• WARNING: Tube/Valve amplifiers use potentially LETHAL HIGH VOLTAGES.
    Building, troubleshooting and testing of these amplifiers should only be
    performed by someone who is thoroughly familiar with
    the safety precautions around high voltages.

rectification and xfmr stress

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
Hi

I was wondering. Does the way you rectify have influence on transformer heating?

I have a tranny that has 2x 2A 7-3.15-0-3.15-7V winding for heating of output tubes.

With the use of the 6.3v ac for heaters it was very fine.

Now i changed the output tubes to 12,6 Volt types.

I made 4 boards with a diode for rect and a regulator with diode for 12,6 volt

The tranny is getting quite warm now.

Ofcoarse i am asking a little more for it, but it seems really exaggerated.

Now dies the "single diode" rect put great stress on the amp? Would it help if i connected two heaters out-of-phase? that way the tranny really sees a full wave rect.

Anybody?
 
beamnet said:
I was wondering. Does the way you rectify have influence on transformer heating?
...
Now dies the "single diode" rect put great stress on the amp? Would it help if i connected two heaters out-of-phase? that way the tranny really sees a full wave rect.

With a single diode rectifier for anything but very low current (eg. bias supply) your transformer sees a net DC current, which saturates the core, increasing considerably the transformer idle current because the effective inductance of the transformer windings becomes quite drastically reduced. You need a full wave rectifier.

You can esily use a full wave rectifier for every tube but as a stop-gap measure, you can put pairs of tubes in antiphase (as long as the tubes in a pair are the same, of course).

Using a transformer with a difference between half wave currents produces core saturation, and also, a much larger stray field, resulting in more hum.
 
Do not forget that, all other things being equal, peak winding current is a limiting factor for a transformer. This is because the current is what creates the magnetic field, and also, when excessive, saturates the core.

Given an ideal rectifier and a constant load, increasing the filter capacitor after the rectifier increases the peak current, and decreases the conduction angle of the rectifier. This is why you cannot simply increase capacitors to any value - the increased peak current means an increased magnetic field, If the increase is enough to saturate the core, during that angle the transformer becomes ver 'bad' - coupling lowers and ditto inductance. The primary and secondary start looking more like their DC resistance instad of inductive and the heat losses increase a lot.

Often, capacitors are increased to provide better filtering. Speciffically in tube circuits, there will often be a resistive component after the first filter cap, followed by another filter cap. In these situations, when flter caps are increased in value, it is advisable to split the resistance intotwo, and insert one part of it between the rectifier and first filter cap. This will considerably reduce the peak current, and also increase the rectifier conduction angle.

A mistake often made with SS rectifiers involving peak currents is that increasing the peak current increases rectifier recovery. The obvious result is an increase in rectifier heat (often quite dramatic) and a not so obvious is an AC component to the capacitor ripple current, producing a 'crest' of sorts on the ripple voltage, which increases the effective ripple. Also, it stresses the capacitor considerably more. Often there are volumes written about the advantages of vacuum rectification but in reality, if the simple mistake above is avoided by adding a resistance in series with the rectifier (much as a vacuum diode adds it intrinsically), most if not all of the 'objectionable' SS rectification problems will dissapear. The remaining advantage of vacuum rectificiation, that being essentially no recovery time, becomes far less of an issue, if not even a non-issue. In any case, it is then possible to get practically vacuum-like results using fast diodes - with a considerable saving in power that would have been used for a vacuum rectifier's heater, not to mention the saving in price. This is really a simple trick which for some reason I don't see that often. It can also be used to great advantage in DC heater supplies, since these often have a CCS or a series resistor in there somewhere.
 
Ok, i'll shovell the ps for the heaters out.

so i have two 7-0-7V windings, 2A rated ( know that's a little underestimated, think it might be able to provide 50% more)

Now i have two issues. Full wave or bridge? Simulating it does not seem to make that much of a difference.

I have 2200uF after rect and a 7812 (with raising diode)

The heater draws 800mA max

I have two heaters.

Is that too much of capacitance?

I'm learning a lot here on topics i have allways seen too easily
 
I am confused...

Are you using rectification for the heaters only? Or for both heaters and B+?

If using for heaters, there should be little to no difference in the power transfer, since 12 volt tubes draw less current than 6 volt tubes, generally.

I think it would be best to use bridge rectfier and not use the center tap at all. Since you are using a 7812 regulator, 2200 uF should be fine. Me, I don't use any regulation and get a nice solid 6.3 volts. I use 10,000 uF for filtering, though.

Besides... this is just simple low voltage AC-DC conversion, done daily by most electronics today. Why would you have trouble there?

However!

For B+, that is another issue. You MUST use a current limiting resistor from the center tap to ground. I use 220 ohms, 20 watts. Without it the current draw from the secondary will not only heat up the transforrmer, but cause it to hum loudly, and eventually burn it out. Otherwise, a 100uF cap for filtering, then a choke, and anything you want after that (I use from 80 to 330 uF) and hum will be just about the same level as noise (about 5mV P-P).

