Does anyone understands how this power supply works ? - Old Philips PA Amp

[kagliostro]

I was looking to some old Philips PA Amp schematics

My attention was captured by the power supply for grids, someone can figure out how it works ?

at the begin I was thinking to a print mistake but I realized that there are more schematics using that circuit

As you can see they use a CT Transformer with a Full Bridge that has the negative connected to ground, the CT is connected to a cap and than to the grids of final tubes (EL36)

as we know EL36 likes low grid voltage, but how is possible to connect the transformer that way ? How can the CT be connected to grids without a diode ?

I assume there is a rectify in some way, but how it acts ?

VERY VERY CONFUSED about that

Thanks for any explanation

Kagliostro

[rzenc]

As you can see they use a CT Transformer with a Full Bridge that has the negative connected to ground, the CT is connected to a cap and than to the grids of final tubes (EL36)

If I got your question right, it's feeding G2 - control grid #2 aka. screen grid - with positive voltage around 150V labelled +1. Control grid #1 gets bias voltage - labelled -1 (B) around -27V.


Hope it helps.
Best Regards

Rzenc

[jjasilli]

Think of this is as a UL tap  set at 50% rather than a "CT".  As such, it does not need its own separate rectification.  All the voltage in the HT winding, and any tap from it, is rectified by the bridge.  Only a separate winding would need to be separately rectified for DC.

[jeff]

So if you used your existing 300-0-300V transformer you'd get 840V(600X1.4) from the bridge rectifier and 420V(300X1.4) from the CT?

My first thought was this would be a cool way to get a tap half your B+, but with the same transformer you'd get an extra tap double your B+, right?

[HotBluePlates]

So if you used your existing 300-0-300V transformer you'd get 840V(600X1.4) from the bridge rectifier and 420V(300X1.4) from the CT?

Stop thinking in guitar-amp terms. Or better-yet, stop thinking in full-wave rectifier terms.

My attention was captured by the power supply for grids, someone can figure out how it works ?

As you can see they use a CT Transformer with a Full Bridge that has the negative connected to ground, the CT is connected to a cap and than to the grids of final tubes (EL36)

Kagliostro

How does a bridge rectifier work?

Forget the tap, look only at the outside ends of the overall winding. See the first picture for a diagram.

At a given instant, one end is positive and the other is negative. For the positive end, the diode which is between it an the + of the filter cap is forward biased, so current will flow if we complete the path. The other end of the winding is negative, and the diode which is connected from it to ground is forward-biased; that diode completes the path for current to the filter cap - so we have a complete path.

The entire winding is used, so we have an output voltage approaching RMS volts * 1.414 (minus diode drop). However, for each direction of current flow, the center-tap is at 106vac above ground (106v * 1.414 = ~150vdc). Therefore, if a cap is connected to ground, half the output voltage appears across this cap, and alway of the same polarity. So you have a half-B+ output option. See the 2nd and 3rd diagrams to see how this happens.

The bridge always does the same thing, whether you see an given outside end as positive or negative, and compared to ground, the center-tap is always at half-B+ output voltage.

The issue is to stop thinking that the output has to be on the far side of a rectifier. The center-tap in this case is always at half of the full-winding voltage. You could place a rectifier diode between the center-tap and the cap, but it would be largely doing nothing.

[HotBluePlates]

So if you used your existing 300-0-300V transformer you'd get 840V(600X1.4) from the bridge rectifier and 420V(300X1.4) from the CT?

My first thought was this would be a cool way to get a tap half your B+, but with the same transformer you'd get an extra tap double your B+, right?

The thing is that a bridge rectifier uses the voltage appearing across the entire power transformer secondary at any moment; using the center-tap to get half of the total voltage is a handy way to get half-B+.

A typical full-wave rectifier circuit using a rectifier tube uses only half of the secondary winding at any given moment. Therefore, the half-secondary has to supply all the voltage needed. If you used the entire secondary winding (as in a bridge), you'd get double the output voltage, which is usually way more than you need.

