My final (?) PSU remote switch and slow startup for bigger transformers

[pacificblue]

Originally posted by akis
What do you mean by "dropout delay"?

When a relay switches off, back-EMF is generated. Without the diode that goes into the circuit and does more or less damage there. The relay falls off immediately. With diode that energy is forced to circulate through the relay coil until it is used up. the relay remains active during that time. A fly-back diode always delays the deactivation time of a relay.

You can reduce that time by adding a Zener or a resistor (R =~Rrelay) in series with the diode.

[akis]

Damn I had this saved for "further reading" but no time. Thanks for pointing out the problem. I will read some more.

[akis]

OK just so I understand.

With a simple diode as relay protection.

When we take power off the coil, it generates an opposite voltage. This opposite voltage would potentially blow something , so we put a diode P at the ground, N at the +Vcc side, which under normal operation is open circuit, but when the coil tries to push its energy back into the circuit the diode conducts and the current flows through it and through the coil, in other words we short the coil.

However this delays the release time of the coil because ( I suppose) it keeps current flowing. But the current is flowing in the opposite direction so the relay plunger should now be pulled completely to the other side, which is exactly what we want, so what's wrong with that????

We then add a resistor/zener at approximately coil voltage, eg in this case 12 Volts. The diode isolates the zener under normal operation. When the coil "discharges" then the diode opens and the zener acts as a auto-variable resistor, so that the voltage drop across it stays below its rated, in this case 12 Volts. If the initial energy of the coil produced, say 20 Volts, the flow would be restricted by the zener which would increase its own resistance (or burn since there is nothing to protect it from unlimited current ? ) and therefore the resistor/zener would restrict current flow, but now the current will simply choose to go through whatever it would have chosen to go before (and thus cause damage), OR in any case current will still flow and the coil will still remail "active" - what am I missing please?

[pacificblue]

Originally posted by akis
But the current is flowing in the opposite direction so the relay plunger should now be pulled completely to the other side, which is exactly what we want, so what's wrong with that????

Not in the opposite direction. Look at the flow direction through the diode.

Originally posted by akis
We then add a resistor/zener at approximately coil voltage, eg in this case 12 Volts. The diode isolates the zener under normal operation. When the coil "discharges" then the diode opens and the zener acts as a auto-variable resistor, so that the voltage drop across it stays below its rated, in this case 12 Volts. If the initial energy of the coil produced, say 20 Volts, the flow would be restricted by the zener which would increase its own resistance (or burn since there is nothing to protect it from unlimited current ? ) and therefore the resistor/zener would restrict current flow, but now the current will simply choose to go through whatever it would have chosen to go before (and thus cause damage), OR in any case current will still flow and the coil will still remail "active" - what am I missing please?

The Zener works the other way round. It conducts, when the voltage is higher than the Zener voltage and blocks, when the voltage is below. Look at the drawing in the above post. The Zener is connected in the opposite direction.
The discharge goes through the Zener until it falls below 12 V. Then the Zener will block it and the relay deactivates. Without that Zener the relay would remain on until the voltage falls below its release voltage, which can be as low as 10 % of its nominal voltage, e. g. 1,2 V for a 12 V relay.
The Zener is protected from unlimited current by the relay's coil impedance. And the time, during which the current is high is very short.
Just try it out. It has been done successfully before.

[akis]

Thanks for your explanation.