Why put a diode parallel to a resistor?

[Artlav]

I was looking over a switching battery charger schematics, and couldn't figure out one detail.
Why put diodes in parallel with resistors?

It appears in several places.
One - across the current limiting resistor in the P-mosfet gate driver:


Other - across the pull-up resistors on the microcontroller's reset pins:



What purpose(-s) do these serve?

When i saw the one on the gate driver, i thought it was something to make the rise time faster than the fall time.
However, when i found the same arrangement on the pull-up, i started to suspect there is something else going on here and i have no clue what.

Maybe it's just a coincidence, and they used the same part and pattern to speed up the rise time in one place and protect against overvoltage in the other place, but i decided to ask just in case it's not.

[Simon]

well using a resistor between driver and mosfet gate seems to defeat the object of having a driver as it's purpose is to source and sink large currents. Maybe your right and they wanted a faster turn off than turn on.

On the reset pin it is a mystery.

[Precipice]

On the reset pin, it'll be to drain the capacitor back to zero when power goes away, otherwise
a) you won't get a reset after a power glitch
b) you'll be driving a pin on an unpowered chip.

[firewalker]

The reset diode is the AVR042 suggestion for ESD protection.

The Schotky diode on the gate is something for Slow-on, Fast-Off for heavy loads.

Alexander.

[Simon]

I thought that too but I've never used a diode and never seen one suggested I thought that the capacitor would drain through the resistor anyway kjust that it would take longer. The diode work on the assumption that turning power off means that Vcc gets connected to GND

[Precipice]

The diode work on the assumption that turning power off means that Vcc gets connected to GND

Well, yes. If power stays at 3.3V, it's not off, and you don't need another reset.
Systems that don't reset properly on a power glitch are a nightmare - which is why most micros have proper brownout detectors, which will work better than the diode hack.
Reset supervisor chips come in SOT23 (so, since you could ditch at least the capacitor, and probably the resistor, they'd be smaller than the diode hack), and have accurate trip points and reset delays. I use them all over the place. They're cheap, reliable, available, small. All sorts of good things! They're also low power, so if you want to bodge a reset button in, you can usually just feed the reset chip through a resistor, and short across its power pins with a switch to start a reset cycle, so you don't need to stock chips with a reset-in pin for when you're prototyping.
Using power supplies' power-good outputs is often also a good way of making a reset, if they come for free.

>The reset diode is the AVR042 suggestion for ESD protection.

Hmm. That sounds unlikely. Protection for the AVR's ESD protection structure, maybe.

[Kjelt]

Reset supervisor chips come in SOT23
They're cheap, reliable, available, small.

Which ones do you use or can you recommend?

[firewalker]

From AVR042:

If not using High-voltage Programming it is recommended to insert an ESD protecting diode from RESET to Vcc, since this is not
internally provided due to High-voltage Programming.

[Precipice]

Fair enough!
BAT54s don't really count as ESD diodes, though - not without a hefty resistance to stop them exploding. 1K seems to make them invincible, but bare, they'll tend to fail.

[Precipice]

Reset supervisor chips come in SOT23
They're cheap, reliable, available, small.

Which ones do you use or can you recommend?

For hacking around, I've got reels of:
Maxim   DS1811   4.28-4.49V trip 150ms 5k5 pullup    SOT23
Fairchild   FM1233ACS3X   2.70-2.97V trip open drain, active low   SOT23
On Semi   NCP301LSN27T1   2.7V reset chip open drain, active low   SOT23-5

Not all of those are current parts any more, I suspect. Microchip make nice ones, if you're hunting. Just pick something suitable for your supply voltage, make sure you fit a pullup resistor if it's open-drain, and get one with a delay if you're feeling paranoid and want to wait for the power the be properly stable before you release nreset.

For customer projects, it depends on what they can source in a bundle deal with other silicon. (Say, if I'm using a load of TI stuff, it'll probably be a TI chip - although a lot of the time, I'm using proper PMICs and carefully crafted power supplies, so the reset control will come for free, or I'll be relying on the brownout and watchdog engines if it's a tiny micro).

[David Hess]

I thought that too but I've never used a diode and never seen one suggested I thought that the capacitor would drain through the resistor anyway kjust that it would take longer. The diode work on the assumption that turning power off means that Vcc gets connected to GND

The capacitor will drain through the resistor when power is lost however if power is only lost for a short period of time like with a glitch, the capacitor may not drain low enough to trigger a reset when power is restored.  The diode prevents this problem by discharging the capacitor immediately.

The parallel connected diode and resistor in series with the MOSFET gate are there to create unequal turn on and turn off times.

[Dago]

A diode in parallel with a gate resistor when driving a FET or an IGBT is usually used to compensate for the usually slower turn-off of the device (so usually the diode allows the gate to drain faster).

[Artlav]

So, a coincidence.
The diodes serve specific purpose in each case, actually making sense both times.

Thanks all.

[rx8pilot]

I have used the diode/resistor arrangement on my high-current switching designs to deal with capacitive loads on the FET. The resistor allows me to control the transition time from OFF to ON to reduce the inrush current as the capacitors charge. It's a delicate setup though - high value resistor will make the FET so slow it can cause damage by dissipating too much heat. A low value resistor allows the FET to turn on so fast that the inrush current of the load causes a problem as well.

In the case of an over current fault, I can pull the gate down very fast through the diode to save the rest of the circuit.