Re: About heater voltage
Ash Small, Sun Jun 14 2015, 06:25PM

If I only had a 12V transformer, I'd rectify and smooth it, then put something like a 6V 21W bulb in series with the heater filament.

Maybe no need for the rectifier and smoothing cap, maybe just a suitable load in series with the heater filament?

(If the current measures a bit low, just add a resistor in parallel with the bulb)

Re: About heater voltage
mister_rf, Sun Jun 14 2015, 11:10PM

1. There is no point in unwinding and rewinding the secondary, I'm thinking you can improvise another tap off the secondary coil to pick up the right voltage.

2. Do you have two identical transformers? Why not wire the primaries in series and just use one of the two secondary? This connection would make the primary voltage to each transformer half of the mains. Therefore each transformer output voltage would be half what it should be and…

3. ...for the ideal solution complete the above idea by connecting the two secondary windings in parallel and in phase.

Re: About heater voltage
Patric, Mon Jun 15 2015, 06:14AM

I can not unwind or make a tap, the transformer is cast (sealed).
No place for 2 transformers, to big.
A bulb in series gives heat and light...
Still wondering if a diode could be trusted, because she cut the sine in half so there is no loss in power and the a811 gets only half of the voltage. My main question is: it it not bad for the filament over longer time? No one with any experience?

PS: it is for a Tesla coil.

Re: About heater voltage
Dr. Slack, Mon Jun 15 2015, 07:42AM

Don't diode load the output of a transformer, the rectified DC will quickly saturate the core and make it do things you didn't want.

The ideal, from all sorts of points of view (cost, convenience, adjustability, self-indicating, rated to get rid of heat) is a low voltage filament lamp (12v auto bulbs are great for this purpose, readily available, cheap, range of wattages) in series with the heater, increase the number in parallel until the requisite current flows.

Bulbs generate heat and light, well ain't that a b1tch!. I guess that as TCs are completely stealthy, you wouldn't want to blow your cover.

One step down in convenience is to use a capacitor in series, no excess heat and light. Either use a very large low voltage cap in series with the output (construct a non-polar one with two anti-series electrolytics, each shunted with a diode to charge it up), or my favourite which I've used on all sorts of projects, a low value mains-rated cap in series with the transformer input. Increase the capacitance until the right current flows.

The series cap and the series lamp configurations are essentially current sources, that will deliver a nice soft-start to your filament. While good, this will mean a longer warmup time, don't draw cathode current until it's fully hot. With a current source supply, you can monitor the heater temperature by observing the heater voltage.

I've just re-read Mr_RF's suggestion of the two transformers. #2 could be very clever, except that it doesn't work. The transformer with the unused secondary looks just like a high value inductor, and collapses the output of the loaded transformer. The only way this will work is to go to #3 and connect the secondaries in parallel, primaries in series. 6v is probably just fine for 6.3v rated.

Do you have two 811's? Two of those in series from 12v would work!

Re: About heater voltage
Signification, Mon Jun 15 2015, 08:04AM


I've seen those series/parallel configurable transformers (made by: Signal Transformer)--with dual primary and secondary windings. Some can be wired in this manner (primarys in series).
Is your reasoning for failure with two identical discreet transformers because there is no common core?
Perhaps a light load on the unused secondary will work?

Re: About heater voltage
Patric, Mon Jun 15 2015, 08:18AM


Are you sure about that? Most transformers secondary are rectified...

Re: About heater voltage
Dr. Slack, Mon Jun 15 2015, 08:46AM

If you draw current in *both* directions from the output of a transformer in a *balanced* way, so getting a net zero DC transformer current, using a bridge rectifier, a voltage doubler, a centre-tapped 'push-pull' type full wave rectifier (which although each winding only has a single diode, the two windings are magnetically coupled so they compensate each other), then all will work fine.

If you try to draw uni-directional current using a single diode in series with a single secondary winding, then it will go pear-shaped real fast.

The two key observations are that

a) a transformer secondary is a short piece of copper wire, so you cannot develop a DC voltage across it

b) the core of a transformer has a high permeability, so only a tiny DC current is needed to saturate it

=> do not do anything with either the primary or the secondary that will cause a net DC on the transformer.

