[medium jim]

For us newbies...

I agree with this statement.  However, the original post is Vacuum tube full rectification, not full rectification for a vacuum tube amp.  As I see it, one would need 2 dual diode tubes if the transformer is not center tapped.  Would this still be cheaper than one tube and twice the windings?

Thanks, Mike

But would result in tube sag, good for guitar amps, not so much for home audio.
To clarify, there will always be some sag in a tube rectified circuit, especially in class a/b.  This can be minimized with SS rectification.
Jim

[Ericus Rex]

The difference is that the full-wave rectifier requires 2 diodes and a centre-tap. The bridge requires 4 diodes and needs no centre tap.

w

There must be some other difference.  Otherwise, why would anyone use a full wave bridge and double the number of parts.  I'm confused.

[cheap-Jack]

Hi.

There must be some other difference.  Otherwise, why would anyone use a full wave bridge and double the number of parts.  I'm confused.

Bridge rectification using 4 diodes will double the 'efficiency' of full wave using only 2 diodes & the power iron can be made smaller due to less winding because of no centre taping needed. It is a win-win situation.

Full wave rectification have been a historic PS topology century back when full wave rectification tubes were only available & hence popular back then. No other better choice until the invention of semi-conductors a few decades back.

I have seen many home audio PS designs of HV & large loading currents use bridge rectification of 4 sand diodes for its high efficiency, low voltage drop across the diodes (vs relatively large voltage drop across vacuum tubes) & less heat & zip LV heater PS needed for tubes. Cheaper to build & cheaper to run for much much longer service life.

For single secondary winding power irons, either half wave rectification using one diode or bridge rectification using 4 diodes can be chosen. For highest efficiency, bridge rectification using 4 sand diodes is the better choice given a sand diode only cost 20 cents a piece!

One drawback of HV bridge rectification using sand diodes in tube amps is the tubes need a short time to heat up while rectification using solid state diode(s) will be ready immediately. So HV will applied to the tube plates well before the tubes are heated up to conduct. This is said to be 'harmful' to the tubes.

So for rectification using solid state diode(s), make sure there will be a time delay circuit for the HV to apply to the tubes after the tubes heated up.

c-J

[mgalusha]

Hammond provides a handy guide showing various connection schemes and their voltage losses. Note they show diodes but most of this still applies.

http://www.hammondmfg.com/pdf/5c007.pdf

mike

[Ericus Rex]

Hi.
Bridge rectification using 4 diodes will double the 'efficiency' of full wave using only 2 diodes & the power iron can be made smaller due to less winding because of no centre taping needed. It is a win-win situation.

c-J

Thanks Cheap-Jack.  That's exactly what I said earlier in the post but that comment was poo-poo'd by Wakibaki.  What really confused me was that he later stated  "The only case where one half of the cycle is unused is half-wave rectification."  Full wave NON-bridge rectifiers are just two half-wave rectifiers in one bottle so there still is an unused half-cycle for each of the diode sections in the bottle.  Right?

Mike, thanks for that link.  Perfect information!  I don't feel quite so dumb anymore.

Any answers for Mark concerning his second question?

[JohnR]

Full wave NON-bridge rectifiers are just two half-wave rectifiers in one bottle so there still is an unused half-cycle for each of the diode sections in the bottle.  Right?

I'm not entirely sure what you mean by "unused" but each diode only conducts for half the cycle in both cases (fullwave center-tapped and fullwave bridge). As far as the output is concerned, it's a rectified sine wave (sans filter) in both cases.

Any answers for Mark concerning his second question?

Which one is that?

[AVnerdguy]

For us newbies...

I agree with this statement.  However, the original post is Vacuum tube full rectification, not full rectification for a vacuum tube amp.  As I see it, one would need 2 dual diode tubes if the transformer is not center tapped.  Would this still be cheaper than one tube and twice the windings?

