How to measure the area under the curve?

[LukasN]

Hi,

I want to measure the current inside a boost converter circuit.
My HP 34401A measures 04.9367 mADC and I want to double check this (because I can ) with my Rigol DS2072A. I never used a oscilloscope for this.

Which function do I have to use to get the right voltage?

The voltage @DSO (screen see below) is measured at the probes of the HP 34401A. The resistance of the probes, the cable and the meter is 5.90 Ohms.


Thank you.


edit: changed title from "Another Current question in switchmode boost converter"

[ovnr]

"Inside" a boost converter? Do you mean the inductor current, the switch current, the diode/active rectifier current, the output current, the input current, the output filter capacitor current, or the input capacitor current?


Because there's plenty of different ones to pick from!


No matter what you do, you will want to insert a low-value low-inductance shunt resistor (SMD preferably) into the current path of interest, then probe that. If you're dealing with high-frequency and/or high-current converters, even a few hundred nH will massively skew your measurements, and your test leads + multimeter present much more than that.


As for "get the right voltage": Measure a DC current with your setup, say 1A, and look at the voltage you get. If you're actually right about the 5.9 ohms, you should see 5.9 volts across the meter, which is hilariously improbable if you're in high-current mode and don't have horrifically bad leads. Just use Ohm's law in any event.

[LukasN]

and thought I should find out how precise it is when I measure only with a meter. I've never measured the area under the curve, the integral with a oscilloscope. I have to learn it .

[MLXXXp]

You may find this video if Dave's useful:
http://youtu.be/Dh0xYu8YvaE

[dannyf]

I want to measure the current inside a boost converter circuit.

Charge up a capacitor with it.

[LukasN]

I want to measure the current inside a boost converter circuit.

Charge up a capacitor with it.

I will try this soon.

Measure a DC current with your setup, say 1A, and look at the voltage you get. If you're actually right about the 5.9 ohms, you should see 5.9 volts across the meter, which is hilariously improbable if you're in high-current mode and don't have horrifically bad leads.

I used 100mA to stay in same range. I'm not spot on 590mV but near enough. 5,9 Ohms was measured with another 15 to 20 year old bench top meter (Fluke).
I've got skills in electronics an know Ohm's law but I'm not what you would call a engineer .

You may find this video if Dave's useful:
http://youtu.be/Dh0xYu8YvaE

OK, I can confirm the 1,73µVs, measured with the "Per.Area" function, with the math expression Intg(CH2). They do both the same as said by Dave in the video.
Now I have to multiply this with the frequency, because I got only one period ... I get 33,68mV.
By using Ohm's law, I get 5,71 mA. OK, this is about 11% more then I measured 24 hours ago with my meter.

I think I have to measure again but follow to ovnr and use a low-value low-inductance shunt resistor instead of a meter shunt .


Thank you all for help!!!