interest check: universal psu boards

[looser101]

Quote:

With #3, you would not populate the bridge rectifier section of the TREAD board and just feed the DC directly to the regulator input.

I would leave the bridge rectifier in. It makes an excellent polarity converter. Then you plug AC, DC (center+), or DC (center-) wallwart and still get the correct polarity to the regulator. Just think, you can plug just about any wallwart into a DC(AC?) input jack and never have to worry about polarity. There are some drawbacks but I think the versatility is worth it.

R

 

[amb]

Quote:

Originally Posted by looser101 I would leave the bridge rectifier in. It makes an excellent polarity converter. Then you plug AC, DC (center+), or DC (center-) wallwart and still get the correct polarity to the regulator. Just think, you can plug just about any wallwart into a DC(AC?) input jack and never have to worry about polarity. There are some drawbacks but I think the versatility is worth it.

True enough. The only downside is two additional diode drops in voltage before the regulator. One needs to take that into account when selecting a DC wallwart.

 

[tangent]

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Originally Posted by Clutz why someone would attach the TREAD into an AC/DC wall-wart

There are several reasons I recommend looking into this: it puts the transformer far away from the regulator, you can find cheap unregulated wall warts for about the same cost as a raw transformer, and it frees you from having to deal with the dangers of AC power. The latter advantage is very important: it addresses the needs of people who want a regulated supply but aren't willing to pay for an Elpac WM080 and aren't willing to go 100% DIY.

Quote:

Is the issue here that most cheap AC/DC wallwarts provide noisy DC

Check out the oscillograms in my power supply tests article. The Radio Shack supply is a typical unregulated unit, and the Elpac WM1224 is a switching-regulated supply. Compare the noise they put out with that of the other supplies shown, which are all linear-regulated.

 

[Clutz]

Quote:

Originally Posted by amb Clutz, most people build cmoys for battery power, so a resistor-divider virtual ground scheme is used. An improved scheme would be to use a TLE2426 rail splitter chip instead of the resistor divider. If you want to power a cmoy with wall power, you could use a single TREAD and retain the virtual ground scheme. Or, you could use two TREADs and eliminate the virtual ground. I see little point in doing the latter for a cmoy, because cmoys are supposed to be small, simple and portable, and it seems silly to have a power supply box several times the size/weight and complexity of the amp itself.

I'm not actually interested in using a TREAD or a STEPS as a power supply unit for a Cmoy. I used the Cmoy as an example, because it is a design I understand and understand where/why there is a V+, Virtual Ground and V-. So it provided a useful example where I could ask a question about implementing a TREAD PSU.

Quote:

Originally Posted by amb There are many amps out there without onboard virtual ground, and require a split power supply. If you were to power such an amp with the TREAD, you'll need two to make the split supply.

Okay.

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Originally Posted by amb The STEPS is a slightly different animal than the TREAD because the power transformer is onboard. You can also build a split supply with two of them, but you'd be connecting the AC mains input of the two together instead of the dual secondary transformer scheme. The output side is then wired in series just like my diagram. As for your wallwart question, since the TREAD doesn't have an onboard power transformer, you could use 1.) a regular power transformer, 2.) a AC wallwart as the transformer, or 3.) an unregulated DC wallwart. With #3, you would not populate the bridge rectifier section of the TREAD board and just feed the DC directly to the regulator input.

Thanks for you help amb and tangent- as usual your posts have been very useful. I really appreciate it.

Cheers,
Clutz

 

[Clutz]

Quote:

Originally Posted by tangent There are several reasons I recommend looking into this: it puts the transformer far away from the regulator, you can find cheap unregulated wall warts for about the same cost as a raw transformer, and it frees you from having to deal with the dangers of AC power. The latter advantage is very important: it addresses the needs of people who want a regulated supply but aren't willing to pay for an Elpac WM080 and aren't willing to go 100% DIY.

I like this solution. It's more interesting than just getting an Elpac and I don't have to live in fear of zapping myself (given that I have Tourette Syndrome which sometimes causes my hands to shake- this is not an entirely unreasonable fear).

I suppose I will be placing an order for the TREAD soon (and some ALPS pots).