One thing I did wrong with the input of a transformer many years ago was to try to sense its input current for a smart HiFi I was building. I put three diodes in series to develop enough voltage to turn on the input of an opto-coupler, but then only put one diode in shunt to let current flow the other way, because I didn't know better at the time. The imbalance of just two diode drops in 240VAC was sufficient to saturate the transformer, and it hummed like mad. Output current will mess it up just as easily as input current.

If you look at the literature for various single-ended converter topologies, many have compensating windings to stop the transformer flux from saturating, which it would with an unbalanced drive or load.

Re: About heater voltage
Dr. Slack, Mon Jun 15 2015, 09:03AM


No. It is because a transformer draws input current to meet the requirements of its load (plus some magnetising current which we can neglect to first order). A single current will flow through both transformers, so the output voltages will be whatever ohm's law makes them. If the other load is light, so high resistance, it will develop most of the voltage across it, 'collapsing' the voltage across the load output.

Re: About heater voltage
Bored Chemist, Mon Jun 15 2015, 07:18PM

While there is some hysteresis in magnetic materials, transformers are not noted for their memory. (They deliberately use materials with low hysteresis)
Imagine, for the sake of discussion, you set up the transformer feeding the filament through a diode and you throw the switch just as the voltage at the o/p of the transformer is zero and it's in the direction so that the diode will conduct.
As far as the transformer is concerned the diode acts like a short length of wire. it conducts.
For that cycle of the mains the transformer doesn't even "know" the diode is there.
After 1/100 or 1/120 of a second the voltage changes direction and, from the point of view of the transformer, it is feeding an open circuit because the diode won't let any current flow.
These two states of affairs- neither of which bothers the transformer, repeat in sequence.

How can running a resistive load via a diode harm the transformer?

( the valve may hum a bit more because the filament temperature is cycling half as fast)

Re: About heater voltage
klugesmith, Mon Jun 15 2015, 07:47PM

Patric: When using AC source of 12 volts RMS, a single diode rectifier would cut the average power (not the voltage) in half. Your a811 heater would get 12/sqrt(2) = 8.5 volts RMS, minus the diode drop, and probably not last very long.
-----
BC: There's more to it than what you said.

To begin with, V=0 time is the worst moment to close the [primary] switch. In steady state operation the positive half-cycle of mains voltage is bucked by positive half-cycle of induced voltage in primary. That requires a flux density swing of 2 Bmax during the half-cycle. Normally that's from -Bmax to +Bmax. When turned on at V=0 and B=0, magnetizing current and B start increasing, but typically the core can't support +2 BMax without saturation, so the primary current is limited only by winding resistance. This saturation dies out after a few cycles, and often causes an audible grunt from the transformer.
-----------------------
As for the possibility of eventual damage to transformer from significant resistive load on a single-diode rectifier ...
Google "rectifier transformer" about derating transformers for rectifier duty. I bet they don't even mention the single-diode case.

One exception is X-ray generators whose Coolidge tube is self-rectifying. High-voltage circuit has DC current in the secondary winding. To avoid core saturation, a parallel diode-resistor network goes in series with the primary winding. It's oriented so primary DC ampere-turns cancel the secondary DC ampere-turns. If you had lots of these running continuously, and in phase with each other, the electric power provider would be very unhappy with DC load on their distribution transformer.

Re: About heater voltage
Patric, Mon Jun 15 2015, 08:14PM


I thought the diode cuts the p to p voltage in half, so the current is half, so the power is 1/4.

Re: About heater voltage
klugesmith, Mon Jun 15 2015, 10:01PM

Wrong. Wish I had the discipline to let this wait until tomorrow.

Here's a picture of your a811's heater on AC power at nominal value, 6.3 volts RMS.

The voltage comes in pulses of alternating polarity and amplitude of 8.91 V. (that's 17.8 volts peak-to-peak). Current waveform is similar, with amplitude of 5.66 A. So the heating power (V x I, at each instant) comes in two pulses, each reaching 8.91 x 5.66 = 50.4 watts. What counts is the _average_ power. That's 25.2 watts, same as you would get with DC 6.3 V x 4 A.

