Inductor, switch, snubber diode(flyback diode)

[Bassalisk]

Hello,

I am designing a simple circuit and I need some guidance over snubber diode.

When we open the switch, when we have a inductor in circuit, we will have that arc over switch which is something that we don't want.

Using this snubber diode, we can avoid this. I feel that we will have HUGE voltage spike, on inductor, when we open that switch.

How can I (safely) measure that voltage spike, in order to determine how much inverse breakdown voltage my diode has to have? I am assuming oscilloscope.

My source is 30V DC.

 

[technician]

When the switch is closed current flows through the inductor and there is a magnetic field.
When you open the switch the current stops instantaneously and the magnetic field collapses very quickly.
This gives a high induced emf which can damage other components in the circuit, can cause a spark across the switch contacts and even cause electric shock to anyone in contact with the terminals of the coil.
The way to reduce this high emf is to arrange for the current to decrease slowly so that the magnetic field decreases slowly.
Before the advent of solid state diodes a common technique was to have a resistor (a lamp bulb) permanently connected in parallel with the coil. When the switch is opened any induced emf tries to maintain the current (lenz's law) and the resistor allows this to happen The current decreases slowly and there is no large induced emf.
The disadvantage is that power is wasted because of the resistor permanently connected.
The modern solution is to replace the resistor with a diode which is connected in reverse across the coil. No current will flow through the diode when the switch is closed and current flows through the coil.
When the switch is opened the polarity of the emf across the coil is to try to keep the current flowing (this is the opposite polarity to the supply voltage for the coil!)
The diode now acts like a low resistance and the current decreases slowly avoiding any large induced emf.
I do not think that reverse breakdown of the diode is an important matter.
This is a common technique used when relays are used in transistor circuits. A diode is connected across the relay coil (in opposite polarity to the power supply).

 

[Bassalisk]

When the switch is closed current flows through the inductor and there is a magnetic field.
When you open the switch the current stops instantaneously and the magnetic field collapses very quickly.
This gives a high induced emf which can damage other components in the circuit, can cause a spark across the switch contacts and even cause electric shock to anyone in contact with the terminals of the coil.
The way to reduce this high emf is to arrange for the current to decrease slowly so that the magnetic field decreases slowly.
Before the advent of solid state diodes a common technique was to have a resistor (a lamp bulb) permanently connected in parallel with the coil. When the switch is opened any induced emf tries to maintain the current (lenz's law) and the resistor allows this to happen The current decreases slowly and there is no large induced emf.
The disadvantage is that power is wasted because of the resistor permanently connected.
The modern solution is to replace the resistor with a diode which is connected in reverse across the coil. No current will flow through the diode when the switch is closed and current flows through the coil.
When the switch is opened the polarity of the emf across the coil is to try to keep the current flowing (this is the opposite polarity to the supply voltage for the coil!)
The diode now acts like a low resistance and the current decreases slowly avoiding any large induced emf.
I do not think that reverse breakdown of the diode is an important matter.
This is a common technique used when relays are used in transistor circuits. A diode is connected across the relay coil (in opposite polarity to the power supply).

I understand. I will now open another thread regarding this matter. I will show what I have to make.