I am trying to understand how an inductive load can pose a threat to its high side driver chip when the load is turned off. When the driver is ON, the load is connected to 12V supply through the driver's MOSFET switch. When the MOSFET switch is turned off, the electromagnetic field collapses which causes a high potential across the inductive load but with the opposite polarity as shown in scenario 2 below. This causes 112V appears at VCC which can potentially damage the chip. Note- I picked 100V as a random high vtg which generates across the inductive load when it is turned off.




Next question- How the MOSFET switch "T2" is working to protect VCC from the high voltage generated by turning off the inductive load?

Note- T1 and T2 are withing the high side driver IC with T1 to connect the load to 12V, whereas T2 to protect VCC from high voltage spikes.

 

Why not connect the protection diode directly across the inductive device, as is usually done? Or possibly slow the turn off just a bit. Consider that the voltage spike is directly related to the rate of current change, and so if that rate is a bit slower then the voltage will be quite a bit less. There will be more dissipation in the control mosfet, so there does need to be some calculation as to how much time in the linear mode is OK. Stretching the switch off time from one microsecond to one millisecond, as an example, will make the dV/dT .001of the original.
Next question: Have you actually measured the transient pulse??

 

This schematic is from the high side driver datasheet. My guess is the HSD manufacturer doesn't want to assume that the application engineer is going to connect a diode across the inductive load so they have provided protection within the driver chip against the voltage spikes generated by turning off the inductive load. But I am not able to understand how T2 exactly protects T1 from such an event.

 

But I am not able to understand how T2 exactly protects T1 from such an event.

Don't see how it can either.
You need a diode from the output (cathode) to ground to provide spike protection.

 

Are you sure T2 is connected to ground and not the output?

 

The T2 works as an active clamping circuit. But to work with an inductive load it has to be connected to output and not to ground (it only clamps Vcc in your circuit).

Here is an article from analog devices how it works:
https://www.analog.com/en/technical...nductive-loads-with-safe-demagnetization.html

 

That circuit would also protect T1 from excessive Vds if the T1 source were pulled excessively low by some transient.

 

If an external diode is tobe used, it should be solidly connected across the inductive portion of the load device. The benefit of the external diode is that usually it can be sized to handle the energy in the device it is connected to.
And still, slowing the switch-off time can reduce the spike energy quite a bit. That fact is never mentioned in the protective-diode religion, though. Why is that???

 

If an external diode is tobe used, it should be solidly connected across the inductive portion of the load device. The benefit of the external diode is that usually it can be sized to handle the energy in the device it is connected to.
And still, slowing the switch-off time can reduce the spike energy quite a bit. That fact is never mentioned in the protective-diode religion, though. Why is that???

It is easier to just put a external diode parallel to the inductive load than to slowly switch off the mosfet. And the mosfet needs to dissipate more energy while turning off

 

Certainly that is true. But pay attention to the units: I suggested changing one MICROSECOND to ONE MILLISECOND, which is still a rather short time. I was not saying "slowly", just a reduction in what the very fast system can do. And consider that relay operating the starter solenoid on my various MOPAR vehicles, which some performed flawlessly for over 20 years.

 

Certainly that is true. But pay attention to the units: I suggested changing one MICROSECOND to ONE MILLISECOND, which is still a rather short time. I was not saying "slowly", just a reduction in what the very fast system can do. And consider that relay operating the starter solenoid on my various MOPAR vehicles, which some performed flawlessly for over 20 years.

I'm in no way suggesting that you are wrong. I just think people avoid it because they have to think more.

And there are possible downsides of using a freewheeling diodes too (https://www.te.com/commerce/Documen...v&DocNm=13C3264_AppNote&DocType=CS&DocLang=EN)

 

I'm in no way suggesting that you are wrong. I just think people avoid it because they have to think more.

