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Purpose of Schottky Diodes

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ssylee

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I've been using this motor driver circuit (see attached) for the motors I have received with **broken link removed**. However, for my motor driver circuit, I did not connect the catch Schottky diodes. While the cheap motors accompanied with the kit could be culprits of them burning out, I'm not sure if the fact that I removed those diodes could cause the burn outs. It would be great to know what you think. Thanks.
 

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I've been using this motor driver circuit (see attached) for the motors I have received with **broken link removed**. However, for my motor driver circuit, I did not connect the catch Schottky diodes. While the cheap motors accompanied with the kit could be culprits of them burning out, I'm not sure if the fact that I removed those diodes could cause the burn outs. It would be great to know what you think. Thanks.

hi,
The motor windings are an inductive load.

When switching the current flowing thru an inductive load a back emf is generated.
formula: Vemf = - L dI/dt. Where I is the interrupted current and the 't' is the speed at which the current is interrupted [ switched OFF].

So if you have 1amp flowing in the motor and you switch it off in say, 1mSec, thats Vemf = [ L *1]/0.001 = -L * 1000V at the FET pins.!!

The diodes will become forward biased by this high voltage and clamp the FET pin to approx Vsupply + Vdiode forward voltage drop.

So the diodes are are essential.
 
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I have read and re-read your message and haven't figured out what burned out.

Is it the motor, the microprocessor, the battery, some MOSFETs, the IXDN404PI, what?
 
I can't see any Schottky diodes in the schematic, they're ordinary silicon diodes.

The diodes are required to protect the MOSFETs from the high voltage generated when the motor is turned off, as said above.
 
I can't see any Schottky diodes in the schematic, they're ordinary silicon diodes.

The diodes are required to protect the MOSFETs from the high voltage generated when the motor is turned off, as said above.

hi hero,
Schottky diodes.:)
 

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hi,
The motor windings are an inductive load.

When switching the current flowing thru an inductive load a back emf is generated.
formula: Vemf = - dI/dt. Where I is the interrupted current and the 't' is the speed at which the current is interrupted [ switched OFF].

I think you forgot the "L" for the inductance.

The equation is Vemf = L dI/dt.

The inductance can be 1 mH or so, but the switching times will be more like 1 µs, so you still get silly peak voltages if you don't have diodes.

The Schottky diodes have a lower voltage drop and faster switching times, which is why they are preferable.

If the circuit is controlling the current by PWM, the supply will only be interrupted for very short times. The current will continue to flow in the windings, through the Schottky diode, until the power is restored. See figure 4.6 of Jones on Stepping Motor Current Limiting. I know that is for stepper motors, but the current control by PWM applies to any inductive load.
 
hi hero,
Schottky diodes.:)


Then why are they drawn as ordinary silicon diodes?

I shouldn't have to look at the datasheet to see what type of diode it is.

The correct symbol for a Schottky diode is posted below.
**broken link removed**
 
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I think you forgot the "L" for the inductance.

The equation is Vemf = L dI/dt.

The inductance can be 1 mH or so, but the switching times will be more like 1 µs, so you still get silly peak voltages if you don't have diodes.

The Schottky diodes have a lower voltage drop and faster switching times, which is why they are preferable.

If the circuit is controlling the current by PWM, the supply will only be interrupted for very short times. The current will continue to flow in the windings, through the Schottky diode, until the power is restored. See figure 4.6 of Jones on Stepping Motor Current Limiting. I know that is for stepper motors, but the current control by PWM applies to any inductive load.

hi Diver,
Thanks for the heads up, silly me, I have now corrected my error.:rolleyes:

I think its time for tea and toast.!

Regards
 
Then why are they drawn as ordinary silicon diodes?

I dont know.;)

I shouldn't have to look at the datasheet to see what type of diode it is.

He did say 'Schottky in his text.

hi,
Hows the job search going.?:)
 
I've applied for a few jobs, perhaps not as many as I should.

The good news is I have a job interview and I've enrolled on a PAT testing course at college.
 
I haven't seen the link but I know that I have to buy the book listed on that site to do the course.
 
I've been using this motor driver circuit (see attached) for the motors I have received with **broken link removed**. However, for my motor driver circuit, I did not connect the catch Schottky diodes. While the cheap motors accompanied with the kit could be culprits of them burning out, I'm not sure if the fact that I removed those diodes could cause the burn outs. It would be great to know what you think. Thanks.

Hi there,


I have a question too: What would make you remove those diodes in the first place?

Second question: Who designed that circuit with 470 ohm gate resistors?
That doesnt look too well designed to me. Because of the way MOSFETs turn on and off the
gate resistors usually have to be more carefully chosen and with asymmetrical drive also.

Also, there are usually built in diodes that could operate similar to the ones you 'removed'
and so there is a chance that they would do almost the same job. I think the whole design
would have to be looked at more carefully, not just the diodes.
 
Where's the shoot-through protection?

I found the datasheet for the MOSFET driver and there doesn't seem to be any.
https://www.electro-tech-online.com/custompdfs/2010/03/99018.pdf

If both the top and bottom MOSFETs turn on simultaneously, they'll short circuit the supply which will probably be fatal to the MOSFETs unless the current limiting is fast enough to protect them.
 
I have a question too: What would make you remove those diodes in the first place?

Yeah, I was wondering that too. It's rather like saying, "I took off all those useless wheel nuts from my car."
 
Then why are they drawn as ordinary silicon diodes?

I shouldn't have to look at the datasheet to see what type of diode it is.

The correct symbol for a Schottky diode is posted below.
**broken link removed**

Schottky diodes are almost always shown with the normal diode symbol, I don't think I've ever seen that symbol used in service manuals?.
 
I've not read many service manuals but every schematic I've seen calling for Schottky diodes uses that symbol.
 
Thanks for the note. It was the motor which died. The symbol for Schottky diode is incorrect as I searched for the part number in the Multisim circuit drawing program.
 
schottky diodes, to answer the question which is in the title of this thread, are just normal diodes with low forward voltages, e.g 0.14V, 0.2V only, whereas silicon or germanium ones are around 0.5-0.7V. They're four times costlier too, in my case. :)

One of the applications is the construction of low-dropout rectifiers (AC to DC converters).
 
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