Fit Cheap DC Ammeter to Power supply output?

Any time you put a sense resistor (or meter shunt) between the supply and load you're increasing the output impedance of the supply. A really great low noise regulator like the Jung has output impedance in the tens of micro-Ohms so it defeats the effort you've one to assure low impedance, low noise.

I'd follow Mark's direction -- just remember that the regulator is going to draw current of its own.
What "low noise" box needs a 10 AMP meter??

If you must: measure the current INTO the regulator. This will include the regulator's current and any wobble in it. Typically this will be small compared to load, and fairly steady.

No won't be doing it - asked here before purchase and got good advice.

Why 10 amps because the power supply is capable of delivering 6 at 24 Volts :)
Agree on it s adding an impedance in series with voltage source, by the way I´d expect an inductive component, but would like to know why it might introduce *noise* , I see no mechanism for that.

You're right, the meter and shunt have DC resistance -- for a 1mA meter (salvaged from an HP 400) it's about 38 Ohms. The shunt would have to be ~0.39 Ohm to measure 100 mA, and the inductance is damped by the small parallel resistance. Noise should be trivial.


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