Thanks, Mike

Oops! You are correct. I was thinking in my generic electronic theory mode. I do both tube and SS frequently and nowadays the main advantage to a full wave bridge is the reduced cost of 2 extra SS diodes vs. the copper windings and a heavier xfmr. The center tapped full wave rectifier was more common in tube equipment with the center tap referenced to the chassis. Today in a SS device a full wave bridge is more common and chassis reference can be non existent due to switching power supplies - another topic for another time.

[Mark Korda]

Hi Ericus and all others helping,thanks.Ericus,don't go breaking your back digging up a book,but if you do ever find it,it is Basic Electronics,Van Valkenburgh...ect Vol.1.,page 45.I don't mean to throw another (wrench) into all of this,but what is tube sag?To me thats what happens if you hang around your computer to long instead of getting off your can,going outside and getting some exersize.Not trying to get off the topic,but I've never come across that term(tube sag)....thanks guys...Mark.

[wakibaki]

Look at what you said:-

The full wave rectifier is still not the most efficient since the negative portion of each half is still wasted, though it certainly is better than a half wave.

'the negative portion of each half is still wasted'.

This is simply not true.

Full wave NON-bridge rectifiers are just two half-wave rectifiers in one bottle so there still is an unused half-cycle for each of the diode sections in the bottle.  Right?

NO!

One half wave rectifier acts on the positive going cycle, the other acts on the negative going half, but because of the centre tap being tied to ground, the two halves of the secondary winding act 180 degrees out of phase, so that one rectifier is active on the positive-going half of the INPUT waveform, the other is active on the negative-going half of the INPUT waveform.

A half wave rectifier only produces output on the positive going half of the INPUT cycle. By INPUT cycle I mean the AC flowing in the primary.

You have to be careful with the use of the word EFFICIENCY too.

Efficiency has a special meaning in engineering. It's the ratio of the input energy to the output energy. Because a half-wave rectifier only admits one half of the input cycle, strictly speaking it has roughly the same EFFICIENCY as a full-wave rectifier. If it was admitting the whole cycle but only using half of it then the efficiency would be ~50%

A bridge rectifier has the same efficiency as a full-wave centre-tapped rectifier if you neglect the energy lost in the diodes, which is in fact negligible in the case of solid state.

Look at the diagrams I have taken the trouble to draw, and try to understand how they work, instead of quibbling.

@Mark Korda

More important, LOOK AT THE ESSENTIAL SAFETY INFORMATION I have provided. You are operating a dangerous device..

Single insulated devices without a safety ground are no longer legal for sale in almost every jurisdiction in the world.

w

If you don't believe me go over and post on diyaudio.com where there are plenty of engineers who will set you straight PDQ.

[wakibaki]

[mboxler]

Hi Ericus and all others helping,thanks.Ericus,don't go breaking your back digging up a book,but if you do ever find it,it is Basic Electronics,Van Valkenburgh...ect Vol.1.,page 45.I don't mean to throw another (wrench) into all of this,but what is tube sag?To me thats what happens if you hang around your computer to long instead of getting off your can,going outside and getting some exersize.Not trying to get off the topic,but I've never come across that term(tube sag)....thanks guys...Mark.

Hi Mark...

This is your thread, your topic.  To paraphrase Neo..."There is no wrench".

See if this link helps.

http://www.aikenamps.com/Sag.html

Also, if you struggle remembering Ohm's law (as I do), I find the Ohm circle chart very helpful. 




Mike

[Mark Korda]

Hi Mike,I have not checked that link you sent me yet,but I will.I never got a tattoo,too short for the NBA,not good enough to QB San Fran,but always wondered if I ever got to the point of taking an electronic test if it would be legal if I had that chart on my arm like Tom Brady has the plays written down on his.I've seen the chart before,and thats the only way I could memorize it.I thank you for thinking of sending that to me.I'll check that site out now ..take care Mike....Mark.

[wakibaki]

Maybe you guys think I don't know what I'm talking about. 

I just don't often spend much time on sites where DB testing is not open to discussion.

A couple of my own design headphone amps.

