B
Ben Lessani
Guest
Hi, I've been dabling with my car central locking, making my attempts
to integrate the alarm into the system. When I lock the doors remotely
with the key, the door pins go down (naturally), but when I press an
additional button on the key, it deadlocks the doors. There is a wire
(lets call it JOE) which is always at 12V unless the doors are
deadlocked, when it is at 6V. I am not sure of the current running
through this circuit, but I know its not powerful enough to switch a
relay. I was thinking I could use a transistor type circuit to achieve
what I need. I can use a relay in its opposite state, so when no
voltage is supplied, the circuit is complete, but when a voltage is
applied - it opens the circuit. I was thinking I could put a resistor
from JOE to the BASE of a transistor, to reduce the voltage down to 6V
as standard, then when the doors are deadlocked, the voltage will drop
6V, taking the voltage at B to be 0V. I've drawn some diagrams below
to try and help explain what I'm trying to say. The wire does not have
enough current to close a relay, so I need a transistor to close the
relay. But I want to resist the voltage input at the resistor using
R1, so that it is normally at 6V, this means that when I deadlock the
doors, and the voltage drops, the input at the transistor is 0V,
leaving the 'switch' open, leaving the relay open.
12V
|
State 1 (JOE AT 12V) |
8 |--|-----
8 | | RELAY CLOSED
8 |--|-----
C |
/
12V 6V |/
----| R1 |----B-|
|\
\
E |
|
|
GND
********************************************
12V
|
State 2 (JOE AT 6V) |
8 |--|-----
8 | \ RELAY OPEN
8 |--|-----
C |
/
6V 0V |/
----| R1 |----B-|
|\
\
E |
|
|
GND
Could you please email me with a response at
Qlessani@nildram.co.ukQ
Delete the Q's if you want to email me, everyone else is doing it, so
I thought I might join in the fun
.
Thanks V. much
Ben Lessani
to integrate the alarm into the system. When I lock the doors remotely
with the key, the door pins go down (naturally), but when I press an
additional button on the key, it deadlocks the doors. There is a wire
(lets call it JOE) which is always at 12V unless the doors are
deadlocked, when it is at 6V. I am not sure of the current running
through this circuit, but I know its not powerful enough to switch a
relay. I was thinking I could use a transistor type circuit to achieve
what I need. I can use a relay in its opposite state, so when no
voltage is supplied, the circuit is complete, but when a voltage is
applied - it opens the circuit. I was thinking I could put a resistor
from JOE to the BASE of a transistor, to reduce the voltage down to 6V
as standard, then when the doors are deadlocked, the voltage will drop
6V, taking the voltage at B to be 0V. I've drawn some diagrams below
to try and help explain what I'm trying to say. The wire does not have
enough current to close a relay, so I need a transistor to close the
relay. But I want to resist the voltage input at the resistor using
R1, so that it is normally at 6V, this means that when I deadlock the
doors, and the voltage drops, the input at the transistor is 0V,
leaving the 'switch' open, leaving the relay open.
12V
|
State 1 (JOE AT 12V) |
8 |--|-----
8 | | RELAY CLOSED
8 |--|-----
C |
/
12V 6V |/
----| R1 |----B-|
|\
\
E |
|
|
GND
********************************************
12V
|
State 2 (JOE AT 6V) |
8 |--|-----
8 | \ RELAY OPEN
8 |--|-----
C |
/
6V 0V |/
----| R1 |----B-|
|\
\
E |
|
|
GND
Could you please email me with a response at
Qlessani@nildram.co.ukQ
Delete the Q's if you want to email me, everyone else is doing it, so
I thought I might join in the fun
Thanks V. much
Ben Lessani