Thanks again,
Clutz

 

[BrokenEnglish]

Quote:

Originally Posted by tangent There are several reasons I recommend looking into this: it puts the transformer far away from the regulator, you can find cheap unregulated wall warts for about the same cost as a raw transformer, and it frees you from having to deal with the dangers of AC power. The latter advantage is very important: it addresses the needs of people who want a regulated supply but aren't willing to pay for an Elpac WM080 and aren't willing to go 100% DIY.

to add another benefit: amp-manufacturers could integrate a tread for their overseas-customers with 230/240v-mains. it's very hard to find proper commercial regulated higher voltage psus for example over here in europe (uk seems to be an exception), but it is no problem to get simple and cheap unregulated dc or ac wallwarts.

 

[individual6891]

Any chance you could add a board layout to the TREAD documents tangent?

 

[bg4533]

Anyone know of any wallwarts that are a little higher voltage? I would like to make a power supply that puts out ~28V.

 

[skyskraper]

ive seen a couple of 30v psu's around but they were pretty pricey iirc. using a transformer would be cheap and easy to get 28v with the tread

 

[BrokenEnglish]

Quote:

Originally Posted by bg4533 Anyone know of any wallwarts that are a little higher voltage? I would like to make a power supply that puts out ~28V.

no problem with an 24v ac-transformer. in fact i'm using one of my 24v pollin-wallwarts (s. above) to get 29v@300mA with one of my diy-psus (lm317) at the moment. without load i measure ~36v dc after the smoothing caps (before the regulator).

rule of thumb:
Voutput[dc] = (sqrt(2)*Vinput[ac]) - 2v [voltage drop rectifier] - 2v [voltage drop regulator]

/edit: oh... also important to note: you need to keep the ripple-voltage low, to get 28v out of 24v ac-transformers. better use 2,2f minimum as smoothing cap.

 

[tangent]

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Originally Posted by individual6891 Any chance you could add a board layout to the TREAD documents tangent?

Done. Direct link.

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Originally Posted by BrokenEnglish Voutput[dc] = (sqrt(2)*Vinput[ac]) - 2v [voltage drop rectifier] - 2v [voltage drop regulator]

True, but only under no load. Under load, Vinput[ac] goes down, until it is exactly the rated voltage when the load current equals the rated current. You can sometimes get a graph of this in the part's datasheet. So, you need the transformer to be rated for much higher current than you intend to draw if you want to take advantage of this effect.

Quote:

better use 2,2f minimum as smoothing cap.

2.2F is a little excessive, don't you think?

 

[BrokenEnglish]

Quote:

Originally Posted by tangent True, but only under no load. Under load, Vinput[ac] goes down, until it is exactly the rated voltage when the load current equals the rated current. You can sometimes get a graph of this in the part's datasheet. So, you need the transformer to be rated for much higher current than you intend to draw if you want to take advantage of this effect.

you're right. let me better it:
Voutput[dc] = (sqrt(2)*Vinput[ac_RATED]) - 2v [voltage drop rectifier] - 2v [voltage drop regulator] - Urip

Quote:

2.2F is a little excessive, don't you think?

no.. i calculated:
Urip[V] = 1.5 * Iload[mA] / C[uF]
worst case: 1.5*1500/2200 = 1v
vs. 1.5*1500/1000 = 2.25v
you'd have to subtract Urip from Voutput - so less Urip is better.
/edit: unless he specifies what current he needs...

 

[tangent]

Quote:

no.. i calculated:

I think we have a units and/or decimal separator problem. 2.2F is 2,200,000uF. In European notation, that's 2,2 farads = 2.200.000 microfarads. It looks like you mean 2200uF.

 

[BrokenEnglish]

Quote:

Originally Posted by tangent I think we have a units and/or decimal separator problem. 2.2F is 2,200,000uF. In European notation, that's 2,2 farads = 2.200.000 microfarads. It looks like you mean 2200uF.

oh... sorry... sometimes i forget to use the . as separator.. and i missed the next smaller step: mf

...you're right: i meant 2200µf

 

[individual6891]

Quote:

Originally Posted by tangent Done. Direct link.

thanks tangent