Now for simplicity, let's double the voltage to 12.6 V_RMS and keep the heater resistance unchanged.

Peak (and p-p) voltage have doubled, to 17.8 (and 35.6) volts. Without rectification, the current doubles to 8 A_RMS (peak 11.3 A). The average and peak power quadruple, to 100.8 and 201.6 watts.

Now let's add your single diode, but for simplicity we'll make it ideal.

Peak output voltage is still 17.8 V. Peak to peak voltage is not a useful measurement in this case. Peak current is still 11.3 A, and peak power is still 201.6 watts. Average power is now 50.4 watts, half of what it was without rectification.

Multimeters on DC would show you the average values of rectified waveforms: 5.67 volts and 3.60 amps. Product is 20.4 watts. You might mistakenly think that the a811 would heat less than normal.
Ordinary analog multimeters on AC range (rectified average reading, sine-calibrated) would indicate 12.6 V or 0 V, depending on probe orientation.
True-RMS meters on AC (if they don't reject the DC component) would correctly indicate 8.91 volts and 5.66 amps, whose product is 50.4 W. Heater temperature would be the same as you would get with 8.91 DC volts.

If we figure 0.5 volt forward drop (e.g. Schottky diode), the peak voltage drops from 17.8 to 17.3 volts. Average now 5.43 V. RMS 8.59 V. Heating power 46.9 W. In the real world, higher temperature would increase the heater resistance and reduce these current and power values, but not enough to stop abusing the a811.

[edit] The numerical values 5.67 (V) and 5.66 (A) are similar by coincidence. 6.3 x 2/pi is about the same as 4.0.

Re: About heater voltage
Electra, Mon Jun 15 2015, 11:38PM

Yes I've seen small transformers burnt out in homemade power supplys because someone thought to try the diode trick, rather than buy one at the right voltage. If the transformer is quite large compared to the current drawn it possibly might tolerate the unbalanced output. Even then you will still have to add some series resistance to reduce the RMS volts, as already explained above it won't be half.

It's only 4A Unless your trying to do it with zero to spend, A 6v transformer can't be that difficult or expensive to obtain. If you have any Toroid transformers with a large enough hole in the center they are quite easy to wind an extra secondary on too.

Re: About heater voltage
Patric, Tue Jun 16 2015, 07:07PM


Wow, thank you! I am still thinking.

Re: About heater voltage
Sulaiman, Tue Jun 16 2015, 10:53PM

Using half-wave rectification of a transformer output works just fine,
there are two considerations;
. you will probably be able to get 1/4 of rated VA as watts due to heating
(google 'transformer utilization factor')

. since 811s have a directly heated cathode,
I think that you will get a lot of 50/60 Hz humm instead of a little 100/120 Hz hum

I would probably use two transformers, primaries in series and secondaries in parallel
two 50VA transformers wired like this will act like a single 50VA transformer
because although the current capacity is the same for each transformer (heating of windings),
hence twice the output current of a single transformer
..... the voltage is halved.

Alternatively use one transformer with a suitable series resistor
(12V-6,3V)/4 A= 5.7V/4A = 1.425 Ohm, power = 5.7V x 4A = 22.8W ...use a 1.5 Ohm 50W resistor.
e.g.
this will also be gentle with your heater ... much lower surge current on initial heating
just be sure to mount the resistor on a large heatsink/panel.

Re: About heater voltage
Bored Chemist, Thu Jun 25 2015, 07:47PM

What is the mechanism by which the transformer "remembers" that it has been running for "a few cycles"?
As I pointed out, transformers are made from materials with a low propensity to "remember" what the flux ever was.
If the transformer is switched on at V=0 then, since there is zero voltage and zero current (because changing it from zero would have required a voltage) there is no energy transfer.
As the voltage rises (rather slowly in terms of electronics- tens of miliseconds or so) the current rises and the magnetic field rises, energy gets stored up in the magnetic field
Then, (again, rather slowly) the voltage and current fall; energy is fed back to the power grid.

A perfect transformer feeding a resistive load looks to the outside world like a resistor (with a resistance equal to the load resistance times the square of the turns ratio).

Obviously, no transformer is perfect, but they are often good, so what's the mechanism for causing harm?

