Hacking a redundant DVD/CD player power supply for low voltage tube power.

[anotherjim]

I'm looking at making a low voltage tube project (I don't know exactly what yet) but thought of this idea to help power it.
I found a schematic of an SMPS that's the same kind of thing as the one I have. Note that mine can take 240vAC power so C1 (no C2) is 400v rated and will have around 320vDC across it! ***BE CAREFUL!***

T1 doesn't have the high tap for a -22v output (display?) and I've already replaced the 5v adjustment resistors R6 & R7 with a trimmer pot. My board had 5k6/5k1 here so I used a 10k trimmer.
The regulation of the +/- 12v (or any other secondary) depends on the shunt regulator U3 setting. Out of the box, it had only 4.7v on the 5v and the others were +/-10v. I don't think it's faulty, just cheap!

Anyway, no problems getting 6.3v heater power. It doesn't show max current anywhere, but I think enough for at least x3 dual-triode tubes.

Now, the other supplies for Tube HT. The board is single-sided, so easy to hack. I could either...
Cut the T1 centre tap track out, ground the -12v end and have a single +24v for tube HT or...
Do the same cut and use the centre tap for an extra +12v output or...
Do the cut and replace the half-wave diode rectifiers with a full-wave bridge and get near 30v out.
Incidentally, my board has (I think) 1N4007 diodes instead of 1N4148.

Can you have a bridge rectifier on SMPS outputs? I've not come across one. They all seem to be halfwave.
Should the diodes be fast recovery types? The control IC U1 can be in either 50kHz or 100Khz versions. Mine is the former.

[anotherjim]

Ok, would totem pole secondaries work?

Bear in mind, the 12v windings only give 12v with the 5v raised to 6v.

[PRR]

> Can you have a bridge rectifier on SMPS outputs? I've not come across one. They all seem to be halfwave.
Should the diodes be fast recovery types?

"SMPS" covers a thousand different topologies.

Your small/simple switcher has one active device so the available energy is in one side of the wave. A bridge is a waste of half the diodes. Also doubles your voltage losses. Yes, anything higher than airplane power (400Hz) will need fast power diodes or the back-swing leakage will smoke them.

Note the the +/-12V are 1N4148 diodes, fast but only 100mA. (Which I guess is ample for you?)

I'm not sure why there are 1N4007 on the transformer primary flyback load, seems wrong?

Remember that cold heaters suck 3 or 4 times their rated running current. Switchers often stall.

[anotherjim]

Very good points Paul, I'll have to have a closer look at the diodes in my example, I could have UF4007 (there are several differences from that schematic, but no major ones).  I did wonder, what with the switcher only giving one power swing per cycle and now I know why bridge rect's don't get seen with them.
Tube heater cold current is another good heads-up. Thanks for that.
Actually, I haven't tested yet with any kind of load other than the fixed resistors on the outputs. 100R on the +5v and 1k on both +/-12v. Have just made a breadboard friendly 9pin tube adapter so get to try some kind of circuit with it.

[anotherjim]

Have checked those diodes. Had to take them out to read them. I have FR107 which are 1A 1000v PIV fast recovery. The only 1N4007s form the AC input bridge rectifier, as you'd expect.

[Rob Strand]

Out of the box, it had only 4.7v on the 5v and the others were +/-10v. I don't think it's faulty, just cheap!

Not sure what is going on there.  Feedback is feedback it should regulate according to the feedback resistor values.   Since the 5V rail is the feedback point you should have minimum load on the 5V rail of about 1/10 the rating to measure any of the rails.

I did wonder, what with the switcher only giving one power swing per cycle and now I know why bridge rect's don't get seen with them.

In a flyback design like this the transformer actually acts as an inductor.   When the switch is on the primary circuit looks like the primary inductance.  In this phase the inductor is charged up.   When the switch is off the secondary circuit like the secondary inductance.  In this phase the stored energy in the inductor is passed to the output.

These things work quite similar to those NE555 boost converters you play with.  The difference is the inductor has two separate windings which lets you set the output voltage.

While the transformer turns ratio still controls the output voltage but one view is it's the inductances (and switch duty) which set the voltage.

For higher powers, say over 80W there are switchmode supplies which work differently and the transformer does work as a transformer, for example the power supply in a desktop computer.

For your unit the rating of the PSU is set by the switch-mode chip.   In particular the peak current of the switch.

https://media.digikey.com/pdf/Data%20Sheets/Fairchild%20PDFs/FSDH321%20Series.pdf

You can see the rating of device depends not only on the thermal conditions but also the range of input supply.   The reason input voltage comes into play is the higher input voltage sets the peak current (primary turns) and the lower input voltage sets the turns ratio - not quite that either since it depends on the chip.

What I'm leading to here is by taking a 5V design and jacking up the output voltage to 6V you will need to back-off the output current as the transformer won't be optimally designed for 6V.  In fact you don't have any info on where the design stands in terms of how stressed the chip is.    These chips do have a habit of blowing up when stressed. 

