Adding the ability to measure ac ammeter in dt830d multimeter

Use a suitable known value "shunt" resistor in line with the circuit, and measure the voltage across that.

eg. A one ohm shunt would give 1V per amp.
A ten milliohm shunt would give ten millivolts per amp.

For in-circuit / PCB use, they are generally just precision low value resistors. For higher currents and stand-alone use, they are often four terminal devices, so the measuring points are separate from the power terminals.

rjenkinsgb said:

Use a suitable known value "shunt" resistor in line with the circuit, and measure the voltage across that.

eg. A one ohm shunt would give 1V per amp.
A ten milliohm shunt would give ten millivolts per amp.

For in-circuit / PCB use, they are generally just precision low value resistors. For higher currents and stand-alone use, they are often four terminal devices, so the measuring points are separate from the power terminals.

https://uk.farnell.com/murata-power-solutions/3020-01103-0/shunt/dp/1498416?cjevent=d6a2e4e5ad5811ee823b00c40a18ba74&cjdata=MXxZfDB8WXww&CMP=AFC-CJ-UK-8989328&gross_price=true&source=CJ

https://www.mouser.co.uk/ProductDetail/Ohmite/CS10FR002E?qs=xZ%2FP%252Ba9zWqZDTHjTl9QFXQ%3D%3D

Click to expand...

Every multimeter has shunt resistance, but this model of multimeter cannot measure amperes and mA ac. The challenge is actually this.

It can measure AC voltage. That is all that is needed when you use an appropriate shunt connected in the circuit being measured!

If you have noticed, the shunt is the same in all multimeters that measure ac and dc amperes, but there is a different selection selector, one selector measures dc and the other selector measures ac...now we wanted to add a circuit to the prop and multimeter. provide ac and mA for us!?

Now I checked the test with a 1n4007 diode, I put the series test lamp in the circuit and the ammeter of the dt832 multimeter, the amperage is measured, but due to the power loss compared to the diode, it does not show 0.39A, instead it shows 0.22A.
Vrms=230v
r filament lamp stat off = 33 ohm
i = 0.39A
i with diode =0.22A
Related video analysis and testing

I did it..
Measurement mode without trick and with trick
Measuring the amperage of the 100w filament series test lamp with a dt832 multimeter
because a diode creates a loss, the ammeter shows 0.22A, by using a diode bridge, the loss was minimized and it showed the same amperage ac 0.39A true rms
Link to the full video of the relevant test

electronium said:

Every multimeter has shunt resistance, but this model of multimeter cannot measure amperes and mA ac. The challenge is actually this.

Click to expand...

That's why you need a separate, EXTERNAL current shunt!

eg.



For low power electronics, you can just use a normal resistor, eg. one ohm or 100 milliohm.

Or you could use an add-on current clamp, eg:


Just using external rectifiers can work under some circumstances, but not where the voltage drop they add will affect the circuit, or if the current is too high.

This time, testing the small 15w 220v refrigerator lamp with the mA section of the dt832 multimeter, with and without tricks, and as you can see, the mA ac section works well, now it is enough to have two diode bridges and input and output terminals for the ampere and milliampere sections. If we consider inside the multimeter or outside the multimeter, the work is done
video test mA ac with dt832

Your link is not doing much for me? What am I supposed to see or hear?

Before we start talking about a shunt and measuring the AC voltage drop across a shunt maybe we should look at the AC Voltage ranges of the meter. The meter only has 200 and 500 VAC ranges with 100 mV and 1.0 volt resolution. It's only an inexpensive meter and adding a shunt sans amplification is not going to get you anything. Typical shunts only output 50 mV or 100 mV full scale.

Next even meters that measure AC voltages and currents unless specified otherwise only actually average and display RMS values. This is fine for pure sine waves but that's as good as it gets, average responding RMS indicating. Good meters are RMS responding RMS indicating. Here nor there but something to consider.

You could use a peak detector which will charge to the peak value of the AC waveform and then just do the basic math.

I would still be curious as to what I am supposed to take away with your link?

Ron