If you need assistance, please send an email to forum at 4hv dot org. To ensure your email is not marked as spam, please include the phrase "4hv help" in the subject line. You can also find assistance via IRC, at irc.shadowworld.net, room #hvcomm.
Support 4hv.org!
Donate:
4hv.org is hosted on a dedicated server. Unfortunately, this server costs and we rely on the help of site members to keep 4hv.org running. Please consider donating. We will place your name on the thanks list and you'll be helping to keep 4hv.org alive and free for everyone. Members whose names appear in red bold have donated recently. Green bold denotes those who have recently donated to keep the server carbon neutral.
Special Thanks To:
Aaron Holmes
Aaron Wheeler
Adam Horden
Alan Scrimgeour
Andre
Andrew Haynes
Anonymous000
asabase
Austin Weil
barney
Barry
Bert Hickman
Bill Kukowski
Blitzorn
Brandon Paradelas
Bruce Bowling
BubeeMike
Byong Park
Cesiumsponge
Chris F.
Chris Hooper
Corey Worthington
Derek Woodroffe
Dalus
Dan Strother
Daniel Davis
Daniel Uhrenholt
datasheetarchive
Dave Billington
Dave Marshall
David F.
Dennis Rogers
drelectrix
Dr. John Gudenas
Dr. Spark
E.TexasTesla
eastvoltresearch
Eirik Taylor
Erik Dyakov
Erlend^SE
Finn Hammer
Firebug24k
GalliumMan
Gary Peterson
George Slade
GhostNull
Gordon Mcknight
Graham Armitage
Grant
GreySoul
Henry H
IamSmooth
In memory of Leo Powning
Jacob Cash
James Howells
James Pawson
Jeff Greenfield
Jeff Thomas
Jesse Frost
Jim Mitchell
jlr134
Joe Mastroianni
John Forcina
John Oberg
John Willcutt
Jon Newcomb
klugesmith
Leslie Wright
Lutz Hoffman
Mads Barnkob
Martin King
Mats Karlsson
Matt Gibson
Matthew Guidry
mbd
Michael D'Angelo
Mikkel
mileswaldron
mister_rf
Neil Foster
Nick de Smith
Nick Soroka
nicklenorp
Nik
Norman Stanley
Patrick Coleman
Paul Brodie
Paul Jordan
Paul Montgomery
Ped
Peter Krogen
Peter Terren
PhilGood
Richard Feldman
Robert Bush
Royce Bailey
Scott Fusare
Scott Newman
smiffy
Stella
Steven Busic
Steve Conner
Steve Jones
Steve Ward
Sulaiman
Thomas Coyle
Thomas A. Wallace
Thomas W
Timo
Torch
Ulf Jonsson
vasil
Vaxian
vladi mazzilli
wastehl
Weston
William Kim
William N.
William Stehl
Wesley Venis
The aforementioned have contributed financially to the continuing triumph of 4hv.org. They are deserving of my most heartfelt thanks.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Yay, I have finally got my new scope, a 4 channel 50MHz Rigol, DS1054Z. I'm still exploring all the things it says it can do, and I was a bit amused that the very first thing I do with it is to connect it to the mains, but, there you go, in at the deep end.
The experiemntal setup is a 600VA transformer, so it will have quite a high magnetisng current, with a roughly 1A lamp load, driven by a 40v rms secondary, with either a direct connection, or with two series silicon diodes. Units and per div are correct.
The first piccy is driving a balanced load. The primary current waveform (yellow) shows two principal features, the load current in phase with the primary voltage (blue), and the magnetising current in quadrature to the input voltage. The magnetising current is low for most of the cycle, but kicks up at the zero crossings where flux is maximum, this is a transformer which pushes the core Vs. The purple trace shows the secondary load voltage. This is not a sinusoidal mains supply, 3rd harmonic-a-go-go.
The second piccy shows the effect of introducing the diode into the secondary in series. As expected, the load voltage is half wave rectified. The primary load-related current follows the output current, half of it has disappeared. What is interesting is the magnetising related current peaks at the zero crossings, one has reduced, the other has increased. That is evidence that the secondary rectification is driving a DC current in the transformer, polarising it, so shifting the primary current level at which saturation occurs.
