Series inductance in a resistor

[onemilimeter]

Power or current sensing resistors are usually labelled as "low inductance", "very low inductance", or "non-inductive". Values of inductance are seldom given. For Vishay parts, "very low inductance" is refer to 0.5 nH to 5 nH. Do you think "non-inductive" resistor can be treated as a resistor having less series inductance than those labelled as "very low inductance"? By the way, for current sensing resistor used in switching power supply (e.g. switching frequency @ 500 kHz), do you think a resistor with 0.12uH will give any problem? Thanks.

[Zero999]

The level of inductance which can be tolerated depends on the current as well as the frequency, the higher the current the more effect the inductance will have.

[Alex]

There is no such thing as 'non-inductive' any real component will have some inductance.

In the case of a current sense resistor, to add to Hero's post, the inductance of a current sense resistor will add to the measured voltage (at a different phase) across the resistor which you are trying to measure to calculate the current. This will be an error in your measurement. You can find out the voltage that will be developed due to the inductance of the part by V(t) = L * dI(t)/dt. The higher the inductance, the higher parasitic voltage. The higher the rate of change of current (directly linked to operating frequency), the higher the parasitic voltage. Parasitic capacitances will also cause ringing and if very severe, it could destroy the circuit, possibly the switching MOSFET or some other MOS part.

[onemilimeter]

... if very severe, it could destroy the circuit, possibly the switching MOSFET or some other MOS part.

That's what I'm worrying at the moment. I'm building a 3-phase inverter now (maximum dc-link voltage is 100V, maximum phase current is 30 A peak-peak, maximum switching frequency is 20kHz). I wish to measure the dc-link current using a current sensing resistor, which will be placed between the dc-link capacitor and the negative rail of the inverter (see attached figure). I heard people saying that the dc-link capacitor must be placed as close to the inverter as possible because longer wire will add more inductance and hence increase the Ldi/dt. I wonder how long is considered "longer" in practice.

[Alex]

Yes, you can follow best practises, but if you want to do this properly you should treat it numerically, as it is a numerical problem specific to your arrangement. It makes sense for the DC side capacitor to be close to the transistors to delivery energy when needed. It is also important to keep leads short to minimise inductance. Also important that your source can provide the energy required between cycles so that the DC rail doesnt droop too much. But how much is too much depends on you application and you cant avoid using the calculator.

[Zero999]

The voltage doesn't make much difference, it's the current which is important here.

I wouldn't worry too much about the parasitic inductance causing voltage spikes and destroying components but it will mess with the measurements. To put it into perspective, unless the parasitic inductance is huge, the amount of energy stored in it will be tiny

Suppose you use a wire wound resistor with a inductance of 1uH

E = ¹/₂I²L = 0.5*30²*10^-6 = 450uJ which is not going to kill anything.

The accuracy is the real problem.
X_L = 2piFL = 2*3.142*500*10³ = 3.142 Ohm, higher at the harmonics which will totally mess up the readings.

Do the calculations for 120nH and you'll find the induced voltage is probably still high enough to mess up the readings but the energy will be too low to cause any harm.