Tube guitar amps using tube rectifier always use a full-wave circuit (unless they use something like Geezer's hybrid rectifier), because a bridge would require 2 additional recitfier tubes, 2 additional sockets and 2 additional 5v heater windings. Due to the cost, it was cheaper in the old days to pay for more secondary high-voltage winding than to use a bridge rectifier which would better utilize the secondary.

[jeff]

I was just thinking, cuz there's been talk about dropping the voltage with a VVR on the power tubes to reduce volume.
Say you take an amp using a 300-0-300V transformer. Could you replace it with a 150-0-150V transformer and wire it this way? Run the preamp off the bridge, then have a switch to run the power tubes off the bridge or the tap to lower the voltages?

As far as the bridge rectifier, could you use the tube rect and two SS diodes? 'Nother words tube rect to B+ and the SS diodes to ground. I think I remember someone talking about this but can't remember if it was possible.

I know, I know. You're thinking there's Jeff with another one of his crazy hypothetical questions, but it's good to talk things out. If I can understand why things won't work it gives me a better understanding of how things do work. Plus if it is possible that's one more thing in our bag of tricks. Maybe one day someone'll come across a 150-0-150V transformer and try it out.

[DummyLoad]

modified delon full wave doubler. like marshall major PS.

http://en.wikipedia.org/wiki/File:Full-wave_voltage_doubler.svg 

[jeff]

That's a noodle frier.
Correct me if I'm wrong but that schematic looks to me like a bridge rect with the center tap connected to the two caps. If you removed the CT connection is that the same as a bridge rect with two caps in series instead of one cap?

So would this give you 4 times the voltage as a full wave, CT grounded transformer would give you?

Full wave get.
Bridge little more complicated.
This'll take some thinkin'

[John]

Would the typical 150-0-150 transformer supply enough amperes though?

[HotBluePlates]

Correct me if I'm wrong but that schematic looks to me like a bridge rect with the center tap connected to the two caps. If you removed the CT connection is that the same as a bridge rect with two caps in series instead of one cap?

Yes. But don't read any more into it.

The intent of the Marshall circuit is to enforce splitting the applied voltage equally between the series caps. This is normally done in a Fender amp by using bleeder resistors across the caps. Often, they are also included on the Marshall, but aren't strictly needed.

It doesn't double the voltage again (to 4x); that is done with a different arrangement of caps and diodes called a voltage doubler. The difference is that in the typical bridge circuit, both caps are being charged at the same time. In the doubler arrangement, one cap is charged to the full winding voltage, then the other cap is charged; the two caps are stacked so the total output is the sum of the two caps, each charged to full winding voltage.

[kagliostro]

Thanks to all you friends

Many thanks HBP

your schematics were really helpful

I think I have understand your explanation

I try to explain what I've understand

in example 01 there is a "normal" full-wave rectifier (two diodes with the CT of transformer connected to ground)

and the "reverse" solution, the center of the diodes is connected to ground so at the CT of the transformer there is the positive voltage

in example 02 I've tried to draw the current path in the two opposite conditions of AC wave

about example 03 considering the other examples I was thinking to the current we can get from the full-wave rectifier in opposition to the Full-Bridge rectify

The CT of the transformer in second circuit in the example 03 is NOT connected in any way

someone can confirm my assumptions are corrects ?

Thanks again

Kagliostro

p.s.: By mistake I loaded some wrong examples - this are the examples I wont you see - SORRY

[jeff]

It doesn't double the voltage again (to 4x); that is done with a different arrangement of caps and diodes called a voltage doubler.

Thanks for the explination.
 I think what threw me was "doubler". But now I think I get it, it's a full wave doubler, it gives you double what a full wave would give you.
Even though a bridge gives you double what a full wave I never thought of it as a doubler. I hear doubler and think the circuit you mentioined earlier.

It's a bridge rect using the CT to equalize the caps.

They should call it a "self equalizing bridge rect" to avoid comfusing me



 

[jeff]

 I love this site. From kagliostro's initial question to DummyLoads schematic to HotBluePlates clarification it's a real trip to understanding.

 I had a schematic for a tube stereo that made no sense at all to me. It's been a long while since I even looked at it or thought about it. Because of this post, it just popped into my head. BAM, now I get it.