Looks like a score draw. Saturation is affected, but with a transformer this size and small load, we are nowhere near to smoke coming out. It may be different with a transformer where the rectified load is a larger fraction of its rated load. I did notice the lamp was rather less bright with the rectified input than the full input, so it works to reduce output power to a resistor.
I was about to start postulating that the transformer flux shifts due to recitification, so shifting the magnitudes of the saturation peaks to maintain zero DC input current. However, there is one big problem with rushing to that conclusion. I am measuring the input current through a current transformer for isolation (don't want to connect my nice new scope straight to mains without thinking carefully), so of course the indicated input current has zero DC under all conditions. While the relative shifts of the load currents and the saturation peaks are suggestive, it's no more than that until I get DC coupled.
So, next measurement is of the components above, but with a DC coupled primary current measurement. The measurement after that is with a smaller, better transformer, so that the magnetising current is lower and less peaky, and it's more representative of what somebody would do in practice.
I was about to start postulating that the transformer flux shifts due to recitification, so shifting the magnitudes of the saturation peaks to maintain zero DC input current. However, there is one big problem with rushing to that conclusion. I am measuring the input current through a current transformer for isolation (don't want to connect my nice new scope straight to mains without thinking carefully), so of course the indicated input current has zero DC under all conditions. While the relative shifts of the load currents and the saturation peaks are suggestive, it's no more than that until I get DC coupled.
I don't think, there can be a permanent DC current component in an inductor if there is no DC voltage component, which is the case if it is driven by mains. A DC component would imply a DC voltage drop along the wire resistance of the inductor. Faradays law states that the flux phi is:
phi = integral V * dt
A DC component in V results in a flux phi getting larger and larger over time. The AC component cancels out in the integral. So you won't find a DC current component in the primary even if you omit the current transformer. You could get a permanent DC primary current if the primary were superconducting
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Finally, after 2 weeks of yak shaving, got some relevant, and DC coupled, pictures
I am using a fairly cheap 50VA transformer, 240v to 24v, so turns ratio N of about 10, and the nominally 50 watt load of a Weller soldering iron on the secondary. The input voltage is measured via a 1000:1 resistive divider. The input current is measured through a 1 ohm shunt in the neutral, into a unity gain diff-amp (that's what took all the time). The secondary load voltage is measured by a standard 10:1 scope probe. All oscillographs show mains input voltage in light blue, primary input current in yellow, and secondary load voltage (and so current as the load is pure resistive) in pink.
The first picture shows the mains primary input voltage and the primary current. The nominal mains voltage is 240v rms, about 330v peak. The core is being run somewhat into saturation, drawing a roughly 100mA blip around the zero crossings, which is where the core runs to maximum flux. This is in quadrature to the input voltage, so little real power is being drawn.
The second picture shows the situation with a normal load connected.
The load voltage follows the input voltage, via the scale factor N. The input current substantially follows the load current, but the core saturation current draw can still clearly be seen as a lump around the zero crossings. I chose polarities such that the input voltage, input current and load current were all in phase. The input current is peaking to about 300mA, 1/Nth of the load current peak as expected.
The third picture shows what happens when a silicon diode is placed in series with the load.
During the first positive half cycle, the load diode is blocking. There is no load voltage, and initially no primary current. While the input voltage is still at its peak, the primary input current starts to grow. This indicates that the core is going into saturation. It is being saturated earlier, at a lower total volt.seconds than the unloaded case, which indicates that the core did not start the cycle at maximum negative flux.
As the input cycle continues, the magnetising current builds to more than 500mA, larger than the normal load current was, but still a long way short from being dangerous. In this particular case, the rms input current when using the rectified load is about 2% higher than for the normal load, the input fuse and primary winding won't even notice. The secondary rms is lower, so overall, the transformer will run cooler.
As the negative half cycle starts, the output voltage goes negative. For a substantial part of this cycle, the primary current is still positive, whereas the load current is negative, an unusual state of affairs for a transformer.
Finally as the cycle continues, the primary current comes into phase with the load current. If you look in detail at the primary current at the end of the load conduction phase, you will see it follows the shape of the secondary current closely. The saturation bulge that was visible in the normal load case is not present here. This confirms that at this point in the cycle, the transformer core has not reached maximum negative flux.
