UNI-T UT61E RMS measurement - Page 1
Forum section: electronics / testgear

author: 4x1md
author_rank: Regular Contributor
timestamp: August 19, 2017, 08:20:56 am
content: This is my first post on this forum.I have a Hakko Presto soldering iron which doesn't have any temperature control. When I use it with a thin tip, in order to reduce the heat, I connect it to AC through a diode. The diode acts as a half-wave rectifier, thus the effective voltage on the soldering iron becomes:Veff= 230 * 1.41 / 2 = 163VI tried to measure the voltage on the soldering iron using my UNI-T UT61E DMM which, as the manufacturer states, is as true RMS multimeter.In DC mode I measured about 100V which is the average value, i.e.Vavg= Vpeak/ 3.14 = 230V * 1.41 / 3.14 = 98.7VIn AC mode I measured about 120V which neither the average value nor the the expected RMS voltage. I measured a similar value with another multimeter in AC mode which is not true RMS.Why did the DMM measure 120V instead of 163V? What I'm doing wrong or don't understand about measuring the RMS voltage in this case?

author: WhichEnt2
author_rank: Regular Contributor
timestamp: August 19, 2017, 09:13:54 am
content: It's about 155 V ac+dc rms.

author: alm
author_rank: Super Contributor
timestamp: August 19, 2017, 09:44:52 am
content: A signal that varies between 163 V and 1 V will clearly have a DC offset. That is the value you should measure in DC mode. In AC mode, it should measure the RMS value of the AC component of the signal. The RMS value of the AC + DC voltage (proportional to the amount of heat an ideal resistor would dissipate) is determined using \$\sqrt{V_{AC}^2+V_{DC}^2}\$, as posted by WhichEnt2. Some DMMs provide an AC+DC mode to save you from having to take two measurements and doing the mental arithmetic.

author: 4x1md
author_rank: Regular Contributor
timestamp: August 25, 2017, 10:38:57 am
content: Ok. And how can I mathematically derive the AC component of the half-wave rectifier output?

author: alm
author_rank: Super Contributor
timestamp: August 26, 2017, 08:53:01 am
content: The way I would do it (assuming the signal is large enough amplitude and low enough frequency so the diode behaves close to ideal):\$V_{0,RMS}\$ is the RMS voltage of the input to the rectifier.T the period of the input signal.AC+DC RMS voltage: For half wave rectification, the RMS voltage would be \$\sqrt{\left.{\left({T\over 2}V_{0,RMS}^2+{T\over 2}0^2\right)\left.\right/ T}\right.}\$, or \$V_{0,RMS}\over \sqrt{2}\$. This follows from the definition of RMS. For lower voltages or higher currents, you would subtract the diode forward drop.DC voltage: The DC voltage is the average of the signal, i.e. the integral over one period divided by the period. For a half-wave rectified signal, this would be the integral over one half period (twice the peak voltage divided by the voltage drop times a half period) divided by the period. The peak voltage is sqrt(2) times the RMS value of the original AC voltage. Again, you could substract the diode forward drop from the function before integration.AC RMS voltage: You can determine the AC RMS voltage from the AC+DC RMS voltage and DC voltage using the equation posted earlier.You can verify your equations using LTspice. It can calculateaverage and RMS valuesof any signal.

author: joeqsmith
author_rank: Super Contributor
timestamp: August 26, 2017, 05:03:35 pm
content: Quote from: 4x1md on August 19, 2017, 08:20:56 amQuote from: 4x1md on August 25, 2017, 10:38:57 amIgnoring the diode and using your 230VTake your first part, 163 and square that or 26,442.Take your second part, 98.7 and square that or 9741.27442 - 9741 = 16702. Take the square root of that, or 129V.If I run the same test here, 4007 diode and 100K,I measure 122VACrms at the line.Pulsed DC across R measures 54VDC.VACrms across R measures 66.5.122VACrms = 172.5VpeakVpeak/2 = RMS = 86.2554^2 = 291666.5^2= 4422.252916+4422.25= 7338.25sqrt(7338.25) = 85.67So we are in the ballpark.If we wanted to calc just the AC, we already have the AC RMS so calc the pulsed DC part as you did172.5Vpeak * 0.3185 = 54.94VDC (again in the ballpark of what we measured)Take both the RMS and DC parts and square them,86.25^2=7439.0654.94^2=3018.54Again, subtract and take the square root.sqrt(7439.06 - 3018.54)sqrt(4420.53)= 66.48. Again in the ballpark of what we measured.I would say you were very close in your measurements.

author: Fungus
author_rank: Super Contributor
timestamp: August 26, 2017, 05:56:30 pm
content: Quote from: 4x1md on August 19, 2017, 08:20:56 amIn AC mode the meter will always center the signal before measuring it.If you draw your half-rectified signal on paper and draw another line along the middle of it, that weird shaped signal is what your meter is trying to measure the RMS of.

author: alm
author_rank: Super Contributor
timestamp: August 26, 2017, 07:21:40 pm
content: Quote from: Fungus on August 26, 2017, 05:56:30 pmFor a (half-)rectified sine wave, the mean value is not at the midpoint between its minimum and maximum, so neither is the DC component (since that is defined as the mean).Obviously a substantial (parasitic) capacitance across the load compared to the load current (\$T\over t_{RC}\$ is not negligible) will have a large effect on the result and can even turn it to mostly DC (c.f. traditional unregulated power supply).