About heater voltage / General Science and Electronics / Forums

Forums

4hv.org :: Forums :: General Science and Electronics

« Previous topic | Next topic »

About heater voltage

previous 1 2 3 4 next

Move Thread

LAN_403

klugesmith

Mon Jun 15 2015, 07:47PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1714

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.

Patric

Mon Jun 15 2015, 08:14PM

Registered Member #2899 Joined: Wed Jun 02 2010, 06:31PM
Location: Deinze, Belgium
Posts: 254

klugesmith wrote ...

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" regarding the derating of 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.

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

klugesmith

Mon Jun 15 2015, 10:01PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1714

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

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.

Electra

Mon Jun 15 2015, 11:38PM

Registered Member #816 Joined: Sun Jun 03 2007, 07:29PM
Location:
Posts: 156

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.

Patric

Tue Jun 16 2015, 07:07PM

Registered Member #2899 Joined: Wed Jun 02 2010, 06:31PM
Location: Deinze, Belgium
Posts: 254

klugesmith wrote ...

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.

Wow, thank you! I am still thinking.

Sulaiman

Tue Jun 16 2015, 10:53PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

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.

Bored Chemist

Thu Jun 25 2015, 07:47PM

Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022

klugesmith wrote ...

-----
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.
-----------------------

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?

klugesmith

Thu Jun 25 2015, 09:49PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1714

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?

Bored Chemist

Sat Jun 27 2015, 10:58AM

Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022

"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.

klugesmith

Sat Jun 27 2015, 03:09PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1714

Bored Chemist wrote ...
"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.
...
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. ...

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 Î¦ = 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 = dÎ¦/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 dÎ¦/dt = k N^2 di/dt = L di/dt. In simple transformer, Î¦ 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).

previous 1 2 3 4 next

Moderator(s): Chris Russell, Noelle, Alex, Tesladownunder, Dave Marshall, Dave Billington, Bjørn, Steve Conner, Wolfram, Kizmo, Mads Barnkob