op amp to mosfet output destroyed

[g2c]

Hello, below circuit is a loV disconnect.
V+ is 12V from a lead battery. It worked nicely for 2 years until I disconnected and reconnected the load (mechanically) after what the op amp o/p stopped doing the job and started following the supply even when v+ < v-
Was this expected? did I have to connect the opamp o/p to the gate via a resistor? value? any other protection?
Thanks for your help!

 

[Matthewjtaylor]

Hello, below circuit is a loV disconnect.
V+ is 12V from a lead battery. It worked nicely for 2 years until I disconnected and reconnected the load (mechanically) after what the op amp o/p stopped doing the job and started following the supply even when v+ < v-
Was this expected? did I have to connect the opamp o/p to the gate via a resistor? value? any other protection?
Thanks for your help!


View attachment 148035

it's possible that disconnecting and reconnecting the load could have caused issues with the op-amp or MOSFET. When connecting the op-amp output directly to the MOSFET gate, it's important to include a gate resistor to limit the current and protect the op-amp from potential damage. A typical value for this resistor is between 100Ω to 1kΩ. This helps dampen any voltage spikes and ensures proper switching.

 

[alec_t]

If your load was inductive there could well have been a big voltage spike generated when you disconnected it. That spike could have caused gate-drain breakdown and hence damaged the op-amp.

 

[ZipZapOuch]

This op amp is ok for automotive use but may suffer if it sees voltage spikes from the unregulated supply (alternator) or inductive kick from a starter motor turning off or even from a relay coil de-energizing because the "absolute maximum" supply voltage should only be 16volts.

Likewise, applying current to a Mosfet gate (which is essentially a capacitor) can easily exceed the maximum source or sink loads of the op amp output of only 18mA on repeated long-term use. A resistor should be there to avoid the initial charge of the gate which is short but can eventually cause a fail.

I assume the circuit is designed to cut flow to your load when battery voltage falls below 10v.

Finally, the mosfet could have failed short circuit and the op amp is working perfectly but the MOSFET is not responding to the op amps instructions.

I would guess the zeners are not too sensitive with the 12k resistors instructions place. I would focus on the op amp and add a gate resistor, next look for the performance of the mosfet.

 

[g2c]

Thank you all for bothering to look into my troubles. I just replaced the quad opamp, not an easy job as it was soldered to the pcb. The circuit works again as it should. I will insert a ~300Ohm per your recommendation.

Btw the loads are a 12V to 12V powering a router and a optical adapter and a 12V to 7V powering two arduino boards with logic. all because we have here many mains shortage, some lasting more than the reserve of a Schneider ups.

 

[Tony Stewart]

Next time, don't "disconnect ... the load (mechanically) " (with the current enabled). Dry contact current breaks are notorious for creating as much voltage as possible in a microsecond before the air ionizes and arcs. Although 0.8A is not much to look at , it doesn't take much to clamp it. Even a 100 mA signal diode can do this in less than a microsec. when applied from Iout and cathode (bar)to V+.

 

[g2c]

Next time, don't "disconnect ... the load (mechanically) " (with the current enabled). Dry contact current breaks are notorious for creating as much voltage as possible in a microsecond before the air ionizes and arcs. Although 0.8A is not much to look at , it doesn't take much to clamp it. Even a 100 mA signal diode can do this in less than a microsec. when applied from Iout and cathode (bar)to V+.

Not sure i understand.. do you mean i should wire a reversed diode on the output of the circuit? Isn't the MOSFET's one enough ?

 

[Nigel Goodwin]

Not sure i understand.. do you mean i should wire a reversed diode on the output of the circuit? Isn't the MOSFET's one enough ?

No, that's not why it's there - it's simply a consequence of the manufacturing process, not to actually be used.

 

[danadak]

There are MOSFETs specifically designed and characterized to use the output
parasitic diode, as well as many its just there with little or no specs. So choose appropriately.
Used in motor control and inductive load switching applications. Its typically slow
so used for low freq applications. Datasheets have been getting better with more
data on the parasitic diode and its characteristics.

https://ww1.microchip.com/downloads/en/Appnotes/901.pdf

• Infineon's "Power MOSFET Basics" discusses the forward voltage and reverse recovery characteristics of the body diode, essential for designing energy-efficient systems
  Infineon
  Infineon
  .
• "Avalanche Design Guidelines" by Infineon explains the role of the body diode in handling energy surges during unclamped inductive switching
  Infineon
  .

Regards, Dana.

 

[crutschow]

Not sure i understand.. do you mean i should wire a reversed diode on the output of the circuit? Isn't the MOSFET's one enough ?

The MOSFET diode is the wrong polarity for protection as the transient voltage is negative going.
The diode would go from the MOSFET source terminal (cathode) to ground.

 

[g2c]

This is something I can do w/o a complete rework of the board. Any remark?
Thanks!

 

[danadak]

What is the frequency of modulation being switched ? Only reason
I ask is 300 ohms in gate can cause possible overrating of MOSFET
due to slowed switch rate, eg. more time spent going thru active region
hence dissipating power.

There are ap notes on web to calculate gate R for optimal operation.


Regards, Dana.

 

[g2c]

Frequency is practically null. The whole thing - lo v disconnect & load - is powered by an SLA battery, charged by a maintainer. The fet will be blocked ~once a month if mains are off for ~10 hours and V+ goes below 10.5V. It will be saturated again after the maintainer charges the battery to more than 11.5V. Given circuit hysteresis, the transition from one state to the other should be fast

 

[g2c]

I think the gate resistor shoul be chosen to limit the opamp outout current to 18mA i e. 15/18=830 Ohms

 

[ZipZapOuch]

I think the gate resistor shoul be chosen to limit the opamp outout current to 18mA i e. 15/18=830 Ohms

It is an odd op amp without short-circuit protection so being in the "completely safe" region with your occasional switching circuit is probably a good idea as long as your mosfet doesn't overheat during each switching cycle. I hope the off-to-on (high-to-low) transition is fast and your mosfet doesn't warm up because of a slow transition. You should be ok

 

[g2c]

Fet's input c=3.4nF. 5rc (>99%xition time) would amount to 3.4*.83*5~14us

 

[g2c]

modified board

 

[Nigel Goodwin]

Where are your decoupling capacitors?.

 

[danadak]

This may have been the casue of failure :

It worked nicely for 2 years until I disconnected and reconnected the load (mechanically) after what the op amp o/p stopped doing the job and started following the supply even when v+ < v-

If the Inv input was taken below its supply rail,. with no current limit, would have fried
the input structure thru the forward biased Zener acting as a simple diode when v+ goes
more negative than a diode drop.....Fix would be a series R to Inv input, say 1K ohm from
junction 12K and zener anode.

 

[g2c]

Where are your decoupling capacitors?.

I didn't think I needed any as it essentially a dc thing with slow ~.5Hz 'pompings ' by the maintainer. Only ac fluctuation when it disrupts / re-establishs power and i didn't see any changes with decoupling of the amp supply