EEVblog #376 - Multimeter Fuse Diode Followup

How diode bridge protection across a shunt resistor in a multimeter works, and why the fuse blows before the diode does on a well designed multimeter like the Fluke.
A followup to the previous video on Multimeter input protection:

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[EEVblog]

The 10A range does not have any diode protection for the (50mOhm) shunt resistor. The diodes are only for the mA jack. The 10A shunt is physically a much beefer shunt so it able to handle much higher peak pulses than the mA shunt. Thus Fluke deemed it was not necessary (or possible) to protect it using semiconductor devices. The 10A fuse would simply blow before any damage is done to the 10A current shunt.

[EEVblog]

But however, the diodes still function overvoltage protection devices for the ADC inputs in the case of the 10A range. In the 10A range, the 5ohm mA shunt resistor taps off the voltage from 10A shunt, with the diode protection after that. So for the 10A range, the protection diodes functionality reduced to that of ADC and not shunt resistor protection.

[bgdwiepp]

if you're wondering why 8.3ms on diodes, it's because that's half a sinewave from a 60hz source.

[MetalPhreakAU]

L1 is just a choke coil on the input (common terminal side). Dave mentioned it in his fluke teardown video. Doesn't really have any effect or importance on the circuit here.

[BaeHat]

As i see it, there is even more "protection" of the protective diodes, as the shunt resistor will also carry some current.
So, in the time from the fault to the fuse blows, not only will the diodes carry some current to blow the fuse, the shunt ressistor will also carry some current, (1-2 amps)? which will make the fuse blow even faster.

[dumyyyyyy]

You just busted the myth that "the fuse blow the last one! "(old Romanian electronics myth). Thank you for this nice video!

[MetalPhreakAU]

The diodes only conduct when there is a fault current. 10A on the 10A input will just go through the fuse and the sense resistor. The diodes are there to protect the sense resistor. Say you had a fault current of 100A through that 0.05ohm sense resistor. There is now a 5v drop over the sense resistor. That's 500w of power dissipated! That sense resistor would fail before the fuse breaks. But the diode bridge conducts at ~3.6V so the power impulse bypassed through them instead.

[viesturssilins]

Link to the the first video on input protection is wrong.

[envisionelec]

Ah yes. I routinely puzzle engineers by spec'ign "piss ant" diodes for surge current. I always point 'em to the datasheet. Never had a problem!

[Coolkeys]

There must be some electricians on here, fault currents on UK domestic mains can be up to 8000 amps god knows about 415v 3 phase. Could inductor L1 slow the rise in fault current to give the protection more time to work?

[BarsMonster]

Wait, how come it's not important? It limits current rise time, so that protection would have time to react...

[MeneGR]

2:24 ".. into the jacks amp". Good one!

~155 videos to catch up!

[rsattahip]

Oops, never did that before. I hate that easy mistake.