And there are possible downsides of using a freewheeling diodes too (https://www.te.com/commerce/Documen...v&DocNm=13C3264_AppNote&DocType=CS&DocLang=EN)

Certainly the diode will extend the time for the magnetic field to collapse, and so depending on the design the contacts may open much slower. For a PWM system with duty cycle controlling force, the control characteristics will be altered. And if a panel builder, or tech altering a panel gets the built in diode connection backward things will not work right. (Been there, done that).
And in 50+ years of designing and building assorted industrial equipment I have NEVER had a failure to to unsurpresed inductive spikes.. Of course, I always design with an adequate safety margin. I never used a 20 volt transistor on a 12 volt circuit, for an example. the half-cent price difference is simply not worth the risk.

 

Certainly the diode will extend the time for the magnetic field to collapse, and so depending on the design the contacts may open much slower. For a PWM system with duty cycle controlling force, the control characteristics will be altered. And if a panel builder, or tech altering a panel gets the built in diode connection backward things will not work right. (Been there, done that).
And in 50+ years of designing and building assorted industrial equipment I have NEVER had a failure to to unsurpresed inductive spikes.. Of course, I always design with an adequate safety margin. I never used a 20 volt transistor on a 12 volt circuit, for an example. the half-cent price difference is simply not worth the risk.

Here is the application note link for your reference. Please check the section Protection against low energy spikes and load dump and let me know if you are able to figure the T2's role in clamping the VCC voltage at a safe value by allowing energy to flow through it.

https://www.st.com/resource/en/appl...drivers-for-automotive-stmicroelectronics.pdf

 

I'm in no way suggesting that you are wrong. I just think people avoid it because they have to think more.

And there are possible downsides of using a freewheeling diodes too (https://www.te.com/commerce/Documen...v&DocNm=13C3264_AppNote&DocType=CS&DocLang=EN)

Here is the application note link for your reference. Please check the section Protection against low energy spikes and load dump and let me know if you are able to figure the T2's role in clamping the VCC voltage at a safe value by allowing energy to flow through it.

https://www.st.com/resource/en/appl...drivers-for-automotive-stmicroelectronics.pdf

 

Here is the application note link for your reference. Please check the section "Protection against low energy spikes and load dump" and let me know if you are able to figure the T2's role in clamping the VCC voltage at a safe value by allowing energy to flow through it.

I appreciate all of you for sharing your thoughts.

https://www.st.com/resource/en/appl...drivers-for-automotive-stmicroelectronics.pdf

 

T2 is an active clamp against load dumps and energy spikes on VCC it does not help with inductive load switching. So it does not protect T1 against inductive load spikes.

 

One thing being ignored is that any protection against any transient produced by ANY device should really be connected across that transient producing device. That is a flaw of the original drawing. It may not be clear to all, but it certainly stands out from where I stand.

 

T2 is an active clamp against load dumps and energy spikes on VCC it does not help with inductive load switching. So it does not protect T1 against inductive load spikes.

don't you think load dump (i.e. high voltage at VCC) can happen if the inductive load is switched off too fast?
Can you list out the reasons causing the load dump?

One thing being ignored is that any protection against any transient produced by ANY device should really be connected across that transient producing device. That is a flaw of the original drawing. It may not be clear to all, but it certainly stands out from where I stand.

do you mean the hand drawing where a flyback diode across the inductor is missing?

 

Consider that we have no information as to the surrounding that this is used in, nor are we given the resistanc or inductance of the solenoid, o t frequency of application. We are given the voltage but no hint of the source. So there are some guesses being made.

 

Consider that we have no information as to the surrounding that this is used in, nor are we given the resistanc or inductance of the solenoid, o t frequency of application. We are given the voltage but no hint of the source. So there are some guesses being made.

This is from this application note (https://www.st.com/resource/en/appl...drivers-for-automotive-stmicroelectronics.pdf) associated with the high side driver VNQ7E100AJTR ( which has 4 internal high side switches used to control 4 different automotive loads e.g. Headlights, cooling pump, etc.). I am reviewing protective features that this switch comes with and one of them is Protection against low energy spikes and load dump .