[PRR]

This is the VERY common way to power transistor op-amps and power amps:



It gives symmetrical + and - voltages from "ground".

The "ground" can be anywhere. If you call the bottom of C4 "ground", then you have all positive voltages, one twice the other.

If the winding is 200V CT (100-0-100), the total output is 282V, the tap is 141V.

The usual run of 560VCT (280-0-280) "tube PTs" are not much use this way unless you have need of 800 volts (large transmitter tubes).

A 230+230 winding will give 310V and 620V DC, perfect for HIGH-power 6550 6146 etc amps.

[jeff]

Looking at it with the CT grounded it looks like a full wave rect plus a negative full wave rect?
Diodes reversed cap reversed voltage reversed.

[WiderGates]

Hi,
for example, Fender used this kind of power supply in TheTwin.
There you have a power switch 100W / 25W, which halves the voltage for the powertubes anode, screengrid and fixed bias.
See attachment.


Ciao
Dieter

[jeff]

Hot diggity, that's the ticket!

[WiderGates]

Hi,
but don't forget you need a PT 170-0-170 Volt with this system.


Ciao

[jeff]

What about current draw. If you have a transformer rated at 100mA as a full wave CT and change it to a bridge what can you draw? 100mA or 50 mA. I'm assuming nothing's free. If you have a "X" watt transformer and you wire it to get twice the voltage will you only get 1/2 the current capability?

[kagliostro]

I understand that so:

if I use the whole winding (not CT connection) with a Full-Rectifier Bridge I can have only the current at which the winding is rated

if I use the Center Tap and a Full-Wave Rectifier is the same to put the two half of the transformer in parallel so the current is the sum of the current at which each winding is rated

Kagliostro

[jeff]

If you have a full wave rect that's rated for 100mA with the CT grounded, as each leg swings positive the current passes through the diode and is blocked when it swings negative. Using this wiring(see below) the diodes are passing the current when the leg swings positive charging the upper cap, just as in the full wave, but the reverse diodes are passing when the leg swings nevative to charge the lower cap. What was unused is now being used.

 If the transformer is rated for 100mA as a full wave and you hooked a 100mA load to the positive AND a 100mA load to the negative, are you pulling more current, or are you pulling current throughout each legs full cycle instead of only the positive half?

Are you working the winding harder by using a 100mA pos and a 100 mA neg load, or using it more efficentlly by using each leg's positive AND negitave swing?

What does this mean you can pull if you ground the lower cap, instead of the CT, to get twice the voltage, instead of a neg and a pos? A tran rated at 100mA full wave is rated at ?mA full wave doubler.

[PRR]

> Are you working the winding harder or using it more efficentlly

More efficient on total power, but very different Volt/Current values.

Bridge needs about 70% of the secondary copper -- we use the 2-diode plan only because diodes are (used to be) more expensive than the copper, and the common-cathode form works exceptionally well in vacuum rectifiers.

When you allow for primary utilization (unchanged) and the pesky 6V windings (unchanged), the Bridge output is maybe 10%-15% higher total VA.

Fender Twin PT rated to deliver 450V 400mA DC could, with Bridge, give 900V 220mA.

So with most of the "usual" tube PTs, the question is moot. Twice the voltage at a bit more than half the amps is NOT a useful output for most tube amps. (SE transmitter tubes could be an exception, for folks more reckless than Buttery or DummyL.)

[HotBluePlates]

I understand that so:

if I use the whole winding (not CT connection) with a Full-Rectifier Bridge I can have only the current at which the winding is rated

if I use the Center Tap and a Full-Wave Rectifier is the same to put the two half of the transformer in parallel so the current is the current at which the winding is rated x 2

Kagliostro

Pay attention to the biasing of the diodes.

In the full-wave bridge (CT not used), each half of the winding sees a diode which is forward-biased, so the total current available is equal to the current rating of the secondary; that means half of the total current is supplied by each half-winding. Each half winding contributes its share of the voltage, so the total voltage is the sum of the half-winding voltages.

In the full-wave circuit (not the bridge, and using the CT), only one of the diodes is forward-biased at any moment. While one half-winding is positive, the other half-winding is negative. The secondary still contributes the total current that it can, which now looks like double current, because it is all coming from a single half-winding. However, only one half-winding is used, so you get only the voltage from that half winding.