Re: About heater voltage
klugesmith, Thu Jun 25 2015, 09:49PM

I didn't say the magnetizing inrush current would be harmful, just that it happens, esp. when primary is switched on at zero voltage crossing. A laboratory demonstration would be in order if this weren't work time. . Look up inrush current, transformer. Wikipedia has this picture: but says that remedies for toroidal cores are different from those for other kinds. That's debatable IMHO, and brings remanent flux into the discussion.

One document covering many kinds of inrush current says, in the Remedies section, "7.1 - Zero Voltage Switching
An easy way to ensure zero voltage switching is to use a 'solid state relay' (aka SSR) [8]. Most of the common SSRs are already designed for zero-crossing switching, and they do not activate unless the voltage across the relay is below around 30 volts or so. Because of this, they are completely unsuited for use with transformers, because transformer inrush current is at its very worst if the power is applied at zero volts. Never use a zero crossing SSR with transformers!"

When transformer lab time comes up, I'd like to do the half-wave-rectifier example before the inrush example. I think the latter can be done fairly well without an oscilloscope.

The oldest microwave ovens I've taken apart had triacs in the primary circuit, but no "dimming" feature. I figured the triacs were to synchronize turn-on time with the AC voltage cycles. Newer MWO control boards have relays. I've always suspected that the relay coil drive is synchronized, taking "operate delay" into account, to avoid zero voltage switching. Experts?

Re: About heater voltage
Bored Chemist, Sat Jun 27 2015, 10:58AM

"I didn't say the magnetizing inrush current would be harmful, just that it happens, esp. when primary is switched on at zero voltage crossing."
that's twice you have said it, but you haven't explained it.
How is there a rush of current?
A current flowing through an inductor implies a stored energy- but there's no voltage to drive a current or to provide any energy for any such current.
How can there be a large current (or indeed, any) with zero voltage at the outset of the experiment?

All microwaves I have seen have had timers so they would need a triac, relay or whatever to cut the power when your baked potato is cooked.
Also, they all have interlocks with the door so that might also involve a relay or triac.

Find a CA3059 or something like it with a delay, and you will actually have evidence of a zero crossing avoidance switch.
On the other hand, there's a really good reason not to use zcs electronic relays in transformer circuits- bit it's not to do with switching on.
A ZCS will also try to wait till the end of the mains voltage cycle before turning off.
But with an inductive load, that means it's not turning off when the current is zero - because the current and voltage are out of phase.
If you try to turn off the current through an inductor you have to dissipate the stored energy somewhere. Also, if you try to turn it off quickly then because v = L dI/dT you get a huge voltage spike.
That's likely to damage the solid state relay because semiconductors are fast.

Re: About heater voltage
klugesmith, Sat Jun 27 2015, 03:09PM

Let's deal with your second point first. That's about turning off ZVS solid-state relays (and anything else using triacs or SCR's). Gate drive can be turned off at any phase, but the power semiconductor inherently stays "on" until the current reaches zero. So no large dI/dt, no extraordinary voltage spike. If the load is inductive then the voltage across the switch steps rapidly up to mains voltage at that phase, and that dV/dt is a design consideration.

The first point can definitely be explained to your satisfaction, by me or someone else. Dr Slack is around these days. Unlike the recent Kirchhoff's Law argument, this should end with one of us changing his mind and going away the better for it. If I came to agree with you then we would both be wrong, IMHO.

This could be a matter of your model being a bit too simple. Linear model of a real transformer (as opposed to an ideal or perfect transformer) has a magnetizing inductance. But for both directions of net* current, there is a (soft) limit beyond which the inductance goes to zero due to core saturation. THe matter before us is the physics underlying that inductance.
* Net current, or magnetizing current, is the difference between primary current and (Ns/Np) times the secondary current. That's why large secondary currents don't push core toward saturation.
Got to run now.

[edit]Which would you prefer, explanations or measurements? A third option would be simulations, e.g. using SwitcherCad (LTSpice) with its nonlinear core models, a skill I have never learned.

