running multiple relay''s from BS-1
--- In basicstamps@yahoogroups.com, "Allan Lane" <allan.lane@h...>
wrote:
> In 'normal' use (while the relay is activated)
> the diode across the relay coil
> should be reverse-biased (Cathode to +5, Anode to
> the Stamp).
>
> The purpose of the diode is that when you release
> the coil, there is a current flowing, maintaining
> a magnetic field, that is holding the contacts
> closed. When you switch off the coil,
> the magnetic field collapses. This generates
> additional current in the coil which wants to go
> somewhere. The diode gives it a place to go.
>
> Without the diode, the current can only go into your
> relay-driver circuitry -- which tends to burn it out.
> Some Darlington transistor IC's have built-in
> diodes for this purpose.
This statement needs a clarification.
The purpose of the diode is not to prevent "additional current".
is to clamp the kickback voltage generated when the semiconductor
switch is turn off.
This effect in math terms is defined as V=L*di/dt
where V is the generated voltage, L is the inductance of the coil in
Hy, di is the instaneous current and dt is the time it takes the
swith to open.
The faster the switch, the larger the generated voltage.
Imagen as dt approaches zero, what is the voltage equal to?
wrote:
> In 'normal' use (while the relay is activated)
> the diode across the relay coil
> should be reverse-biased (Cathode to +5, Anode to
> the Stamp).
>
> The purpose of the diode is that when you release
> the coil, there is a current flowing, maintaining
> a magnetic field, that is holding the contacts
> closed. When you switch off the coil,
> the magnetic field collapses. This generates
> additional current in the coil which wants to go
> somewhere. The diode gives it a place to go.
>
> Without the diode, the current can only go into your
> relay-driver circuitry -- which tends to burn it out.
> Some Darlington transistor IC's have built-in
> diodes for this purpose.
This statement needs a clarification.
The purpose of the diode is not to prevent "additional current".
is to clamp the kickback voltage generated when the semiconductor
switch is turn off.
This effect in math terms is defined as V=L*di/dt
where V is the generated voltage, L is the inductance of the coil in
Hy, di is the instaneous current and dt is the time it takes the
swith to open.
The faster the switch, the larger the generated voltage.
Imagen as dt approaches zero, what is the voltage equal to?