alex36
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Why reverse diode is used in relay ?? In my book its written that ,to protect op amp from damage but how does it protects?
Why does a reverse diode protect an op amp in a relay circuit?
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If you try and switch off the current to an inductor/relay coil instantly you get a high voltage spike that can damage the driving circuit. This is because the equation for the voltage across an inductor is...
V = L* dI/dt
where L is the inductance and dI/dt is the "rate of change of current".
If the current changes rapidly (eg when switching off) then dI/dt is very large and so the voltage becomes very large. This effect can be useful and is frequently used in voltage booster circuits (eg when you want to make a step up DC to DC converter).
In your relay circuit the high voltage can cause damage. The diode prevents the voltage going very high by limiting it ("clamping it") to around Vd = 0.7V. The diode is sometimes called a clamp diode, a freewheeling diode or a flyback diode...
https://en.wikipedia.org/wiki/Flyback_diode
Sometimes relays are used to switch inductive loads like motors so there might be two diodes. One on the coil side to protect the driver and one on the switch side to help stop arcing across the relay contacts.
V = L* dI/dt
where L is the inductance and dI/dt is the "rate of change of current".
If the current changes rapidly (eg when switching off) then dI/dt is very large and so the voltage becomes very large. This effect can be useful and is frequently used in voltage booster circuits (eg when you want to make a step up DC to DC converter).
In your relay circuit the high voltage can cause damage. The diode prevents the voltage going very high by limiting it ("clamping it") to around Vd = 0.7V. The diode is sometimes called a clamp diode, a freewheeling diode or a flyback diode...
https://en.wikipedia.org/wiki/Flyback_diode
Sometimes relays are used to switch inductive loads like motors so there might be two diodes. One on the coil side to protect the driver and one on the switch side to help stop arcing across the relay contacts.
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alex36 said:
the part being protected could be the output of an IC, opamp or otherwise
or it could be as in my circuit below, a single transistor
CWatters said:
The way you described that infers that the diode just clamps ( stops) the back EMF. This isn't really what happens.
That energy has to go somewhere and it doesn't/isn't just stopped at the or by the diode.
So, better to look at it that any spike over around 0.7V ( the conduction voltage of the diode) is conducted back through the diode to the positive rail ( the +5V in my circuit above) and away from the transistorDave