[edit] (break time) OK, BC, let's start with lesson 1. Physics behind v = L di/dt as well as transformer equations. Consider N turns of wire carrying current i, wound on a ferromagnetic core. The core material is permeable enough, compared to other materials in the system, that practically all of the magnetic flux from Ampere's law runs through the core material and links all of the turns.
At any instant, total flux Φ = kNi where k depends on core geometry and permeability.
According to Faraday's law, changing flux induces a voltage in any loop linked by the flux. Ignoring petty details of sign, v = dΦ/dt. No scaling factor if we use SI units; 1 volt is the same as 1 weber per second. The N turns are wired in series so total v = N dΦ/dt = k N^2 di/dt = L di/dt. In simple transformer, Φ is common to all windings and is equal to k(N1i1 + N2i2 + ...). To show magnetizing current inrush, it's sufficient to use a 2-terminal inductor with saturation behavior (next lesson).

Re: About heater voltage
Dr. Slack, Sun Jun 28 2015, 09:53AM


Thanks for volunteering me Klugesmith. However, while I am still around these days, my oscilloscope isn't, it's hors de combat after a stupid accident with a freezer and some hand cleanser (don't ask). After 40 years of always have a scope at my elbow, it's like losing a limb. While I am investigating the alternatives, I'm not going to rush into a purchase. I won't make any substantive posts here until I can post some actual measurements. I'm not sure I trust simulations to be convincing enough. At only 50Hz and some harmonics, I could hook up a sound card just for this, but the one thing they don't do well is DC.

Anybody else care to scope input and output currents while drawing a normal and an unbalanced rectified load from a transformer, at peak currents of <50% and ~100% rated load, and post the pictures?

Re: About heater voltage
Uspring, Sun Jun 28 2015, 11:18AM

klugesmith wrote:

* Net current, or magnetizing current, is the difference between primary current and (Ns/Np) times the secondary current. That's why large secondary currents don't push core toward saturation.

Yes, and under heavy loads, magnetizing current can be a small difference between large currents. This holds only for AC currents, though. DC currents on e.g. the secondary side will cause extra flux in the core, without it being compensated on the primary side. If during normal operation the flux is close to the saturation, then there will definitely be a problem by a diode causing DC currents near max load secondary AC currents.

Re: About heater voltage
Bored Chemist, Sun Jun 28 2015, 12:19PM

I re-read the OP and I think we are at crossed purposes
There are two ways to look at the OP's question
"Can I use such a transformer with a diode in series with the tube without risk? "

The uncertainty is where the diode is ; which of there two interpretations do you mean?
Can I feed half-wave rectified mains electricity to a transformer whose output drives a heater?

and "Can I connect a transformer to the mains and then half wave rectify the output and connect it to a heater ?"

The second one is well known to not work properly.
I was wondering about the first i.e
"Can I use such (a transformer with a diode in series) with the tube without risk? "
rather than "Can I use such a transformer with (a diode in series with the tube) without risk? "

Re: About heater voltage
Uspring, Sun Jun 28 2015, 01:25PM

Thanks for the clarification.

Can I feed half-wave rectified mains electricity to a transformer whose output drives a heater?

The DC component of rectified mains will see only the DC resistance of the primary. That will likely blow the fuse before the transformer is harmed

Re: About heater voltage
Bored Chemist, Tue Jun 30 2015, 08:45PM

That seems to make sense, but, as I said before, how does it "know" that?

For one half of the cycle the mains simply isn't connected to the transformer (because there's a diode in the way).
For the other half, it behaves exactly the same way as it normally would without the diode, because the diode acts (nearly) like just a bit of wire.

Re: About heater voltage
Uspring, Thu Jul 02 2015, 12:01PM

You're right, I overlooked the role of the diode, when it is not turned on. My previous remark is only true if the voltage supplied to the transformer looks like a half wave rectified voltage. That is not the case.

Say we begin with a positive half cycle where the diode conducts. After this half cycle a current will have built up and it will keep on flowing until the end of the negative half cycle. At this point it will be zero again. Exactly the same will happen on the next cycles. Input current will have a sinusoidal shape on top of a DC offset. This means, that the max flux will be twice as high as in the case without a diode.

When the secondary is loaded, input current and offset will increase. The effect is similar to that as if there is a diode on the secondary. It is somewhat worse, since also the magnetizing current will contribute to additional DC flux. Putting a diode on the secondary side is thus less harmful than on the primary but still dangerous unless the transformer has a lot of spare flux capability before going into saturation.

