Re: [EVGRAY] J ² = 1

Database ID: 110714
2018-11-22T18:03:56+00:00
onielsen2000 <[email protected]>

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Now the image is attached!

Regards
Ole
 

---In [email protected], <onielsen@...> wrote :

 Image missing. Here it is again.

Hi Andreas,

Both ways can be used. For low inductance of a current shunt let the return wire go close to and in parallel with the shunt resistor to minimize any area for the magnetic field.

Here the return wire is twisted around the current shunt.

Also the 50 Ohm feed through is necessary. Without it a lot of reflections will happen unless the scope itself has a 50 Ohm input. As scope inputs usually are 1M Ohm the signal will be reflected due to the great impedance mismatch from 50m Ohm to 1M Ohm and the signal will drown in noise. 50m Ohm to 50 Ohm reduces the reflections a lot.

Regards
Ole

 

---In [email protected], <onielsen@...> wrote :

 Hi Andreas,

Both ways can be used. For low inductance of a current shunt let the return wire go close to and in parallel with the shunt resistor to minimize any area for the magnetic field.

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