My :2c:

Gabe
 
we are talking only heaters here.

i used 4xkt88. 6.3 volts AC, 1,6 amps per tube-> no excess heat

switched to GU50 tubes, 12,6V 800ma

the xformer has 7-3.15-0-3.15-7volt windings, so 7-0-7 yoelds 14 volts ac. Rectified with one diode it drives the xformer incredibly warm.

So which is kind of logical, as the xformer has to produce all it's powe in a half wave. Plus it gets all these nasty spikes from the rect.

The question left was: which is better? full wave or bridge rectification?

what would be the pro's and cons for both systems?

I've build lots of small power supplies, but now i have to consider the max rating of the xformer
 
the xformer has 7-3.15-0-3.15-7volt windings, so 7-0-7 yoelds 14 volts ac. Rectified with one diode it drives the xformer incredibly warm.

So which is kind of logical, as the xformer has to produce all it's power in a half wave. Plus it gets all these nasty spikes from the rect.

Hmmmmmm. This seems wrong to me. Half wave rectification will give you a final voltage of if I recall about 20-30 percent less than RMS, simply because of the recovery rate. When the other half wave is going the there is no current flow so the load draws current. Voltage goes down until the next positive peak reaches the same level and returns the voltage to the mean level (average, which is less than RMS?), whatever that is. But it is definitely less than the RMS value of 14, unless you have a ridiculously large capacitor that maintains much of its current.

That is why a full wave (bridge or standard) gives closer to the full wave (rather than just RMS), because the cap has double the chance to charge up, since now the half waves are happening 120 times a second and are one right after the other, versus only 60 half waves with 1/120th of a second in between. A big enough cap and you can get to about 90% or more of peak.

So, I quess you are correct, since at half wave there needs to be more current flow to charge the rectifier than the full wave. But I am looking at it from a voltage times current aspect, if that makes any sense.

On the other hand, once the cap is full it takes less effort for the full wave to keep it charged.

As for nasty spikes, a couple of small value caps (0.1 µF or so) will take care of those.

Gabe
 
You could go at this 2 basic ways:

1) You could use the 14VAC with a resistor in series to get the required 12.6V/800mA. This gives you AC heating, but is very simple. Actually, if your 7V winding gave you 6.3V at the heaters with 1.6A, using the 7+7=14V would give you 12.6V at the heaters at 800mA.

2) DC heating - here you have a choice of 1 common rectifier or one rectifier per tube. You do need one 7812 per tube as 1A is the rated current (1.5A max).
For a common rectifier you really have no other choice but to use a bridge. This will give you DC with ripple peaks at (14 * SQRT(2))-1.2V, ~=18.6V. The 7812 needs a minimum of 15V on the input to function properly, so your 'sawtooth on top of DC' at the filter cap must never drop below 15V at full load. This is what gives you the size of the filter cap - if you have a scope handy, it will be a simple matter of trial and error to figure out the right value.
In the mean time, here is a good approximation: Given that the voltage on the heaters is kept constant, so is the heater current. In effect, your power supply is loaded by a constant current sink of 4x0.8=3.2A. Every 10ms, the caps are charged to 18.6V approximately, and after that, discharged with a constant current of 3.2A. In order for the voltage never to fall below 15V, they must not discharge by more than 3.6V. Since C=delta(U) * I * t, putting that into the formula yields 3.6 * 3.2 * 0.01 = 115200uF! The actual capacitance will be lower because for a short time every 10ms, the rectifier feeds the cap and the load. Still, this is a lot of capacitance, and you also need some quite humongous diodes to pull it off, with very high peak currents through the winding, meaning voltage loss on the diodes, and increased transformer losses.
An alternative solution would be to use 4 smaller rectifiers and caps, one for each tube. Using a half-wave rectifier does have a SLIGHT advantage that you only lose one diode drop on each peak, but you have to account with twice as long a time between replenishments of the filter cap. So, you end up with some 56000uF PER TUBE. Still not good. IMHO, I'd go with the AC heating...

given that delta(V)=t*I/C, and your I=4x800mA, and delta(V)=18.6-15=3.6V, and t=20ms, you get C=dV*I*t=3.6 *3.2 * 0.02 ~=230000uF!
 
230,000 uF!!!!!!??????????:eek:

Yikes, man! Too much!

I have 10,000 uF after a full wave (yeah, I opted for a 25 amp) bridge feeding (6) 6.3 volt 300mA tubes... very clean and very quiet. They draw 1.8 amps total. I use the same bridge/cap for another set of power output tubes that draw 3 amps total, again very clean. Also, I get the exact voltage rated.

According to your calculations, he might as well get one of those 1 or 2 farad caps for automotive audio... they're at 12 volts...

"Why calculate what you can measure?"

Ah, thoery is a wondeful thing, isn't it?
 
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.