Re: About heater voltage
klugesmith, Thu Jul 02 2015, 07:45PM

Yup, but not the way BC explained it. See example where diode behaves like a wire for more than 90% of each cycle.

The answer came as a surprise. Starting today I know that putting a diode in series with primary of an unloaded transformer will, if the core does not saturate, be harmless! Just simulated that. Will try it for real at next opportunity, applying 115 V through diode (and small fuse ) to a nominal 230 V primary winding. Maybe one could win a bar bet at a EE convention.

An intuitive but wrong idea is that a half-wave rectifier rectifies the input voltage. It does if the load is resistive, since the current is being rectified. With inductive load, when the input voltage reverses the load continues to draw current which keeps the diode on, so the voltage on primary follows the input voltage in negative half-cycle -- until the current wants to go negative.

For clarity I simulated with 100V peak, at 50 Hz, and inductance to give 0.1A peak normal magnetizing current. (Same L as 120V RMS, 60 Hz, 0.1A RMS). Set winding resistance to 40 ohms so the loss effects are plainly visible. Offset the rectified traces by 0.2 ms so they don't eclipse the unrectified ones.





By switching on at zero voltage, the absolute magnetizing current goes to twice its normal steady-state level. Then, in unrectified case, the DC component dies out with L/R time constant.
With real materials the absolute magnetic flux would have to go to twice its normal level, which is often impossible with any amount of magnetizing current. That situation persists in the rectified case.

Now suppose we set up the unloaded rectified case for real, with 115 V on a small 230 V transformer. I bet we can make the transformer lose its smoke by putting a heavy resistive load in parallel with the primary, after the rectifier.

Re: About heater voltage
Bored Chemist, Thu Jul 02 2015, 08:16PM

A voltage doesn't "know" it is rectified.
It's just a voltage across a component.

Re: About heater voltage
Ash Small, Thu Jul 02 2015, 08:29PM

If it's a transformer with 'EI' laminates and you can take it apart, you can re-assemble it with all the 'E's stacked together, and all the 'I's stacked together, then put it together with a sheet of papar between the 'E's and the 'I's.

This will introduce an air gap which will prevent it saturating. Varying the thickness/number of layers of paper varies the air gap.

Inductance will be reduced, but it won't saturate.

You might also then be able to use it as a crude single ended output transformer as well

Re: About heater voltage
Uspring, Fri Jul 03 2015, 08:51AM

klugesmith wrote:

The answer came as a surprise. Starting today I know that putting a diode in series with primary of an unloaded transformer will, if the core does not saturate, be harmless!

Yes, I realized that a bit late, as you can see in my last post. Another thing to note is, that secondary voltage does not decrease with a diode on the primary side. So there is no point in doing that wrt to the OPs intention. I am just wondering if putting diodes on both sides of the transformer so that the additional DC effects due to load are opposite to each other, will help.

Re: About heater voltage
Dr. Slack, Fri Jul 03 2015, 11:19PM

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.

Re: About heater voltage
Uspring, Tue Jul 07 2015, 10:46AM

Dr. Slack wrote:

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

Re: About heater voltage
Patric, Wed Jul 15 2015, 03:24PM

Maybe I do not understand it all, but when the diode is in the secondary, the voltage is cut in halve, so anyhow the power is 1/4 by a resistive load?

Re: About heater voltage
Dr. Slack, Wed Jul 15 2015, 05:00PM


A diode will drop the voltage by 0.7v (so not much) and make the load operate for half the time (so 50% duty cycle) so half of the power

Re: About heater voltage
Dr. Slack, Thu Jul 16 2015, 04:39PM

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.

Re: About heater voltage
klugesmith, Thu Jul 16 2015, 07:04PM

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.

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.

Re: About heater voltage
Bored Chemist, Sat Jul 18 2015, 11:11AM

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.

Re: About heater voltage
Uspring, Sat Jul 18 2015, 05:53PM

Some comments to Dr. Slacks last post:

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.

Re: About heater voltage
Patric, Sun Jul 19 2015, 08:07AM

This is the infamous transformer: