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 = 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.
Vavg = Vpeak / 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.