a schematic of these transistor switches?

You mean you want the relay to be activated when one of the transistors goes off.
 

Thread Starter

Rickm

Joined Jun 24, 2009
20
prefer relay to go off when a switch is opened, but eithe would work, and i cant get it small enough with bmp
 

mik3

Joined Feb 4, 2008
4,843
If you connect the emitters and the collectors of the transistors together the relay will be activated if anyone of the transistors turns on.
 

Thread Starter

Rickm

Joined Jun 24, 2009
20
I need for the relay to go off with only one switch open, this is to be used in a alarm system, each switch is a window, door, ect.... each switch is considered N/C and the leds give a status of the switch
 
Last edited:

Thread Starter

Rickm

Joined Jun 24, 2009
20
with 20 switches, and 20 leds at 2v and 35mA, in series that means at least 40v In parallel I would be looking at 2.5-3.5 watt resistors to be safe (at 5 volts)
 

Thread Starter

Rickm

Joined Jun 24, 2009
20
this is all good but the switches are door switches, 2 wires and a open reed switch, when a magnet comes close to them they close (SPST)
 

eblc1388

Joined Nov 28, 2008
1,542
I need for the relay to go off with only one switch open, this is to be used in a alarm system, each switch is a window, door, ect.... each switch is considered N/C and the leds give a status of the switch
So you want the LEDs to light when the windows, door etc.. are closed.

The following circuit does that. It shows only one channel. You can extend it to whatever channels you wants by adding CD4078B. Each cd4078B would give another 8-channel. You can control/reduce the brightness of the LED by increasing the 1K resistor to any value below 10K.

The combined gate OUTPUT will goes LOW if any switch opens.

This is how it works. If the switch remains closed, the LED will light. The voltage drop across the LED is about 2~3V which is lower than the logic HIGH input level(~6V) of the 8-input NOR gate. Thus the gate output remains HIGH.

If any switch is open, the LED will not light. The 1K resistor would pullup the gate input voltage to 12V and the output of the NOR gate will go LOW.

Note: Unused CD4078B inputs should be tied to GROUND or 0V.

 

Attachments

BillB3857

Joined Feb 28, 2009
2,516
If you put all your Normally Closed (N/C) switches in series and put a resistor/LED series combination across each switch. Resistor value would need to be high enough to allow relay to drop out when switch opens.With all OK, the relay is picked up. Any switch opening will drop the relay and light the associated LED.
 

Attachments

Thread Starter

Rickm

Joined Jun 24, 2009
20
Eblc1388, I think I'll give your thoughts a try, the only thing is where does it leave me on the output voltage, 12v? I don't have a splice model to try the IC so can't really test before building. I ordered the the CD4078B (5 of them) this morning, whats $10 if it doesn't work $3 for the chips and $7 shipping. Thanks for the advice from all and any further advice is welcome.
 

eblc1388

Joined Nov 28, 2008
1,542
where does it leave me on the output voltage, 12v?
Yes, 12V when all of the switches are in closed state(normally closed switches). Dropping to about 0.6V if one or more switches become open.

You can drive a relay using a PNP transistor with this type of output. The relay will energise if the PNP base voltage is less than 11V. It is just the upside down version of a NPN version.

Do you need assistance on a schematic of how to drive a relay using PNP transistor?
 
Top