If you compare the two, it looks like full-wave gives double current and half voltage, while the full-wave bridge gives double voltage and half current. Really, the transformer supplies largely the same power either way; how that power is split between voltage and current is changed.

We tend to learn/understand one style first, and compare the other to that style. Really, if you want to understand them you need to learn the features and functions of each on their own.

[jjasilli]

Yes, I think people tend to think in terms of DC.  But an analysis valid to DC may be invalid for AC.  AC changes the picture, because Time enters the equation. 

[kagliostro]

thanks HBP for your attempt to explain  better the matter

The secondary still contributes the total current that it can, which now looks like double current, because it is all coming from a single half-winding. However, only one half-winding is used, so you get only the voltage from that half winding.

am I correct if I say that the transformer is (as example and assuming there are not losses) 21,2W

so

when we use the full winding voltage (no CT use) we have 100mA available current
(21.2W/212v=0.1A)

when we use half winding voltage (because we use the CT) we have 200mA available current
(21.2W/106v=0.2A)

that result is because in the single time unit (with the full-Wave rectify) we have a fast switch of the use of the two halves of the entire winding so in that unit of time we are taking current from each winding giving the effect to double the current

something like if we have two different torch and switch between each other very fast, if we do it at a certain frequency (faster than the persistence of the image on the retina) we see both torch lighted at the same time

I'm not sure of that and at the moment that is the way I understand this aspect of PS rectify

is this correct or totally wrong ??   ??

or all the power is in one torch at a time so one torch give the doube of light ???


Very confused 

Kagliostro

[HotBluePlates]

am I correct if I say that the transformer is (as example and assuming there are not losses) 21,2W

so

when we use the full winding voltage (no CT use) we have 100mA available current
(21.2W/212v=0.1A)

when we use half winding voltage (because we use the CT) we have 200mA available current
(21.2W/106v=0.2A)

Yes, exactly.

Your math is the proof of what I described. The same VA (volt-amperes, or power) will be available from the transformer in any circuit. If your circuit provides double the voltage, it will also provide half the current.

that result is because in the single time unit (with the full-Wave rectify) we have a fast switch of the use of the two halves of the entire winding so in that unit of time we are taking current from each winding giving the effect to double the current

something like if we have two different torch and switch between each other very fast, if we do it at a certain frequency (faster than the persistence of the image on the retina) we see both torch lighted at the same time

No, not like that.

Draw a diagram like I did. Assume a voltage on the secondary, including any point used as a ground reference. If you have a grounded center-tap, remember that if one end is positive, and the center-tap is zero, then the other end is negative.

Note polarities during that half-cycle throughout the diagram. I say "half-cycle" because during the next half-cycle, the polarities will be reversed.

Note the biasing of the diodes (which ones will conduct, which ones will not conduct). Your diagram will be correct for the full-wave if only 1 diode is conducting during any half-cycle; during the next half-cycle, that diode is off and the other diode is on. Your diagram will be correct for the bridge if you have 2 diodes (one for each end of the secondary) conducting during a half-cycle; during the next half-cycle, the first 2 diodes will turn off, and the diodes which were off will now turn on.

[kagliostro]

MANY THANKS HBP

I think I've understand

is correct to say that the thickness of the wire used for the secondary winding can sustain a certain power (the 21.2W of my example)

so in the unit of time the current that is possible to take from the winding is equal to the power / voltage

if I take 212v is 100ma (21.2W/212v=0.1A)

if I take 106v is 200ma (21.2W/106v=0.2A)

same power less voltage more Ampere or more voltage less Ampere

so the same thickness of wire can afford different currents depending on voltage

is that right ?

Kagliostro

[PRR]

> so the same thickness of wire can afford different currents depending on voltage

No.

In the CT 2-diode plan, the two secondary windings are "parallel" to the output.

In the bridge they are series.

[kagliostro]

OK thanks PRR

I think I've understand what I can aspect from a PT with different use of the CT and rectify

I'm not very deep on the reasons but sure about results

Thanks to all

Kagliostro