Another thing missing in the design info is the relative amounts of total power assigned to -12V, +12V and 5V windings.  This comes down to how much winding space as assigned to each winding.  If you can accurately measure the (usually very low) DC resistances of the windings, then scale the DC resistances of the 5V winding to correspond to 12V (n^2 scaling), you can estimate the proportions of current allowed in the transformer design.   Even that is extremely rough because the heating depends on the AC resistance and the difference between AC an DC resistance goes up (by a large factor) when you have thicker wire.

Unfortunately switchmodes are complicated.  You don't know how much margins for power and stability of feedback loop.   For example, up'ing the output voltage could upset the safety margins in the feedback loop ; possibly meaning it will go unstable sooner when the caps get old.

Sweeping all these problems under the carpet.  It would at least be wise to not load the PSU too close to the power limits in the datasheet.   

As far modifying the winding arrangements go, I suspect you should be able to do it.   It's slightly safer from a feedback stability perspective to keep the 12V windings separate from the 5V winding.

A good indicator of something going wrong is to check the idle current, and low-load current, before and after the mods.

I'm not sure why there are 1N4007 on the transformer primary flyback load, seems wrong?

The D8 and D9 stuff is normal.  D8 is part of a voltage clamp which protects the switch from over voltage due to glitches cause by the transformer's leakage inductance.   D9 provider power to the chip when running.  Start-up power is supplied via R1 then when the chip powers up it gets a more beefy supply via D9.

[anotherjim]

Well, a pair of 6.3v heaters make a difference. Adjusted to 5v and the 12v rails are much closer to 12v. Taken up to 6v they reach about 13.5v. This makes sense though doesn't it? The harder the switcher works to keep the "5v" output happy, the other secondaries are going to get boosted along with it.

Lashed up a Valvecaster thing with an ECC82 (12AU7) running plate load from the +12v. Works ok, nothing to shout about though. only one tube, but the switcher survived the Mullard flash start heaters (which always scares me!).

[Rob Strand]

Well, a pair of 6.3v heaters make a difference. Adjusted to 5v and the 12v rails are much closer to 12v. Taken up to 6v they reach about 13.5v. This makes sense though doesn't it? The harder the switcher works to keep the "5v" output happy, the other secondaries are going to get boosted along with it.

Looks reasonable to me.   Those power supplies are designed to produce the nominal voltages with specified loads on *each* rail.   When things are loaded differently from the specified loads the non-feedback outputs can be all over the place.  With light loads you can even get peak detecting effects which can produce quite high output voltages compared to the nominal.

Bumping the 5V to 6V we would expect the 12V rails to rise.  Trying to predict the rise isn' t so easy.  The simplest estimate would be 12*(6/5) = 14.4V.  We could try to account for the diodes (6+0.35)/(5+0.35)*(12+0.7)-0.7 = 14.4V But in reality it's much more difficult because you have two coupled windings, leakage inductances, winding resistances.   The 13.5V doesn't look unreasonable.

Re output current ratings.  Since the 12V rails use low current diodes you would only expect 50mA to 100mA from those output.  That's less than 1.2W per 12V rail so less than 2.4W for the two, say 2W.  The maximum rating for the chip is somewhere in the 8W to 17W zone depending on unknown specifics but the point is that leaves about 6W to 15W for the 5V rail.   The diode on the 5V rail is 2A, so we would expect perhaps 1A load on the 5V or 5W which is consistent with the upper limits for the chip.   So the 5V rail (bumped to 6V) should be able to drive the heaters.

[anotherjim]

As I mentioned earlier, I don't have the exact same board, it's only in the same general arrangement as that schematic I found on the web. The 12v outputs have FR107 1A diodes. However, the higher rating will be moot if I exceed the total power rating of the switcher. I don't expect I'll need to draw even 10mA from the 12v supplies!

It also has better output filtering, 220uF + 100n disc compared to that schematic. Those 220uF are only 25v rated so I'll have to keep an eye on the voltage rise or swap them for 35v or better if I stack the secondaries.

Another difference, the control chip Startup input is directly connected to the switcher Drain pins - no series resistor other than the primary winding. I think the chip data said something about that being ok.

[Rob Strand]

As I mentioned earlier, I don't have the exact same board, it's only in the same general arrangement as that schematic I found on the web. The 12v outputs have FR107 1A diodes. However, the higher rating will be moot if I exceed the total power rating of the switcher. I don't expect I'll need to draw even 10mA from the 12v supplies!

The angle I was taking was the 12V only used up a small proportion of the rating so that left a considerable proportion of the power rating to the 5V rail.   The 1A diodes on the 12V completely stuffs-up that argument!

Another difference, the control chip Startup input is directly connected to the switcher Drain pins - no series resistor other than the primary winding. I think the chip data said something about that being ok.

Ah, OK.   Yes, my reading of the datasheet is it's OK too.

There's an example at the end of the datasheet *with* the resistor, which is used for some trickery.  The 56k on your schematic might help protect the chip against glitches as the current source on the Vstr pins sometimes gets fried (at least on other similar chips).

Anyway it looks like it's all working fine.