# Siglent SDS1104X-E Integrate Function - Units of Measure

I am using a Siglent SDS1104X-E four channel oscilloscope which has a variety of math functions, one of which is to integrate a signal.

Currently I am measuring a voltage pulse across a gate resistor which invariably feeds the gate of a MOSFET. I want to determine the area under the voltage curve and use it to calculate the equivalent current (in terms of area, not peak value) using the measured resistance value.

Here is the wave form for which I would like to integrate (section isolated using the gate function).

When I set up my scope I am guessing that the resulting integrated waveform is representative of the area under the measured curve versus time, is this correct? I am assuming that the peak value of this curve is the total area under the measured waveform. What confuses me is the unit of measure for this resultant waveform... uWb. After some searching I can only assume this is micro-Webers, which apparently converts 1:1 to microvolt-seconds. Is this a correct assumption? If so, why is the scope measuring in uWb and not uv-s?

Next, how do I make use of this value to calculate current if the resistor is 36.18 Ohms?

Source

As you see Wb=Vs, so it doesn't make any difference.

Next, how do I make use of this value to calculate current if the resistor is 36.18 Ohms?

$$I=\frac{V}{R}$$

• Weber is a measurement of magnetic flux. While wikipedia says it has a 1:1 va
– mike
Apr 25, 2021 at 20:26
• Ok, let's try this again....while wikipedia says a Webber equals a volt-second, a volt-second is not equal to a volt. Why is a scope displaying a Weber, a measure of magnetic flux, instead of volt-seconds. How did you derive that you can simply apply ohms law to calculate current?
– mike
Apr 25, 2021 at 20:29
• Can you explain what the scope is doing?
– mike
Apr 25, 2021 at 20:30
• @mike Someone who wrote the firmware of the scope choose Wb as a result of integrating Volts over time. What other law do you know to calculate current from resistance and voltage? Apr 25, 2021 at 20:37

Strange unit to choose, $$\\text{Wb}\$$, but $$\\rm{Wb =V\cdot s}\$$, so you're good. Perhaps they want to avoid confusion for people thinking $$\\rm{Vs}\$$ means volts.

That white trace looks exactly like what I would expect when integrating the yellow trace over that interval. The "peak" is 4V, roughly triangular over an interval of about 4μs, so I would expect the area to be about:

$$A = \frac{1}{2}xy = \frac{1}{2}\times 4\mu \times 4 = 8\mu$$

I'd say that $$\8.9\rm{\ \mu V\cdot s}\$$ is correct.

Assuming the resistance remains constant:

$$\int{I\cdot dt} = \int{\frac{V}{R}\cdot dt} = \frac{1}{R}\int{V\cdot dt}$$

Essentially, if you have 8.9μWb, just divide by $$\R\$$:

$$\int{I\cdot dt} = \frac{8.9\rm{\ \mu Wb}}{36.18\rm{\ \Omega}} = 246\rm{\ nA\cdot s}$$