UNI-T UT61E RMS measurement

[4x1md]

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:
V_eff = 230 * 1.41 / 2 = 163V

I 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.
V_avg = V_peak / 3.14 = 230V * 1.41 / 3.14 = 98.7V

In 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?

[WhichEnt2]

It's about 155 V ac+dc rms.

[alm]

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.

[4x1md]

Ok. And how can I mathematically derive the AC component of the half-wave rectifier output?

[alm]

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 calculate average and RMS values of any signal.

[joeqsmith]

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:
V_eff = 230 * 1.41 / 2 = 163V

I 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.
V_avg = V_peak / 3.14 = 230V * 1.41 / 3.14 = 98.7V

In 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?

Ok. And how can I mathematically derive the AC component of the half-wave rectifier output?

Ignoring the diode and using your 230V
Take 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 = 172.5Vpeak   
Vpeak/2 = RMS = 86.25

54^2 = 2916
66.5^2= 4422.25
2916+4422.25= 7338.25
sqrt(7338.25) = 85.67
So 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 did
172.5Vpeak * 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=7439.06 
54.94^2=3018.54
Again, 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. 

[Fungus]

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?

In 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.

[alm]

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.

For 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).