- #1
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?
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SUMMARY
The discussion centers on the use of reverse diodes, specifically flyback diodes, in relay circuits to protect operational amplifiers and other components from high voltage spikes generated by inductive loads. When the current to an inductor or relay coil is switched off rapidly, a high voltage spike occurs due to the formula V = L * dI/dt, which can damage the driving circuit. The flyback diode clamps this voltage spike to approximately 0.7V, redirecting excess energy back to the power supply, thereby preventing damage to sensitive components such as transistors and op-amps.
PREREQUISITES- Understanding of inductive loads and their behavior in electrical circuits
- Familiarity with operational amplifiers and their applications
- Knowledge of diode types, specifically flyback diodes
- Basic grasp of voltage, current, and the relationship defined by V = L * dI/dt
- Research the function and application of flyback diodes in relay circuits
- Study the principles of inductive kickback and its implications in circuit design
- Explore the use of clamp diodes in protecting various electronic components
- Learn about designing circuits with inductive loads and the necessary protective measures
Electrical engineers, circuit designers, and hobbyists working with relay circuits and inductive loads who need to understand protection mechanisms for sensitive components.
- #2
CWatters
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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.
- #3
davenn
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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
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