I want to understand how to select the diode rates

I found this circuit in YT video i can not found it again

Batteries 400v - Capacitor 450v 1000uf - load 3.6 ohm - diode



This diode is placed to prevent the back emf on the switch

How diode rate is selected for a circuit to prevent back emf?

 

1n4007/UN4007

 

1n4007/UN4007

I want to know how to select the rate, what is the principle to select the diode

 

The diode will not be needed, because when the button is pressed about 120 amps will flow and thus the wiring will melt as the button evaporates. And if it is in a simulator there will be no arcing because all simulator components are perfect.
Plus, the component L1 has no inductance, only a resistance of 3.6 "m". Possibly short for ohms? or millihenries?

 

The diode will not be needed, because when the button is pressed about 120 amps will flow and thus the wiring will melt as the button evaporates.

The diode is reverse biased?????????????????

 

The diode will not be needed, because when the button is pressed about 120 amps will flow and thus the wiring will melt as the button evaporates. And if it is in a simulator there will be no arcing because all simulator components are perfect.
Plus, the component L1 has no inductance, only a resistance of 3.6 "m". Possibly short for ohms? or millihenries?

I doubt this circuit need a thyristor not just a switch
But the diode should be work as back emf suppression

 

Is that microfarad and millihenry?

 

That 120+ amps will not flow in the diode, but in the rest of the circuit, until something melts. The caption describes the coil as 3.6 ohms. Amps =400V/3.6 ohms
And the circuit is a good example of what is not to to be trusted on YT.

 

Is that microfarad and millihenry?

1 microfarad capacitor
3.5 milihenry inductance - 3.6 ohm coil

 

That 120+ amps will not flow in the diode, but in the rest of the circuit, until something melts. The caption describes the coil as 3.6 ohms. Amps =400V/3.6 ohms
And the circuit is a good example of what is not to to be trusted on YT.

It is push button means this current will pass for few milliseconds, i think nothing will melt as this fast (expect the switch so it need thyristor)
This current will start from capacitor so the second circuit only will get this big current
After the switch open their will be a back emf in this case the diode should do the work

 

The voltage developed is inductance multiplied by the rate of current change. Wit 400 volts and over 120 amps there will be an arc as the switch contacts open and thus the rate of change will not be fast.
And if this is a simulation then the voltage source has zero internal resistance and the capacitor has not much effect.
If this is an actual circuit what is the purpose?

 

It was a demonstration circuit
I want to learn how to select a diode ( diode rate) for a circuit like this

I think if we add a diode like 1n4007 which is 1kv and 1A to supress the bemf will not work or damage

 

Probably that is true. But given that we have no clue as to how fast the current changes with the non-realistic values, we have no idea as to the voltage or the power associated with that circuit operation. Do those contacts open in a nanosecond?? Or do they take a millisecond to cut off the current??
If a transistor oof some type is used, what is it's time to transition from full conduction down to 10% conduction?? Microseconds or picoseconds? Or milliseconds???The rate of change is what makes the difference.

 

The diode must be rated as follows:
Voltage: equivalent to supply voltage
Current: equivalent to load current

AS it does not state how long the switch is closed for, the current is unknown.

 

The diode must be rated as follows:
Voltage: equivalent to supply voltage
Current: equivalent to load current

AS it does not state how long the switch is closed for, the current is unknown.

It is normal open push switch means it will take very few milliseconds, just a press

So diode should be 120A, 400v!
Is there a diode has rate like this?

 

1 microfarad capacitor
3.5 milihenry inductance - 3.6 ohm coil

So diode should be 120A, 400v!
Is there a diode has rate like this?

See diagram:

ADDED:
One more (may be better) circuit:

ADDED:
Final circuit:

_____

 

It is normal open push switch means it will take very few milliseconds, just a press

So diode should be 120A, 400v!
Is there a diode has rate like this?

I wouldn't consider a switch that will switch a 120A inductive load to be "normal". That's a pretty rare (and expensive) switch.
But diodes that take more than 120A are quite common.
https://www.littelfuse.com/products...tifier/rectifier-capsule-type/w0507yh420.aspx