I do measure a slight net DC current on the input with the rectified load, but it is very small, well below the noise level of the scope. I also measure the same net DC when using the normal load, so I will put it down to scope offsets or non-linearity. I am not best pleased with the displayed noise level of the scope, it seems to be too many LSB, even with only 8 bits vertically, but I guess you get what you pay for, and I didn't want to afford Agilent. I will play with averaging traces offline to see if that gives me more confidence. As any input DC would be expected to change with a load rectification, I can conclude that there is no DC drawn at the input as a result of rectifying the output.
Summary.
a) Rectifying the load on a transformer does shift the flux in the core so that one saturation peak is increased, and the other is diminished
b) Even when the load is at 100% of the transformer rating, it doesn't appear to endanger the transformer, or at least this size of transformer. If you are contemplating repeating this on a much larger transformer, it would be worth doing this sort of check again.
So I stand corrected. Go for it. To halve the output power into a resistive load, it appears to be OK to half-wave rectify the secondary of a transformer.
Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1714
Thank you for the excellent pictures and explanation there, Dr Slack. You beat me to it. Are you retired?
To carry the flag a bit farther, I want to dig out a 12 volt 50 VA transformer and make a Slackian picture using 3 ohm (4 amp) resistive load, already at hand. Also measure the temperature rise, at least with fingertips.
Then repeat with the transformer in a beginner's regulated DC power supply: half-wave rectifier, filter capacitor, 3-terminal VR, 3 ohm load. Let's see what happens with 4 amps DC in the secondary. And a DC current according to transformer Utilization Factor as pointed to by Sulaiman(?) .
The margins may depend on transformer's design application, and consequent trade-offs on core loss and copper loss. In a 12 V lighting transformer, such as in a desk lamp, we care only about losses at full load. In a 12 V doorbell transformer, we care mostly about losses at no load.
Dr. Slack wrote ... ... I do measure a slight net DC current on the input with the rectified load, but it is very small, well below the noise level of the scope. I also measure the same net DC when using the normal load, so I will put it down to scope offsets or non-linearity. ...
Agreed! Another way to check for DC current is to measure across sense resistor with handheld meter on DC mV, both ways. If the meter's dual-slope ADC fails to properly reject the strong 50 Hz signal, or the meter's auto-ranging is confounded, put a passive RC filter between sense R and the meter.
If you can get the Rigol captured waveforms into a spreadsheet, try numerically integrating the no-load secondary voltage. Result can be charted WRT the magnetizing current to see a BH plot. If it's not a closed figure, add a spreadsheet knob for scope offset voltage in the channel being integrated. If the spreadsheet is problematic, put the scope waveforms into a text file and let somebody else take a shot at it.
Registered Member #193
Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
Another thought has just occurred to me. I don't think it will matter much because valve heaters are generally designed to warm and cool relatively slowly. Imagine that the mains frequency was very low. In effect the lamp would be flashing. Each cycle would expose it to the "shock" of heating up again. I don't think that would be good for it.
a) I do agree, that the current bumps in the first diagram are due to beginning saturation. Hysteresis would show up as current difference at the points of max primary voltage. Little of this is visible.
b) The positive saturation bump in the third diagram is considerably higher, maybe 5 times as high, as in the unloaded case. Here it is not something to worry about, but the height of the bump depends on how the core goes into saturation. That is a gradual process and might be different for other transformers. For complete saturation, i.e. ur=1 current might be much larger.
c) In the rectified case, RMS primary current is 2% higher. That does not seem much, but at the same time the power throughput has been halved.
d) As I've proved in my last post, a DC component in the primary current is impossible in a steady state situation. It is possible as a transient phenomenon.
e) Nice measurement Would be interesting to measure at 100% power, i.e. half the load resistance.
This site is powered by e107, which is released under the GNU GPL License. All work on this site, except where otherwise noted, is licensed under a Creative Commons Attribution-ShareAlike 2.5 License. By submitting any information to this site, you agree that anything submitted will be so licensed. Please read our Disclaimer and Policies page for information on your rights and responsibilities regarding this site.
About heater voltage