6N6P









LME49600









If you need any more convincing, take a look at my website - http://www.wakibaki.com

w

[Ericus Rex]

I must not be communicating this very well.  My apologies to all.  I don't think, nor do I speak, EE.

Waki, when I look at your "center tapped full wave rectifier" diagram from 4PM today this it what I see:  Two full sine waves representing AC current 180 degrees out of phase with each other.  There are two diodes which only let the current going in one direction through to the filter.  Therefore, 1/2 of each of those complete sine waves is blocked at the diode.  Yes, I know.  The two add up to form the complete rippled DC, also in your diagram.  But 1/2 of each of those out-of-phase currents is not allowed past the diode (perhaps "wasted" was not the right word).  With the full wave bridge rectifier, two more diodes are added to the circuit but are backwards compared to the two present in the NON-bridged full wave rectifier.  These two new diodes allow the formerly blocked current to flow into the filter circuit and be utilized.  Did you see the Hammond link that Mgalusha posted?  It clearly states that the full wave bridged rectifier (both choke and cap input) has a higher VDC than the full-wave resistive load.  I can only account for this by the utilization of the current that is blocked in the NON-bridged full wave rectifier.  Have I misinterpreted those diagrams?

[JohnR]

Have I misinterpreted those diagrams?

With regard to the diagrams, Hammond specify the voltage across the full center-tapped winding. The output voltage compared to the full-wave bridge is therefore half. To put it another way, to get the same output voltage from a fullwave rectifier as a full-wave bridge, the output voltage of the center-tapped transformer needs to be double. This is not (electrical) efficiency though.

If you want to use vacuum-tube rectification, the simple way is to use a standard fullwave circuit with a center-tapped transformer and a heater winding. These transformers are available off-the-shelf in a range of voltages and current capabilities that will work for most common types of amp and preamp (Hammond, Edcor, others). If (electrical) efficiency is a concern, you probably shouldn't be using tubes

FWIW I haven't seen wakibaki post any incorrect information...

[wakibaki]

It clearly states that the full wave bridged rectifier (both choke and cap input) has a higher VDC than the full-wave resistive load.  I can only account for this by the utilization of the current that is blocked in the NON-bridged full wave rectifier.  Have I misinterpreted those diagrams?

The short answer is yes, but this is hardly your fault.

The Hammond .pdf is unfortunately misleading.

Let's look at just one of these diagrams:-



The legend states that the peak voltage is 1.41 * the secondary RMS voltage. It also states that the average value of the DC (a sine wave) is 0.9* the RMS voltage. A sine wave, however, does NOT appear across the load.

It is true that the DC value of a sine wave is 0.9  the RMS value, but in the circuit shown the voltage across the load, the resistance, unless the current being drawn is very high, is 1.41 * the RMS value. This is because the capacitor charges to the PEAK voltage, and unless the current drawn is very high, maintains the voltage being supplied at close to the peak value. The voltage does decline slightly, but it is topped up every full cycle. The variation in the output voltage is called ripple and this is the ripple often mentioned in connection with the design of PSU filters. If the resistance were reduced to the point where it was very low, then the average value of the DC being supplied to it would decline, as the capacitor would approach fully discharged within the cycle time. Since the whole point of having the capacitor there is to ensure a virtually continuous DC voltage, it would be nonsensical to design with a resistance so low (or a capacitor so small) that the capacitor routinely discharged to that point.

Why exactly Hammond have chosen to present this information in this misleading way is a mystery to me, unless it is because they expect it to be read only by experienced engineers, in which case they might be expected to know the average DC value of a sine wave anyway. The information is irrelevant in this context.

Without entering into a detailed discussion of all the diagrams, the Hammond document is, IMO, best ignored. There are many other, better references for the understanding of rectifier circuits.

allaboutcircuits.com has a section dealing with half- and full-wave rectifiers, both centre-tap and bridge and also voltage multipleiers.

A very good book (although dealing only with solid state) is The Art of Electronics by Paul Horowitz and Winfield Hill.

A good modern book on tubes is Valve Amplifiers by Morgan Jones. It has a section on valve PSU design.

w